chapter-9-作业参考答案

信号与系统奥本海姆英文版课后答案chapter9

Chapter 9 Answers9.1 （a ）The given integral may be written as(5)0t j t e e dt σω∞-+⎰If σ<-5 ,then the function (5)te σ-+ grows towards ∞ with increasing t and the given integral does not converge .but if >-5,then the integral does converge (b) The given integral may be written as0(5)t j tee d σω-+-∞⎰t If σ>-5 ,then the function (5)te σ-+ grows towards ∞as t decreases towards -∞and the given integral doesnot converge .but if σ<-5,then the integral does converge (c) The given integral may be written as5(5)5t j t e e d σω-+-⎰t Clearly this integral has a finite value for all finite values of σ. (d) The given integral may be written as(5)t j t e e d σω∞-+-∞⎰tIfσ>-5 ,then the function (5)t e σ-+ grows towards ∞as t decreases towards -∞and the given integraldoes not converge If σ<-5, ,then function (5)te σ-+ grows towards ∞ with increasing t and the given integral does not converge If σ=5, then the integral stilldoes not have a finite value. therefore, the integral does not converge for any value of σ. (e) The given integral may be written as0(5)t j tee d σω-+-∞⎰t+ (5)0t j t e e d σω∞-+⎰t The first integral converges for σ<-5, the second internal converges if σ>-5,therefore, the given internal converges whenσ<5.(f) The given integral may be written as0(5)t j t e e d σω-+-∞⎰tIf σ>5 ,then the function (5)te σ--+grows towards ∞ as t decrease towards -∞ and the given integral does not converge .but if σ<5,then the integral does converge. 9.2 (a)X(s)= 5(1)t dt eu t e dt ∞---∞⎰- =(5)0s tedt ∞-+⎰ =(5)5s es -++As shown in Example 9.1 the ROC will be {}Re s >-5. (b) By using eg.(9.3), we can easily show that g(t)=A 5te-u(-t-0t ) has the Laplace transformG(s)= 0(5)5s t Ae s ++The ROC is specified as {}Re s <-5 . Therefore ,A=1 and 0t =-19.3 Using an analysis similar to that used in Example 9.3 we known that given signal has a Laplace transform of the formX(s)115s s β+++The corresponding ROC is {}Re s >max(-5,Re{β}). Since we are given that the ROC isRe{s}>-3, we know that Re{β}=3 . there are no constraints on the imaginary part of β. 9.4 We know form Table 9.2 that111()sin(2)()()()Lt x t e t u t X s X s -=-←−→=-, Re{s}>-1 We also know form Table 9.1 thatx(t)= 1()Lx t -←−→X(s)= 1()X s -The ROC of X(s) is such that if 0s was in the ROC of 1()X s , then -0s will be in the ROC of X(s). Putting the two above equations together ,we havex(t)= 1x (-t) =sin(2)()t e t u t --X(s)= =-222(1)2s -+, {}Re s <1the denominator of the form 2s -2s+5. Therefore, the poles of X(s) are 1+2j and 1-2j.9.5 (a) the given Laplace transform may be written as ()X s =24(1)(3)s s s +++.Clearly ,X(s) has a zero at s=-2 .since in X(s) the order of the denominator polynomial exceeds the order of the numerator polynomial by 1 ,X(s) has a zero at ∞. Therefore ,X(s) has one zero in finite s-plane and one zero at infinity.(b) The given Laplance transform may be written asX(s)=1(1)(1)s s s +-+= 11s -Clearly ,X(s) has no zero in the finite s-plane .Since in X(s) the of the denominator polynomial exceedsthe order the numerator polynomial by 1,X(s) has a zero at .therefore X(s) has no zero in the finite s-plane and one zero at infinity.(c) The given Laplace transform may be written as22(1)(1)()1(1)s s s X s s s s -++==-++ Clearly ，X （s ）has a zero at s=1.since in X(s) the order of the numerator polynomial exceeds the order of the denominator polynomial by 1,X(s) has zeros at ∞ .therefore , X(s) has one zero in the s-plane and no zero at infinity .9.6 (a) No. From property 3 in Section 9.2 we know that for a finite-length signal .the ROC is the entire s-plane .therefore .there can be no poles in the finite s-plane for a finite length signal . Clearly in this problem this not the case.(b) Yes. Since the signal is absolutely integrable, The ROC must include, the j ω-axis . Furthermore ,X(s) has a pole at s=2 .therefore, one valid ROC for the signal would be Re{s}<2. From property 5 in section 9.2 we know that this would correspond to a left-sided signal(C) No . Since the signal is absolutely integrable, The ROC must include , the j ω-axis . Furthermore ,X(s) has a pole at s=2. therefore ,we can never have an ROC of the form Re{s}> α. From property 5 in section 9.2 we knew that x(t) can not be a right-side signal(d) Yes . Since the signal is absolutely integrable, The ROC must include , the j ω-axis . Furthermore ,X(s) has a pole at s=2 .therefore, one valid ROC for the signal could be α<Re{s}<2 such that α<0 .From property 6 in section 9.2 ,we know that this would correspond to a two side signal9.7 We may find different signal with the given Laplace transform by choosing different regions of02s =- 13s =- 212s j =- 312s j =-Based on the locations of the locations of these poles , we my choose form the following regions of convergence: (i) Re{s}>- 12(ii)-2< Re{s}<- 12(iii)-3<Re{s}<-2 (iv)Re{s}<-3Therefore ,we may find four different signals the given Laplace transform. 9.8 From Table 9.1,we know thatG(t)= 2()()(2)L te x t G s X s ←−→=-. The ROC of G(s) is the ROC of X(s) shifted to the right by 2We are also given that X(s) has exactly 2 poles at s=-1 and s=-3. since G(s)=X(s-2), G(s)also has exactly two poles ,located at s=-1+2=1 and s=-3+2=-1 since we are given G(j ω) exists , we may infer that j ω-axis lies in the ROC of G(s). Given this fact and the locations of the poles ,we may conclude that g(t) is a two sidesequence .Obviously x(t)= 2te g(t) will also be two sided9.9 Using partial fraction expansion X(s)= 4243s s -++Taking the inverse Laplace transform, X(t)=443()2()t t e u t e u t ---9.10 The pole-zero plots for each of the three Laplace transforms is as shown in Figure S9.10(a) form Section 9.4 we knew that the magnitude of the Fourier transform may be expressed aswe se that the right-hand side of the above expression is maximum for ω=0 and decreases as ω becomesincreasing more positive or more negative . Therefore 1()H j ω is approximately lowpass(b) From Section 9.4 we know that the magnitude of the Fourier transform may be express aswe see that the right-hand side of the above expression is zero for ω=0.It then increams withincreasing |ω| until |ω| reach 1/2. Then it starts decreasing as |ω| increase even further. Therefore | 2H (j )ω| is approximately bandpass.(c) From Section 9.4 we know that the magnitude of the Fourier transform may be express as2We see that the right-hand side of the above expression is zero for ω=0. It then increases withincreasing |ω| until |ω| reaches 21. Then |ω| increases,| 3()H j ω| decreases towards a value of1(because all the vector lengths became almost identical and the ratio become 1) .Therefore |3()H j ω| is approximately highpass.9.11 X(s) has poles ats=1-2and 1-2.X(s) has zeros ats=1212.From Section 9.4 we know that |X(j ω)| is11(Length of vector from toto ωωThe terms in the numerator and denominator of the right-band side of above expression cancel ourgiving us |X(j ω)|=1.9.12 (a) If X(s) has only one pole, then x(t) would be of the form A ate -.Clearly such a signal violates condition 2. Therefore , this statement is inconsistent with the given information.(b) If X(s) has only two poles, then x(t) would be of the form A 0sin()ate t ω- .Clearly such a signal could be made to satisfy all three conditions(Example:0ω=80π,α=19200). Therefore, this statement is consistent with the given information. (c) If X(s) has more than two poles (say 4 poles), then x(t) could be assumed to be of theform 00sin()sin()at btAe t Be t ωω--+. Clearly such a signal could still be made to satisfy all three conditions. Therefore, this statement is consistent with the given information. 9.13 We have1}Re{,1)(->+=s s s X β.Also,(Length of vector form ω to -1)(Length of vector form ω to 11}Re{1),()()(<<--+=s s X s X s G αTherefore, ].11[)(2s s s s G -++-=ααβComparing with the given equation for G(s), ,1-=α .21=β9.14. Since X(s) has 4 poles and no zero in the finite s-plane, we many assume that X(s) is of the form .))()()(()(d s c s b s a s As X ----=Since x(t) is real ,the poles of X(s) must occur in conjugate reciprocal pairs. Therefore, we mayassume that b=*a and d=*c . This result in .))()()(()(**c s c s a s a s As X ----=Since the signal x (t) is also even , the Laplace transform X(s) must also be even . This implies thatthe poles have to be symmetric about the j ω-axis. Therefore, we may assume that c=*a -. This results in .))()()(()(**a S a s a s a s As X ++--=We are given that the location of one of the poles is (1/2)4πj e . If we assume that this pole is a, we have 4444AX(s)=.1111(s-)(s-)(s+)(s+)2222j j j j e e e e ππππ--This gives us22().11()()44AX s s s s s =Also ,we are give that()(0)4x t dt X ∞-∞==⎰Substituting in the above expression for X(s), we have A=1/4. Therefore,221/4().11()()44X s s s s s =9.15. Taking the Laplace transform of both sides of the two differential equations, we haves X(s)=1)(2+-s Y and s Y(s)=2X(s) . Solving for X(s) and Y(s), we obtain4)(2+=s s s X and Y(s)= 22s 4+.The region of convergence for both X(s) and Y(s) is Re{s}>0 because both are right-hand signals. 9.16. Taking the Laplace transform of both sides of the given differential equations ,we obtain ).(])1()1()[(223s X s s s s Y =+++++αααα therefore,.)1()1(1)()()(223αααα+++++==s s s s X s Y s H(a) Taking the Laplace transform of both sides of the given equation, we haveG(s) = s H(s)+ H(s). Substituting for H(s) from above,.1)1()1()1()(22223αααααα++=++++++=s s s s s s s GTherefore, G(s) has 2 poles.(b) we know that H(s) =.))(1(122αα+++s s s Therefore, H(s) has poles at and j ),2321(,1+--α ).2321(j --α If the system has to be stable,then the real part of the poles has to be less than zero. For this to be true, we require that ,02/<-α i.e.,0>α.9.17 The overall system show in Figure 9.17 may be treated as two feedback system of the form shown in figure 9.31 connected in parallel. By carrying out an analysis similar to that described in Section 9.8.1, we find the system function of the upper feedback system to be.82)/2(41/2)(1+=+=s s s s HSimilarly, the system function of the lower feedback system is .21)2/1(21/1)(2+=+=s s s HThe system function of the overall system is now.1610123)()()(221+++=+=s s s s H s H s HSince H(s)=Y(s)/X(s), we may write]123)[(]1610)[(2+=++s s X s s s Y . Taking the inverse Laplace transform, we obtaindtt dx t x t y dt t dy dt t y d )(3)(12)(16)(10)(2+=++9.18. ( a) From problem 3.20, we know that differential equation relating the input and output of the RLC circuit is2()()()().d y t dy t y t x t dtdt++=Taking the Laplace transform of this (while nothing that the system is causal and stable), we obtain 2()[1]().Y s s s X s ++= Therefore ,2()1(),()1Y s H s X s s s ==++ 1{}.2e s ℜ>-(b) We note that H(s) has two poles at12s =--12s =-+From Section 9.4 we know that the magnitude of the Fourier transform may be expressed asWe see that the right hand side of the above expression Increases with increasing |ω| until |ω| reaches 12. Then it starts decreasing as |ω| increasing even further. It finally reaches 0 for |ω|=∞.Therefore 2|()|H j ω is approximately lowpass.(c) By repeating the analysis carried out in Problem 3.20 and part (a) of this problem with R =310-Ω,we can show that2()1(),()1Y s H s X s s s ==++ {}0.0005.e s ℜ>-(d) We haveWe see that when |ω| is in he vicinity 0.0005, the right-hand side of the above equation takes onextremely large value. On either side of this value of |ω| the value of |H (j ω)| rolls off rapidly. Therefore, H(s) may be considered to be approximately bandpass. 9.19. (a) The unilateral Laplace transform isX(s) = 20(1)t st e u t e dt -∞--+⎰= 20t st e e dt -∞--⎰=21+s {} 2.e s ℜ>-(b) The unilateral Laplace transform is2(3)0()[(1)()(1)]t st X s t t e u t e dt δδ-∞-+-=++++⎰2(3)0[()]t st t e e dt δ-∞-+-=+⎰612e s -=++ {} 2.e s ℜ>- (c) The unilateral Laplace transform is240()[()()]t t st X s e u t e u t e dt -∞---=⎰240[]t t st e e e dt -∞---=+⎰1124s s =+++ {} 2.e s ℜ>-9.20. In Problem 3.29, we know that the input of the RL circuit are related by ).()()(t x t y dtt dy =+Applying the unilateral Laplace transform to this equation, we have ).()()0()(s x s y y s sy =+--(a) For the zero-state response, set (0)0y -=.Also we have u s x =)(L{)(2t u et-}=21+s .Therefore,y(s)(s+1)=.21+sComputing the partial fraction expansion of the right-hand side of the above equation and then taking its inverse unilateral Laplace transform, we have ).()()(2t u e t u e t y t t ---=(b) For the zero-state response, assume that x(t) = 0.Since we are given that (0)1y -=,.11)(0)(1)(+=⇒=+-s s y s y s sy Taking the inverse unilateral Laplace transform, we have ()().t y t e u t -=Figure S9.212()2()().t t y t e u t e u t --=-9.21. The pole zero plots for all the subparts are shown in figure S9.21. (a) The Laplace transform of x(t) isX(s)= 230()t t st e e e dt ∞---+⎰= (2)(3)00[/(2)]|[/(3)]|s t s te s e s -+∞-+∞-++-+ =211252356s s s s s ++=++++(b) Using an approach similar to that show in part (a), we have41(),4L t e u t s -←−→+ {} 4.e s ℜ>-Also,551(),55L t j t e e u t s j -←−→+-and(){}551,555LT t j t e e u t e s s j --←−→ℜ>-++.From this we obtain()()()()55555215sin 52525LTt t j t t j t e t u t e e e e u t js ----⎡⎤=-←−→⎣⎦++ ,where {}5e s ℜ>- .Therefore,()()(){}245321570sin 5,51490100LTt t s s e u t e t u t e s s s s --+++←−→ℜ>-+++. R b Im(c)The Laplace transform of ()x t is ()()023t t st X s e e e dt --∞=+⎰()()()()2300/2|/3|s t s t e s e s ----∞-∞⎡⎤⎡⎤=--+--⎣⎦⎣⎦ 211252356s s s s s -=+=---+.The region of convergence (ROC) is {}2e s ℜ<.(d)Using an approach along the lines of part (a),we obtain(){}21,22LT t e u t e s s -←−→ℜ>-+. (S9.21-1) Using an approach along the lines of part (c) ,we obtain(){}21,22LT t e u t e s s -←−→ℜ<-. (S9.21-2)From these we obtain()()222224t LT t t s e e u t e u t s --=+-←−→-, {}22e s -<ℜ<. Using the differentiation in the s-domain property , we obtain(){}22222228,2244t LT d s s te e s ds s s -+⎡⎤←−→-=--<ℜ<⎢⎥-⎣⎦-. (e)Using the differentiation in the s-domain property on eq.(S9.21-1),we get()(){}2211,222LT t d te u t e s ds s s -⎡⎤←−→-=ℜ>-⎢⎥+⎣⎦+.Using the differentiation in the s-domain property on eq (S9.21-2),we get ()(){}2211,222LT t d te u t e s ds s s ⎡⎤--←−→=-ℜ<⎢⎥-⎣⎦-.Therefore,()()()(){}222224,2222t LT t t st e te u t te u t e s s s ---=--←−→-<ℜ<+-.(f)From the previous part ,we have ()()(){}2221,22LT t t t e u t te u t e s s -=--←−→-ℜ<-.(g)Note that the given signal may be written as ()()()1x t u t u t =-- .Note that (){}1,0LTu t e s s←−→ℜ>.Using the time shifting property ,we get(){}1,0s LT e u t e s s--←−→ℜ>.Therefore ,()1x t()()11,sLT e u t u t s----←−→ All s . Note that in this case ,since the signal is finite duration ,the ROC is the entire s-plane.(h)Consider the signal ()()()11x t t u t u t =--⎡⎤⎣⎦.Note that the signal ()x t may beexpressed as ()()()112x t x t x t =+-+ . We have from the previous part()()11sLT e u t u t s----←−→, All s . Using the differentiation in s-domain property ,we have()()()12111s s s LT d e se e x t t u t u t ds ss ---⎡⎤--+=--←−→=⎡⎤⎢⎥⎣⎦⎣⎦, All s . Using the time-scaling property ,we obtain()121s s LT se e x t s --+-←−→, All s .Then ,using the shift property ,we have()21212s sLT s se e x t es ---+-+←−→ ,All s . Therefore ,()()()21122112s s s sLT s se e se e x t x t x t e s s----+--+=+-+←−→+, All s. (i) The Laplace transform of ()()()x t t u t δ=+ is (){}11/,0X s s e s =+ℜ>.(j) Note that ()()()()33t u t t u t δδ+=+.Therefore ,the Laplace transform is the same as the result of the previous part.9.22 (a)From Table 9.2,we have()()()1sin 33x t t u t =.(b)From Table 9.2 we know that()(){}2cos 3,09LT st u t e s s ←−→ℜ>+. Using the time scaling property ,we obtain()(){}2cos 3,09LT s t u t e s s -←−→-ℜ<+Therefore ,the inverse Laplace transform of ()X s is()()()cos 3x t t u t =--.(c)From Table 9.2 we know that ()()(){}21cos 3,119LTt s e t u t e s s -←−→ℜ>-+. Using the time scaling property ,we obtain ()()(){}21cos 3,119LTt s e t u t e s s -+-←−→-ℜ<-++. Therefore ,the inverse Laplace transform of ()X s is ()()()cos 3t x t e t u t -=--.(d)Using partial fraction expansion on ()X s ,we obtain ()2143X s s s =-++ .From the given ROC ,we know that ()x t must be a two-sided signal .Therefore ()()()432t t x t e u t e u t --=+-.(e)Using partial fraction expansion on ()X s ,we obtain()2132X s s s =-++. From the given ROC ,we know that ()x t must be a two-sided signal ,Therefore,()()()332ttx t e u t e u t --=+-.(f)We may rewrite ()X s as ()2311s X s s s =+-+1=1=+Using Table 9.2 ,we obtain()())())()/2/23cos /2sin/2t t x t t e u t u t δ--=+.(g)We may rewrite ()X s as ()()2311s X s s =-+.From Table 9.2,we know that(){}21,0LT tu t e s s ←−→ℜ>.Using the shifting property ,we obtain()(){}21,11LT t e tu t e s s -←−→ℜ>-+.Using the differentiation property ,()()()(){}2,11LT t t t d s e tu t e u t te u t e s dt s ---⎡⎤=-←−→ℜ>-⎣⎦+. Therefore,()()()()33t t x t t e u t te u t δ--=--.9.23.The four pole-zero plots shown may have the following possible ROCs:·Plot (a): {}2e s ℜ<- or {}22e s -<ℜ< or {}2e s ℜ>.·Plot (b): {}2e s ℜ<- or {}2e s ℜ>-. ·Plot (c): {}2e s ℜ< or {}2e s ℜ>. ·Plot (d): Entire s-plane.Also, suppose that the signal ()x t has a Laplace transform ()X s with ROC R . (1).We know from Table 9.1 that()()33LT te x t X s -←−→+.The ROC 1R of this new Laplace transform is R shifted by 3 to the left .If ()3t x t e - is absolutely integrable, then 1R must include the jw -axis.·For plot (a), this is possible only if R was {}2e s ℜ> . ·For plot (b), this is possible only if R was {}2e s ℜ>-. ·For plot (c), this is possible only if R was {}2e s ℜ> . ·For plot (d),R is the entire s-plane. (2)We know from Table 9.2 that(){}1,11LT t e u t e s s -←−→ℜ>-+.Also ,from Table 9.1 we obtain()()(){}2,11LT t X s x t e u t R R e s s -⎡⎤*←−→=ℜ>-⎡⎤⎣⎦⎣⎦+I If ()()te u t x t -*is absolutely integrable, then 2R must include the jw -axis.·For plot (a), this is possible only if R was {}22e s -<ℜ<. ·For plot (b), this is possible only if R was {}2e s ℜ>-. ·For plot (c), this is possible only if R was {}2e s ℜ< . ·For plot (d),R is the entire s-plane.(3)If ()0x t = for 1t > ,then the signal is a left-sided signal or a finite-duration signal . ·For plot (a), this is possible only if R was {}2e s ℜ<-. ·For plot (b), this is possible only if R was {}2e s ℜ<-. ·For plot (c), this is possible only if R was {}2e s ℜ< . ·For plot (d),R is the entire s-plane.(4)If ()0x t =for 1t <-,then the signal is a right-sided signal or a finite-duration signal ·For plot (a), this is possible only if R was {}2e s ℜ>.·For plot (b), this is possible only if R was {}2e s ℜ>- . ·For plot (c), this is possible only if R was {}2e s ℜ>.·For plot (d),R is the entire s-plane.9.24.(a)The pole-zero diagram with the appropriate markings is shown Figure S9.24.(b)By inspecting the pole-zero diagram of part (a), it is clear that the pole-zero diagram shown in Figure S9.24 will also result in the same ()X jw .This would correspond to the Laplace transform()112X s s =-, {}12e s ℜ<.(c)≮()X jw π=-≮()1X jw .(d)()2X s with the pole-zero diagram shown below in Figure S9.24 would have the property that ≮()2X jw =≮()X jw .Here ,()211/2X s s -=-. (e) ()()21/X jw X jw =.(f)From the result of part (b),it is clear that ()1X s may be obtained by reflecting the poles and zeros in the right-half of the s-plane to the left-half of the s-plane .Therefore, ()11/22s X s s +=+.From part (d),it is clear that ()2X s may be obtained by reflecting the poles (zeros) in the right-half of the s-plane to the left-half and simultaneously changing them to zeros (poles).Therefore,()()()()2211/22s X s s s +=++9.25.The plots are as shown in Figure S9.25. 9.26.From Table 9.2 we have()()(){}2111,22LT t x t e u t X s e s s -=←−→=ℜ>-+and()()(){}3111,33LTt x t e u t X s e s s -=←−→=ℜ>-+.Using the time-shifting time-scaling properties from Table 9.1,we obtain()(){}22112,22s LT s e x t e X s e s s ---←−→=ℜ>-+and()(){}33223,33s LT s e x t e X s e s s---+←−→-=ℜ>--.Therefore, using the convolution property we obtain ()()()()23122323s s LTe e y t x t x t Y s s s --⎡⎤⎡⎤=-*-+←−→=⎢⎥⎢⎥+-⎣⎦⎣⎦. 9.27.From clues 1 and 2,we know that ()X s is of the form()()()AX s s a s b =++. Furthermore , we are given that one of the poles of ()X s is 1j -+.Since ()x t is real, the poles of ()X s must occur in conjugate reciprocal pairs .Therefore, 1a j =-and 1b j =+and ()()()11AH s s j s j =+-++. From clue 5,we know that ()08X =.Therefore, we may deduce that 16A = and ()21622H s s s =++ .Let R denote the ROC of ()X s .From the pole locations we know that there are two possible choices of R .R may either be {}1e s ℜ<-or {}1e s ℜ>-.We will now useclue 4 to pick one .Note that()()()()22LTt y t e x t Y s X s =←−→=-.The ROC of ()Y s is R shifted by 2 to the right .Since it is given that ()y t is not absolutely integrable ,the ROC of ()Y s should not include the jw axis -.This is possible only ofR is {}1e s ℜ>-.9.28.(a) The possible ROCs are(i) {}2e s ℜ<-.(ii) {}21e s -<ℜ<-. (iii) {}11e s -<ℜ<.( iv) {}1e s ℜ>.(b)(i)Unstable and anticausal. (ii) Unstable and non causal. （iii ）Stable and non causal. (iv) Unstable and causal. 9.29.(a)Using Table 9.2,we obtain (){}1,11X s e s s =ℜ>-+and(){}1, 2.2H s e s s =ℜ>-+(b) Since ()()()y t x t h t =*,we may use the convolution property to obtain()()()()()112Y s X s H s s s ==++.The ROC of ()Y s is {}1e s ℜ>-.(c) Performing partial fraction expansion on ()Y s ,we obtain . ()1112Y s s s =-++.Taking the inverse Laplace transform, we get()()()2t t y t e u t e u t --=-. (d)Explicit convolution of ()x t and ()h t gives us()()()y t h x t d τττ∞-∞=-⎰()()20t e e u t d ττττ∞---=-⎰t t e e d ττ--=⎰ for0t >()2.t t e e u t --⎡⎤=-⎣⎦ 9.30.For the input ()()x t u t =, the Laplace transform is (){}1,0.X s e s s=ℜ>The corresponding output ()()1t t y t e te u t --⎡⎤=--⎣⎦ has the Laplace transform()()(){}221111,0111Y s e s s s s s s =--=ℜ>+++. Therefore,()()()(){}21,0.1Y s H s e s X s s ==ℜ>+ Now ,the output ()()3123t t y t e e u t --⎡⎤=-+⎣⎦has the Laplace transform()()(){}12316,0.1313Y s e s s s s s s s =-+=ℜ>++++ Therefore , the Laplace transform of the corresponding input will be()()()()(){}1161,0.3Y s s X s e s H s s s +==ℜ>+ Taking the inverse Laplace transform of the partial fraction expansion of ()1,X s we obtain ()()()3124.t x t u t e u t -=+9.31.(a).Taking the Laplace transform of both sides of the given differential equation and simplifying, weobtain()()()212Y s H s X s s s ==--.b).The partial fraction expansion of ()H s is()1/31/321H s s s =--+. (i).If the system is stable ,the ROC for ()H s has to be {}12e s -<ℜ< . Therefore ()()()21133t t h t e u t e u t -=---.(ii).If the system is causal, the ROC for ()H s has to be {}2e s ℜ> .Therefore()()()21133t t h t e u t e u t -=-.(iii)If the system is neither stable nor causal ,the ROC for ()H s has to be {}1e s ℜ<-.Therefore ,()()()21133t t h t e u t e u t -=--+-9.32. If ()2t x t e =produces ()()21/6t y t e =,then ()()21/6H =. Also, by taking the Laplace transform of both sidesof the given differential equation we get ()()()()442s b s H s s s s ++=++.Since ()21/6H = ,we may deduce that 1b = .Therefore()()()()()222424s H s s s s s s +==+++. 9.33.Since ()()()t t t x t e e u t e u t --==+-,()()(){}112,111111X s e s s s s s -=-=-<ℜ<+-+-. We are also given that ()2122s H s s s +=++.Since the poles of ()H s are at 1j -±, and since ()h t is causal ,we may conclude that the ROC of()H s is {}1e s ℜ>-.Now()()()()()22221Y s H s X s s s s -==++-. The ROC of ()Y s will be the intersection of the ROCs of ()X s and ()H s .This is {}11e s -<ℜ<. We may obtain the following partial fraction expansion for ()Y s :()22/52/56/5122s Y s s s s +=-+-++. We may rewrite this as ()()()222/521411551111s Y s s s s ⎡⎤⎡⎤+=-++⎢⎥⎢⎥-++++⎢⎥⎢⎥⎣⎦⎣⎦.Nothing that the ROC of ()Y s is {}11e s -<ℜ<and using Table9.2,we obtain ()()()()224cos sin 555t t t y t e u t e tu t e tu t --=-++9.34.We know that()()(){}111,0LTx t u t X s e s s=←−→=ℜ> Therefore,()1X s has a pole at0s =.Now ,the Laplace transform of the output()1y t of the system with()1x t as the input is()()()11Y s H s X s =Since in clue 2, ()1Y s is given to be absolutely integrable ,()H s must have a zero at 0s =whichcancels out the pole of ()1X s at 0s =.We also know that()()(){}2221,0LT x t tu t X s e s s=←−→=ℜ> Therefore , ()2x s has two poles at 0s =.Now ,the Laplace transform of the output ()2y t of the system with ()2x t as the input is()()()22Y s H s X s =Since in clue 3, ()2Y s is given to be not absolutely integrable ,()H s does not have two zeros at0s =.Therefore ,we conclude that ()H s has exactly one zero at 0s =. From clue 4 we know that the signal ()()()()2222d h t dh t p t h t dt dt=++is finite duration .Taking the Laplace transform of both sides of the above equation ,we get ()()()()222P s s H s sH s H s =++. Therefore,()()222P s H s s s =++.Since ()p t is of finite duration, we know that ()P s will have no poles in the finite s-plane .Therefore, ()H s is of the form()()1222Ni i A s z H s s s =-=++∏,where i z ,1,2,....,i N =represent the zeros of ()P s .Here ,A is some constant.From clue 5 we know that the denominator polynomial of ()H s has to have a degree which is exactly one greater than the degree of the numerator polynomial .Therefore, ()()1222A s s H s s s -=++.Since we already know that ()H s has a zero at 0s = ,we may rewrite this as ()222As H s s s =++ From clue 1 we know that ()1H is0.2.From this ,we may easily show that 1A = .Therefore,()222s H s s s =++. Since the poles of ()H s are at 1j -± and since ()h t is causal and stable ,the ROC of ()H s is {}1e s ℜ>-.9.35.(a) We may redraw the given block diagram as shown in Figure S9.35. From the figure ,it is clear that()()1F s Y s s=. Therefore, ()()1/f t dy t dt =. Similarly, ()()/e t df t dt =.Therefore, ()()221/e t d y t dt =.From the block diagram it is clear that()()()()()()()21111266d y t dy t y t e t f t y t y t dtdt=--=--.。

(含答案)九年级英语人教全一册课后作业Unit 9 I like music that I can

答卷时应注意事项１、拿到试卷，要认真仔细的先填好自己的考生信息。

２、拿到试卷不要提笔就写，先大致的浏览一遍，有多少大题，每个大题里有几个小题，有什么题型，哪些容易，哪些难，做到心里有底；３、审题，每个题目都要多读几遍，不仅要读大题，还要读小题，不放过每一个字，遇到暂时弄不懂题意的题目，手指点读，多读几遍题目，就能理解题意了；容易混乱的地方也应该多读几遍，比如从小到大，从左到右这样的题；４、每个题目做完了以后，把自己的手从试卷上完全移开，好好的看看有没有被自己的手臂挡住而遗漏的题；试卷第１页和第２页上下衔接的地方一定要注意，仔细看看有没有遗漏的小题；５、中途遇到真的解决不了的难题，注意安排好时间，先把后面会做的做完，再来重新读题，结合平时课堂上所学的知识，解答难题；一定要镇定，不能因此慌了手脚，影响下面的答题；６、卷面要清洁，字迹要清工整，非常重要；７、做完的试卷要检查，这样可以发现刚才可能留下的错误或是可以检查是否有漏题，检查的时候，用手指点读题目，不要管自己的答案，重新分析题意，所有计算题重新计算，判断题重新判断，填空题重新填空，之后把检查的结果与先前做的结果进行对比分析。

亲爱的小朋友，你们好!经过两个月的学习，你们一定有不小的收获吧，用你的自信和智慧，认真答题，相信你一定会闯关成功。

相信你是最棒的！Unit9I like music that I can dance toSection AI.根据句意和汉语提示写出所缺的单词1.The play begins with a short___________(对话)between father and son.2.They want to know more about___________(澳大利亚的)history.3.In my mind,Feng Xiaogang is an excellent___________(导演).4.Sometimes we___________(闭上)off our eyes to the facts.refusing to face the truth.5.This picture wasn't true because it was___________(粘贴)from the Internet.6.We love peace.but we are not afraid of___________(战争).7.I can’t___________(抽出)the time for a holiday at the moment.8.The trousers are very expensive.The cloth feels___________(平滑的).9.When I am feeling_________(沮丧)，he will always listen to me and try to help me.10.This dictionary is available in___________(电子的)form.Ⅱ.根据句意和首字母提示写出所缺的单词11.The Internet is very useful and it can offer p___________of information to us.12.He used to like treating friends to dinner in restaurants,but now he p___________to invite them to his home.13.You can't make a noise in class.in that c___________you'll be punished.14.I s___________we'll go to the park to have a picnic next week.15.We went to see him o___________in a while when he was in Beijing.Ⅲ.从方框中选出合适的短语，并用其正确形式填空16.I don’t___________walking very much today.17.The crops grow well this year because there is___________rain.18.He was so tired that he___________his brain to have a rest.19.I want to get home___________to watch the football match.20.-Are you going climbing tomorrow,Jack?-I’m not sure.It___________the weat her.Ⅳ.单项选择()21.-What makes you___________that I'm against the rule?-Look!You're late again.A.supposeB.imagineC.predictD.accept()22.Please remember to_________the electricity and water before you leave t he laboratory.A.take offB.shut offC.go offD.put off()23.You can’t come here soon.___________,we decide not to wait for you an y longer.A.HoweverB.In that caseC.So farD.After that()24.-Which invention do you like best?-Wechat.It is an invention__________can help us communicate with others online fre ely.A.whoB.thatC.what()25.Some people prefer tea__________milk.However,I like drinking tea wit hout anything in it.A.toB.withC.ofV.根据汉语意思完成句子。

Chapter9反译法课后练习答案

Chapter9反译法课后练习答案Chapter 9 反译法Drills 9.1.14. 我认为，安理会不能也不应对这些极其严重的侵犯人权的行为漠不关心（袖手旁观）。

5．我们认为，他今天上午发言的要旨和口气都无助于在中东建立持久和平。

6．我觉得再也不能忍受下去了。

7．我看你可不必担心。

8．我想不是这样。

（我想并非如此）Drills 9.1.26.那天他碰巧不在那里。

7.她不是乘公共汽车来上学的。

8.我来这里不是要逗弄你。

9.我们战斗了这么多年，不是为了在最后向敌人的走狗屈服。

10.她把信留在这里，并不是有意想让你看。

11.人民摆脱了殖民主义者的奴役，绝不是为了套上霸权主义者的枷锁。

12.你如果拒绝应邀参加会议，那就不礼貌了。

13.我们要到那里去，并不是代表我国政府去进行谈判。

14.不要因为我使唤了她就认为我不好。

15.他这样做，并不是因为没有校对员，而是他不愿在报纸上无意中出现任何差错。

16.我离开了家，但并不是因为我怕我父亲。

/我不是怕我父亲才离开家的。

/ cf. 我没有离开家，因为我怕我父亲。

Drills 9.1.36.她没有回答我。

7.禁止招贴。

8.滚石不生苔，专业不聚财。

/ 见异思迁，一事无成。

9.在任何情况下，我们都不会放弃原则。

10.她发现她确实不能指望谁了。

11.当然，林登·约翰逊在这方面是别人望尘莫及的。

12.我走开了，不去参观名胜了。

Drills 9.2a1.焦急不安2.不出所料3.不及格、不履行、没做到4.无上权威、至高权威5.不失体面、不失尊严6.一大片不毛之地7.困惑不解、茫然不知所措8.心不在焉9.不负所望10.安然无恙11.一点也不差12.得不治之症13.一股不可小看的力量、一股力量，决不能掉以轻心14.不迟、不晚15.油漆未干16.无期徒刑17.放心不下18.不为过、没有扩大化、没有超出范围、没有越过雷池一步Drills 9.2b12.不要客气。

13.她临处死刑而不屈。

Chapter 9参考译文及答案

Chapter 9Learning Aims学完本章，学生应能：1．解释客户投诉索赔的一般原因；2．学会如何进行投诉和索赔；3．学会如何对投诉和索赔作出答复；4．掌握投诉理赔信件的撰写原则、结构等有关知识；5．熟悉一些化工产品的名称。

Background Information在国际贸易中，一家公司向另一家公司提出投诉或索赔是很常见的。

在执行合同的过程中，买卖双方都应遵守合同条款并严格履行各自的义务。

任何一方如果不能严格履约，就会给另一方带来麻烦，有时还会使另一方遭受巨大损失。

在这种情况下，受损失的一方有权根据合同规定要求责任方保证不再发生此类事件，这种行为叫“投诉”，或要求责任方赔偿损失，这种行为叫“索赔”。

相关的信件就是投诉信或索赔信。

通常有两种情况促使买方提出投诉或索赔：一是由于货物错发、质量不良、货物损坏、短重或迟发而引起的真正的投诉或索赔；二则是因为买方不想收到货物或找到了更低价的货源，所以吹毛求疵以此为借口解除合同。

撰写投诉索赔信件需要注意以下几点原则：①第一时间迅速、清楚告诉对方问题所在；②在信件中注明发票号、产品信息等情况，叙述问题清晰有条理；③提供足够的依据；④明确说明投诉或索赔想要得到的具体要求，如果未得到满意回应则礼貌地加强措辞；⑤结尾语气友好但必须坚决。

投诉信往往包含以下要点：①信的开头应对提出的抱怨表示遗憾；②详尽陈述事实，说明出现了什么差错；③说明不满的原因，并要求予以解释；④提及由此而造成的不便；⑤提出如何纠正的建议。

每一项投诉不管看起来多么微不足道，对写信方来讲都是很重要的。

所以收到这类信后，应立即作答。

答复信应说明事实，礼貌而公平，并遵循以下要点：①对来信表示感谢。

②对给对方造成的不便表示歉意。

③说明你将采取什么行动。

④期待进一步合作。

写索赔信应遵循以下要点：①告知收信人有关索赔货物的详细情况。

②清楚地表达你的索赔要求。

③要求对方采取什么行动：换货，维修，退款或赔偿。

chapter9投资作业-chapter15习题

习题Chapter 93. You are a consultant to a large manufacturing corporation that is considering a project with the following net after-tax cash flows (in millions of dollars):The project’s beta is 1.8. Assuming that, what is the net present value of the project? What is the highest possible beta estimate for the project before its NPV becomes negative?4. Are the following true or false? Explain.a. Stocks with a beta of zero offer an expected rate of return of zero.b. The CAPM implies that investors require a higher return to hold highly volatile securities.c. You can construct a portfolio with beta of .75 by investing .75 of the investment budget in T-bills and the remainder in the market portfolio.In problems 13 to 15 assume that the risk-free rate of interest is 6% and the expected rate of return on the market is 16%.13. Ashare of stock sells for $50 today. It will pay a dividend of $6 per share at the end of the year. Its beta is 1.2. What do investors expect the stock to sell for at the end of the year?14. I am buying a firm with an expected perpetual cash flow of $1,000 but am unsure of its risk. If I think the beta of the firm is .5, when in fact the beta is really 1, how much more will I offer for the firm than it is truly worth?15. A stock has an expected rate of return of 4%. What is its beta?17. Suppose the rate of return on short-term government securities (perceived to be riskfree) is about 5%. Suppose also that the expected rate of return required by the market for a portfolio with a beta of 1 is 12%. According to the capital asset pricing model (security market line):a. What is the expected rate of return on the market portfolio?b. What would be the expected rate of return on a stock with beta＝0?c. Suppose you consider buying a share of stock at $40. The stock is expected to pay $3 dividends next year and you expect it to sell then for $41. The stock risk has been evaluated at beta＝－0.5. Is the stock overpriced or underpriced?21. The security market line depicts:a. A security’s expected return as a function of its systematic risk.b. The market portfolio as the optimal portfolio of risky securities.c. The relationship between a security’s return and the return on an index.d. The complete portfolio as a combination of the market portfolio and the risk-free asset.22. Within the context of the capital asset pricing model (CAPM), assume:• Expected return on the market ＝15%.• Risk-free rate _ 8%.• Expected rate of return on XYZ security ＝17%.• Beta of XYZ security ＝1.25.Which one of the following is correct?a. XYZ is overpriced.b. XYZ is fairly priced.c. XYZ’s alpha is＝-0.25%.d. XYZ’s alpha is=0 .25%.The following table shows risk and return measures for two portfolios. answer question 26 and 2726. When plotting portfolio R on the preceding table relative to the SML, portfolio R lies:a. On the SML.b. Below the SML.c. Above the SML.d. Insufficient data given.27. When plotting portfolio R relative to the capital market line, portfolio R lies:a. On the CML.b. Below the CML.c. Above the CML.d. Insufficient data given.31. Karen Kay, a portfolio manager at Collins Asset Management, is using the capital asset pricing model for making recommendations to her clients. Her research department has developed the information shown in the following exhibit.a. Calculate expected return and alpha for each stock.b. Identify and justify which stock would be more appropriate for an investor who wants toi. add this stock to a well-diversified equity portfolio.ii. hold this stock as a single-stock portfolio.Chapter 111. A portfolio management organization analyzes 60 stocks and constructs a meanvariance efficient portfolio using only these 60 securities.a. How many estimates of expected returns, variances, and covariances are needed to optimize this portfolio?b. If one could safely assume that stock market returns closely resemble a single index structure, how many estimates would be needed?2. The following are estimates for two of the stocks in problem 1.The market index has a standard deviation of 22% and the risk-free rate is 8%.a. What is the standard deviation of stocks A and B?b. Suppose that we were to construct a portfolio with proportions:Stock A: .30Stock B: .45T-bills: .25Compute the expected return, standard deviation, beta, and nonsystematic standard deviation of the portfolio.4. Consider the two (excess return) index model regression results for A and B:a. Which stock has more firm-specific risk?b. Which has greater market risk?c. For which stock does market movement explain a greater fraction of return variability?d. Which stock had an average return in excess of that predicted by the CAPM?e. If rf were constant at 6% and the regression had been run using total rather than excess returns, what would have been the regression intercept for stock A ?Use the following data for problems 5 through 9. Suppose that the index model for stocks A and B is estimated from excess returns with the following results:5. What is the standard deviation of each stock?6. Break down the variance of each stock to the systematic and firm-specific components.7. What are the covariance and correlation coefficient between the two stocks?8. What is the covariance between each stock and the market index?9. Are the intercepts of the two regressions consistent with the CAPM? Interpret their values.15. Based on current dividend yields and expected growth rates, the expected rates of return on stocks A and B are 11% and 14%, respectively. The beta of stock A is .8, while that of stock B is 1.5. The T-bill rate is currently 6%, while the expected rate of return on the S&P 500 index is 12%. The standard deviation of stock A is 10% annually, while that of stock B is 11%.a. If you currently hold a well-diversified portfolio, would you choose to add either of these stocks to your holdings?b. If instead you could invest only in bills and one of these stocks, which stock would you choose? Explain your answer using either a graph or a quantitative measure of the attractiveness of the stocks.16. Assume the correlation coefficient between Baker Fund and the S&P 500 Stock Index is .70. What percentage of B aker Fund’s total risk is specific (i.e., nonsystematic)?a. 35%b. 49%c. 51%d. 70%17. The correlation between the Charlottesville International Fund and the EAFE Market Index is 1.0. The expected return on the EAFE Index is 11%, the expected return on Charlottesville International Fund is 9%, and the risk-free return in EAFE countries is 3%. Based on this analysis, the implied beta of Charlottesville International is:a. Negativeb. .75c. .82d. 1.00chapter 122. Which of the following most appears to contradict the proposition that the stock market is weakly efficient? Explain.a. Over 25% of mutual funds outperform the market on average.b. Insiders earn abnormal trading profits.c. Every January, the stock market earns abnormal returns.3. Suppose that, after conducting an analysis of past stock prices, you come up with the following observations. Which would appear to contradict the weak form of the efficient market hypothesis? Explain.a. The average rate of return is significantly greater than zero.b. The correlation between the return during a given week and the return during the following week is zero.c. One could have made superior returns by buying stock after a 10% rise in price and selling after a 10% fall.d. One could have made higher-than-average capital gains by holding stocks with low dividend yields.4. Which of the following statements are true if the efficient market hypothesis holds?a. It implies that future events can be forecast with perfect accuracy.b. It implies that prices reflect all available information.c. It implies that security prices change for no discernible reason.d. It implies that prices do not fluctuate.5. Which of the following observations would provide evidence against the semistrong form of the efficient market theory? Explain.a. Mutual fund managers do not on average make superior returns.b. You cannot make superior profits by buying (or selling) stocks after the announcement of an abnormal rise in dividends.c. Low P/E stocks tend to have positive abnormal returns.d. In any year approximately 50% of pension funds outperform the market.6. The semistrong form of the efficient market hypothesis asserts that stock prices:a. Fully reflect all historical price information.b. Fully reflect all publicly available information.c. Fully reflect all relevant information including insider information.d. May be predictable.7. Assume that a company announces an unexpectedly large cash dividend to its shareholders.In an efficient market without information leakage, one might expect:a. An abnormal price change at the announcement.b. An abnormal price increase before the announcement.c. An abnormal price decrease after the announcement.d. No abnormal price change before or after the announcement.8. Which one of the following would provide evidence against the semistrong form of the efficient market theory?a. About 50% of pension funds outperform the market in any year.b. All investors have learned to exploit signals about future performance.c. Trend analysis is worthless in determining stock prices.d. Low P/E stocks tend to have positive abnormal returns over the long run.Chapter 141. Which security has a higher effective annual interest rate?a. A 3-month T-bill selling at $97,645 with par value $100,000.b. A coupon bond selling at par and paying a 10% coupon semiannually.2. Treasury bonds paying an 8% coupon rate with semiannual payments currently sell at par value. What coupon rate would they have to pay in order to sell at par if they paid their coupons annually? (Hint: what is the effective annual yield on the bond?)3. Two bonds have identical times to maturity and coupon rates. One is callable at 105, the other at 110. Which should have the higher yield to maturity? Why?4. Consider a bond with a 10% coupon and with yield to maturity _ 8%. If the bond’s yield to maturity remains constant, then in 1 year, will the bond price be higher, lower, or unchanged? Why?5. Consider an 8% coupon bond selling for $953.10 with 3 years until maturity making annual coupon payments. The interest rates in the next 3 years will be, withcertainty, . Calculate the yield to maturity and realized compound yield of the bond.6. Philip Morris may issue a 10-year maturity fixed-income security, which might include a sinking fund provision and either refunding or call protection.a. Describe a sinking fund provision.b. Explain the impact of a sinking-fund provision on:i. The expected average life of the proposed security.ii. Total principal and interest payments over the life of the proposed security.c. From the investor’s point of view, explain the rationale for demanding a sinking fund provision.7. Bonds of Zello Corporation with a par value of $1,000 sell for $960, mature in 5 years, and have a 7% annual coupon rate paid semiannually.a. Calculate the:i. Current yield.ii. Yield to maturity (to the nearest whole percent, i.e., 3%, 4%, 5%, etc.).iii. Realized compound yield for an investor with a 3-year holding period and a reinvestment rate of 6% over the period. At the end of 3 years the 7% coupon bonds with 2 years remaining will sell to yield 7%.b. Cite one major shortcoming for each of the following fixed-income yield measures:i. Current yield.ii. Yield to maturity.iii. Realized compound yield.9. A 20-year maturity bond with par value of $1,000 makes semiannual coupon payments at a coupon rate of 8%. Find the bond equivalent and effective annual yield to maturity of the bond if the bond price is:a. $950.b. $1,000.c. $1,050.12. Consider a bond paying a coupon rate of 10% per year semiannually when the market interest rate is only 4% per half year. The bond has 3 years until maturity.a. Find the bond’s price today and 6 months from now after the next coupon is paid.b. What is the total (6 month) rate of return on the bond?14. A bond with a coupon rate of 7% makes semiannual coupon payments on January15 and July 15 of each year. The Wall Street Journal reports the asked price for the bond on January 30 at 100:02. What is the invoice price of the bond? The coupon period has 182 days.20. A 30-year maturity, 8% coupon bond paying coupons semiannually is callable in 5 years at a call price of $1,100. The bond currently sells at a yield to maturity of 7% (3.5% per half-year).a. What is the yield to call?b. What is the yield to call if the call price is only $1,050?c. What is the yield to call if the call price is $1,100, but the bond can be called in 2 years instead of 5 years?22. A 2-year bond with par value $1,000 making annual coupon payments of $100 is priced at $1,000. What is the yield to maturity of the bond? What will be the realized compound yield to maturity if the 1-year interest rate next year turns out to be (a) 8%,(b) 10%26. Alarge corporation issued both fixed and floating-rate notes 5 years ago, with terms given in the following table:a. Why is the price range greater for the 9% coupon bond than the floatingrate note?b. What factors could explain why the floating-rate note is not always sold at par value?c. Why is the call price for the floating-rate note not of great importance to investors?d. Is the probability of call for the fixed-rate note high or low?e. If the firm were to issue a fixed-rate note with a 15-year maturity, what coupon rate would it need to offer to issue the bond at par value?f. Why is an entry for yield to maturity for the floating-rate note not appropriate?27. On May 30, 1999, Janice Kerr is considering one of the newly issued 10-year AAA corporate bonds shown in the following exhibit.a. Suppose that market interest rates decline by 100 basis points (i.e., 1%). Contrast the effect of this decline on the price of each bond.b. Should Kerr prefer the Colina over the Sentinal bond when rates are expected to rise or to fall?c. What would be the effect, if any, of an increase in the volatility of interest rates on the prices of each bond?Chapter 152. Which one of the following statements about the term structure of interest rates is true?a. The expectations hypothesis indicates a flat yield curve if anticipated futureshort-term rates exceed current short-term rates.b. The expectations hypothesis contends that the long-term rate is equal to the anticipated short-term rate.c. The liquidity premium theory indicates that, all else being equal, longer maturities will have lower yields.d. The liquidity preference theory contends that lenders prefer to buy securities at the short end of the yield curve.8. Suppose the following table shows yields to maturity of zero coupon U.S. Treasury securities as of January 1, 1996:a. Based on the data in the table, calculate the implied forward 1-year rate of interest at January 1, 1999.b. Describe the conditions under which the calculated forward rate would be an unbiased estimate of the 1-year spot rate of interest at January 1, 1999.c. Assume that 1 year earlier, at January 1, 1995, the prevailing term structure for U.S. Treasury securities was such that the implied forward 1-year rate of interest at January 1, 1999, was significantly higher than the corresponding rate implied by the term structure at January 1, 1996. On the basis of the pure expectations theory of the term structure, briefly discuss two factors that could account for such a decline in the implied forward rate.16. Below is a list of prices for zero-coupon bonds of various maturities.a. An 8.5% coupon $1,000 par bond pays an annual coupon and will mature in 3years. What should the yield to maturity on the bond be?b. If at the end of the first year the yield curve flattens out at 8%, what will be the1-year holding-period return on the coupon bond?21. The yield to maturity (YTM) on 1-year zero-coupon bonds is 5% and the YTM on 2-year zeros is 6%. The yield to maturity on 2-year-maturity coupon bonds with coupon rates of 12% (paid annually) is 5.8%. What arbitrage opportunity is available for an investment banking firm? What is the profit on the activity?22. Suppose that a 1-year zero-coupon bond with face value $100 currently sells at $94.34, while a 2-year zero sells at $84.99. You are considering the purchase of a2-year-maturity bond making annual coupon payments. The face value of the bond is $100, and the coupon rate is 12% per year.a. What is the yield to maturity of the 2-year zero? The 2-year coupon bond?b. What is the forward rate for the second year?c. If the expectations hypothesis is accepted, what are (1) the expected price of the coupon bond at the end of the first year and (2) the expected holding-period return on the coupon bond over the first year?d. Will the expected rate of return be higher or lower if you accept the liquidity preference hypothesis?。

操作系统第九版部分课后作业习题答案解析

CHAPTER 9 Virtual Memory Practice Exercises9.1 Under what circumstances do page faults occur? Describe the actions taken by the operating system when a page fault occurs.Answer:A page fault occurs when an access to a page that has not beenbrought into main memory takes place. The operating system veriﬁesthe memory access, aborting the program if it is invalid. If it is valid, a free frame is located and I/O is requested to read the needed page into the free frame. Upon completion of I/O, the process table and page table are updated and the instruction is restarted.9.2 Assume that you have a page-reference string for a process with m frames (initially all empty). The page-reference string has length p;n distinct page numbers occur in it. Answer these questions for any page-replacement algorithms:a. What is a lower bound on the number of page faults?b. What is an upper bound on the number of page faults?Answer:a. nb. p9.3 Consider the page table shown in Figure 9.30 for a system with 12-bit virtual and physical addresses and with 256-byte pages. The list of freepage frames is D, E, F (that is, D is at the head of the list, E is second, and F is last).Convert the following virtual addresses to their equivalent physical addresses in hexadecimal. All numbers are given in hexadecimal. (A dash for a page frame indicates that the page is not in memory.)• 9EF• 1112930 Chapter 9 Virtual Memory• 700• 0FFAnswer:• 9E F - 0E F• 111 - 211• 700 - D00• 0F F - EFF9.4 Consider the following page-replacement algorithms. Rank these algorithms on a ﬁve-point scale from “bad” to “perfect” according to their page-fault rate. Separate those algorithms that suffer from Belady’s anomaly from those that do not.a. LRU replacementb. FIFO replacementc. Optimal replacementd. Second-chance replacementAnswer:Rank Algorithm Suffer from Belady’s anomaly1 Optimal no2 LRU no3 Second-chance yes4 FIFO yes9.5 Discuss the hardware support required to support demand paging. Answer:For every memory-access operation, the page table needs to be consulted to check whether the corresponding page is resident or not and whether the program has read or write privileges for accessing the page. These checks have to be performed in hardware. A TLB could serve as a cache and improve the performance of the lookup operation.9.6 An operating system supports a paged virtual memory, using a central processor with a cycle time of 1 microsecond. It costs an additional 1 microsecond to access a page other than the current one. Pages have 1000 words, and the paging device is a drum that rotates at 3000 revolutions per minute and transfers 1 million words per second. The following statistical measurements were obtained from the system:• 1 percent of all instructions executed accessed a page other than the current page.•Of the instructions that accessed another page, 80 percent accesseda page already in memory.Practice Exercises 31•When a new page was required, the replaced page was modiﬁed 50 percent of the time.Calculate the effective instruction time on this system, assuming that the system is running one process only and that the processor is idle during drum transfers.Answer:effective access time = 0.99 × (1 sec + 0.008 × (2 sec)+ 0.002 × (10,000 sec + 1,000 sec)+ 0.001 × (10,000 sec + 1,000 sec)= (0.99 + 0.016 + 22.0 + 11.0) sec= 34.0 sec9.7 Consider the two-dimensional array A:int A[][] = new int[100][100];where A[0][0] is at location 200 in a paged memory system with pages of size 200. A small process that manipulates the matrix resides in page 0 (locations 0 to 199). Thus, every instruction fetch will be from page 0. For three page frames, how many page faults are generated bythe following array-initialization loops, using LRU replacement andassuming that page frame 1 contains the process and the other twoare initially empty?a. for (int j = 0; j < 100; j++)for (int i = 0; i < 100; i++)A[i][j] = 0;b. for (int i = 0; i < 100; i++)for (int j = 0; j < 100; j++)A[i][j] = 0;Answer:a. 5,000b. 509.8 Consider the following page reference string:1, 2, 3, 4, 2, 1, 5, 6, 2, 1, 2, 3, 7, 6, 3, 2, 1, 2, 3, 6.How many page faults would occur for the following replacement algorithms, assuming one, two, three, four, ﬁve, six, or seven frames? Remember all frames are initially empty, so your ﬁrst unique pages will all cost one fault each.•LRU replacement• FIFO replacement•Optimal replacement32 Chapter 9 Virtual MemoryAnswer:Number of frames LRU FIFO Optimal1 20 20 202 18 18 153 15 16 114 10 14 85 8 10 76 7 10 77 77 79.9 Suppose that you want to use a paging algorithm that requires a referencebit (such as second-chance replacement or working-set model), butthe hardware does not provide one. Sketch how you could simulate a reference bit even if one were not provided by the hardware, or explain why it is not possible to do so. If it is possible, calculate what the cost would be.Answer:You can use the valid/invalid bit supported in hardware to simulate the reference bit. Initially set the bit to invalid. O n ﬁrst reference a trap to the operating system is generated. The operating system will set a software bit to 1 and reset the valid/invalid bit to valid.9.10 You have devised a new page-replacement algorithm that you thinkmaybe optimal. In some contorte d test cases, Belady’s anomaly occurs. Is the new algorithm optimal? Explain your answer.Answer:No. An optimal algorithm will not suffer from Belady’s anomaly because —by deﬁnition—an optimal algorithm replaces the page that will notbe used for the long est time. Belady’s anomaly occurs when a pagereplacement algorithm evicts a page that will be needed in the immediatefuture. An optimal algorithm would not have selected such a page.9.11 Segmentation is similar to paging but uses variable-sized“pages.”Deﬁnetwo segment-replacement algorithms based on FIFO and LRU pagereplacement schemes. Remember that since segments are not the samesize, the segment that is chosen to be replaced may not be big enoughto leave enough consecutive locations for the needed segment. Consider strategies for systems where segments cannot be relocated, and thosefor systems where they can.Answer:a. FIFO. Find the ﬁrst segment large enough to accommodate the incoming segment. If relocation is not possible and no one segmentis large enough, select a combination of segments whose memoriesare contiguous, which are “closest to the ﬁrst of the list” andwhich can accommodate the new segment. If relocation is possible, rearrange the memory so that the ﬁrstNsegments large enough forthe incoming segment are contiguous in memory. Add any leftover space to the free-space list in both cases.Practice Exercises 33b. LRU. Select the segment that has not been used for the longestperiod of time and that is large enough, adding any leftover spaceto the free space list. If no one segment is large enough, selecta combination of the “oldest” segments that are contiguous inmemory (if relocation is not available) and that are large enough.If relocation is available, rearrange the oldest N segments to be contiguous in memory and replace those with the new segment.9.12 Consider a demand-paged computer system where the degree of multiprogramming is currently ﬁxed at four. The system was recently measured to determine utilization of CPU and the paging disk. The results are one of the following alternatives. For each case, what is happening? Can the degree of multiprogramming be increased to increase the CPU utilization? Is the paging helping?a. CPU utilization 13 percent; disk utilization 97 percentb. CPU utilization 87 percent; disk utilization 3 percentc. CPU utilization 13 percent; disk utilization 3 percentAnswer:a. Thrashing is occurring.b. CPU utilization is sufﬁciently high to leave things alone, and increase degree of multiprogramming.c. Increase the degree of multiprogramming.9.13 We have an operating system for a machine that uses base and limit registers, but we have modiﬁed the ma chine to provide a page table.Can the page tables be set up to simulate base and limit registers? How can they be, or why can they not be?Answer:The page table can be set up to simulate base and limit registers provided that the memory is allocated in ﬁxed-size segments. In this way, the base of a segment can be entered into the page table and the valid/invalid bit used to indicate that portion of the segment as resident in the memory. There will be some problem with internal fragmentation.9.27.Consider a demand-paging system with the following time-measured utilizations:CPU utilization 20%Paging disk 97.7%Other I/O devices 5%Which (if any) of the following will (probably) improve CPU utilization? Explain your answer.a. Install a faster CPU.b. Install a bigger paging disk.c. Increase the degree of multiprogramming.d. Decrease the degree of multiprogramming.e. Install more main memory.f. Install a faster hard disk or multiple controllers with multiple hard disks.g. Add prepaging to the page fetch algorithms.h. Increase the page size.Answer: The system obviously is spending most of its time paging, indicating over-allocationof memory. If the level of multiprogramming is reduced resident processeswould page fault less frequently and the CPU utilization would improve. Another way toimprove performance would be to get more physical memory or a faster paging drum.a. Get a faster CPU—No.b. Get a bigger paging drum—No.c. Increase the degree of multiprogramming—No.d. Decrease the degree of multiprogramming—Yes.e. Install more main memory—Likely to improve CPU utilization as more pages canremain resident and not require paging to or from the disks.f. Install a faster hard disk, or multiple controllers with multiple hard disks—Also animprovement, for as the disk bottleneck is removed by faster response and morethroughput to the disks, the CPU will get more data more quickly.g. Add prepaging to the page fetch algorithms—Again, the CPU will get more datafaster, so it will be more in use. This is only the case if the paging action is amenableto prefetching (i.e., some of the access is sequential).h. Increase the page size—Increasing the page size will result in fewer page faults ifdata is being accessed sequentially. If data access is more or less random, morepaging action could ensue because fewer pages can be kept in memory and moredata is transferred per page fault. So this change is as likely to decrease utilizationas it is to increase it.10.1、Is disk scheduling, other than FCFS scheduling, useful in asingle-userenvironment? Explain your answer.Answer: In a single-user environment, the I/O queue usually is empty. Requests generally arrive from a single process for one block or for a sequence of consecutive blocks. In these cases, FCFS is an economical method of disk scheduling. But LOOK is nearly as easy to program and will give much better performance when multiple processes are performing concurrent I/O, such as when aWeb browser retrieves data in the background while the operating system is paging and another application is active in the foreground.10.2.Explain why SSTF scheduling tends to favor middle cylindersover theinnermost and outermost cylinders.The center of the disk is the location having the smallest average distance to all other tracks.Thus the disk head tends to move away from the edges of the disk.Here is another way to think of it.The current location of the head divides the cylinders into two groups.If the head is not in the center of the disk and a new request arrives,the new request is more likely to be in the group that includes the center of the disk;thus,the head is more likely to move in that direction.10.11、Suppose that a disk drive has 5000 cylinders, numbered 0 to 4999. The drive is currently serving a request at cylinder 143, and the previous request was at cylinder 125. The queue of pending requests, in FIFO order, is86, 1470, 913, 1774, 948, 1509, 1022, 1750, 130Starting from the current head position, what is the total distance (in cylinders) that the disk arm moves to satisfy all the pending requests, for each of the following disk-scheduling algorithms?a. FCFSb. SSTFc. SCANd. LOOKe. C-SCANAnswer:a. The FCFS schedule is 143, 86, 1470, 913, 1774, 948, 1509, 1022, 1750, 130. The total seek distance is 7081.b. The SSTF schedule is 143, 130, 86, 913, 948, 1022, 1470, 1509, 1750, 1774. The total seek distance is 1745.c. The SCAN schedule is 143, 913, 948, 1022, 1470, 1509, 1750, 1774, 4999, 130, 86. The total seek distance is 9769.d. The LOOK schedule is 143, 913, 948, 1022, 1470, 1509, 1750, 1774, 130, 86. The total seek distance is 3319.e. The C-SCAN schedule is 143, 913, 948, 1022, 1470, 1509, 1750, 1774, 4999, 86, 130. The total seek distance is 9813.f. (Bonus.) The C-LOOK schedule is 143, 913, 948, 1022, 1470, 1509, 1750, 1774, 86, 130. The total seek distance is 3363.12CHAPTERFile-SystemImplementationPractice Exercises12.1 Consider a ﬁle currently consisting of 100 blocks. Assume that the ﬁlecontrol block (and the index block, in the case of indexed allocation)is already in memory. Calculate how many disk I/O operations are required for contiguous, linked, and indexed (single-level) allocation strategies, if, for one block, the following conditions hold. In the contiguous-allocation case, assume that there is no room to grow atthe beginning but there is room to grow at the end. Also assume thatthe block information to be added is stored in memory.a. The block is added at the beginning.b. The block is added in the middle.c. The block is added at the end.d. The block is removed from the beginning.e. The block is removed from the middle.f. The block is removed from the end.Answer:The results are:Contiguous Linked Indexeda. 201 1 1b. 101 52 1c. 1 3 1d. 198 1 0e. 98 52 0f. 0 100 012.2 What problems could occur if a system allowed a ﬁle system to be mounted simultaneously at more than one location?Answer:4344 Chapter 12 File-System ImplementationThere would be multiple paths to the same ﬁle, which could confuse users or encourage mistakes (deleting a ﬁle with one path deletes theﬁle in all the other paths).12.3 Why must the bit map for ﬁle allocation be kept on mass storage, ratherthan in main memory?Answer:In case of system crash (memory failure) the free-space list would notbe lost as it would be if the bit map had been stored in main memory.12.4 Consider a system that supports the strategies of contiguous, linked, and indexed allocation. What criteria should be used in deciding which strategy is best utilized for a particular ﬁle?Answer:•Contiguous—if ﬁle is usually accessed sequentially, if ﬁle isrelatively small.•Linked—if ﬁle is large and usually accessed sequentially.• Indexed—if ﬁle is large and usually accessed randomly.12.5 One problem with contiguous allocation is that the user must preallocate enough space for each ﬁle. If the ﬁle grows to be larger than thespace allocated for it, special actions must be taken. One solution to this problem is to deﬁne a ﬁle structure consisting of an initial contiguous area (of a speciﬁed size). If this area is ﬁlled, the operating system automatically deﬁnes an overﬂow area that is linked to the initialc ontiguous area. If the overﬂow area is ﬁlled, another overﬂow areais allocated. Compare this implementation of a ﬁle with the standard contiguous and linked implementations.Answer:This method requires more overhead then the standard contiguousallocation. It requires less overheadthan the standard linked allocation. 12.6 How do caches help improve performance? Why do systems not use more or larger caches if they are so useful?Answer:Caches allow components of differing speeds to communicate moreefﬁcie ntly by storing data from the slower device, temporarily, ina faster device (the cache). Caches are, almost by deﬁnition, more expensive than the device they are caching for, so increasing the number or size of caches would increase system cost.12.7 Why is it advantageous for the user for an operating system to dynamically allocate its internal tables? What are the penalties to the operating system for doing so?Answer:Dynamic tables allow more ﬂexibility in system use growth — tablesare never exceeded, avoiding artiﬁcial use limits. Unfortunately, kernel structures and code are more complicated, so there is more potentialfor bugs. The use of one resource can take away more system resources (by growing to accommodate the requests) than with static tables.Practice Exercises 4512.8 Explain how the VFS layer allows an operating system to support multiple types of ﬁle systems easily.Answer:VFS introduces a layer of indirection in the ﬁle system implementation. In many ways, it is similar to object-oriented programming techniques. System calls can be made generically (independent of ﬁle system type). Each ﬁle system type provides its function calls and data structuresto the VFS layer. A system call is translated into the proper speciﬁc functions for the ta rget ﬁle system at the VFS layer. The calling program has no ﬁle-system-speciﬁc code, and the upper levels of the system call structures likewise are ﬁle system-independent. The translation at the VFS layer turns these generic calls into ﬁle-system-speciﬁc operations.。

Chapter-9_NP完全问题

南京理工大学
9 NP完全问题 NP Complete Problem
南京理工大学
本章内容提要
• • • • • • • 易解问题与难解问题 P类问题和类问题类问题和NP类问题类问题和 NP完全问题完全问题 co-NP类问题类问题 NPI类问题类问题 P、NP、co-NP、NPI类之间的区别与联系、、、类之间的区别与联系 NP完全问题的计算机处理技术简介完全问题的计算机处理技术简介
南京理工大学
9.1.2 易解问题与难解问题的主要区别
在学术界已达成这样的共识：在学术界已达成这样的共识：把多项式时间复杂性作为易解问题与难解问题的分界线，主要原因有：易解问题与难解问题的分界线，主要原因有： 1) 多项式函数与指数函数的增长率有本质差别
问题规模 n logn n
1 10 20 50 100 0 3.3 4.3 5.6 6.6 1 10 20 50 100
特点：对同一输入实例，运行算法，所得结果是一样的。特点：对同一输入实例，运行算法A，所得结果是一样的。
• 定义定义9.2(P类问题如果对于某个判定问题，存在一个非类问题): 类问题如果对于某个判定问题∏，负整数k，对于输入规模为n的实例能够以O(nk)的时间运的实例，负整数，对于输入规模为的实例，能够以的时间运行一个确定性算法，得到yes或no的答案，则称该判定问题的答案，行一个确定性算法，得到或的答案 ∏是一个是一个P(Polynomial)类问题。类问题。是一个类问题
3) 多项式时间复杂性忽略了系数，不影响易解问题与难解多项式时间复杂性忽略了系数，
问题的划分
问题规模n 问题规模 5 10 100 1000 多项式函数 n8 390625 108 1016 1024 108n 5×108 × 109 1010 1011 n1000 51000 101000 102000 103000 1.1n 1.611 2.594 13780.6 指数函数 20.01n 1.035 1.072 2

(完整word版)国际财务管理课后习题答案chapter9

CHAPTER 9 MANAGEMENT OF ECONOMIC EXPOSURESUGGESTED ANSWERS AND SOLUTIONS TO END-OF-CHAPTERQUESTIONS AND PROBLEMSQUESTIONS1. How would you define economic exposure to exchange risk?Answer: Economic exposure can be defined as the possibility that the firm’s cash flows and thus its market value may be affected by the unexpected exchange rate changes.2. Explain the following statement: “Exposure is the regression coefficient.”Answer: Exposure to currency risk can be appropriately measured by th e sensitivity of the firm’s future cash flows and the market value to random changes in exchange rates. Statistically, this sensitivity can be estimated by the regression coefficient. Thus, exposure can be said to be the regression coefficient.3. Suppose that your company has an equity position in a French firm. Discuss the condition under which the dollar/franc exchange rate uncertainty does not constitute exchange exposure for your company.Answer: Mere changes in exchange rates do not necessarily constitute currency exposure. If the French franc value of the equity moves in the opposite direction as much as the dollar value of the franc changes, then the dollar value of the equity position will be insensitive to exchange rate movements. As a result, your company will not be exposed to currency risk.4. Explain the competitive and conversion effects of exchange rate changes on the firm’s operating cash flow.Answer: The competitive effect: exchange rate changes may affect operating cash flows by altering the firm’s competitive position.The conversion effect: A given operating cash flows in terms of a foreign currency will be converted into higher or lower dollar (home currency)amounts as the exchange rate changes.5. Discuss the determinants of operating exposure.Answer: The main determinants of a firm’s operating exposure are (1) the structure of the markets in which the firm sources its inputs, such as labor and materials, and sells its products, and (2) the firm’s ability to mitigate the effect of exchange rate changes by adjusting its markets, product mix, and sourcing.6. Discuss the implications of purchasing power parity for operating exposure.Answer: If the exchange rate changes are matched by the inflation rate differential between countries, firms’ competitive positions will not be altered by exchange rate changes. Firms are not subject to operating exposure.7. General Motors exports cars to Spain but the strong dollar against the peseta hurts sales of GM cars in Spain. In the Spanish market, GM faces competition from the Italian and French car makers, such as Fiat and Renault, whose currencies remain stable relative to the peseta. What kind of measures would you recommend so that GM can maintain its market share in Spain.Answer: Possible measures that GM can take include: (1) diversify the market; try to market the cars not just in Spain and other European countries but also in, say, Asia; (2) locate production facilities in Spain and source inputs locally; (3) locate production facilities, say, in Mexico where production costs are low and export to Spain from Mexico.8. What are the advantages and disadvantages of financial hedging of the firm’s operating exposure vis-à-vis operational hedges (such as relocating manufacturing site)?Answer: Financial hedging can be implemented quickly with relatively low costs, but it is difficult to hedge against long-term, real exposure with financial contracts. On the other hand, operational hedges are costly, time-consuming, and not easily reversible.9. Discuss the advantages and disadvantages of maintaining multiple manufacturing sites as a hedge against exchange rate exposure.Answer: To establish multiple manufacturing sites can be effective in managing exchange risk exposure, but it can be costly because the firm may not be able to take advantage of the economy of scale.10. Evaluate the following statement: “A firm can reduce its currency exposure by diversifying across different business lines.”Answer: Conglomerate expansion may be too costly as a means of hedging exchange risk exposure. Investment in a different line of business must be made based on its own merit.11. The exchange rate uncertainty may not necessarily mean that firms face exchange risk exposure. Explain why this may be the case.Answer: A firm can have a natural hedging position due to, for example, diversified markets, flexible sourcing capabilities, etc. In addition, to the extent that the PPP holds, nominal exchange rate changes do not influenc e firms’ competitive positions. Under these circumstances, firms do not need to worry about exchange risk exposure.PROBLEMS1. Suppose that you hold a piece of land in the City of London that you may want to sell in one year. As a U.S. resident, you are concerned with the dollar value of the land. Assume that, if the British economy booms in the future, the land will be worth £2,000 and one British pound will be worth $1.40. If the British economy slows down, on the other hand, the land will be worth less, i.e., £1,500, but the pound will be stronger, i.e., $1.50/£. You feel that the British economy will experience a boom with a 60% probability and a slow-down with a 40% probability.(a) Estimate your exposure b to the exchange risk.(b) Compute the variance of the dollar value of your property that is attributable to the exchange rate uncertainty.(c) Discuss how you can hedge your exchange risk exposure and also examine the consequences of hedging.Solution: (a) Let us compute the necessary parameter values:E(P) = (.6)($2800)+(.4)($2250) = $1680+$900 = $2,580E(S) = (.6)(1.40)+(.4)(1.5) = 0.84+0.60 = $1.44Var(S) = (.6)(1.40-1.44)2 + (.4)(1.50-1.44)2= .00096+.00144 = .0024.Cov(P,S) = (.6)(2800-2580)(1.4-1.44)+(.4)(2250-2580)(1.5-1.44)= -5.28-7.92 = -13.20b = Cov(P,S)/Var(S) = -13.20/.0024 = -£5,500.You have a negative exposure! As the pound gets stronger (weaker) against the dollar, the dollar value of your British holding goes down (up).(b) b2Var(S) = (-5500)2(.0024) =72,600($)2(c) Buy £5,500 forward. By doing so, you can eliminate the volatility of the dollar value of your British asset that is due to the exchange rate volatility.2. A U.S. firm holds an asset in France and faces the following scenario:In the above table, P* is the euro price of the asset held by the U.S. firm and P is the dollar price of the asset.(a) Compute the exchange exposure faced by the U.S. firm.(b) What is the variance of the dollar price of this asset if the U.S. firm remains unhedged against thisexposure?(c) If the U.S. firm hedges against this exposure using the forward contract, what is the variance of thedollar value of the hedged position?Solution: (a)E(S) = .25(1.20 +1.10+1.00+0.90) = $1.05/€E(P) = .25(1,800+1,540+1,300 +1,080) = $1,430Var(S) = .25[(1.20-1.05)2 +(1.10-1.05)2+(1.00-1.05)2+(0.90-1.05)2]= .0125Cov(P,S) = .25[(1,800-1,430)(1.20-1.05) + (1,540-1,430)(1.10-1.05)(1,300-1,430)(1.00-1.05) + (1,080-1,430)(0.90-1.05)]= 30b = Cov(P,S)/Var(S) = 30/0.0125 = €2,400.(b) Var(P) = .25[(1,800-1,430)2+(1,540-1,430)2+(1,300-1,430)2+(1,080-1,430)2]= 72,100($)2.(c) Var(P) - b2Var(S) = 72,100 - (2,400)2(0.0125) = 100($)2.This means that most of the volatility of the dollar value of the French asset can be removed by hedging exchange risk. The hedging can be achieved by selling €2,400 forward.MINI CASE: ECONOMIC EXPOSURE OF ALBION COMPUTERS PLCConsider Case 3 of Albion Computers PLC discussed in the chapter. Now, assume that the pound is expected to depreciate to $1.50 from the current level of $1.60 per pound. This implies that the pound cost of the imported part, i.e., Intel’s microprocessors, is £341 (=$512/$1.50). Other variables, such as the unit sales volume and the U.K. inflation rate, remain the same as in Case 3.(a) Compute the projected annual cash flow in dollars.(b) Compute the projected operating gains/losses over the four-year horizon as the discounted present value of change in cash flows, which is due to the pound depreciation, from the benchmark case presented in Exhibit 12.4.(c) What actions, if any, can Albion take to mitigate the projected operating losses due to the pound depreciation?Suggested Solution to Economic Exposure of Albion Computers PLCa) The projected annual cash flow can be computed as follows:______________________________________________________Sales (40,000 units at £1,080/unit) £43,200,000Variable costs (40,000 units at £697/unit) £27,880,000Fixed overhead costs 4,000,000Depreciation allowances 1,000,000Net profit before tax £15,315,000Income tax (50%) 7,657,500Profit after tax 7,657,500Add back depreciation 1,000,000Operating cash flow in pounds £8,657,500Operating cash flow in dollars $12,986,250______________________________________________________b) ______________________________________________________Benchmark CurrentVariables Case Case______________________________________________________Exchange rate ($/£) 1.60 1.50Unit variable cost (£) 650 697Unit sales price (£) 1,000 1,080Sales volume (units) 50,000 40,000Annual cash flow (£) 7,250,000 8,657,500Annual cash flow ($) 11,600,000 12,986,250Four-year present value ($) 33,118,000 37,076,946Operating gains/losses ($) 3,958,946______________________________________________________c) In this case, Albion actually can expect to realize exchange gains, rather than losses. This is mainly due to the fact that while the selling price appreciates by 8% in the U.K. market, the variable cost of imported input increased by about 6.25%. Albion may choose not to do anything.。

旅游英语视听说第二版课后答案chapter9

旅游英语视听说第二版课后答案chapter91、4．—Alice’s never late for school.—________. [单选题] *A．So am I.B．So was I.C．Neither am I. (正确答案)D．Neither have I.2、The relationship between employers and employees has been studied（）. [单选题] *A. originallyB. extremelyC. violentlyD. intensively(正确答案)3、Betty works as a waitress to earn money for her education. [单选题] *A. 服务员(正确答案)B. 打字员C. 秘书D. 演员4、( ) It tells what is going on ___the county and all____the world. [单选题] *A. across; over(正确答案)B. all; acrossC. in; inD.to; for5、In the closet（）a pair of trousers his parents bought for his birthday. [单选题] *A. lyingB. lies(正确答案)c. lieD. is lain6、It’s windy outside. _______ your jacket, Bob. [单选题] *A. Try onB. Put on(正确答案)C. Take offD. Wear7、You can buy some pieces of bread from "_______". [单选题] *A. Bakery(正确答案)B. Travel AgencyC. LaundryD. Ticket Office8、My brother is _______ actor. He works very hard. [单选题] *A. aB. an(正确答案)C. theD. one9、--How is your friend coming?--I’m not sure. He _______ drive here. [单选题] *A. may(正确答案)B. canC. mustD. will10、32．There are about __________ women doctors in this hospital. [单选题] * A．two hundred ofB．two hundreds ofC．two hundredsD．two hundred (正确答案)11、We’re proud that China _______ stronger and stronger these years. [单选题] *A. will becomeB. becameC. is becoming(正确答案)D. was becoming12、The reason I didn't attend the lecture was simply _____ I got a bad cold that day. [单选题] *A. becauseB. asC. that(正确答案)D. for13、Location is the first thing customers consider when_____to buy a house. [单选题] *A.planning(正确答案)B.plannedC.having plannedD.to plan14、--What’s your _______, Jim Green?--Jim. [单选题] *A. full nameB. first name(正确答案)C. last nameD. family name15、—Excuse me, how long does it ______ to walk to the library? —About 15 minutes, I’m afraid.（）[单选题] *A. take(正确答案)B. spendC. costD. pay16、16．Lily is a lovely girl. We all want to ________ friends with her. [单选题] *A．haveB．make(正确答案)C．doD．take17、One thousand dollars a month is not a fortune but at least can help cover my living（）. [单选题] *A. billsB. expenses(正确答案)C. pricesD. charges18、—______ you speak French?—Yes, I can.（）[单选题] *A. NeedB. Can(正确答案)C. MightD. Must19、( ) The Great Wall was listed by the UNESCO as ___ World Heritage Site. [单选题]*A. a(正确答案)B. theC.\D.an20、Sitting at the back of the room（）a very shy girl with two bright eyes. [单选题] *A. is(正确答案)B. areC. hasD. there was21、The secretary was asked to_____of the waste paper on the desk. [单选题] *A.disappearB.dispose(正确答案)C.declareD.got rid22、Mary _______ Math. [单选题] *A. is good at(正确答案)B. do well inC. is good forD. is good with23、The scenery is so beautiful. Let’s _______. [单选题] *A. take photos(正确答案)B. take mapsC. take busD. take exams24、In order to find the missing child, villagers _______ all they can over the past five hours. [单选题] *A. didB. doC. had doneD. have been doing(正确答案)25、Look! There are some boats ______ the river.（）[单选题] *A. on(正确答案)B. overC. betweenD. in26、The three guests come from different _______. [单选题] *A. countryB. countrysC. countryesD. countries(正确答案)27、--Miss Li, could you please help me _______ math problem？--OK. Let me try. [单选题] *A. look upB. work out(正确答案)C. set upD. put up28、Ships can carry more goods than _____ means of transport. [单选题] *A. the otherB. anotherC. any other(正确答案)D. any29、She found her wallet（）she lost it. [单选题] *A. where(正确答案)B. whenC. in whichD.that30、36．The students will go to the Summer Palace if it __________ tomorrow. [单选题] * A．won’t rainB．isn’t rainingC．doesn’t rain (正确答案)D．isn’t rain。

chapter-9-Lesson-Planning

第14页，共30页。
案例一
PEP BOOK3 Unit Two ( Pa从rt B教: L师et’的s角tal度k) 出发，“使学生…” “发
展学生的…” “训练学生的…” “培养
Teaching aims: 1. Aims on the knowledge
学生的…”，目标主体均为教师
(1) To enable the children to grasp the usage of the single and plural forms
5) By lesson planning, teachers also become aware of the teaching aids that are needed.
6) Lesson planning is a good practice and a sign of
professionalism. 7) Lesson planning will enable the teacher to improve
The time and date of the lesson;
The time duration of the lesson
第12页，共30页。
5.2 Teaching Aims
1. to learn the use of comparative form of adjectives
Able to use a number of comparative adjectives to compare things or people
Micro planning is planning for a specific unit or a lesson. Teachers may have different ways of writing their own lesson plans.

信号与系统-奥本海姆-中文答案-chapter-9

第九章9.6 解：(a) 若是有限持续期信号Roc 为整个s 平面，故存在极点不可能，故不可能为有限持续期。

(b) 可能是左边的。

(c) 不可能是右边的，若是右边信号，它并不是绝对可积的。

(d) x(t)可能为双边的。

9.8 解：因为te t x t g 2)()(=的傅氏变换，)(ωj G 收敛所以)(t x 绝对可积若)(t x 为左边或者右边信号，则)(t x 不绝对可积故)(t x 为双边信号9.10 解：(a) 低通(b) 带通(c) 高通9.14 解：dt e t x s X st ⎰∞∞--=)()(，由)(t x 是偶函数可得)()()(t d e t x s X st --=⎰-∞∞ dt e t x t s ⎰∞∞----=)()(dt e t x t s ⎰∞∞---=)()( )(s X -=421πj e s =为极点，故421πj e s -=也为极点，由)(t x 是实信号可知其极点成对出现，故421πj e s -=与421πj e s --=也为极点。

)21)(21)(21)(21()(4444ππππj j j j e s e s e s e s Ms X --++--=由⎰∞∞-=4)(dt t x 得 4)0(=x所以，M ＝1/4 即，42}Re{42<<-s 9.21 解： (a) 3121)(+++=s s s X 2}Re{->s (b) 25)5(541)(2++++=s s s X 4}Re{->s (c) 3121)(----=s s s X 2}Re{<s (d) 22)2(1)2(1)(--+=s s s X 2}Re{2<<-s (e) 22)2(1)2(1)(-++-=s s s X 2}Re{2<<-s (f) 2)2(1)(-=s s X 2}Re{<s (g) )1(1)(s e ss X --= 0}Re{>s (h) 22)1()(se s X s -=- 0}Re{>s(i) ss X 11)(+= 0}Re{>s (j) ss X 131)(+= 0}Re{>s9.23 解：1． Roc 包括 Re{s}=32． Roc 包括 Re{s}=03． Roc 在最左边极点的左边4． Roc 在最右边极点的右边图1：1，2}Re{>s2，2}Re{2<<-s3，2}Re{-<s4，2}Re{>s图2： 1，2}Re{->s2，2}Re{->s3，2}Re{-<s4，2}Re{->s图3： 1，2}Re{>s2，2}Re{<s3，2}Re{<s4，2}Re{>s图4： 1，S 为整个平面2，S 为整个平面3，S 为整个平面4，S 为整个平面9.25 解：图略9.27 解：)(t x Θ为实信号，)(s X 有一个极点为j s +-=1)(s X ∴另一个极点为j s --=1)1)(1()(j s j s M s X ++-+=∴ 又Θ8)0(=X16=∴M 则，)1(8)1(8)(j s j j s j s X -+-++= 1}Re{->s 或者1}Re{-<s 之一使其成立又 )(2t x e tΘ不是绝对可积的 ∴ 对任一个s ，右移2，不一定在Roc 中因此，1}Re{-<s9.35 解： (a) )(1)(*)(s X st u t x L −→−Θ 那么方框图表示的方程为)(*)(*)(6)(*)()()(*)(*)()(*)(2)(t u t u t y t u t y t y t u t u t x t u t x t x --=++即 ⎰⎰⎰⎰⎰⎰∞-∞-∞-∞-∞-∞---=++t tt t t t dt d y d y t y dt d x d x t x ττττττττ)(6)()()()(2)( 对两边求导可得)(6)()()()()(2222t x dt t dx dtt x d t y dt t dy dt t y d --=++ (b) 126)(22++--=s s s s s H 121-==s s 是)(s H 的二重极点，由于系统是因果的所以 1}Re{->sRoc 包含虚轴，所以系统是稳定的。

CreatingBrandEquity习题及答案详解

Chapter 9:Creating Brand EquityGENERAL CONCEPT QUESTIONSMultiple Choice1.At the heart of a successful brand is ________, backed by creatively designed andexecuted marketing.a.priceb.promotionc. a great product or serviced. a great slogane. a brand conceptAnswer: c Page: 273 Level of difficulty: Easy2.The strategic branch management process involves four main steps. Which of thefollowing would NOT be among those steps?a.Measuring consumer brand knowledge.b.Identifying and establishing brand positioning.c.Planning and implementing brand marketing.d.Measuring and interpreting brand performance.e.Growing and sustaining brand value.Answer: a Page: 274 Level of difficulty: Hard3.The American Marketing Association defines a ________ as “a name, term, sign,symbol, or design, or a combination of them, intended to identify the goods or services of one seller or group of sellers and to differentiate them from those of competitors.”a.holistic product conceptb.product conceptc.service conceptd.brande.brand imageAnswer: d Page: 274 Level of difficulty: Medium4.The earliest signs of branding in Europe were medieval ________ requirement thatcraftspeople put trademarks on their products to protect themselves and consumers against inferior quality.a.kings’b.churches’c.consumers’ernments’e.guilds’Answer: e Page: 274 Level of difficulty: Medium5.Consumers learn about brands through ________ and product marketing programs.a.the mass mediab.past experiences with the productc.the sales forced.shopping botse.independent information sourcesAnswer: b Page: 274 Level of difficulty: Medium6.The world’s strongest brands share common attributes. Which of the following wouldNOT be among those common attributes?a.The brand that spends the most is the most respected and valued.b.The company monitors sources of brand equity.c.The pricing strategy is based on consumer perceptions of value.d.The brand stays relevant.e.The brand excels at delivering the benefits consumers truly desire.Answer: a Page: 275 Level of difficulty: Hard7.________ is endowing products and services with the power of a brand.a.Brand imageb.The branding conceptc.Brandingd.Brand positioninge.Brand partitioningAnswer: c Page: 275 Level of difficulty: Easy8.Brand ________ is the added value endowed to products and services.a.loyaltyb.equityc.preferenced.satisfactione.benefitsAnswer: b Page: 276 Level of difficulty: Medium9.The premise of ________ models is that the power of a brand lies in what customershave seen, read, learned, thought, and felt about the brand over time.a.product-based brand equityb.service-based brand equityc.functional-based brand equityd.mission-driven brand equitye.consumer-based brand equityAnswer: e Page: 276 Level of difficulty: Hard10. ________ can be defined as the differential effect that brand knowledge has onconsumer response to the marketing of that brand.a.Mission-driven brand equityb.Consumer-based brand equityc.Product-driven brand equityd.Service-driven brand equitye.Function-based brand equityAnswer: b Page: 277 Level of difficulty: Medium11.When a consumer expresses thoughts, feelings, images, experiences, beliefs, and soon that become associated with the brand, the consumer is expressing brand ________.a.knowledgeb.loyaltyc.behaviord.preferencee.equityAnswer: a Page: 277 Level of difficulty: Easy12.________ is what drives the differences that manifest themselves in brand equity.a.Brand imageb.Consumer incomec.Consumer purchasing powerd.Consumer knowledgee.Brand perceptionAnswer: d Page: 277 Level of difficulty: Medium13.Strong brands possess all of the following marketing advantages EXCEPT ________.a.greater loyaltyrger marginsc.guaranteed profitsd.improved perceptions of product performancee.more elastic consumer response to price decreasesAnswer: c Page: 277 Level of difficulty: Medium14.When a marketer expresses his or her vision of what the brand must be and do forconsumers, they are expressing what is called ________.a. a brand promiseb. a brand missionc.brand equityd. a brand positione. a brand conceptAnswer: a Page: 278 Level of difficulty: Hard15.There are four key components—or pillars—of brand equity. Which of thosecomponents or pillars measures the breadth of a brand’s appeal?a.Differentiationb.Relevancec.Esteemd.Knowledgee.ValueAnswer: b Page: 278 Level of difficulty: Hard16.Two pillars that point to the brand’s future value, rather than just reflecting its past,are differentiation and relevance. Differentiation and relevance combine to determine what is called brand ________.a.positionb.imagec.depthd.knowledgee.strengthAnswer: e Page: 279 Level of difficulty: Medium17.David Aaker views brand equity as a set of five categories of brand assets andliabilities linked to a brand that add or subtract from the value provided by a product or service to a firm and/or that firm’s customers. All of the following would be among Aaker’s five categories EXCEPT ________.a.brand loyaltyb.brand awarenessc.perceived qualityd.brand pricee.brand associationsAnswer: d Page: 279 Level of difficulty: Hard18.According to Aaker, a particularly important concept for building brand equity is________—the unique set of brand associations that represent what the brand stands for and promises to consumers.a.brand knowledgeb.brand preferencec.brand identityd.brand visione.brandable differencesAnswer: c Page: 279 Level of difficulty: Medium19.General Motors states that its ________ is a “world class car with employees whotreat customers with respect and as friends.”a.core identityb.perceived identityc.extended identityd.defensible identitypetitive identityAnswer: a Pages: 280 Level of difficulty: Medium20.According to the BRANDZ model of brand strength, brand building involves asequential series of steps. Which of these steps would address or answer the question “Do I know about it?”a.Relevanceb.Presencec.Performanced.Advantagee.BondingAnswer: b Page: 280 Level of difficulty: Medium21.The MasterCard “priceless”ad campaign is a good example of brand duality. Twoadvantages of the brand are demonstrated in this campaign. What are those advantages?a.Positive and negative advantages.b.Local and global advantages.c.Rational and emotional advantages.d.Segmented and differentiated advantages.e.Price and promotional advantages.Answer: c Page: 280 Level of difficulty: Hard22.All of the following are considered to be among the “six brand building blocks”EXCEPT ________.a.brand salienceb.brand performancec.brand imageryd.brand feelingse.brand prideAnswer: e Page: 280 Level of difficulty: Medium23.With respect to the “six brand building blocks,” ________ focus on customers’ ownpersonal opinions and evaluations.a.brand salienceb.brand performancec.brand imageryd.brand judgmentse.brand resonanceAnswer: d Page: 280 Level of difficulty: Medium24.With respect to the brand resonance pyramid, at which of the following “buildingblock levels”would we expect the consumer to have developed an intense, active loyalty?a.Salienceb.Imageryc.Feelingsd.Judgmentse.ResonanceAnswer: e age: 281 Level of difficulty: Hard25.From a marketing management perspective, there are three main sets of brand equitydrivers. Which of these drivers was most applicable when McDonald’s decided to use the “golden arches” and Ronald McDonald as symbols of their brand?a.The product and all accompanying marketing activities and supporting marketingprograms.b.The service and all accompanying marketing activities and supporting marketingprograms.c.The initial choices for the brand elements or identities making up the brand.d.Associations indirectly transferred to the brand by linking it to some other entity.e.The profitability associated with brand development.Answer: c Page: 281 Level of difficulty: Medium26.________ are those trademarkable devices that serve to identify and differentiate thebrand.a.Brand elementsb.Brand valuec.Brand perceptiond.Brand imagee.Brand tracksAnswer: a Page: 281 Level of difficulty: Medium27.Six brand elements assist in brand building. Which of the following would NOT beamong those preferred brand elements?a.Adaptableb.Protectablec.Memorabled.Likeabilitye.Subliminal natureAnswer: e Page: 282 Level of difficulty: Medium28.If a brand element can be used to introduce new products in the same or differentcategories, the brand element is said to be ________.a.memorableb.meaningfulc.likeabled.transferablee.adaptableAnswer: d Page: 282 Level of difficulty: Medium29.Before selecting a brand name, marketers must be assured that the brand name will bewell received and understood by the consumers. In order to do this several tests may need to be performed. A(n) ________ asks such questions as “How easily is the name pronounced?”a.association testb.learning testc.memory testd.preference teste.design testAnswer: b Page: 282 Level of difficulty: Medium30.Brand names are not only important brand element. Often, ________, the moreimportant it is that brand elements capture the brand’s intangible characteristics.a.the less concrete brand benefits areb.the more concrete brand benefits arec.the more varied brand perceptions ared.the less varied brand perceptions aree.the more sophisticated brand benefits areAnswer: a Pages: 282–283 Level of difficulty: Hard31.With respect to powerful brand elements, a ________ is an extremely efficient meansto build brand equity. The element can function as useful “hooks”or “handles”to help consumers grasp what the brand is and what makes it special.a.spokespersonb.product shapec.slogand.patente.promotional descriptorAnswer: c Page: 283 Level of difficulty: Medium32.Brand elements and secondary associations can make important contributions tobuilding brand equity; however, the primary input comes from ________.a.the product or service and supporting marketing activitiesb.marketing researchc.the consumerd.the distributorse.the organizational mission statementAnswer: a Page: 284 Level of difficulty: Medium33.A ________ can be defined as any information-bearing experience a customer orprospect has with the brand, the product category, or the market that relates to the marketer’s product or service.a.brand valueb.brand elementc.brand traitd.brand charactere.brand contactAnswer: e Page: 284 Level of difficulty: Medium34.Chupa Chups has the distinction of being the world’s largest maker of ________ andhas begun to diversify its brand beyond its original market (children). Today, Chupa Chups is targeting teens and youth through “nonendorser endorsers”such as Jerry Seinfeld.a.veggie chipsb.apple juicec.roller skatesd.lollipopse.granolaAnswer: d Page: 284 Level of difficulty: Hard35.Considering holistic marketing activities, the rapid expansion of ________ hascreated opportunities to personalize marketing.a.target marketingb.globalizationc.the Internetd.standardizatione. CD technologyAnswer: c Page: 284 Level of difficulty: Hard36.________ is about making sure that the brand and its marketing are as relevant aspossible to as many customers as possible—a challenge, given that no two customers are identical.a.Personalizing marketingb.Segmenting marketingc.Brand imageryd.Emotionalizing brandse.Rationalizing brandsAnswer: a Page: 284 Level of difficulty: Medium37.The traditional “marketing-mix”concept and the notion of the “4 Ps”may notadequately describe modern marketing programs. ________ is about mixing and matching marketing activities to maximize their individual and collective effects.a.Personalizing marketingb.Individualizing marketingc.Globalizing marketingd.Institutionalizing marketinge. Integrating marketingAnswer: e Page: 285 Level of difficulty: Medium38.Godin identifies five specific steps to enhance effective permission marketing. Whichof the following would NOT be among those steps?a.Offer the prospect an incentive to volunteer.b.Offer the interested prospect a curriculum over time that teaches the consumerabout the product or service.c.Reinforce the incentive to guarantee that the prospect maintains the permission.d.Always offer a monetary incentive to demonstrate sincerity of the offer.e. Over time, leverage the permission to change consumer behavior toward profits. Answer: d Page: 285 Level of difficulty: Hard39.________ is the consumers’ ability to identify the brand under different conditions, asreflected by their brand recognition or recall performance.a.Brand awarenessb.Brand imagec.Brand alternationd.Brand perceptione.Brand reflectionAnswer: a Page: 286 Level of difficulty: Medium40.A marketing manager has decided to study the perceptions and beliefs held byconsumers, as reflected in the associations held in consumer memory. Which of the following terms is most closely associated with the marketing manager’s objective of study?a.Brand awarenessb.Brand imagec.Brand elementd.Brand concepte.Brand traitAnswer: b Page: 286 Level of difficulty: Medium41.For a brand to succeed, marketers must “walk the walk” and ensure that employeesand marketing partners do the same. Marketers often must use ________ to motivate those groups to support the brand.a.global brandingb.retro-brandingc.internal brandingd.external brandinge.dual brandingAnswer: c Page: 286 Level of difficulty: Medium42.Mark Thomas has observed that Shell delivers on its promises to supply the bestgasoline possible to the driving public. Shell promotions, employees, and distributors send a common and consistent message about delivering on Shell promises to Mr.Thomas. A good term that describes what occurs when customers experience the company as delivering on its brand promise is the term ________.a.brand imageb.brand enhancementc.brand beliefd.brand attitudee.brand bondingAnswer: e Page: 286 Level of difficulty: Medium43.The main secondary sources of brand knowledge come from all of the followingEXCEPT ________.a.other brandsb.peoplec.thingsd.local, state, and federal governmentse.placesAnswer: d Page: 287 Level of difficulty: Easy44.Brand equity can be measured in two ways. Which of the following would be a goodrepresentation of one of those ways?a.Statistical analysis of demographics.b.Secondary evaluation of governmental statistics.c.Directly assessing the actual impact of brand knowledge on consumer response todifferent aspects of marketing.d.Evaluating published statistics of competitors.e.Hiring independent evaluators.Answer: c Page: 288 Level of difficulty: Medium45.A structured approach to assessing the sources and outcomes of brand equity and themanner in which marketing activities create brand value is called ________.a.the brand value chainb.the brand portfolioc.the brand life cycled.brand partitioninge.brand positioningAnswer: a Page: 288 Level of difficulty: Medium46.A brand manager is concerned that his organization’s brand image and physical salesare slipping in the marketplace. The manager has decided to query consumers about the health of the brand and try to discover ways to leverage the brand’s equity. Which of the following terms will most likely provide the structure and process for the manager’s investigation?a. A brand demographic matrix analysis.b. A secondary search of good brand characteristics.c. A brand audit.d.An organizational audit.e. A brand positioning study.Answer: c Page: 289 Level of difficulty: Medium47.Within a brand audit, the ________ has as its purpose to provide a current,comprehensive profile of how all the products and services sold by a company are marketed and branded.a.cost curve calculationb.product balance sheetc.strategic pland.brand inventorye.product assessmentAnswer: d Page: 289 Level of difficulty: Medium48.Many firms use ________ to supplement traditional focus groups. They studyconsumers in their everyday habitats at home, at work, at play, or shopping.a.ethnographyb.demographyc.geodemographyd.psychographicse.product profilesAnswer: a Page: 290 Level of difficulty: Hard49.Irene and Dave Washburn are part of a long-term ________. The Washburn’s supplya national research firm with information about their brand habits, preferences,dislikes, and beliefs on a monthly basis for a period of two years.a.image studyb.demographic studyc.psychological profile studyd.promotion management studye.tracking studyAnswer: e Page: 290 Level of difficulty: Medium50.________ has the job of estimating the total financial value of the brand.a.Brand trackingb.Brand auditingc.Brand equityd.Brand valuatione.Brand partitioningAnswer: d Page: 290 Level of difficulty: Medium51.According to 2004 Brand Value estimates, ________ was ranked number-one in theworld with a brand value of $67.39 billion.a.Microsoftb.IBMc.Coca-Colad.GEe.McDonald’sAnswer: c Page: 291 Level of difficulty: Medium52.A company’s major enduring asset is ________.a.its leadershipb.its brandc.its cultured.its shareholderse.its workersAnswer: b Page: 291 Level of difficulty: Easy53.According to Interbrand’s Brand Strength formula, the most important factor in theequation is ________ with a 25 percent weight.a.stabilityb.marketc.geographic spreadd.trende.leadershipAnswer: e Page: 292 Level of difficulty: Hard54.According to Scott Bedbury’s book, A New Brand World, all of the following areconsidered to be important principles for twenty-first century branding EXCEPT ________.a.consumers will tell you what your brand image should beb.relying on brand awareness has become marketing fool’s goldc.you have to know it before you can grow itd.great brands establish enduring customer relationshipse.all brands need good parentsAnswer: a Page: 294 Level of difficulty: Hard55.When a firm uses an established brand to introduce a new product, it is called a(n)________.a.sub-brandb.brand valuec.brand extensiond.brand mixe. brand postureAnswer: c Page: 296 Level of difficulty: Medium56.All of the following would be examples of companies that use family branding as oneof their branding strategies EXCEPT ________.a.Campbell Soupb.Searsc.Heinzd.General Electrice.Hunt’sAnswer: b Page: 297 Level of difficulty: Hard57.According to Ries and Trout, Cadbury suffered from the ________ when thecompany allowed its brand to become diluted by putting their name on such variants as mashed potatoes, powdered milk, soups, beverages, as well as chocolates and candies.a.“greed is good” trapb.“pyramid principle”c.“branding fallout” conceptd.“image syndrome”e.“line-extension” trapAnswer: e Page: 299 Level of difficulty: Hard58.A ________ is the set of all brands and brand lines a particular firm offers for sale tobuyers in a particular category.a.brand partitionb.brand positionc.brand portfoliod.brand concepte.brand imageAnswer: c Page: 301 Level of difficulty: Medium59.________ brands are positioned with respect to competitors’brands so that moreimportant (and more profitable) flagship brands can retain their desired positioning.a.Flankerb.Attackerc.Defenderd.Blitzkriege.IndividualAnswer: a Page: 302 Level of difficulty: Medium60.The role of ________ in the brand portfolio often may be to attract customers to thebrand franchise. Trading up will often occur with this type of brand.a.the cash cowb.flanker brandc.fighting brandd.high-end prestige brande.low-end entry-level brandAnswer: e Page: 303 Level of difficulty: MediumTrue/False61.At the heart of a successful brand are the successful people that run the organization. Answer: False Page: 273 Level of difficulty: Medium62.Google’s successful brand name was derived from the word googol—the numberrepresented by a 1 followed by 100 zeroes.Answer: True Page: 273 Level of difficulty: Medium63.The American Marketing Association defines a brand as “a name, term, sign, symbol,or design, or a combination of them, intended to identify the goods or services of one seller or group of sellers and to differentiate them from those of competitors.”Answer: True Page: 274 Level of difficulty: Easy64.One of the attributes shared by the world’s strongest brands is that the brand becomesinvulnerable to attack from competitors due to the support of the marketplace. Answer: False Page: 275 Level of difficulty: Hard65.The added value endowed to products and services is called brand equity. Answer: True Page: 276 Level of difficulty: Medium66.Brand knowledge is indicated when the consumer refuses to purchase competitivebrands.Answer: False Page: 277 Level of difficulty: Medium67.One of the advantages of having a strong brand is the ability to have a more elasticconsumer response to price decreases of the brand.Answer: True Page: 277 Level of difficulty: Medium68.The marketer’s vision of what the brand must be and do for consumers is called abrand promise.Answer: True Page: 278 Level of difficulty: Medium69.There are four key components—or pillars—of brand equity; one of thesecomponents is called knowledge.Answer: True Page: 278 Level of difficulty: Medium70.An illustration of the extended brand identity of GM’s Saturn brand is a “world-classcar with employees who treat customers with respect and as friends.”Answer: False Page: 280 Level of difficulty: Hard71.Under the BRANDZ model of brand strength, the brand objective of bonding istranslated to mean “nothing else beats it.”Answer: True Page: 280 Level of difficulty: Easy72.Brand resonance is associated with how a brand name sounds when said byspokespersons or consumers.Answer: False Page: 280 Level of difficulty: Hard73. Brand salience relates to how often and easily the brand is evoked under variouspurchase or consumption situations.Answer: True Page: 280 Level of difficulty: Medium74. Brand imagery is the consumers’ emotional responses and reactions with respect tothe brand.Answer: False Page: 280 Level of difficulty: Medium75. In the brand resonance pyramid, the lowest level (e.g., identity—Who are you?) isassociated with what is called salience.Answer: True Page: 281 Level of difficulty: Hard76. A few good illustrations of power brand elements are Nike’s “swoosh”logo, theempowering “Just Do It”slogan, and the mythological “Nike”name based on the winged goddess of victory.Answer: True Page: 281 Level of difficulty: Medium77. One of the selection criteria for creating a successful brand element is that it shouldbe projectable.Answer: True Page: 282 Level of difficulty: Medium78. If a brand element has the characteristic of being memorable, it is said to mean thatthe brand is credible and suggestive of its corresponding category.Answer: False Page: 282 Level of difficulty: Hard79. One of the ways for testing a brand element is to ask consumers how well theyremember the brand name; this is called a learning test.Answer: False Page: 282 Level of difficulty: Medium80. A powerful—but sometimes overlooked—brand element is slogans.Answer: True Page: 283 Level of difficulty: Easy81. A classic case of a company using a slogan to build brand equity is that of Avis’s 41-year-old “We Try Harder” ad campaign.Answer: True Page: 283 Level of difficulty: Easy82. A brand contract is defined as any information-bearing experience a customer orprospect has with a brand.Answer: True Page: 284 Level of difficulty: Medium83. Personalizing marketing is based on the fact that all customers are identical in severalways.Answer: False Page: 284 Level of difficulty: Medium84. Gillette sent 1,000 young males a new, innovative razor product. In return, thecompany asked the young males to supply their e-mail addresses so future contacts could be initiated. This would be an example of what is called permission marketing. Answer: True Page: 285 Level of difficulty: Medium85. Brand awareness can be perceived as being the consumers’ability to identify thebrand under different conditions, as reflected by their brand recognition or recall performance.Answer: True Page: 286 Level of difficulty: Medium86. Internal branding is the perceptions and beliefs held by consumers, as reflected in theassociations held in consumer memory.Answer: False Page: 286 Level of difficulty: Hard87. Marketers must now “walk the walk”to deliver the brand promise. Often internalbranding is necessary to make sure that all employees assist in meeting the brand promise.Answer: True Page: 286 Level of difficulty: Medium88. The brand promise will not be delivered unless everyone in the company lives thebrand.Answer: True Page: 286 Level of difficulty: Easy89.The indirect approach to assessing brand equity assesses the actual impact of brandknowledge on consumer response to different aspects of marketing.Answer: False Page: 288 Level of difficulty: Medium90. A brand value chain often does not exist because of an organization’s commitment toits service value chain.Answer: False Page: 288 Level of difficulty: Medium91.The brand audit can be used to set strategic direction for the brand.Answer: True Page: 289 Level of difficulty: Easy92. The purpose of brand strength(an assessment tool) is to provide a current,comprehensive profile of how all the products and services sold by a company are marketed and branded.Answer: False Page: 289 Level of difficulty: Hard93.In order to explore more detailed information about brands, many firms are now usingethnography to supplement traditional focus groups.Answer: True Page: 290 Level of difficulty: Medium94.Tracking studies collect information from consumers on a routine basis over time. Answer: True Page: 290 Level of difficulty: Easy95. In essence, brand equity is basically the same concept as brand valuation. Answer: False Page: 290 Level of difficulty: Medium96. At present, the most valuable brand in the world is Microsoft.Answer: False Page: 291 Level of difficulty: Medium97. A company’s most enduring asset is its intellectual capital generated by the topofficers of the organization.Answer: False Page: 291 Level of difficulty: Medium98. With respect to brand revitalization, the primary strategy for recovery is to hire apublic relations firm to think of a new image for the firm and its products. Answer: False Page: 294 Level of difficulty: Hard99.When Honda expanded its brand into such areas as automobiles, motorcycles,snowblowers, lawnmowers, marine engines, and snowmobiles, it was pursuing a strategy called category extension of its brand.Answer: True Page: 296 Level of difficulty: Medium100. Brand dilution occurs when consumers no longer associate a brand with a specific product or highly similar products and start thinking less of the brand.Answer: True Page: 299 Level of difficulty: MediumEssay101. The world’s strongest brands share a list of common attributes. List and briefly characterize five of those attributes.Suggested Answer: According to information presented in the text, the world’s strongest brands possess ten common characteristics. Those characteristics appear as: (1) the brand excels at delivering the benefits consumers truly desire; (2) the brand stays relevant; (3) the pricing strategy is based on consumer perceptions of value; (4) the brand is properly positioned; (5) the brand is consistent; (6) the brand portfolio and hierarchy makes sense; (7) the brand makes use of and coordinates a full repertoire of marketing activities to build equity; (8) the brand’s managers understand what the brand means to consumers; (9) the brand is given proper, sustained support; and, (10) the company monitors sources of brand equity. The students may choose any five of the above ten for their discussion.See chapter material for addition discussion material.Page: 275 Level of difficulty: Hard。

2019年精选牛津版初中英语九年级上册课后练习第九篇

2019年精选牛津版初中英语九年级上册课后练习第九篇第1题【单选题】When she knew her son was out of danger, she was relaxed.A、not sadB、not worriedC、not excited【答案】：【解析】：第2题【单选题】—What would you like to drink, tea or ________?—Neither. I just want a glass of water.A、breadB、cakeC、coffee【答案】：【解析】：第3题【单选题】Tom _______ Jack 1:0 at the 17th Tennis School Cup last night.A、beatedB、beatC、winedD、won【答案】：【解析】：第4题【单选题】—There are few mistakes in Mike"s homework.—I think he is the most careful one ________ our classmates.A、aboutB、amongC、between【答案】：【解析】：第5题【单选题】You needn"t _________. There"s enough time for you to go to the airport.A、sayB、hurryC、lie【答案】：【解析】：第6题【单选题】There is a white house _______ many trees.A、amongB、betweenC、duringD、on【答案】：【解析】：第7题【单选题】Let"s move the books onto the desk in our room.A、toB、overC、with【答案】：【解析】：第8题【单选题】You must remember ________.A、what your father saidB、what did your father sayC、your father said whatD、what has your father said【答案】：【解析】：第9题【单选题】She is confident of victory in Saturday"s match, because she is well prepared.A、practiceB、successC、energy【答案】：【解析】：第10题【单选题】The soup is not enough . Add some salt, please.A、sweetB、sourC、saltyD、crispy【答案】：【解析】：第11题【单选题】_______ he is ill, he is not here today.A、BecauseB、Because ofC、SoD、Though【答案】：【解析】：第12题【单选题】—The doctor said that Molly had better stay away from fast food. —But it"s hard for her to give it up.A、not eatB、not cookC、not sell【答案】：【解析】：第13题【单选题】— There have only been five ______ in the shop today.— Don"t worry. It will become better and better.A、scientistsB、customersC、doctors【答案】：【解析】：第14题【单选题】We have to work on weekdays and have no chance to see each other. But we often spend the weekends together.A、on timeB、from Monday to FridayC、every day I【答案】：【解析】：第15题【单选题】Shenzhen was not prepared for the heavy rains which caused deaths and disasters in May, 2014.A、helpfulB、usefulC、ready【答案】：【解析】：第16题【填空题】【答案】：【解析】：第17题【完形填空】阅读下面短文，掌握其大意，然后从每小题所给的A、B、C、D四个选项中，选出一个最佳选项。

Chapter 9 课后答案

新编语言学教程Chapter 9 答案Psycholinguistics1.Define the following terms briefly.(1)psycholinguistics: the study of the relation between language and mind: themental structures and processes which are involved in the acquisition, comprehensionand production of language.(2) language production: the process involved in creating and expressing meaningthrough language, such as the four successive stages provided by Levelt(1989): conceptualization, formulation, articulation andself-regulation.(3) language comprehension: From a psycholinguistic point of view, we store agreat deal of information about the properties of the language, and retrievethis information when we understand language. Besides, language comprehensioncan be treated in four levels: sound, word, sentence and text comprehensions.(4)Sapir-Whorf Hypothesis: It refers to the view that the language system couldinfluence or even determine one’s thought, and a particular language imposesparticular ideas of nature or beliefs of one’s culture.(5)linguistic determinism: One’s language structure determines his cognitivestructure. That is, learning a language may change the way a person thinks orperceives the objective world.(6)linguistic relativity: As one’s language influences one’s cognitive system,speakers of different languages perceive the world differently.2.Psycholinguistics is the study of psychological aspects of language; it usually investigatesthe psychological states and mental activities associated with the use oflanguage. Most problems in psycholinguistics are comparatively more concrete,involving the study of language acquisition especially in children and linguisticperformance such as producing and comprehending utterances or sentencesamong adults. However, theoretical linguistics is more objective. It usually investigatesthe existing phenomena about languages and its investigations are usuallycarried out in the branches of microlinguistics: phonetics, phonology, morphology,syntax and semantics. Psycholinguistics is an interdisciplinary study of languageand psychology, with structural linguistics and cognitive psychology as itsroots while theoretical linguistics solely focuses on aspects of language.3.(1) The correct form is “They swam across the lake”, which is caus ed by exchange.(2)The correct form is “The spy was bound and gagged” , which is caused by exchange.(3)The correct form is “I will see you in the park”, which originates from substitution.4.The slip-of-the-tongue phenomenon described above can be explained by theparallel distributed processing (PDP) approach in word comprehension. ThePDP approach holds that people use several separate and parallel processes at thesame time to understand spoken or written language. In theslip-of-the-tonguephenomenon, people have already conceptualized his/her idea (thought), butcan not find a proper word to express the idea. This shows that thought precedeslanguage. According to linguistic determinism, language shapes one’s thought.If there isn’t language, there sho uld be no thought. Thus, this phenomenon goesagainst linguistic determinism and shows that thought can exist with or withoutlanguage.5.The fact mentioned here flies at the face of linguistic determinism which says thatone’s language structure determines one’s cognitive structure. That’s to say, a particularlanguage can not shape one’s world view. Language changes along socialchanges. And social changes can lead to the changes of people’s view. At the sametime, one’s world view can affect a particula r language. For example, Xiaojie wasused to refer to the daughter of rich and important families before 1949. Then,since 1949, great changes have taken place in China. The world view of Chinesepeople has changed radically but the language has changed little. During the CulturalRevolution, Xiaojie became very much culturally loaded — young womennot belonging to ‘the revolutionary rank’ and people not to be politically trusted.After 1979, it gradually became popular again, and now it has taken on a derogatorymeaning (hooker). As it is mentioned above, it is social changes that shapeone’s world view, and it is cognitive structure that affects language.。

chapter-9 大学物理习题精选-电磁感应

i= R =
2分 1分 1分
μ 0 Qa 2ω 0
2 RLt 0
i 的流向与圆筒转向一致．
14.21 两个半径分别为 R 和 r 的同轴圆形线圈相距 x，且 R >>r，x >>R．若大线圈通有电流 I 而小线圈沿 x 轴方向以速率 v 运动，试求 x =NR 时(N 为正数)小线圈回路中产生的感应电动势的大小．
K
K
O
ω
b a
K B
O′
(A) (C) (E) 2abB | cosω t |． (B) (D)
ω abB ω abB | cosω t |．
［ D ］
1 ω abB cos ω t ． 2
ω abB | sinω t |．
14.3 如图，长度为 l 的直导线 ab 在均匀磁场 B 中以速度 v 移动，直导线 ab 中的电动势为 (A) Blv． (B) Blv sinα． (D) 0．［ D ］ (C) Blv cosα．
5
x r I R
v
x
解：由题意，大线圈中的电流Ｉ在小线圈回路处产生的磁场可视为均匀的．
B=
故穿过小回路的磁通量为
μ0
μ 0 IR 2 2πIR 2 = 4π ( R 2 + x 2 ) 3 / 2 2( R 2 + x 2 ) 3 / 2
3分
K K μ0 μ 0 πr 2 RI 2 IR 2 2 Φ = B⋅S = πr ≈ 2 (R 2 + x 2 )3 / 2 2x 3
解：(1) 由法拉第电磁感应定律： Φ = B
i
K
K
1 xy 2
y = tg θ x
x = vt
2分

曼昆-宏观经济经济学第九版-英文原版答案9

Answers to Textbook Questions and ProblemsCHAPTER 9 Economic Growth II: Technology, Empirics, and PolicyQuestions for Review1. In the Solow model, we find that only technological progress can affect the steady-state rate of growthin income per worker. Growth in the capital stock (through high saving) has no effect on the steady-state growth rate of income per worker; neither does population growth. But technological progress can lead to sustained growth.2. In the steady state, output per person in the Solow model grows at the rate of technological progress g.Capital per person also grows at rate g. Note that this implies that output and capital per effectiveworker are constant in steady state. In the U.S. data, output and capital per worker have both grown at about 2 percent per year for the past half-century.3. To decide whether an economy has more or less capital than the Golden Rule, we need to compare themarginal product of capital net of depreciation (MPK –δ) with the growth rate of total output (n + g).The growth rate of GDP is readily available. Estimating the net marginal product of capital requires a little more work but, as shown in the text, can be backed out of available data on the capital stock relative to GDP, the total amount o f depreciation relative to GDP, and capital’s share in GDP.4. Economic policy can influence the saving rate by either increasing public saving or providingincentives to stimulate private saving. Public saving is the difference between government revenue and government spending. If spending exceeds revenue, the government runs a budget deficit, which is negative saving. Policies that decrease the deficit (such as reductions in government purchases or increases in taxes) increase public saving, whereas policies that increase the deficit decrease saving. A variety of government policies affect private saving. The decision by a household to save may depend on the rate of return; the greater the return to saving, the more attractive saving becomes. Taxincentives such as tax-exempt retirement accounts for individuals and investment tax credits forcorporations increase the rate of return and encourage private saving.5. The legal system is an example of an institutional difference between countries that might explaindifferences in income per person. Countries that have adopted the English style common law system tend to have better developed capital markets, and this leads to more rapid growth because it is easier for businesses to obtain financing. The quality of government is also important. Countries with more government corruption tend to have lower levels of income per person.6. Endogenous growth theories attempt to explain the rate of technological progress by explaining thedecisions that determine the creation of knowledge through research and development. By contrast, the Solow model simply took this rate as exogenous. In the Solow model, the saving rate affects growth temporarily, but diminishing returns to capital eventually force the economy to approach a steady state in which growth depends only on exogenous technological progress. By contrast, many endogenous growth models in essence assume that there are constant (rather than diminishing) returns to capital, interpreted to include knowledge. Hence, changes in the saving rate can lead to persistent growth. Problems and Applications1. a. In the Solow model with technological progress, y is defined as output per effective worker, and kis defined as capital per effective worker. The number of effective workers is defined as L E (or LE), where L is the number of workers, and E measures the efficiency of each worker. To findoutput per effective worker y, divide total output by the number of effective workers:Y LE =K12(LE)12LEY LE =K12L12E12LEY LE =K12 L1E1Y LE =KLE æèççöø÷÷12y=k12b. To solve for the steady-state value of y as a function of s, n, g, and δ, we begin with the equationfor the change in the capital stock in the steady state:Δk = sf(k) –(δ + n + g)k = 0.The production function ycan also be rewritten as y2 = k. Plugging this production functioninto the equation for the change in the capital stock, we find that in the steady state:sy –(δ + n + g)y2 = 0.Solving this, we find the steady-state value of y:y* = s/(δ + n + g).c. The question provides us with the following information about each country:Atlantis: s = 0.28 Xanadu: s = 0.10n = 0.01 n = 0.04g = 0.02 g = 0.02δ = 0.04δ = 0.04Using the equation for y* that we derived in part (a), we can calculate the steady-state values of yfor each country.Developed country: y* = 0.28/(0.04 + 0.01 + 0.02) = 4Less-developed country: y* = 0.10/(0.04 + 0.04 + 0.02) = 12. a. In the steady state, capital per effective worker is constant, and this leads to a constant level ofoutput per effective worker. Given that the growth rate of output per effective worker is zero, this means the growth rate of output is equal to the growth rate of effective workers (LE). We know labor grows at the rate of population growth n and the efficiency of labor (E) grows at rate g. Therefore, output grows at rate n+g. Given output grows at rate n+g and labor grows at rate n, output perworker must grow at rate g. This follows from the rule that the growth rate of Y/L is equal to thegrowth rate of Y minus the growth rate of L.b. First find the output per effective worker production function by dividing both sides of theproduction function by the number of effective workers LE:Y LE =K 13(LE )23LE YLE =K 13L 23E 23LEY LE =K 13L 13E 13Y LE =K LE æèçöø÷13y =k 13To solve for capital per effective worker, we start with the steady state condition:Δk = sf (k ) – (δ + n + g )k = 0.Now substitute in the given parameter values and solve for capital per effective worker (k ):Substitute the value for k back into the per effective worker production function to find output per effective worker is equal to 2. The marginal product of capital is given bySubstitute the value for capital per effective worker to find the marginal product of capital is equal to 1/12.c. According to the Golden Rule, the marginal product of capital is equal to (δ + n + g) or 0.06. In the current steady state, the marginal product of capital is equal to 1/12 or 0.083. Therefore, we have less capital per effective worker in comparison to the Golden Rule. As the level of capital per effective worker rises, the marginal product of capital will fall until it is equal to 0.06. To increase capital per effective worker, there must be an increase in the saving rate.d. During the transition to the Golden Rule steady state, the growth rate of output per worker will increase. In the steady state, output per worker grows at rate g . The increase in the saving rate will increase output per effective worker, and this will increase output per effective worker. In the new steady state, output per effective worker is constant at a new higher level, and output per worker is growing at rate g . During the transition, the growth rate of output per worker jumps up, and then transitions back down to rate g .3. To solve this problem, it is useful to establish what we know about the U.S. economy: • A Cobb –Douglas production function has the form y = k α, where α is capital’s share of income.The question tells us that α = 0.3, so we know that the production function is y = k 0.3.• In the steady state, we know that the growth rate of output equals 3 percent, so we know that (n +g ) = 0.03.• The deprec iation rate δ = 0.04. • The capital –output ratio K/Y = 2.5. Because k/y = [K /(LE )]/[Y /(LE )] = K/Y , we also know that k/y =2.5. (That is, the capital –output ratio is the same in terms of effective workers as it is in levels.)a. Begin with the steady-state condition, sy = (δ + n + g)k. Rewriting this equation leads to a formulafor saving in the steady state:s = (δ + n + g)(k/y).Plugging in the values established above:s = (0.04 + 0.03)(2.5) = 0.175.The initial saving rate is 17.5 percent.b. We know from Chapter 3 that with a Cobb–Douglas production function, capital’s share ofincome α = MPK(K/Y). Rewriting, we haveMPK = α/(K/Y).Plugging in the values established above, we findMPK = 0.3/2.5 = 0.12.c. We know that at the Golden Rule steady state:MPK = (n + g + δ).Plugging in the values established above:MPK = (0.03 + 0.04) = 0.07.At the Golden Rule steady state, the marginal product of capital is 7 percent, whereas it is 12 percent in the initial steady state. Hence, from the initial steady state we need to increase k to achieve the Golden Rule steady state.d. We know from Chapter 3 that for a Cobb–Douglas production function, MPK = α (Y/K). Solvingthis for the capital–output ratio, we findK/Y = α/MPK.We can solve for the Golden Rule capital–output ratio using this equation. If we plug in the value0.07 for the Golden Rule steady-state marginal product of capital, and the value 0.3 for α, we findK/Y = 0.3/0.07 = 4.29.In the Golden Rule steady state, the capital–output ratio equals 4.29, compared to the current capital–output ratio of 2.5.e. We know from part (a) that in the steady states = (δ + n + g)(k/y),where k/y is the steady-state capital–output ratio. In the introduction to this answer, we showed that k/y = K/Y, and in part (d) we found that the Golden Rule K/Y = 4.29. Plugging in this value and those established above:s = (0.04 + 0.03)(4.29) = 0.30.To reach the Golden Rule steady state, the saving rate must rise from 17.5 to 30 percent. Thisresult implies that if we set the saving rate equal to the share going to capital (30 percent), we will achieve the Golden Rule steady state.4. a. In the steady state, we know that sy = (δ + n + g)k. This implies thatk/y = s/(δ + n + g).Since s, δ, n, and g are constant, this means that the ratio k/y is also constant. Since k/y =[K/(LE)]/[Y/(LE)] = K/Y, we can conclude that in the steady state, the capital–output ratio isconstant.b. We know that capital’s share of income = MPK ⨯ (K/Y). In the steady state, we know from part (a)that the capital–output ratio K/Y is constant. We also know from the hint that the MPK is afunction of k, which is constant in the steady state; therefore the MPK itself must be constant.Thus, capital’s share of income is constant. Labor’s share of income is 1 – [C apital’s Share].Hence, if capital’s share is constant, we see that labor’s share of income is also constant.c. We know that in the steady state, total income grows at n + g, defined as the rate of populationgrowth plus the rate of technological change. In part (b) we showed that labor’s and capital’s share of income is constant. If the shares are constant, and total income grows at the rate n + g, thenlabor income and capital income must also grow at the rate n + g.d. Define the real rental price of capital R asR = Total Capital Income/Capital Stock= (MPK ⨯K)/K= MPK.We know that in the steady state, the MPK is constant because capital per effective worker k isconstant. Therefore, we can conclude that the real rental price of capital is constant in the steadystate.To show that the real wage w grows at the rate of technological progress g, defineTLI = Total Labor IncomeL = Labor ForceUsing the hint that the real wage equals total labor income divided by the labor force:w = TLI/L.Equivalently,wL = TLI.In terms of percentage changes, we can write this asΔw/w + ΔL/L = ΔTLI/TLI.This equation says that the growth rate of the real wage plus the growth rate of the labor forceequals the growth rate of total labor income. We know that the labor force grows at rate n, and,from part (c), we know that total labor income grows at rate n + g. We, therefore, conclude that the real wage grows at rate g.5. a. The per worker production function is F (K, L )/L = AK α L 1–α/L = A (K/L )α = Ak α b. In the steady state, Δk = sf (k ) – (δ + n + g )k = 0. Hence, sAk α = (δ + n + g )k , or, after rearranging:k *=sA d +n +g éëêêùûúúa 1-a æèççöø÷÷.Plugging into the per-worker production function from part (a) givesy *=A a 1-a æèççöø÷÷s d +n +g éëêêùûúúa 1-a æèççöø÷÷.Thus, the ratio of steady-state income per worker in Richland to Poorland isy *Richland/y *Poorland ()=s Richland d +n Richland +g /s Poorlandd +n Poorland +g éëêêùûúúa1-a =0.320.05+0.01+0.02/0.100.05+0.03+0.02éëêêùûúúa1-c. If α equals 1/3, then Richland should be 41/2, or two times, richer than Poorland.d. If 4a 1-a æèççöø÷÷= 16, then it must be the case that a 1-a æèççöø÷÷, which in turn requires that α equals 2/3.Hence, if the Cobb –Douglas production function puts 2/3 of the weight on capital and only 1/3 on labor, then we can explain a 16-fold difference in levels of income per worker. One way to justify this might be to think about capital more broadly to include human capital —which must also be accumulated through investment, much in the way one accumulates physical capital.6. How do differences in education across countries affect the Solow model? Education is one factoraffecting the efficiency of labor , which we denoted by E . (Other factors affecting the efficiency of labor include levels of health, skill, and knowledge.) Since country 1 has a more highly educated labor force than country 2, each worker in country 1 is more efficient. That is, E 1 > E 2. We will assume that both countries are in steady state. a. In the Solow growth model, the rate of growth of total income is equal to n + g , which isindependent of the work force’s level of education. The two countries will, thus, have the same rate of growth of total income because they have the same rate of population growth and the same rate of technological progress.b. Because both countries have the same saving rate, the same population growth rate, and the samerate of technological progress, we know that the two countries will converge to the same steady-state level of capital per effective worker k *. This is shown in Figure 9-1.Hence, output per effective worker in the steady state, which is y* = f(k*), is the same in bothcountries. But y* = Y/(L E) or Y/L = y*E. We know that y* will be the same in both countries, but that E1 > E2. Therefore, y*E1 > y*E2. This implies that (Y/L)1 > (Y/L)2. Thus, the level of incomeper worker will be higher in the country with the more educated labor force.c. We know that the real rental price of capital R equals the marginal product of capital (MPK). Butthe MPK depends on the capital stock per efficiency unit of labor. In the steady state, bothcountries have k*1= k*2= k* because both countries have the same saving rate, the same population growth rate, and the same rate of technological progress. Therefore, it must be true that R1 = R2 = MPK. Thus, the real rental price of capital is identical in both countries.d. Output is divided between capital income and labor income. Therefore, the wage per effectiveworker can be expressed asw = f(k) –MPK • k.As discussed in parts (b) and (c), both countries have the same steady-state capital stock k and the same MPK. Therefore, the wage per effective worker in the two countries is equal.Workers, however, care about the wage per unit of labor, not the wage per effective worker.Also, we can observe the wage per unit of labor but not the wage per effective worker. The wageper unit of labor is related to the wage per effective worker by the equationWage per Unit of L = wE.Thus, the wage per unit of labor is higher in the country with the more educated labor force.7. a. In the two-sector endogenous growth model in the text, the production function for manufacturedgoods isY = F [K,(1 –u) EL].We assumed in this model that this function has constant returns to scale. As in Section 3-1,constant returns means that for any positive number z, zY = F(zK, z(1 –u) EL). Setting z = 1/EL,we obtainY EL =FKEL,(1-u)æèççöø÷÷.Using our standard definitions of y as output per effective worker and k as capital per effective worker, we can write this asy = F[k,(1 –u)]b. To begin, note that from the production function in research universities, the growth rate of laborefficiency, ΔE/E, equals g(u). We can now follow the logic of Section 9-1, substituting thefunction g(u) for the constant growth rate g. In order to keep capital per effective worker (K/EL) constant, break-even investment includes three terms: δk is needed to replace depreciating capital, nk is needed to provide capital for new workers, and g(u) is needed to provide capital for thegreater stock of knowledge E created by research universities. That is, break-even investment is [δ + n + g(u)]k.c. Again following the logic of Section 9-1, the growth of capital per effective worker is thedifference between saving per effective worker and break-even investment per effective worker.We now substitute the per-effective-worker production function from part (a) and the function g(u) for the constant growth rate g, to obtainΔk = sF [k,(1 –u)] – [δ + n + g(u)]kIn the steady state, Δk = 0, so we can rewrite the equation above assF [k,(1 –u)] = [δ + n + g(u)]k.As in our analysis of the Solow model, for a given value of u, we can plot the left and right sides of this equationThe steady state is given by the intersection of the two curves.d. The steady state has constant capital per effective worker k as given by Figure 9-2 above. We alsoassume that in the steady state, there is a constant share of time spent in research universities, so u is constant. (After all, if u were not constant, it wouldn’t be a ―steady‖ state!). Hence, output per effective worker y is also constant. Output per worker equals yE, and E grows at rate g(u).Therefore, output per worker grows at rate g(u). The saving rate does not affect this growth rate.However, the amount of time spent in research universities does affect this rate: as more time is spent in research universities, the steady-state growth rate rises.e. An increase in u shifts both lines in our figure. Output per effective worker falls for any givenlevel of capital per effective worker, since less of each worker’s time is spent producingmanufactured goods. This is the immediate effect of the change, since at the time u rises, thecapital stock K and the efficiency of each worker E are constant. Since output per effective worker falls, the curve showing saving per effective worker shifts down.At the same time, the increase in time spent in research universities increases the growth rate of labor efficiency g(u). Hence, break-even investment [which we found above in part (b)] rises at any given level of k, so the line showing breakeven investment also shifts up.Figure 9-3 shows these shifts.In the new steady state, capital per effective worker falls from k1 to k2. Output per effective worker also falls.f. In the short run, the increase in u unambiguously decreases consumption. After all, we argued inpart (e) that the immediate effect is to decrease output, since workers spend less time producingmanufacturing goods and more time in research universities expanding the stock of knowledge.For a given saving rate, the decrease in output implies a decrease in consumption.The long-run steady-state effect is more subtle. We found in part (e) that output per effective worker falls in the steady state. But welfare depends on output (and consumption) per worker, not per effective worker. The increase in time spent in research universities implies that E grows faster.That is, output per worker equals yE. Although steady-state y falls, in the long run the fastergrowth rate of E necessarily dominates. That is, in the long run, consumption unambiguously rises.Nevertheless, because of the initial decline in consumption, the increase in u is not unambiguously a good thing. That is, a policymaker who cares more about current generationsthan about future generations may decide not to pursue a policy of increasing u. (This is analogous to the question considered in Chapter 8 of whether a policymaker should try to reach the GoldenRule level of capital per effective worker if k is currently below the Golden Rule level.)8. On the World Bank Web site (), click on the data tab and then the indicators tab.This brings up a large list of data indicators that allows you to compare the level of growth anddevelopment across countries. To explain differences in income per person across countries, you might look at gross saving as a percentage of GDP, gross capital formation as a percentage of GDP, literacy rate, life expectancy, and population growth rate. From the Solow model, we learned that (all else the same) a higher rate of saving will lead to higher income per person, a lower population growth rate will lead to higher income per person, a higher level of capital per worker will lead to a higher level of income per person, and more efficient or productive labor will lead to higher income per person. The selected data indicators offer explanations as to why one country might have a higher level of income per person. However, although we might speculate about which factor is most responsible for thedifference in income per person across countries, it is not possible to say for certain given the largenumber of other variables that also affect income per person. For example, some countries may have more developed capital markets, less government corruption, and better access to foreign directinvestment. The Solow model allows us to understand some of the reasons why income per person differs across countries, but given it is a simplified model, it cannot explain all of the reasons why income per person may differ.More Problems and Applications to Chapter 91. a. The growth in total output (Y) depends on the growth rates of labor (L), capital (K), and totalfactor productivity (A), as summarized by the equationΔY/Y = αΔK/K + (1 –α)ΔL/L + ΔA/A,where α is capital’s share of output. We can look at the effect on output of a 5-percent increase in labor by setting ΔK/K = ΔA/A = 0. Since α = 2/3, this gives usΔY/Y = (1/3)(5%)= 1.67%.A 5-percent increase in labor input increases output by 1.67 percent.Labor productivity is Y/L. We can write the growth rate in labor productivity asD Y Y =D(Y/L)Y/L-D LL.Substituting for the growth in output and the growth in labor, we findΔ(Y/L)/(Y/L) = 1.67% – 5.0%= –3.34%.Labor productivity falls by 3.34 percent.To find the change in total factor productivity, we use the equationΔA/A = ΔY/Y –αΔK/K – (1 –α)ΔL/L.For this problem, we findΔA/A = 1.67% – 0 – (1/3)(5%)= 0.Total factor productivity is the amount of output growth that remains after we have accounted for the determinants of growth that we can measure. In this case, there is no change in technology, so all of the output growth is attributable to measured input growth. That is, total factorproductivity growth is zero, as expected.b. Between years 1 and 2, the capital stock grows by 1/6, labor input grows by 1/3, and output growsby 1/6. We know that the growth in total factor productivity is given byΔA/A = ΔY/Y –αΔK/K – (1 –α)ΔL/L.Substituting the numbers above, and setting α = 2/3, we findΔA/A = (1/6) – (2/3)(1/6) – (1/3)(1/3)= 3/18 – 2/18 – 2/18= – 1/18= –0.056.Total factor productivity falls by 1/18, or approximately 5.6 percent.2. By definition, output Y equals labor productivity Y/L multiplied by the labor force L:Y = (Y/L)L.Using the mathematical trick in the hint, we can rewrite this asD Y Y =D(Y/L)Y/L+D LL.We can rearrange this asD Y Y =D YY-D LL.Substituting for ΔY/Y from the text, we findD(Y/L) Y/L =D AA+aD KK+(1-a)D LL-D LL =D AA+aD KK-aD LL=D AA+aD KK-D LLéëêêùûúúUsing the same trick we used above, we can express the term in brackets asΔK/K –ΔL/L = Δ(K/L)/(K/L)Making this substitution in the equation for labor productivity growth, we conclude thatD(Y/L) Y/L =D AA+aD(K/L)K/L.3. We know the following:ΔY/Y = n + g = 3.6%ΔK/K = n + g = 3.6%ΔL/L = n = 1.8%Capital’s Share = α = 1/3Labor’s Share = 1 –α = 2/3Using these facts, we can easily find the contributions of each of the factors, and then find the contribution of total factor productivity growth, using the following equations:Output = Capital’s+ Labor’s+ Total FactorGrowth Contribution Contribution ProductivityD Y Y =aD KK+(1-a)D LL+D AA3.6% = (1/3)(3.6%) + (2/3)(1.8%) + ΔA/A.We can easily solve this for ΔA/A, to find that3.6% = 1.2% + 1.2% + 1.2%Chapter 9—Economic Growth II: Technology, Empirics, and Policy 81We conclude that the contribution of capital is 1.2 percent per year, the contribution of labor is 1.2 percent per year, and the contribution of total factor productivity growth is 1.2 percent per year. These numbers match the ones in Table 9-3 in the text for the United States from 1948–2002.Chapter 9—Economic Growth II: Technology, Empirics, and Policy 82。

操作系统第九版部分课后作业习题答案

CHAPTER 9 Virtual Memory Practice Exercises9.1 Under what circumstances do page faults occur? Describe the actions taken by the operating system when a page fault occurs.Answer:A page fault occurs when an access to a page that has not beenbrought into main memory takes place. The operating system veriﬁesthe memory access, aborting the program if it is invalid. If it is valid, a free frame is located and I/O is requested to read the needed page into the free frame. Upon completion of I/O, the process table and page table are updated and the instruction is restarted.9.2 Assume that you have a page-reference string for a process with m frames (initially all empty). The page-reference string has length p;n distinct page numbers occur in it. Answer these questions for any page-replacement algorithms:a. What is a lower bound on the number of page faults?b. What is an upper bound on the number of page faults?Answer:a. nb. p9.3 Consider the page table shown in Figure 9.30 for a system with 12-bit virtual and physical addresses and with 256-byte pages. The list of freepage frames is D, E, F (that is, D is at the head of the list, E is second, and F is last).Convert the following virtual addresses to their equivalent physical addresses in hexadecimal. All numbers are given in hexadecimal. (A dash for a page frame indicates that the page is not in memory.)• 9EF• 1112930 Chapter 9 Virtual Memory• 700• 0FFAnswer:• 9E F - 0E F• 111 - 211• 700 - D00• 0F F - EFF9.4 Consider the following page-replacement algorithms. Rank these algorithms on a ﬁve-point scale from “bad” to “perfect” according to their page-fault rate. Separate those algorithms that suffer from Belady’s anomaly from those that do not.a. LRU replacementb. FIFO replacementc. Optimal replacementd. Second-chance replacementAnswer:Rank Algorithm Suffer from Belady’s anomaly1 Optimal no2 LRU no3 Second-chance yes4 FIFO yes9.5 Discuss the hardware support required to support demand paging. Answer:For every memory-access operation, the page table needs to be consulted to check whether the corresponding page is resident or not and whether the program has read or write privileges for accessing the page. These checks have to be performed in hardware. A TLB could serve as a cache and improve the performance of the lookup operation.9.6 An operating system supports a paged virtual memory, using a central processor with a cycle time of 1 microsecond. It costs an additional 1 microsecond to access a page other than the current one. Pages have 1000 words, and the paging device is a drum that rotates at 3000 revolutions per minute and transfers 1 million words per second. The following statistical measurements were obtained from the system:• 1 percent of all instructions executed accessed a page other than the current page.•Of the instructions that accessed another page, 80 percent accesseda page already in memory.Practice Exercises 31•When a new page was required, the replaced page was modiﬁed 50 percent of the time.Calculate the effective instruction time on this system, assuming that the system is running one process only and that the processor is idle during drum transfers.Answer:effective access time = 0.99 × (1 sec + 0.008 × (2 sec)+ 0.002 × (10,000 sec + 1,000 sec)+ 0.001 × (10,000 sec + 1,000 sec)= (0.99 + 0.016 + 22.0 + 11.0) sec= 34.0 sec9.7 Consider the two-dimensional array A:int A[][] = new int[100][100];where A[0][0] is at location 200 in a paged memory system with pages of size 200. A small process that manipulates the matrix resides in page 0 (locations 0 to 199). Thus, every instruction fetch will be from page 0. For three page frames, how many page faults are generated bythe following array-initialization loops, using LRU replacement andassuming that page frame 1 contains the process and the other twoare initially empty?a. for (int j = 0; j < 100; j++)for (int i = 0; i < 100; i++)A[i][j] = 0;b. for (int i = 0; i < 100; i++)for (int j = 0; j < 100; j++)A[i][j] = 0;Answer:a. 5,000b. 509.8 Consider the following page reference string:1, 2, 3, 4, 2, 1, 5, 6, 2, 1, 2, 3, 7, 6, 3, 2, 1, 2, 3, 6.How many page faults would occur for the following replacement algorithms, assuming one, two, three, four, ﬁve, six, or seven frames? Remember all frames are initially empty, so your ﬁrst unique pages will all cost one fault each.•LRU replacement• FIFO replacement•Optimal replacement32 Chapter 9 Virtual MemoryAnswer:Number of frames LRU FIFO Optimal1 20 20 202 18 18 153 15 16 114 10 14 85 8 10 76 7 10 77 77 79.9 Suppose that you want to use a paging algorithm that requires a referencebit (such as second-chance replacement or working-set model), butthe hardware does not provide one. Sketch how you could simulate a reference bit even if one were not provided by the hardware, or explain why it is not possible to do so. If it is possible, calculate what the cost would be.Answer:You can use the valid/invalid bit supported in hardware to simulate the reference bit. Initially set the bit to invalid. O n ﬁrst reference a trap to the operating system is generated. The operating system will set a software bit to 1 and reset the valid/invalid bit to valid.9.10 You have devised a new page-replacement algorithm that you thinkmaybe optimal. In some contorte d test cases, Belady’s anomaly occurs. Is the new algorithm optimal? Explain your answer.Answer:No. An optimal algorithm will not suffer from Belady’s anomaly because —by deﬁnition—an optimal algorithm replaces the page that will notbe used for the long est time. Belady’s anomaly occurs when a pagereplacement algorithm evicts a page that will be needed in the immediatefuture. An optimal algorithm would not have selected such a page.9.11 Segmentation is similar to paging but uses variable-sized“pages.”Deﬁnetwo segment-replacement algorithms based on FIFO and LRU pagereplacement schemes. Remember that since segments are not the samesize, the segment that is chosen to be replaced may not be big enoughto leave enough consecutive locations for the needed segment. Consider strategies for systems where segments cannot be relocated, and thosefor systems where they can.Answer:a. FIFO. Find the ﬁrst segment large enough to accommodate the incoming segment. If relocation is not possible and no one segmentis large enough, select a combination of segments whose memoriesare contiguous, which are “closest to the ﬁrst of the list” andwhich can accommodate the new segment. If relocation is possible, rearrange the memory so that the ﬁrstNsegments large enough forthe incoming segment are contiguous in memory. Add any leftover space to the free-space list in both cases.Practice Exercises 33b. LRU. Select the segment that has not been used for the longestperiod of time and that is large enough, adding any leftover spaceto the free space list. If no one segment is large enough, selecta combination of the “oldest” segments that are contiguous inmemory (if relocation is not available) and that are large enough.If relocation is available, rearrange the oldest N segments to be contiguous in memory and replace those with the new segment.9.12 Consider a demand-paged computer system where the degree of multiprogramming is currently ﬁxed at four. The system was recently measured to determine utilization of CPU and the paging disk. The results are one of the following alternatives. For each case, what is happening? Can the degree of multiprogramming be increased to increase the CPU utilization? Is the paging helping?a. CPU utilization 13 percent; disk utilization 97 percentb. CPU utilization 87 percent; disk utilization 3 percentc. CPU utilization 13 percent; disk utilization 3 percentAnswer:a. Thrashing is occurring.b. CPU utilization is sufﬁciently high to leave things alone, and increase degree of multiprogramming.c. Increase the degree of multiprogramming.9.13 We have an operating system for a machine that uses base and limit registers, but we have modiﬁed the ma chine to provide a page table.Can the page tables be set up to simulate base and limit registers? How can they be, or why can they not be?Answer:The page table can be set up to simulate base and limit registers provided that the memory is allocated in ﬁxed-size segments. In this way, the base of a segment can be entered into the page table and the valid/invalid bit used to indicate that portion of the segment as resident in the memory. There will be some problem with internal fragmentation.9.27.Consider a demand-paging system with the following time-measured utilizations:CPU utilization 20%Paging disk 97.7%Other I/O devices 5%Which (if any) of the following will (probably) improve CPU utilization? Explain your answer.a. Install a faster CPU.b. Install a bigger paging disk.c. Increase the degree of multiprogramming.d. Decrease the degree of multiprogramming.e. Install more main memory.f. Install a faster hard disk or multiple controllers with multiple hard disks.g. Add prepaging to the page fetch algorithms.h. Increase the page size.Answer: The system obviously is spending most of its time paging, indicating over-allocationof memory. If the level of multiprogramming is reduced resident processeswould page fault less frequently and the CPU utilization would improve. Another way toimprove performance would be to get more physical memory or a faster paging drum.a. Get a faster CPU—No.b. Get a bigger paging drum—No.c. Increase the degree of multiprogramming—No.d. Decrease the degree of multiprogramming—Yes.e. Install more main memory—Likely to improve CPU utilization as more pages canremain resident and not require paging to or from the disks.f. Install a faster hard disk, or multiple controllers with multiple hard disks—Also animprovement, for as the disk bottleneck is removed by faster response and morethroughput to the disks, the CPU will get more data more quickly.g. Add prepaging to the page fetch algorithms—Again, the CPU will get more datafaster, so it will be more in use. This is only the case if the paging action is amenableto prefetching (i.e., some of the access is sequential).h. Increase the page size—Increasing the page size will result in fewer page faults ifdata is being accessed sequentially. If data access is more or less random, morepaging action could ensue because fewer pages can be kept in memory and moredata is transferred per page fault. So this change is as likely to decrease utilizationas it is to increase it.10.1、Is disk scheduling, other than FCFS scheduling, useful in asingle-userenvironment? Explain your answer.Answer: In a single-user environment, the I/O queue usually is empty. Requests generally arrive from a single process for one block or for a sequence of consecutive blocks. In these cases, FCFS is an economical method of disk scheduling. But LOOK is nearly as easy to program and will give much better performance when multiple processes are performing concurrent I/O, such as when aWeb browser retrieves data in the background while the operating system is paging and another application is active in the foreground.10.2.Explain why SSTF scheduling tends to favor middle cylindersover theinnermost and outermost cylinders.The center of the disk is the location having the smallest average distance to all other tracks.Thus the disk head tends to move away from the edges of the disk.Here is another way to think of it.The current location of the head divides the cylinders into two groups.If the head is not in the center of the disk and a new request arrives,the new request is more likely to be in the group that includes the center of the disk;thus,the head is more likely to move in that direction.10.11、Suppose that a disk drive has 5000 cylinders, numbered 0 to 4999. The drive is currently serving a request at cylinder 143, and the previous request was at cylinder 125. The queue of pending requests, in FIFO order, is86, 1470, 913, 1774, 948, 1509, 1022, 1750, 130Starting from the current head position, what is the total distance (in cylinders) that the disk arm moves to satisfy all the pending requests, for each of the following disk-scheduling algorithms?a. FCFSb. SSTFc. SCANd. LOOKe. C-SCANAnswer:a. The FCFS schedule is 143, 86, 1470, 913, 1774, 948, 1509, 1022, 1750, 130. The total seek distance is 7081.b. The SSTF schedule is 143, 130, 86, 913, 948, 1022, 1470, 1509, 1750, 1774. The total seek distance is 1745.c. The SCAN schedule is 143, 913, 948, 1022, 1470, 1509, 1750, 1774, 4999, 130, 86. The total seek distance is 9769.d. The LOOK schedule is 143, 913, 948, 1022, 1470, 1509, 1750, 1774, 130, 86. The total seek distance is 3319.e. The C-SCAN schedule is 143, 913, 948, 1022, 1470, 1509, 1750, 1774, 4999, 86, 130. The total seek distance is 9813.f. (Bonus.) The C-LOOK schedule is 143, 913, 948, 1022, 1470, 1509, 1750, 1774, 86, 130. The total seek distance is 3363.12CHAPTERFile-SystemImplementationPractice Exercises12.1 Consider a ﬁle currently consisting of 100 blocks. Assume that the ﬁlecontrol block (and the index block, in the case of indexed allocation)is already in memory. Calculate how many disk I/O operations are required for contiguous, linked, and indexed (single-level) allocation strategies, if, for one block, the following conditions hold. In the contiguous-allocation case, assume that there is no room to grow atthe beginning but there is room to grow at the end. Also assume thatthe block information to be added is stored in memory.a. The block is added at the beginning.b. The block is added in the middle.c. The block is added at the end.d. The block is removed from the beginning.e. The block is removed from the middle.f. The block is removed from the end.Answer:The results are:Contiguous Linked Indexeda. 201 1 1b. 101 52 1c. 1 3 1d. 198 1 0e. 98 52 0f. 0 100 012.2 What problems could occur if a system allowed a ﬁle system to be mounted simultaneously at more than one location?Answer:4344 Chapter 12 File-System ImplementationThere would be multiple paths to the same ﬁle, which could confuse users or encourage mistakes (deleting a ﬁle with one path deletes theﬁle in all the other paths).12.3 Why must the bit map for ﬁle allocation be kept on mass storage, ratherthan in main memory?Answer:In case of system crash (memory failure) the free-space list would notbe lost as it would be if the bit map had been stored in main memory.12.4 Consider a system that supports the strategies of contiguous, linked, and indexed allocation. What criteria should be used in deciding which strategy is best utilized for a particular ﬁle?Answer:•Contiguous—if ﬁle is usually accessed sequentially, if ﬁle isrelatively small.•Linked—if ﬁle is large and usually accessed sequentially.• Indexed—if ﬁle is large and usually accessed randomly.12.5 One problem with contiguous allocation is that the user must preallocate enough space for each ﬁle. If the ﬁle grows to be larger than thespace allocated for it, special actions must be taken. One solution to this problem is to deﬁne a ﬁle structure consisting of an initial contiguous area (of a speciﬁed size). If this area is ﬁlled, the operating system automatically deﬁnes an overﬂow area that is linked to the initialc ontiguous area. If the overﬂow area is ﬁlled, another overﬂow areais allocated. Compare this implementation of a ﬁle with the standard contiguous and linked implementations.Answer:This method requires more overhead then the standard contiguousallocation. It requires less overheadthan the standard linked allocation. 12.6 How do caches help improve performance? Why do systems not use more or larger caches if they are so useful?Answer:Caches allow components of differing speeds to communicate moreefﬁcie ntly by storing data from the slower device, temporarily, ina faster device (the cache). Caches are, almost by deﬁnition, more expensive than the device they are caching for, so increasing the number or size of caches would increase system cost.12.7 Why is it advantageous for the user for an operating system to dynamically allocate its internal tables? What are the penalties to the operating system for doing so?Answer:Dynamic tables allow more ﬂexibility in system use growth — tablesare never exceeded, avoiding artiﬁcial use limits. Unfortunately, kernel structures and code are more complicated, so there is more potentialfor bugs. The use of one resource can take away more system resources (by growing to accommodate the requests) than with static tables.Practice Exercises 4512.8 Explain how the VFS layer allows an operating system to support multiple types of ﬁle systems easily.Answer:VFS introduces a layer of indirection in the ﬁle system implementation. In many ways, it is similar to object-oriented programming techniques. System calls can be made generically (independent of ﬁle system type). Each ﬁle system type provides its function calls and data structuresto the VFS layer. A system call is translated into the proper speciﬁc functions for the ta rget ﬁle system at the VFS layer. The calling program has no ﬁle-system-speciﬁc code, and the upper levels of the system call structures likewise are ﬁle system-independent. The translation at the VFS layer turns these generic calls into ﬁle-system-speciﬁc operations.。

Chapter 9 参考答案

第九章练习题(参考答案)I. Define the following terms (名词解释) ：标准答案：略II. Multiple Choice Questions (单项选择)：III. Fill in the Blanks Questions (填空题)IV. True/False Questions (判断题)标准答案：1-5: TTTFT 6-10: FTFTF 11-15: TTTFTV. Essay Questions(问答题)1.参考答案: When large market segments can be identified, economies of scale inproduction and marketing can be important competitive advantages of global companies. Other benefits include: (a) a transfer of experience and know-how across countries through improved coordination and integration of marketing activities, (b) ensures access to the toughest customers, and (c) diversity of markets served carries with it additional financial benefits.2.参考答案：Corporate planning is essentially long term, incorporating generalizedgoals for the enterprise as a whole. Strategic planning is conducted at the highest levels of management and deals with products, capital, and research, and long- and short-term goals of the company. Tactical planning pertains to specific actions and to the allocation of resources used to implement strategic planning goals in specific markets. Tactical plans are made at the local level and address marketing and advertising questions.3.参考答案：The four phases are: (a) Phase 1--Preliminary analysis and screening(matching company/country needs); (b) Phase 2--Adapting the marketing mix totarget markets; (c) Phase 3--Developing the marketing plan; and, (d) Phase 4--Implementation and control.4.参考答案：Contractual agreements generally involve the transfer of technology,processes, trademarks, or human skills. The two basic forms of contractual agreements are licensing and franchising. Licensing is associated with patent rights, trademark rights, and the rights to use technological processes in foreign markets. It is a favorite strategy for small and medium-sized companies.Franchising involves offering a standard package of products, systems, and management services.5.参考答案：The three structures are product, geographic, and a matrix approach.Students could select any of the three options, however, the text suggests that the matrix form is preferable in today's market place. A matrix structure permits management to respond to the conflicts that arise between functional activity, product, and geography. Since the new venture will be a joint venture, the matrix structure might allow both of the companies to bring separate expertise to the table. Since a matrix structure encourages sharing of experience, resources, expertise, technology, and information, it seem to be a natural in this situation.。

(完整版)投资学第10版习题答案09

CHAPTER 9: THE CAPITAL ASSET PRICING MODELPROBLEM SETS1.2.If the security’s correlation coefficient with the market portfolio doubles (with all other variables such as variances unchanged), then beta, and therefore the risk premium, will also double. The current risk premium is: 14% – 6% = 8%The new risk premium would be 16%, and the new discount rate for the security would be: 16% + 6% = 22%If the stock pays a constant perpetual dividend, then we know from the original data that the dividend (D) must satisfy the equation for the present value of a perpetuity:Price = Dividend/Discount rate50 = D /0.14 ⇒ D = 50 ⨯ 0.14 = $7.00At the new discount rate of 22%, the stock would be worth: $7/0.22 = $31.82The increase in stock risk has lowered its value by 36.36%.3. a.False. β = 0 implies E (r ) = r f , not zero.b.False. Investors require a risk premium only for bearing systematic(undiversifiable or market) risk. Total volatility, as measured by the standarddeviation, includes diversifiable risk.c.False. Your portfolio should be invested 75% in the market portfolio and 25% in T-bills. Then:β(0.751)(0.250)0.75P =⨯+⨯=4.The expected return is the return predicted by the CAPM for a given level ofsystematic risk.$1$5()β[()]().04 1.5(.10.04).13,or 13%().04 1.0(.10.04).10,or 10%i f i M f Discount Everything E r r E r r E r E r =+⨯-=+⨯-==+⨯-=()β[()].12.18.06β[.14.06]β 1.5.08P f P M f P P E r r E r r =+⨯-=+⨯-→==5.According to the CAPM, $1 Discount Stores requires a return of 13% based on its systematic risk level of β = 1.5. However, the forecasted return is only 12%. Therefore, the security is currently overvalued.Everything $5 requires a return of 10% based on its systematic risk level of β = 1.0. However, the forecasted return is 11%. Therefore, the security is currently undervalued.6.Correct answer is choice a. The expected return of a stock with a β = 1.0 must, onaverage, be the same as the expected return of the market which also has a β = 1.0.7.Correct answer is choice a. Beta is a measure of systematic risk. Since onlysystematic risk is rewarded, it is safe to conclude that the expected return will be higher for Kaskin’s stock than for Quinn’s stock.8.The appropriate discount rate for the project is:r f + β × [E (r M ) – r f ] = .08 + [1.8 ⨯ (.16 – .08)] = .224, or 22.4%Using this discount rate:Annuity factor (22.4%, 10 years)] = $18.09101$15NPV $40$40[$151.224t t ==-+=-+⨯∑The internal rate of return (IRR) for the project is 35.73%. Recall from your introductory finance class that NPV is positive if IRR > discount rate (or, equivalently, hurdle rate). The highest value that beta can take before the hurdle rate exceeds the IRR is determined by:.3573 = .08 + β × (.16 – .08) ⇒ β = .2773/.08 = 3.479. a.Call the aggressive stock A and the defensive stock D. Beta is the sensitivityof the stock’s return to the market return, i.e., the change in the stock returnper unit change in the market return. Therefore, we compute each stock’s betaby calculating the difference in its return across the two scenarios divided by the difference in the market return:.02.38.06.12β 2.00β0.30.05.25.05.25A D ---====--b.With the two scenarios equally likely, the expected return is an average of thetwo possible outcomes:E (r A ) = 0.5 ⨯ (–.02 + .38) = .18 = 18%E (r D ) = 0.5 ⨯ (.06 + .12) = .09 = 9%c.The SML is determined by the market expected return of [0.5 × (.25 + .05)] = 15%, with βM = 1, and r f = 6% (which has βf =0). See the following graph:The equation for the security market line is:E (r ) = .06 + β × (.15 – .06)d.Based on its risk, the aggressive stock has a required expected return of:E (r A ) = .06 + 2.0 × (.15 – .06) = .24 = 24%The analyst’s forecast of expected return is only 18%. Thus the stock’s alpha is:αA = actually expected return – required return (given risk)= 18% – 24% = –6%Similarly, the required return for the defensive stock is:E (r D ) = .06 + 0.3 × (.15 – .06) = 8.7%The analyst’s forecast of expected return for D is 9%, and hence, the stock has a positive alpha:αD = Actually expected return – Required return (given risk)= .09 – .087 = +0.003 = +0.3%The points for each stock plot on the graph as indicated above.e.The hurdle rate is determined by the project beta (0.3), not the firm’s beta. The correct discount rate is 8.7%, the fair rate of return for stock D.10.Not possible. Portfolio A has a higher beta than Portfolio B, but the expected returnfor Portfolio A is lower than the expected return for Portfolio B. Thus, these two portfolios cannot exist in equilibrium.11.Possible. If the CAPM is valid, the expected rate of return compensates only for systematic (market) risk, represented by beta, rather than for the standard deviation, which includes nonsystematic risk. Thus, Portfolio A’s lower rate of return can be paired with a higher standard deviation, as long as A’s beta is less than B’s.12.Not possible. The reward-to-variability ratio for Portfolio A is better than that of the market. This scenario is impossible according to the CAPM because the CAPM predicts that the market is the most efficient portfolio. Using the numbers supplied:.16.10.18.100.50.33.12.24A M S S --====Portfolio A provides a better risk-reward trade-off than the market portfolio.13.Not possible. Portfolio A clearly dominates the market portfolio. Portfolio A hasboth a lower standard deviation and a higher expected return.14.Not possible. The SML for this scenario is: E(r ) = 10 + β × (18 – 10)Portfolios with beta equal to 1.5 have an expected return equal to:E (r ) = 10 + [1.5 × (18 – 10)] = 22%The expected return for Portfolio A is 16%; that is, Portfolio A plots below the SML (α A = –6%) and, hence, is an overpriced portfolio. This is inconsistent with the CAPM.15.Not possible. The SML is the same as in Problem 14. Here, Portfolio A’s required return is: .10 + (.9 × .08) = 17.2%This is greater than 16%. Portfolio A is overpriced with a negative alpha:α A = –1.2%16.Possible. The CML is the same as in Problem 12. Portfolio A plots below the CML, as any asset is expected to. This scenario is not inconsistent with the CAPM.17.Since the stock’s beta is equal to 1.2, its expected rate of return is:.06 + [1.2 ⨯ (.16 – .06)] = 18%110110$50$6()0.18$53$50D P P PE r P P -+-+=→=→=18.The series of $1,000 payments is a perpetuity. If beta is 0.5, the cash flow should be discounted at the rate:.06 + [0.5 × (.16 – .06)] = .11 = 11%PV = $1,000/0.11 = $9,090.91If, however, beta is equal to 1, then the investment should yield 16%, and the price paid for the firm should be:PV = $1,000/0.16 = $6,250The difference, $2,840.91, is the amount you will overpay if you erroneously assume that beta is 0.5 rather than 1.ing the SML: .04 = .06 + β × (.16 – .06) ⇒ β = –.02/.10 = –0.220.r 1 = 19%; r 2 = 16%; β1 = 1.5; β2 = 1a.To determine which investor was a better selector of individual stocks we look at abnormal return, which is the ex-post alpha; that is, the abnormal return is the difference between the actual return and that predicted by the SML. Without information about the parameters of this equation (risk-free rate and market rate of return) we cannot determine which investor was more accurate.b.If r f = 6% and r M = 14%, then (using the notation alpha for the abnormal return):α1 = .19 – [.06 + 1.5 × (.14 – .06)] = .19 – .18 = 1%α 2 = .16 – [.06 + 1 × (.14 – .06)] = .16 – .14 = 2%Here, the second investor has the larger abnormal return and thus appears to be the superior stock selector. By making better predictions, the secondinvestor appears to have tilted his portfolio toward underpriced stocks.c.If r f = 3% and r M = 15%, then:α1 = .19 – [.03 + 1.5 × (.15 – .03)] = .19 – .21 = –2%α2 = .16 – [.03+ 1 × (.15 – .03)] = .16 – .15 = 1%Here, not only does the second investor appear to be the superior stock selector, but the first investor’s predictions appear valueless (or worse).21. a.Since the market portfolio, by definition, has a beta of 1, its expected rate of return is 12%.b.β = 0 means no systematic risk. Hence, the stock’s expected rate of return in market equilibrium is the risk-free rate, 5%ing the SML, the fair expected rate of return for a stock with β = –0.5 is:()0.05[(0.5)(0.120.05)] 1.5%E r =+-⨯-=The actually expected rate of return, using the expected price and dividend for next year is:$41$3()10.1010%$40E r +=-==Because the actually expected return exceeds the fair return, the stock is underpriced.22.In the zero-beta CAPM the zero-beta portfolio replaces the risk-free rate, and thus:E (r ) = 8 + 0.6(17 – 8) = 13.4%23. a.E (r P ) = r f + βP × [E (r M ) – r f ] = 5% + 0.8 (15% − 5%) = 13%α = 14% - 13% = 1%You should invest in this fund because alpha is positive.b.The passive portfolio with the same beta as the fund should be invested 80% in the market-index portfolio and 20% in the money market account. For this portfolio:E (r P ) = (0.8 × 15%) + (0.2 × 5%) = 13%14% − 13% = 1% = α24. a.We would incorporate liquidity into the CCAPM in a manner analogous to theway in which liquidity is incorporated into the conventional CAPM. In thelatter case, in addition to the market risk premium, expected return is alsodependent on the expected cost of illiquidity and three liquidity-related betaswhich measure the sensitivity of: (1) the security’s illiquidity to marketilliquidity; (2) the security’s return to market illiquidity; and, (3) the security’silliquidity to the market return. A similar approach can be used for theCCAPM, except that the liquidity betas would be measured relative toconsumption growth rather than the usual market index.b.As in part (a), nontraded assets would be incorporated into the CCAPM in afashion similar to part (a). Replace the market portfolio with consumptiongrowth. The issue of liquidity is more acute with nontraded assets such asprivately held businesses and labor income.While ownership of a privately held business is analogous to ownership of anilliquid stock, expect a greater degree of illiquidity for the typical privatebusiness. If the owner of a privately held business is satisfied with thedividends paid out from the business, then the lack of liquidity is not an issue.If the owner seeks to realize income greater than the business can pay out,then selling ownership, in full or part, typically entails a substantial liquiditydiscount. The illiquidity correction should be treated as suggested in part (a).The same general considerations apply to labor income, although it is probablethat the lack of liquidity for labor income has an even greater impact onsecurity market equilibrium values. Labor income has a major impact onportfolio decisions. While it is possible to borrow against labor income tosome degree, and some of the risk associated with labor income can beameliorated with insurance, it is plausible that the liquidity betas ofconsumption streams are quite significant, as the need to borrow against laborincome is likely cyclical.CFA PROBLEMS1. a.Agree; Regan’s conclusion is correct. By definition, the market portfolio lies onthe capital market line (CML). Under the assumptions of capital market theory, allportfolios on the CML dominate, in a risk-return sense, portfolios that lie on theMarkowitz efficient frontier because, given that leverage is allowed, the CMLcreates a portfolio possibility line that is higher than all points on the efficientfrontier except for the market portfolio, which is Rainbow’s portfolio. BecauseEagle’s portfolio lies on the Markowitz efficient frontier at a point other than themarket portfolio, Rainbow’s portfolio dominates Eagle’s portfolio.b.Nonsystematic risk is the unique risk of individual stocks in a portfolio that isdiversified away by holding a well-diversified portfolio. Total risk is composed ofsystematic (market) risk and nonsystematic (firm-specific) risk.Disagree; Wilson’s remark is incorrect. Because both portfolios lie on theMarkowitz efficient frontier, neither Eagle nor Rainbow has any nonsystematicrisk. Therefore, nonsystematic risk does not explain the different expected returns.The determining factor is that Rainbow lies on the (straight) line (the CML)connecting the risk-free asset and the market portfolio (Rainbow), at the point oftangency to the Markowitz efficient frontier having the highest return per unit ofrisk. Wilson’s remark is also countered by the fact that, since nonsystematic riskcan be eliminated by diversification, the expected return for bearingnonsystematic risk is zero. This is a result of the fact that well-diversifiedinvestors bid up the price of every asset to the point where only systematic riskearns a positive return (nonsystematic risk earns no return).2.E(r) = r f + β × [E(r M) −r f]Furhman Labs: E(r) = .05 + 1.5 × [.115 − .05] = 14.75%Garten Testing: E(r) = .05 + 0.8 × [.115 − .05] = 10.20%If the forecast rate of return is less than (greater than) the required rate of return,then the security is overvalued (undervalued).Furhman Labs: Forecast return – Required return = 13.25% − 14.75% = −1.50%Garten Testing: Forecast return – Required return = 11.25% − 10.20% = 1.05%Therefore, Furhman Labs is overvalued and Garten Testing is undervalued.3. a.4. d.From CAPM, the fair expected return = 8 + 1.25 × (15 - 8) = 16.75%Actually expected return = 17%α = 17 - 16.75 = 0.25%5. d.6. c.7. d.8. d.[You need to know the risk-free rate]9. d.[You need to know the risk-free rate]10.Under the CAPM, the only risk that investors are compensated for bearing is therisk that cannot be diversified away (systematic risk). Because systematic risk(measured by beta) is equal to 1.0 for both portfolios, an investor would expect the same rate of return from both portfolios A and B. Moreover, since both portfolios are well diversified, it doesn’t matter if the specific risk of the individual securities is high or low. The firm-specific risk has been diversified away for both portfolios.11. a.McKay should borrow funds and invest those funds proportionately inMurray’s existing portfolio (i.e., buy more risky assets on margin). In additionto increased expected return, the alternative portfolio on the capital marketline will also have increased risk, which is caused by the higher proportion ofrisky assets in the total portfolio.b.McKay should substitute low-beta stocks for high-beta stocks in order toreduce the overall beta of York’s portfolio. By reducing the overall portfoliobeta, McKay will reduce the systematic risk of the portfolio and, therefore,reduce its volatility relative to the market. The security market line (SML)suggests such action (i.e., moving down the SML), even though reducing betamay result in a slight loss of portfolio efficiency unless full diversification ismaintained. York’s primary objective, however, is not to maintain efficiencybut to reduce risk exposure; reducing portfolio beta meets that objective.Because York does not want to engage in borrowing or lending, McKaycannot reduce risk by selling equities and using the proceeds to buy risk-freeassets (i.e., lending part of the portfolio).12. a.Expected Return AlphaStock X5% + 0.8 × (14% - 5%) = 12.2%14.0% - 12.2% = 1.8%Stock Y5% + 1.5 × (14% - 5%) = 18.5%17.0% - 18.5% = -1.5%b.i. Kay should recommend Stock X because of its positive alpha, compared toStock Y, which has a negative alpha. In graphical terms, the expectedreturn/risk profile for Stock X plots above the security market line (SML),while the profile for Stock Y plots below the SML. Also, depending on theindividual risk preferences of Kay’s clients, the lower beta for Stock X mayhave a beneficial effect on overall portfolio risk.ii. Kay should recommend Stock Y because it has higher forecasted return andlower standard deviation than Stock X. The respective Sharpe ratios for StocksX and Y and the market index are:Stock X:(14% - 5%)/36% = 0.25Stock Y:(17% - 5%)/25% = 0.48Market index:(14% - 5%)/15% = 0.60The market index has an even more attractive Sharpe ratio than either of the individual stocks, but, given the choice between Stock X and Stock Y, Stock Y is the superior alternative.When a stock is held as a single stock portfolio, standard deviation is the relevant risk measure. For such a portfolio, beta as a risk measure is irrelevant. Although holding a single asset is not a typically recommended investment strategy, some investors may hold what is essentially a single-asset portfolio when they hold the stock of their employer company. For such investors, the relevance of standard deviation versus beta is an important issue.。