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Datavideo SE-200 视频编辑器说明书

Datavideo SE-200 视频编辑器说明书

Table of ContentsWarnings and Precautions (3)Disposal (4)Warranty (5)Standard Warranty (5)Two Year Warranty (5)Packing List (6)Introduction (6)Features (6)Connections & Controls (7)Front Panel (7)Rear Panel (9)Firmware Update (10)Frequently-Asked Questions (11)Specifications (12)Service & Support (16)Disclaimer of Product and ServicesThe information offered in this instruction manual is intended as a guide only. At all times, Datavideo Technologies will try to give correct, complete and suitable information. However, Datavideo Technologies cannot exclude that some information in this manual, from time to time, may not be correct or may be incomplete. This manual may contain typing errors, omissions or incorrect information. Datavideo Technologies always recommend that you double check the information in this document for accuracy before making any purchase decision or using the product. Datavideo Technologies is not responsible for any omissions or errors, or for any subsequent loss or damage caused by using the information contained within this manual. Further advice on the content of this manual or on the product can be obtained by contacting your local Datavideo Office or dealer.Warnings and Precautions1. Read all of these warnings and save them for later reference.2. Follow all warnings and instructions marked on this unit.3. Unplug this unit from the wall outlet before cleaning. Do not useliquid or aerosol cleaners. Use a damp cloth for cleaning.4. Do not use this unit in or near water.5. Do not place this unit on an unstable cart, stand, or table. Theunit may fall, causing serious damage.6. Slots and openings on the cabinet top, back, and bottom areprovided for ventilation. To ensure safe and reliable operation of this unit, and to protect it from overheating, do not block or cover these openings. Do not place this unit on a bed, sofa, rug, or similar surface, as the ventilation openings on the bottom of the cabinet will be blocked. This unit should never be placed near or over a heat register or radiator. This unit should not be placed ina built-in installation unless proper ventilation is provided.7. This product should only be operated from the type of powersource indicated on the marking label of the AC adapter. If you are not sure of the type of power available, consult your Datavideo dealer or your local power company.8. Do not allow anything to rest on the power cord. Do not locatethis unit where the power cord will be walked on, rolled over, or otherwise stressed.9. If an extension cord must be used with this unit, make sure thatthe total of the ampere ratings on the products plugged into the extension cord do not exceed the extension cord’s rating.10. Make sure that the total amperes of all the units that are pluggedinto a single wall outlet do not exceed 15 amperes.11. Never push objects of any kind into this unit through the cabinetventilation slots, as they may touch dangerous voltage points or short out parts that could result in risk of fire or electric shock.Never spill liquid of any kind onto or into this unit.12. Except as specifically explained elsewhere in this manual, do notattempt to service this product yourself. Opening or removingcovers that are marked “Do Not Remove” may expose you todangerous voltage points or other risks, and will void your warranty. Refer all service issues to qualified service personnel. 13. Unplug this product from the wall outlet and refer to qualifiedservice personnel under the following conditions:a. When the power cord is damaged or frayed;b. When liquid has spilled into the unit;c. When the product has been exposed to rain or water;d. When the product does not operate normally undernormal operating conditions. Adjust only thosecontrols that are covered by the operatinginstructions in this manual; improper adjustment ofother controls may result in damage to the unit andmay often require extensive work by a qualifiedtechnician to restore the unit to normal operation;e. When the product has been dropped or the cabinethas been damaged;f. When the product exhibits a distinct change inperformance, indicating a need for service. DisposalFor EU Customers only - WEEE MarkingThis symbol on the product indicates that it should notbe treated as household waste. It must be handedover to the applicable take-back scheme for therecycling of Waste Electrical and Electronic Equipment. For more detailed information about the recycling of this product, please contact your local Datavideo office.WarrantyStandard Warranty• Datavideo equipment is guaranteed against any manufacturing defects for one year from the date of purchase.• The original purchase invoice or other documentary evidence should be supplied at the time of any request for repair underwarranty.• Damage caused by accident, misuse, unauthorized repairs, sand, grit or water is not covered by this warranty.• All mail or transportation costs including insurance are at the expense of the owner.• All other claims of any nature are not covered.• Cables & batteries are not covered under warranty.• Warranty only valid within the country or region of purchase.• Your statutory rights are not affected.Two Year WarrantyAll Datavideo products purchased after 01-Oct.-2008 qualify for a free one year extension to the standard Warranty, providing the product is registered with Datavideo within 30 days of purchase. For information on how to register please visit or contact your local Datavideo office or authorized DistributorsCertain parts with limited lifetime expectancy such as LCD Panels, DVD Drives and Hard Drives are only covered for the first 10,000 hours, or 1 year (whichever comes first).Any second year warranty claims must be made to your local Datavideo office or one of its authorized Distributors before the extended warranty expires.Packing ListItems Description Q’ty1 DAC-50S Unit 12 Accessory List 1 IntroductionThe DAC-50S converts the digital HD-SDI video input to one of the following analog output combinations:one component (YUV) output + one composite (CV) output;one S-Video (Y/C) output + two composite (CV) outputs,with both of which paired with two analog audio outputs (R and L).The outputs can be easily connected to analog video monitor or deck devices such as BataCAM and VHS. The DAC-50S is also able to scale down the signal to connect analog output ports to SD analog equipment.The DAC-50S provides an SDI Loop-thru output, which makes it suitable for more different applications. It further outputs one pair of analog audio channels selected from 16 de-embedded SDI audio channels (Selection is determined by DIP switch settings).The DAC-50S is housed in a rugged aluminum frame, making it durable enough to endure even in the roughest conditions. It is also suitable for 2U rack mount kit of RMK-2 on standard rack, and fits the battery holder of MB-4.Features•Converts 3G/HD-SDI input to analog video and 2 unbalance analog audio channels•Two available output combinations:o one analog component (YUV) output + one composite (CV) outputo one analog S-Video (Y/C) output + two composite (CV) outputs •Supports NTSC / PAL output•Define de-embedded audio channels sent to analog audio outputs via a DIP switch•Solid aluminum housing that fits the 2U rack mount kit of RMK-2 and battery holder of MB-4Connections & Controls Front PanelDIP Switch Setting 0Rear PanelFirmware Update1. Set DIP SW8 to ON2. Connect DAC-50S to the PC using USB cable (DAC-50S is powered byUSB port so external power source is not required)3. USB device connection prompt will appear on the PC screen (DAC-50SUSB) as shown in the diagram below.4. Copy the new firmware.bin and replace the file in the USB device.5. Safely remove the USB device (VP733 Datavideo Tech USB Device) asshown in the diagram below.6.Update is complete; unplug the USB cable and set DIP SW8 to OFF.Frequently-Asked QuestionsThis section describes problems that you may encounter while using DAC-50S. If you have questions, please refer to related sections and follow all the suggested solutions. If problem still exists, please contact your distributor or the service center.No. Problems Solutions1. Serious color bar noise isseen on the output end ofthe DAC-50S if the SDI-INsource is a 100% color bar. DAC-50S does not support 100% color bar and downgrading it to 75% will solve this issue.SpecificationsInputInterface HD-SDI x 1 (BNC)Output Interface Component YPbPr x 3 (BNC)Composite x 2 (one is the same connector as U/Pb) (BNC) S-Video Y/C x 2 (the same connector as Y and V/Pr) (BNC) Audio L/R x 2 (RCA)HD-SDI loop-through x 1 (BNC)Input Format 1080p 50/59.94 Hz 1080i 50/59.94 Hz 720p 50/59.94 HzOutputFormat 625/25 PAL, 525/29.97 NTSC F/W Upgrade mini USB x 1DIP Switches 8-PIN DIP Switch1: SDI audio de-embed2: SDI audio de-embed3: SDI audio de-embed4: IRE (7.5 IRE & 0 IRE)5: Aspect Ratio (16:9 & 4:3)6: Video Output (YUV + CV OUT or Y/C + TWO CV OUT)7. Audio Level (SMPTE & EBU)8. Firmware UpgradePower Output DC 12V / 0.5A (6W)Service & Support。

HullFund8eCh06ProblemSolutions【范本模板】

HullFund8eCh06ProblemSolutions【范本模板】

CHAPTER 6Interest Rate FuturesPractice QuestionsProblem 6.8.The price of a 90—day Treasury bill is quoted as 10.00。

What continuously compounded return (on an actual/365 basis) does an investor earn on the Treasury bill for the 90—day period?The cash price of the Treasury bill is 90100109750360$-⨯=.The annualized continuously compounded return is 36525ln 1102790975%.⎛⎫+=. ⎪.⎝⎭Problem 6.9.It is May 5, 2013。

The quoted price of a government bond with a 12% coupon that matures on July 27, 2024, is 110-17. What is the cash price ?The number of days between January 27, 2013 and May 5, 2013 is 98. The number of days between January 27, 2013 and July 27, 2013 is 181. The accrued interest is therefore 98632486181⨯=. The quoted price is 110。

5312。

The cash price is therefore1105312324861137798.+.=.or $113.78.Problem 6.10。

ch06partSolution

ch06partSolution

Selected Solutions for Exercises inNumerical Methods with Matlab:Implementations and ApplicationsGerald W.RecktenwaldChapter6Finding the Roots of f(x)=0The following pages contain solutions to selected end-of-chapter Exercisesfrom the book Numerical Methods with Matlab:Implementations andApplications,by Gerald W.Recktenwald,c 2000,Prentice-Hall,Upper Saddle River,NJ.The solutions are c 2000Gerald W.Recktenwald.ThePDF version of the solutions may be downloaded or stored or printed onlyfor noncommercial,educational use.Repackaging and sale of these solutionsin any form,without the written consent of the author,is prohibited.The latest version of this PDFfile,along with other supplemental material for the book,can be found at /recktenwald.2Finding the Roots of f(x)=0 6–2The function f(x)=sin(x2)+x2−2x−0.09has four roots in the interval−1≤x≤3.Given the m-file fx.m,which containsfunction f=fx(x)f=sin(x.^2)+x.^2-2*x-0.09;the statement>>brackPlot(’fx’,-1,3)produces only two brackets.Is this result due to a bug in brackPlot or fx?What needs to be changed so that all four roots are found?Demonstrate that your solution works.Partial Solution:The statement>>Xb=brackPlot(’fx’,-1,3)Xb=-0.15790.05262.1579 2.3684returns two brackets.A close inspection of the plot of f(x)reveals that f(x)crosses the x-axis twice near x=1.3.These two roots are missed by brackPlot because there default search interval is too coarse.There is no bug in brackPlot.Implementing a solution using afiner search interval is left as an exercise.6–11Use the bisect function to evaluate the root of the Colebrook equation(see Exercise8) for /D=0.02and Re=105.Do not modify bisect.m.This requires that you write an appropriate function m-file to evaluate the Colebrook equation.Partial Solution:Using bisect requires writing an auxiliary function to evaluate the Cole-brook equation in the form F(f)=0,where f is the friction factor.The following form of F(f)is used in the colebrkz function listed below.F(f)=1√f+2log10/D3.7+2.51Re D√fMany other forms of F(f)will work.function ff=colebrkz(f)%COLEBRKZ Evaluates the Colebrook equation in the form F(f)=0%for use with root-finding routines.%%Input:f=the current guess at the friction factor%%Global Variables:%EPSDIA=ratio of relative roughness to pipe diameter%REYNOLDS=Reynolds number based on pipe diameter%%Output:ff=the"value"of the Colebrook function written y=F(f)%Global variables allow EPSDIA and REYNOLDS to be passed into%colebrkz while bypassing the bisect.m or fzero functionglobal EPSDIA REYNOLDSff=1.0/sqrt(f)+2.0*log10(EPSDIA/3.7+2.51/(REYNOLDS*sqrt(f)));Because the bisect function(unlike fzero)does not allow additional parameters to be passed through to the F(f)function,the values of /D and Re are passed to colebrkz via global variables.Running bisect with colebrkz is left to the reader.For Re=1×105and /D=0.02the solution is f=0.0490.Copyright c 2000,Gerald W.Recktenwald.Photocopying is permitted only for non-commercial educational purposes.Chapter6:Finding the Roots of f(x)=03 6–13Derive the g3(x)functions in Example6.4and Example6.5.(Hint:What is thefixed-pointformula for Newton’s method?)Partial Solution:Thefixed point iteration formulas designated as g3(x)in Example6.4 and Example6.5are obtained by applying Newton’s method.The general form of Newton’smethod for a scalar variable isx k+1=x k−f(x k) f (x k)Example6.4:The f(x)function and its derivative aref(x)=x−x1/3−2f (x)=1−13x−2/3Substituting these expressions into the formula for Newton’s method and simplifying givesx k+1=x k−x k−x1/3k−21−(1/3)x−2/3k=x k(1−(1/3)x−2/3k)−(x k−x1/3k−2)1−(1/3)x−2/3k=x k−(1/3)x1/3k−x k+x1/3k+21−(1/3)x−2/3k=(2/3)x1/3k+21−(1/3)x k=2x1/3k+63−x−2/3kRepeating this analysis for Example6.5is left as an exercise.Copyright c 2000,Gerald W.Recktenwald.Photocopying is permitted only for non-commercial educational purposes.4Finding the Roots of f(x)=0 6–17K.Wark and D.E.Richards(Thermodynamics,6th ed.,1999,McGraw-Hill,Boston,Example 14-2,pp.768–769)compute the equilibrium composition of a mixture of carbon monoxide and oxygen gas at one atmosphere.Determining thefinal composition requires solving3.06=(1−x)(3+x)1/2 x(1+x)1/2for x.Obtain afixed-point iteration formula forfinding the roots of this equation.Implement your formula in a Matlab function and use your function tofind x.If your formula does not converge,develop one that does.Partial Solution:Onefixed point iteration formula is obtained by isolating the factor of (3+x)in the numerator.3.06x(1+x)1/21−x =(3+x)1/2=⇒x=3.06x(1+x)1/21−x2−3=⇒g1(x)=3.06x(1+x)1/21−x2−3Anotherfixed point iteration formula is obtained by solving for the isolated x in the denomi-nator to getx=(1−x)(3+x)1/23.06(1+x)=⇒g2(x)=(1−x)(3+x)1/23.06(1+x)Performing10fixed point iterations with g1(x)givesit xnew1-7.6420163e-012-2.5857113e+003-1.0721050e+014-7.9154865e+015-7.1666488e+026-6.6855377e+037-6.2575617e+048-5.8590795e+059-5.4861826e+0610-5.1370394e+07Thus,g1(x)does not converge.The g2(x)function does converge to the true root of x= 0.340327....Matlab implementations of thefixed point iterations are left as an Exercise. Copyright c 2000,Gerald W.Recktenwald.Photocopying is permitted only for non-commercial educational purposes.Chapter6:Finding the Roots of f(x)=05 6–24Create a modified newton function(say,newtonb)that takes a bracket interval as input instead of a single initial guess.From the bracket limits take one bisection step to determine x0,the initial guess for Newton e the bracket limits to develop relative tolerances on x and f(x)as in the bisect function in Listing6.4.Solution:The newtonb function is listed below.The demoNewtonb function,also listed below, repeats the calculations in Example6.8with the original newton function and with the new newtonb function.Running demoNewtonb gives>>demoNewtonbOriginal newton function:Newton iterations for fx3n.mk f(x)dfdx x(k+1)1-4.422e-018.398e-01 3.526644293139032 4.507e-038.561e-01 3.521380147397333 3.771e-078.560e-01 3.521379706804574 2.665e-158.560e-01 3.5213797068045750.000e+008.560e-01 3.52137970680457newtonb function:Newton iterations for fx3n.mk f(x)dfdx x(k+1)1-4.422e-018.398e-01 3.526644293139032 4.507e-038.561e-01 3.521380147397333 3.771e-078.560e-01 3.521379706804574 2.665e-158.560e-01 3.5213797068045750.000e+008.560e-01 3.52137970680457The two implementations of Newton’s method give identical results because the input to newtonb is the bracket[2,4].This causes the initial bisection step to produce the same initial guess for the Newton iterations that is used in the call to newton.function demoNewtonb%demoNewtonb Use newton and newtonb to find the root of f(x)=x-x^(1/3)-2%%Synopsis:demoNewton%%Input:none%%Output print out of convergence history,and comparison of methodsfprintf(’\nOriginal newton function:\n’);r=newton(’fx3n’,3,5e-16,5e-16,1);fprintf(’\nnewtonb function:\n’);rb=newtonb(’fx3n’,[24],5e-16,5e-16,1);Copyright c 2000,Gerald W.Recktenwald.Photocopying is permitted only for non-commercial educational purposes.6Finding the Roots of f(x)=0 function r=newtonb(fun,x0,xtol,ftol,verbose)%newtonb Newton’s method to find a root of the scalar equation f(x)=0%Initial guess is a bracket interval%%Synopsis:r=newtonb(fun,x0)%r=newtonb(fun,x0,xtol)%r=newtonb(fun,x0,xtol,ftol)%r=newtonb(fun,x0,xtol,ftol,verbose)%%Input:fun=(string)name of mfile that returns f(x)and f’(x).%x0=2-element vector providing an initial bracket for the root%xtol=(optional)absolute tolerance on x.Default:xtol=5*eps%ftol=(optional)absolute tolerance on f(x).Default:ftol=5*eps%verbose=(optional)flag.Default:verbose=0,no printing.%%Output:r=the root of the functionif nargin<3,xtol=5*eps;endif nargin<4,ftol=5*eps;endif nargin<5,verbose=0;endxeps=max(xtol,5*eps);feps=max(ftol,5*eps);%Smallest tols are5*epsif verbosefprintf(’\nNewton iterations for%s.m\n’,fun);fprintf(’k f(x)dfdx x(k+1)\n’);endxref=abs(x0(2)-x0(1));%Use initial bracket in convergence testfa=feval(fun,x0(1));fb=feval(fun,x0(2));fref=max([abs(fa)abs(fb)]);%Use max f in convergence testx=x0(1)+0.5*(x0(2)-x0(1));%One bisection step for initial guessk=0;maxit=15;%Current and max iterationswhile k<=maxitk=k+1;[f,dfdx]=feval(fun,x);%Returns f(x(k-1))and f’(x(k-1))dx=f/dfdx;x=x-dx;if verbose,fprintf(’%3d%12.3e%12.3e%18.14f\n’,k,f,dfdx,x);endif(abs(f/fref)<feps)|(abs(dx/xref)<xeps),r=x;return;endendwarning(sprintf(’root not found within tolerance after%d iterations\n’,k));Copyright c 2000,Gerald W.Recktenwald.Photocopying is permitted only for non-commercial educational purposes.Chapter6:Finding the Roots of f(x)=07 6–27Implement the secant method using Algorithm6.5and Equation(6.13).Test your program by re-creating the results in Example6.10.What happens if10iterations are performed?Replace the formula in Equation(6.13)withx k+1=x k−f(x k)(x k−x k−1)f(x k k−1,whereεis a small number on the order ofεm.How and why does this change the results? Partial Solution:The demoSecant function listed below implements Algorithm(6.5)using Equation(6.13).The f(x)function,Equation6.3,is hard-coded into demoSecant.Note also that demoSecant performs ten iterations without checking for convergence.function demoSecant(a,b);%demoSecant Secant method for finding the root of f(x)=x-x^(1/3)-2=0%Implement Algorithm6.5,using Equation(6.13)%%Synopsis:demoSecant(a,b)%%Input:a,b=initial guesses for the iterations%%Output:print out of iterations;no return values.%copy initial guesses to local variablesxk=b;%x(k)xkm1=a;%x(k-1)fk=fx3(b);%f(x(k))fkm1=fx3(a);%f(x(k-1))fprintf(’\nSecant method:Algorithm6.5,Equation(6.13)\n’);fprintf(’n x(k-1)x(k)f(x(k))\n’);fprintf(’%3d%12.8f%12.8f%12.5e\n’,0,xkm1,xk,fk);for n=1:10x=xk-fk*(xk-xkm1)/(fk-fkm1);%secant formula for updating the rootf=fx3(x);fprintf(’%3d%12.8f%12.8f%12.5e\n’,n,xk,x,f);xkm1=xk;xk=x;%set-up for next iterationfkm1=fk;fk=f;endCopyright c 2000,Gerald W.Recktenwald.Photocopying is permitted only for non-commercial educational purposes.8Finding the Roots of f(x)=0 Running demoSecant with an initial bracket of[3,4](the same bracket used in Example6.10) gives>>demoSecant(3,4)Secant method:Algorithm6.5,Equation(6.13)n x(k-1)x(k)f(x(k))0 3.00000000 4.00000000 4.12599e-011 4.00000000 3.51734262-3.45547e-032 3.51734262 3.52135125-2.43598e-053 3.52135125 3.52137971 1.56730e-094 3.52137971 3.52137971-8.88178e-165 3.52137971 3.52137971-2.22045e-166 3.52137971 3.521379710.00000e+007 3.52137971 3.521379710.00000e+00Warning:Divide by zero.>In/werk/MATLAB_Book/SolutionManual/roots/mfiles/demoSecant.m at line228 3.52137971NaN NaN9NaN NaN NaN10NaN NaN NaNThe secant method has fully converged in6iterations.Continuing the calculations beyond convergence gives afloating point exception because f(x k)−f(x k−1)=0in the denominator of Equation(6.13).In general,it is possible to have f(x k)−f(x k−1)=0before the secant iterations reach convergence.Thus,thefloating point exception exposed by demoSecant should be guarded against in any implementation of the secant method.Implementing thefix suggested in the problem statement is left as an exercise for the reader.Copyright c 2000,Gerald W.Recktenwald.Photocopying is permitted only for non-commercial educational purposes.Chapter6:Finding the Roots of f(x)=09 6–33Write an m-file function to compute h,the depth to which a sphere of radius r,and specific gravity s,floats.(See Example6.12on page281.)The inputs are r and s,and the output ish.Only compute h when s<0.5.The s≥0.5case is dealt with in the following Exercise.If s≥0.5is input,have your function print an error message and stop.(The built-in error function will be useful.)Your function needs to include logic to select the correct root from the list of values returned by the built-in roots function.Partial Solution:The floata function listed below performs the desired computations.We briefly discuss three of the key statements in floata The coefficients of the polynomial are stored in the p vector.Thenc=getreal(roots(p));finds the real roots of the polynomial.The getreal subfunction returns only the real elements of a ing getreal is a defensive programming strategy.The sample calculation in Example6.12obtained only real roots of the polynomial,so getreal would not be necessary in that case.Thek=find(c>0&c<r);statement extracts the indices in the c vector satisfying the criteria0≤c k≤r.Then h=c(k);copies those roots satisfying the criteria to the h vector.No assumption is made that only one root meets the criteria.If more than one root is found a warning message is issued before leaving floata.Testing of floata is left to the reader.Copyright c 2000,Gerald W.Recktenwald.Photocopying is permitted only for non-commercial educational purposes.10Finding the Roots of f(x)=0 function h=floata(r,s)%float Find water depth on a floating,solid sphere with specific gravity<0.5%%Synopsis:h=floata(r,s)%%Input:r=radius of the sphere%s=specific gravity of the sphere(0<s<1)%%Output:h=depth of the sphereif s>=0.5error(’s<0.5required in this version’)elsep=[1-3*r04*s*r^3];%h^3-3*r*h+4*s*r^3=0c=getreal(roots(p));k=find(c>0&c<r);%indices of elements in c such that0<c(k)<rh=c(k);%value of elements in c satisfying above criterionendif length(h)>1,warning(’More than one root found’);end%==============================function cr=getreal(c)%getreal Copy all real elements of input vector to output vector%%Synopsis:cr=getreal(c)%%Input:c=vector of numerical values%%Output cr=vector of only the real elements of c%cr=[]if c has only imaginary elementsn=0;for k=1:length(c)if isreal(c(k))n=n+1;cr(n)=c(k);endendif n==0,cr=[];warning(’No real elements in the input vector’);endCopyright c 2000,Gerald W.Recktenwald.Photocopying is permitted only for non-commercial educational purposes.。

Omega HX15高温相对湿度传感器用户指南说明书

Omega HX15高温相对湿度传感器用户指南说明书

HX15High Temperature Relative Humidity/Temperature®e-mail:**************For latest product manuals:Shop online at®User’sGuideRoHS 2 CompliantIt is the policy of OMEGA Engineering, Inc. to comply with all worldwide safety and EMC/EMI regulations that apply. OMEGA is constantly pursuing certification of its products to the European New Approach Directives. OMEGA will add the mark to every appropriate device upon certification.The information contained in this document is believed to be correct, but OMEGA accepts no liability for anyTABLE OF CONTENTSA. General Description 1.B. Unpacking 1.C. Theory of Operation 2.D. Terminal Connections 3.E. Wiring Example 3.F. Mounting 4.G. Calculations 5.H. R H Calibration 6.I. Temp. Calibration 7.J. Maintenance 7.K. Specifications 8.A. GENERAL DESCRIPTIONThe stainless steel probe provides relative humidity as well astemperature out puts. A thin film polymer capacitor senses relative humidity, while temperature is monitored by a 1000 Ohm platinum RTD. The sensors are protected by a stainless steel filter cap easily removed for cleaning. The probe is connected to an electronicsenclosure with a 40 inch (1m) Teflon cable. The enclosure contains the calibration trimmers, signal and power connections via twoterminal blocks. The probe is available in two configurations, as a2.5” (64 mm) probe with a wall mounting clip, and as an 8.5” (216mm) probe with an adjustable duct flange.B. UNPACKINGVerify that the following parts have been received.1.Remove probe with cable and electronics enclosure.2.Instruction manual.3.Wall mounting clip [for 2.5” (64 mm) probe].4. 2 piece duct flange, with o-ring, (3) screws, and a gasket.[for 8.5” (216 mm) probe]C. THEORY OF OPERATIONA 4 to 20 milliamp loop is a series current loop in which atransmitter will vary the current flow depending upon the parameter being measured (Relative Humidity or Temperature). Advantages ofa current output over a voltage output is that is less susceptible tonoise interference and allows the connection of more than one meter or recorder to the loop as long as the maximum resistance is notexceeded.The typical current loop will consist of a power supply, transmitter, and a meter to measure the current flow. The loop resistance is the sum of the impedance of the meter(s) and the lead wire. Themaximum allowable loop impedance of the probe is found by the Formula:Rmax = (power supply voltage-7 volts) /.02 ampsExample: when using a 24 VDC power supply:Rmax= (24-7) /.02=850 ohms (for total wire length to andfrom the transmitter).AWG WIRE SIZE RESISTANCE PER 1000 FEETIf the meter or recorder being used accepts only voltage, convert the current to voltage by installing a 250 ohm resistor across the input terminals of the recorder to obtain a 1 to 5 volts input.D. TERMINAL CONNECTIONS AND TRIM POTSE. WIRING EXAMPLESTYPICAL CURRENT HOOKUP Wires R1 and R2 can be combined into one single wire with a jumper at pins (4) and (2). This will result in 3 wires instead of 4. TRIMPOTS TYPE A. RH GAIN B. RH ZERO C. RH OFFSET D. TEMP GAIN E. TEMP ZEROPROBE INSTALLATIONF. MOUNTINGA.DUCT MOUNTINGSTEPS1.Slide flange holder onto probe with countersinkhole facing front of probe as shown.2.Position o-ring on probe at desired position (for depth intoduct).3.Slide duct flange onto probe as shown.4.Fasten with (3) 6/32 screws and tighten evenly until secure.5.Position gasket between duct flange and duct wall and fastenassembly to duct with (4) #6 sheet metal screws (notincluded).6.Loosening the 6/32 screws allows for repositioning orremoval of the probe without having to remove the ductflange from the wall.The d uct wall requires a 11/16” (.684” or 17.5 mm) hole forprobe, with (4) mounting holes (for #6 sheet metal screws)evenly spaced on a 2.0” (51 mm) circle. Use duct flange astemplate.B.WALL MOUNTING1.Fasten metal clip to wall.2.Snap probe into clip.G. RH AND TEMPERATURE CALCULATIONS1.Maximum current loop impedance for RH or temperature.Rmax = (V supply – 7 volts) /.02 amps2.RH current output (i = current output in milliamperes)3.Temperature current output.oC = (i-4) x 220/16) – 40 iC = (oC + 40) x (16/220) + 4oF = (i-4) x (396/16) – 40 iF = (oF + 40) x (16/396) + 4The upper limit of the humidity measurement range decreases based on temperature. The humidity measurement limit decreases as follows:2.20% per °C from 95 to 120°C (1.11% per °F from 203 to 248°F)1.00% per °C from 120 to 140°C (0.56% per °F from 248 to 284°F) 0.50% per °C from 140 to 160°C (0.28% per °F from 284 to 320°F) 0.25% per °C from 160 to 180°C (0.14% per °F from 320 to 356°F)For example, at 120°C the upper humidity limit decreases to about 45% ((120-95)(2.20%) = 55% decrease which is 45%), at 180°C it is reduced to about 10%.To calculate the RH accuracy, we’ll use the accuracy specs given in the specifications section:+/-2%RH at 25°C and -40° to 150°C at 0.05 % RH/°C.To clarify lets take an example,RH accuracy at 120°C,=+/- (2% + (120-25)0.05%) = +/- (2% + 4.75%)=+/-6.75%So the accuracy at 120°C is +/- 6.75% and the upper humidity measurement limit is 45% RH.H. RH CALIBRATIONRefer to Section D for the location of trim pots A and B.Note: The HX92-CAL Relative Humidity Calibration Kit is for providing the “low” and “high” RH environments for this procedure. The salt solutions in this kit are prepared according to ASTM standard E104-85 to provide 11.3% and 75.3% relative humidity environments. The containers provided in the kit are designed to fit with these instruments.1.Turn the span (trim pot A) all the way up (clockwise).2.Turn the zero (trim pot B) all the way down (counter-clockwise).3.Place the sensor in the low (11.3%) RH environment. Allowat least one hour for stabilization or until the output stopschanging.4.Verify the output is 4 +/-.02 mA. If it is not, return the unitto Omega for evaluation and repair.5.Adjust the zero (trim pot B) to the point where it just starts tocause a change in the output.6.Place the sensor in the high (75.3%) RH environment. Allowat least one hour for stabilization or until the output stopschanging.7.Adjust the span (trim pot A) so the output is equivalent tothe difference between low and high RH environments.Example: 75.3%-11.3% = 64% which is equivalent to 14.24mA.8.Adjust the sensor in the low RH environment and allow atleast one hour for stabilization or until the output stopschanging. Verify the output is equivalent to the low RHenvironment. Example: 11.3% is equivalent to 5.81 mA.I. TEMPERATURE CALIBRATIONTemperature is factory calibrated only, and does not requireany further calibrations.J. MAINTENANCEIf the probe is operated in a dusty environment, theprotective sensor filter, if clogged, may be removed for cleaning.Unscrew filter and gently blow compressed air through screen.If necessary, use a soft brush to remove lint from sensors.If the sensors are subjected to 100% condensation, they must bedried to obtain correct readings. There is no permanentcalibration shift, nor is recalibration necessary if 100%condensation occurs.The instrument should not be exposed to highConcentrations of ammonia or alcohol vapors. However,any environment that is breathable under normal HVACapplications should not affect the sensors. To maintain originalspecifications, it is generally recommended that the RH sensorbe recalibrated on an annual basis depending upon operatingconditions. The temperature sensor does not require calibration.K. SPECIFICATIONS1.Relative Humidity: Thin film polymer capacitor.Input Voltage Range: 7 to 30 VDC (polarity protected).Range, Accuracy: 3% RH to 95% RH (@-40 to 95°C)*, ±2%RH at 25°CTypical Temp. Characteristics: -40°C to 150°C at .05%RH/°COutput: 4 to 20 mA. For 0%RH to 100%RH.Time Constant: Under 30 seconds, 90% response at 25°C in 1M/sec air.* See RH and Temperature Calculations section for full range.2.Temperature: Thin film 1000 ohm platinum RTD.Input Voltage Range: 7 to 30 VDC (polarity protected).Range, Accuracy: -40°C to 180°C (-40°F to 356°F), ±0.5°C (±1°F)Output: 4 to 20 ma. For -40°C to 180°C (-40°F to 356°F)Time Constant: Under 4 seconds, 60% response in 1m/sec air.3.MechanicalStandard Probe: Stainless steel, 2.5” (64 mm) x .625” (16 mm)diameter. 40” (1m) Teflon cable, metal wall mounting clip.Duct Probe: Stainless steel, 8.5” (216 mm) x .625” (16 mm)diameter. 40” (1m) Teflon cable.Duct Flange: 2.75” (70mm) dia., duct hole 11/16 (.684”,17.5 mm) dia. With 4 mounting holes .156(4 mm)diameter (for #6 sheet metal screws), on 2.00”(51mm) circle.Electronics: Operating temp. -20°C to 70°C(-4°F to 158°F)ABS housing 4.72” (120mm) x 3.14” (80 mm) x2.16” (55 mm) H meets NEMA 1,2,3,4,4X,5,12and 13 specifications.Connectors: Liquid-tight with neoprene gland for .09” to .265”diameter cable.4-pin plug in screw terminal block for output connections.5-pin screw terminal block for cable wire inputconnections, accepts #14 to #22 AWG wires.Weight: 2.5” (64 mm) probe with housing 14 oz. (397 grams).8.5” (216 mm) probe with housing and flange 20 oz. (567grams).WARRANTY/DISCLAIMEROMEGA ENGINEERING, INC. warrants this unit to be free of defects in materials and workmanship for a period of 37 months from date of purchase. OMEGA’s WARRANTY adds an additional one (1) month grace period to the normal three (3) year product warranty to cover handling and shipping time. This ensures that OMEGA’s customers receive maximum coverage on each product.If the unit malfunctions, it must be returned to the factory for evaluation. OMEGA’s Customer Service Department will issue an Authorized Return (AR) number immediately upon phone or written request. Upon examination by OMEGA, if the unit is found to be defective, it will be repaired or replaced at no charge. OMEGA’s WARRANTY does not apply to defects resulting from any action of the purchaser, including but not limited to mishandling, improper interfacing, operation outside of design limits, improper repair, or unauthorized modification.This WARRANTY is VOID if the unit shows evidence of having been tampered with or shows evidence of having been damaged as a result of excessive corrosion; or current, heat, moisture or vibration; improper specification; misapplication; misuse or other operating conditions outside of OMEGA’s control. Components in which wear is not warranted, include but are not limited to contact points, fuses, and triacs.OMEGA is pleased to offer suggestions on the use of its various products. However, OMEGA neither assumes responsibility for any omissions or errors nor assumes liability for any damages that result from the use of its products in accordance with information provided by OMEGA, either verbal or written. OMEGA warrants only that the parts manufactured by the company will be as specified and free of defects. OMEGA MAKES NO OTHER WARRANTIES OR REPRESENTATIONS OF ANY KIND WHATSOEVER, EXPRESSED OR IMPLIED, EXCEPT THAT OF TITLE, AND ALL IMPLIED WARRANTIES INCLUDING ANY WARRANTY OF MERCHANTABILITY AND FITNESS FOR A PARTICULAR PURPOSE ARE HEREBY DISCLAIMED. LIMITATION OF LIABILITY: The remedies of purchaser set forth herein are exclusive, and the total liability of OMEGA with respect to this order, whether based on contract, warranty,negligence, indemnification, strict liability or otherwise, shall not exceed the purchase price of the component upon which liability is based. In no event shall OMEGA be liable for consequential, incidental or special damages.CONDITIONS: Equipment sold by OMEGA is not intended to be used, nor shall it be used: (1)as a “Basic Component” under 10 CFR 21 (NRC), used in or with any nuclear installation or activity; or (2) in medical applications or used on humans. Should any Product(s) be used in or with any nuclear installation or activity, medical application, used on humans, or misused in any way, OMEGA assumes no responsibility as set forth in our basic WARRANTY/ DISCLAIMER language, and, additionally, purchaser will indemnify OMEGA and hold OMEGA harmless from any liability or damage whatsoever arising out of the use of the Product(s) in such a manner.RETURN REQUESTS/INQUIRIESDirect all warranty and repair requests/inquiries to the OMEGA Customer Service Department.BEFORE RETURNING ANY PRODUCT(S) TO OMEGA, PURCHASER MUST OBTAIN AN AUTHORIZED RETURN (AR) NUMBER FROM OMEGA’S CUSTOMER SERVICE DEPARTMENT (IN ORDER TO AVOID PROCESSING DELAYS). The assigned AR number should then be marked on the outside of the return package and on any correspondence.The purchaser is responsible for shipping charges, freight, insurance and proper packaging to prevent breakage in transit.FOR WARRANTY RETURNS, please havethe following information available BEFOREcontacting OMEGA:1.Purchase Order number under whichthe product was PURCHASED,2.Model and serial number of the productunder warranty, and3.Repair instructions and/or specificproblems relative to the product.FOR NON-WARRANTY REPAIRS,consult OMEGA for current repair charges. Have the following information available BEFORE contacting OMEGA:1. Purchase Order number to cover the COST of the repair,2.Model and serial number of theproduct, and 3.Repair instructions and/or specific problemsrelative to the product.OMEGA’s policy is to make running changes, not model changes, whenever an improvement is possible. This affords our customers the latest in technology and engineering.OMEGA is a registered trademark of OMEGA ENGINEERING, INC.© Copyright 2013 OMEGA ENGINEERING, INC. All rights reserved. This document may not be copied, photocopied,reproduced, translated, or reduced to any electronic medium or machine-readable form, in whole or in part, withoutthe prior written consent of OMEGA ENGINEERING, INC.Where Do I Find Everything I Need for Process Measurement and Control?OMEGA…Of Course!Shop online at SMTEMPERATUREⅪߜThermocouple, RTD & Thermistor Probes, Connectors, Panels & AssembliesⅪߜWire: Thermocouple, RTD & ThermistorⅪߜCalibrators & Ice Point ReferencesⅪߜRecorders, Controllers & Process MonitorsⅪߜInfrared PyrometersPRESSURE, STRAIN AND FORCEⅪߜTransducers & Strain GagesⅪߜLoad Cells & Pressure GagesⅪߜDisplacement TransducersⅪߜInstrumentation & AccessoriesFLOW/LEVELⅪߜRotameters, Gas Mass Flowmeters & Flow ComputersⅪߜAir Velocity IndicatorsⅪߜTurbine/Paddlewheel SystemsⅪߜTotalizers & Batch ControllerspH/CONDUCTIVITYⅪߜpH Electrodes, Testers & AccessoriesⅪߜBenchtop/Laboratory MetersⅪߜControllers, Calibrators, Simulators & PumpsⅪߜIndustrial pH & Conductivity EquipmentDATA ACQUISITIONⅪߜData Acquisition & Engineering SoftwareⅪߜCommunications-Based Acquisition SystemsⅪߜPlug-in Cards for Apple, IBM & CompatiblesⅪߜData Logging SystemsⅪߜRecorders, Printers & PlottersHEATERSⅪߜHeating CableⅪߜCartridge & Strip HeatersⅪߜImmersion & Band HeatersⅪߜFlexible HeatersⅪߜLaboratory HeatersENVIRONMENTALMONITORING AND CONTROLⅪߜMetering & Control InstrumentationⅪߜRefractometersⅪߜPumps & TubingⅪߜAir, Soil & Water MonitorsⅪߜIndustrial Water & Wastewater TreatmentⅪߜpH, Conductivity & Dissolved Oxygen Instruments M3199/0313。

CHLORIDE产品解决方案

CHLORIDE产品解决方案


Active A 系列 UPS 的突出客户价值(3 的突出客户价值(
-如何为您省钱? 如何为您省钱?
高达0.9的输出功率因数,较一般UPS可多挂接10-30%以上的服务器负载 采用软线并联方式,省去复杂的并联机框,大幅节省系统成本 系统效率高(up to 94%),省电,运行成本低 功率密度高(如20K只有3U),占用机架空间小,节省机架数量
•输入 输出 输入/输出 输入 输出:
输入:三相四线, 输入:三相四线,380/400/415Vac 50/60Hz
CH-0160L , CH-0200L
输出:三相 输出:三相380/400/415Vac,兼容单相 ,兼容单相220/230/240Vac
•型号 型号: 型号
主机模块: 主机模块 CH-0160L/ CH-0200L
如何实现方案的灵活性? -如何实现方案的灵活性?
5-10K 单单和三单兼容,16/20K三三/三单兼容 支持1+1并机,系统容量随意扩充 支持同步双母线系统(16/20K),配置方案更多选择 共用电池组设计,可通过级联电池模块方便地延长后备时间

Active A 系列 UPS 的突出客户价值(6 的突出客户价值(
欧洲UPS市场排名第一 欧洲UPS市场排名第一 UPS
---来自 来自Frost &Sullivan的奖项 的奖项 来自
2008年度全球UPS市场分布
欧洲200kVA以上UPS市场分布
全球服务网络
葡萄牙 西班牙 土耳其 法国里昂 工业电源
英国 南安普敦
Chloride Power Protection
输入功率因数近似为1100linear100nonlinearthdia23532630thdib1941242040kvainputthdiandpfwith400vratingvoltagevoltagecurrentthdic24392030pfa0999309998pfb0999809999pfc09997099994高过载能力适应各类负载?逆变器1001101hour11012510min1251501min?旁路100135longterm13517010mins1000100ms170bypassload1501301101hr10min1mintimeacac?可带满额高峰值比负载?宽广的输出功率因数pf范围5全面的智能电池管理tbm?电池管理功能恒流均充恒压均充浮充4003001002002002040600currentvoltagetime500t1t2t3t4t1?温度补偿?放电中止电压自动调节?电池自诊断和自检?电池容量预测?放电时间预估?低电压预报警6灵活的冗余解决方案?并联冗余多达6台直接并联以及总共12台的双母线工作能力并联冗余多达6台直接并联以及总共12台的双母线工作能力maintenancebypassswitchbypassswitchinputswitchrectifierinverterchargerdischargeroutputswitch1upsbatteryj10parallelboardx1bypassstaticswitchstaticswitchbcbnbpnmainnupsmaintenancebypassswitch2upsbypassstaticswitchnbinputdistributiondistributionofloadbypassswitchinputswitchrectifierinverterchargerdischargeroutputswitchbatteryj10parallelboardx1staticswitchbcbnbpnmainnupsmaintenancebypassswitchbypassswitchinputswitchrectifierinverterchargerdischargeroutputswitch3upsbatteryj10parallelboardx1bypassstaticswitchstatic

Endress+Hauser TR15和TC15模块化温度计说明书

Endress+Hauser TR15和TC15模块化温度计说明书

TR15热电阻(RTD)温度计TC15热电偶(TC)温度计应用•应用范围广泛•特别适用于高过程压力和高过程温度的蒸汽和气体测量•测量范围:•热电阻(RTD)铠装芯子:–200 … 600 °C (–328 … 1 112 °F)•热电偶(TC)铠装芯子:–40 … 1 100 °C (–40 … 2 012 °F)•静压力可达400 bar (5 800 psi)•最高防护等级:IP68模块化变送器相比于不经过温度变送器而直接接线的测量方法,Endress+Hauser 能为用户提供高测量精度、高测量可靠性的温度变送器。

根据实际工况条件,选择下列信号输出和通信方式:•4 … 20 mA 模拟量输出•HART ®•PROFIBUS ® PA•基金会现场总线(FF)™优势•高灵活性:一体式结构设计,DIN EN 50446标准接线盒和用户自定义插入深度•高兼容性设计,符合DIN 43772标准•延长颈可以防止模块化变送器过热•缩径型或锥管型热保护套管的响应时间短•防爆认证,可在危险区中使用:•本安型(Ex ia)•无火花型(Ex nA)Products Solutions Services技术资料TR15, TC15模块化温度计,带延长颈和棒材保护套管,可选法兰连接或焊接安装TI01100T/28/ZH/04.23-00716209582023-04-26TR15, TC15目录功能与系统设计 (3)测量原理 (3)测量系统 (3)设计 (4)测量范围 (4)输入 (4)测量变量 (4)测量范围 (4)输出 (5)输出信号 (5)温度变送器系列 (5)电源 (5)电缆入口 (8)过电压保护单元 (9)性能参数 (9)操作条件 (9)测量精度 (11)响应时间 (12)绝缘电阻 (13)绝缘强度 (13)自热 (13)标定 (13)材质 (14)安装 (15)安装方向 (15)安装指南 (15)延长颈长度 (16)机械结构 (16)接线盒 (16)设计 (20)铠装芯子 (21)重量 (21)过程连接 (22)备件 (22)证书和认证 (22)其他标准和准则 (22)材料证书 (22)保护套管测试 (22)订购信息 (22)附件 (23)服务专用附件 (23)补充文档资料 (24)2Endress+HauserTR15, TC15Endress+Hauser 3功能与系统设计测量原理热电阻(RTD)采用符合IEC 60751标准的Pt100作为温度传感器。

尼库尔森《微观经济学》课后答案ch18

尼库尔森《微观经济学》课后答案ch18

CHAPTER 18UNCERTAINTY AND RISK AVERSIONMost of the problems in this chapter focus on illustrating the concept of risk aversion. That is, they assume that individuals have concave utility of wealth functions and therefore dislike variance in their wealth. A difficulty with this focus is that, in general, students will not have been exposed to the statistical concepts of a random variable and its moments (mean, variance, etc.). Most of the problems here do not assume such knowledge, but the Extensions do show how understanding statistical concepts is crucial to reading applications on this topic.Comments on Problems18.1 Reverses the risk-aversion logic to show that observed behavior can be used to placebounds on subjective probability estimates.18.2 This problem provides a graphical introduction to the idea of risk-taking behavior. TheFriedman-Savage analysis of coexisting insurance purchases and gambling could bepresented here.18.3 This is a nice, homey problem about diversification. Can be done graphically althoughinstructors could introduce variances into the problem if desired.18.4 A graphical introduction to the economics of health insurance that examines cost-sharingprovisions. The problem is extended in Problem 19.3.18.5 Problem provides some simple numerical calculations involving risk aversion andinsurance. The problem is extended to consider moral hazard in Problem 19.2.18.6 This is a rather difficult problem as written. It can be simplified by using a particularutility function (e.g., U(W) = ln W). With the logarithmic utility function, one cannot use the Taylor approximation until after differentiation, however. If the approximation isapplied before differentiation, concavity (and risk aversion) is lost. This problem can,with specific numbers, also be done graphically, if desired. The notion that fines aremore effective can be contrasted with the criminologist’s view that apprehension of law-breakers is more effective and some shortcomings of the economic argument (i.e., nodisutility from apprehension) might be mentioned.18.7 This is another illustration of diversification. Also shows how insurance provisions canaffect diversification.18.8 This problem stresses the close connection between the relative risk-aversion parameterand the elasticity of substitution. It is a good problem for building an intuitive9798 Solutions Manualunderstanding of risk-aversion in the state preference model. Part d uses the CRRAutility function to examine the “equity-premium puzzle.”18.9 Provides an illustration of investment theory in the state preference framework.18.10 A continuation of Problem 18.9 that analyzes the effect of taxation on risk-takingbehavior.Solutions18.1 p must be large enough so that expected utility with bet is greater than or equal to thatwithout bet: p ln(1,100,000) + (1 –p)ln(900,000) > ln(1,000,000)13.9108p +13.7102(1 –p) > 13.8155, .2006p > .1053 p > .52518.2This would be limited by the individual’s resources: he or she could run out of wealth since unfair bets are continually being accepted.18.3 a.Strategy One Outcome Probability12 Eggs .50 Eggs .5Expected Value = .5∙12 + .5∙0 = 6Strategy Two Outcome Probability12 Eggs .256 Eggs .50 Eggs .25Expected Value = .25 ∙12 + .5 ∙ 6 + .25∙ 0= 3 + 3 = 6Chapter 18/Uncertainty and Risk Aversion 99b.18.4 a. E(L) = .50(10,000) = $5,000, soWealth = $15,000 with insurance, $10,000 or $20,000 without.b. Cost of policy is .5(5000) = 2500. Hence, wealth is 17,500 with no illness, 12,500with the illness.18.5 a. E(U) = .75ln(10,000) + .25ln(9,000) = 9.1840b. E(U) = ln(9,750) = 9.1850 Insurance is preferable.c. ln(10,000 – p ) = 9.184010,000 – p = e 9.1840 = 9,740p = 26018.6 Expected utility = pU(W – f) + (1 – p)U(W). ,[()()]/U p U pU W f U W p U e p U∂=⋅=--⋅∂,()/U f U fp U W f f U e f U∂'=⋅=-⋅-⋅∂100 Solutions Manual,,()()1()U p U fU W f U W e f U W f e --=<'-- by Taylor expansion,So, fine is more effective.If U (W ) = ln W then Expected Utility = p ln (W – f ) + (1 – p ) ln W .,/[ln ()ln ]U pp p f WW f W e U U -=--⋅≈ ,/()/()U ff p f W f p U W f e U U--=--⋅= ,,1U p U fW fe W e -=<18.7 a. U (wheat) = .5 ln(28,000) + .5 ln(10,000) = 9.7251 U (corn) = .5 ln(19,000) + .5 ln(15,000) = 9.7340 Plant corn. b. With half in eachY NR = 23,500Y R = 12,500U = .5 ln(23,500) + .5 ln(12,500) = 9.7491Should plant a mixed crop. Diversification yields an increased variance relative to corn only, but takes advantage of wheat’s high yield.c. Let α = percent in wheat. U = .5 ln[ (28,000) + (1 – α )(19,000)] + .5 ln[α (10,000) + (1 – α )(15,000)] = .5 ln(19,000 + 9,000α) + .5 ln(15,000 – 5,000α)45002500019,0009,00015,0005,000dU d ααα=-=+- 45(150 – 50α) = 25(190 + 90α) α = .444 U = .5 ln(22,996) + .5 ln(12,780) = 9.7494. This is a slight improvement over the 50-50 mix. d. If the farmer plants only wheat,Y NR = 24,000Y R = 14,000U = .5 ln(24,000) + .5 ln(14,000) = 9.8163so availability of this insurance will cause the farmer to forego diversification.Chapter 18/Uncertainty and Risk Aversion 10118.8 a. A high value for 1 – R implies a low elasticity of substitution between states of theworld. A very risk-averse individual is not willing to make trades away from the certainty line except at very favorable terms.b. R = 1 implies the individual is risk-neutral. The elasticity of substitution between wealth in various states of the world is infinite. Indifference curves are linear with slopes of –1. If R =-∞, then the individual has an infinite relative risk-aversion parameter. His or her indifference curves are L-shaped implying an unwillingness to trade away from the certainty line at any price.c. A rise in b p rotates the budget constraint counterclockwise about the W g intercept. Both substitution and income effects cause W b to fall. There is a substitution effect favoring an increase in W g but an income effect favoring a decline. The substitution effect will be larger the larger is the elasticity of substitution between states (the smaller is the degree of risk-aversion).d. i. Need to find R that solves the equation:R R R W W W )955.0(5.0)055.1(5.0)(000+=This yields an approximate value for R of –3, a number consistent with some empirical studies.ii. A 2 percent premium roughly compensates for a ±10 percent gamble. That is:303030)12.1()92(.)(---+≈W W W .The “puzzle” is that the premium rate of return provided by equities seems to be much higher than this.18.9 a. See graph.Risk free option is R , risk option is R '. b. Locus RR' represents mixed portfolios.c. Risk-aversion as represented by curvature of indifference curves will determine equilibrium in RR' (say E ).102 Solutions Manuald. With constant relative risk-aversion, indifference curve map is homothetic so locus ofoptimal points for changing values of W will be along OE.18.10 a. Because of homothetic indifference map, a wealth tax will cause movement along OE(see Problem 18.9).b. A tax on risk-free assets shifts R inward to Rt (see figure below). A flatter RR t'provides incentives to increase proportion of wealth held in risk assets, especially for individuals with lower relative risk-aversion parameters. Still, as the “note” implies, it is important to differentiate between the after tax optimum and the before taxchoices that yield that optimum. In the figure below, the no-tax choice is E on RR'.* EW represents the locus of points along which the fraction of wealth held in risky assets is constant. With the constraint RR t' choices are even more likely to be to the right of EW* implying greater investment in risky assets.c. With a tax on both assets, budget constraint shifts in a parallel way to RR tt'. Even in this case (with constant relative risk aversion) the proportion of wealth devoted to risky assets will increase since the new optimum will lie along OE whereas a constant proportion of risky asset holding lies along EW O.103。

加里森第十四版管理会计课后题答案CH06

加里森第十四版管理会计课后题答案CH06

加里森第十四版管理会计课后题答案CH06Chapter 6Variable Costing and Segment Reporting: Tools for Management Solutions to Questions6-1 Absorption and variable costing differ in how they handle fixed manufacturing overhead. Under absorption costing, fixed manufacturing overhead is treated as a product cost and hence is an asset until products are sold. Undervariable costing, fixed manufacturing overhead is treated as a period cost and is expensed on the current period’s income statement. 6-2 Selling and administrative expenses are treated as period costs under both variable costing and absorption costing.6-3 Under absorption costing, fixedmanufacturing overhead costs are included in product costs, along with direct materials, direct labor, and variable manufacturing overhead. If some of the units are not sold by the end of the period, then they are carried into the next period as inventory. When the units are finally sold, the fixed manufacturing overhead cost that has been carried over with the units is included as part of that period’s cost of goods sold. 6-4 Absorption costing advocates argue that absorption costing does a better job of matching costs with revenues than variable costing. They argue that all manufacturing costs must be assigned to products to properly match the costs of producing units of product with the revenues from the units when they are sold. They believe that no distinction should be made between variable and fixed manufacturing costs for the purposes of matching costs and revenues. 6-5 Advocates of variable costing argue that fixedmanufacturing costs are not really the cost of any particular unit of product. If a unit ismade or not, the total fixed manufacturing costs will be exactly the same. Therefore, how can one say that these costs are part of the costs ofthe products? These costs are incurred to have the capacity to make products during aparticular period and should be charged against that period as period costs according to the matching principle.6-6 If production and sales are equal, net operating income should be the same under absorption and variable costing. When production equals sales, inventories do not increase or decrease and therefore underabsorption costing fixed manufacturing overhead cost cannot be deferred in inventory or released from inventory.6-7 If production exceeds sales, absorption costing will usually show higher net operating income than variable costing. When production exceeds sales, inventories increase and under absorption costing part of the fixedmanufacturing overhead cost of the currentperiod is deferred in inventory to the next period. In contrast, all of the fixed manufacturing overhead cost of the current period is immediately expensed under variable costing. 6-8 If fixed manufacturing overhead cost is released from inventory, then inventory levels must have decreased and therefore production must have been less than sales.6-9 Under absorption costing net operating income can be increased by simply increasing the level of production without any increase in sales. If production exceeds sales, units of product areadded to inventory. These units carry a portion of the current period’s fixedmanufacturing overhead costs into the inventory account, reducing the current period’s reported expenses and causing net operating income to increase.? The McGraw-Hill Companies, Inc., 2012. All rights reserved. Solutions Manual, Chapter 62716-10 Differences in reported net operating income between absorption and variable costing arise because of changing levels of inventory. In lean production, goods are produced strictly to customers’ orders. With production geared to sales, inventories are largely (or entirely) eliminated. If inventories are completely eliminated, they cannot change from one period to another and absorption costing and variable costing will report the same net operating income.6-11 A segment is any part or activity of an organization about which a manager seeks cost, revenue, or profit data. Examples of segments include departments, operations, sales territories, divisions, and product lines.6-12 Under the contribution approach, costs are assigned to a segment if and only if the costs are traceable to the segment (i.e., could be avoided if the segment were eliminated). Common costs are not allocated to segments under the contribution approach.6-13 A traceable cost of a segment is a cost that arises specifically because of the existence of that segment. If the segment were eliminated, the cost would disappear. A common cost, by contrast, is a cost that supports more than one segment, but is not traceable in whole or in part to any one of the segments. If the departments of acompany are treated as segments, then examples of the traceable costs of a department would include the salary of the department’s supervisor, depreciation of machines usedexclusively by the department, and the costs of supplies used by the department. Examples ofcommon costs would include the salary of the general counsel of the entire company, the lease cost of the headquarters building, corporate image advertising, and periodic depreciation of machines shared by several departments. 6-14 The contribution margin is the difference between sales revenue and variable expenses. The segment margin is the amount remaining after deducting traceable fixed expenses from the contribution margin. The contribution margin is useful as a planning tool for many decisions, particularly those in which fixed costs don’t change. The segment margin is useful in assessing the overall profitability of a segment. 6-15 If common costs were allocated tosegments, then the costs of segments would be overstated and their margins would beunderstated. As a consequence, some segments may appear to be unprofitable and managers may be tempted to eliminate them. If a segment were eliminated because of the existence of arbitrarily allocated common costs, the overall profit of the company would decline and the common cost that had been allocated to the segment would be reallocated to the remaining segments―making them appear less profitable. 6-16 There are often limits to how far down an organization a cost can be traced. Therefore, costs that are traceable to a segment maybecome common as that segment is divided into smaller segment units. For example, the costs of national TV and printadvertising might be traceable to a specific product line, but be a common cost of the geographic sales territories in which that product line is sold.? The McGraw-Hill Companies, Inc., 2012. All rights reserved. 272 Managerial Accounting, 14th EditionExercise 6-1 (15 minutes)1. Under absorption costing, all manufacturing costs (variable and fixed) are included in product costs. (All currency values are in thousands of rupees, denoted by R.)Direct materials .................................................................. R120 Direct labor ........................................................................ 140 Variable manufacturing overhead ........................................ 50 Fixed manufacturing o verhead (R600,000 ÷ 10,000 units) .... 60 Absorption costing unit product cost .................................... R370 2. Under variable costing, only the variable manufacturing costs are included in product costs. (All currency values are in thousands of rupees, denoted by R.) Direct materials .................................. R120 Direct labor ........................................ 140 Variable manufacturing overhead ........50 Variable costing unit product cost ........ R310Note that selling and administrative expenses are not treated as product costs under either absorption or variable costing. These expenses are always treated as period costs and are charged against the current period’s revenue.? The McGraw-Hill Companies, Inc., 2012. All rights reserved. Solutions Manual, Chapter 6273Exercise 6-2 (20 minutes)1. 2,000 units in ending inventory × R60 fixed manufacturing overhead per unit = R120,000.2. The variable costing income statement appears below:Sales ................................................. R4,000,000 Variable expenses: Variable cost of goods sold (8,000 units × R310 per unit) ........ R2,480,000 Variable selling and administrative (8,000 units × R20 per unit) .......... 160,000 2,640,000 Contribution margin ............................ 1,360,000 Fixed expenses: Fixed manufacturing overhead .......... 600,000 Fixed selling and administrative ........ 400,000 1,000,000 Net operating income ......................... R 360,000The difference in net operating income between variable and absorption costing can be explained by the deferral of fixed manufacturingoverhead cost in inventory that has taken place under the absorption costing approach. Note from part (1) that R120,000 of fixed manufacturing overhead cost has been deferred in inventory to the next period. Thus, net operating income under the absorption costing approach is R120,000 higher than it is under variable costing.? The McGraw-Hill Companies, Inc., 2012. All rights reserved. 274 Managerial Accounting, 14th EditionExercise 6-3 (20 minutes)1. Year 1 Year 2 Year 3 Beginning inventories .......... 180 150 160 Ending inventories ............... 150 160 200 Change in inventories ..........(30) 10 40 Fixed manufacturing overhead in beginning inventories (@$450 per unit) ................................. $ 81,000 $ 67,500 $72,000 Fixed manufacturing overhead in ending inventories (@$450 per unit) ................................. 67,500 72,000 90,000 Fixed manufacturing overhead deferred in (released from)inventories (@$450 per unit) ................................. $(13,500) $ 4,500 $ 18,000 Variable costing net operating income .............. $292,400 $269,200 $251,800 Add (deduct) fixedmanufacturing overhead cost deferred in (released from) inventory under absorption costing ............ (13,500) 4,500 18,000 Absorption costing net operating income .............. $278,900 $273,700 $269,800 2. Because absorption costing net operating income was greater than variable costing net operating income in Year 4, inventories must have increased during the year and hence, fixed manufacturing overhead was deferred in inventories. The amount of the deferral is just the difference between the two net operating incomes or $27,000 = $267,200 C $240,200.? The McGraw-Hill Companies, Inc., 2012. All rights reserved. Solutions Manual, Chapter 6275。

HullOFOD9eSolutionsCh12第九版期权、期货及其他衍生品课后答案

HullOFOD9eSolutionsCh12第九版期权、期货及其他衍生品课后答案

CHAPTER 12Trading Strategies Involving OptionsPractice QuestionsProblem 12.1.What is meant by a protective put? What position in call options is equivalent to a protective put?A protective put consists of a long position in a put option combined with a long position in the underlying shares. It is equivalent to a long position in a call option plus a certain amount of cash. This follows from put –call parity: 0rT p S c Ke D -+=++Problem 12.2.Explain two ways in which a bear spread can be created.A bear spread can be created using two call options with the same maturity and different strike prices. The investor shorts the call option with the lower strike price and buys the call option with the higher strike price. A bear spread can also be created using two put options with the same maturity and different strike prices. In this case, the investor shorts the put option with the lower strike price and buys the put option with the higher strike price.Problem 12.3.When is it appropriate for an investor to purchase a butterfly spread?A butterfly spread involves a position in options with three different strike prices (12K K ,, and 3K ). A butterfly spread should be purchased when the investor considers that the price of the underlying stock is likely to stay close to the central strike price, 2K .Problem 12.4.Call options on a stock are available with strike prices of $15,$17.5 , and $20 and expiration dates in three months. Their prices are $4, $2, and,$0.5 respectively. Explain how the options can be used to create a butterfly spread. Construct a table showing how profit varies with stock price for the butterfly spread.An investor can create a butterfly spread by buying call options with strike prices of $15 and $20 and selling two call options with strike prices of $1712. The initial investment is1122422$+-⨯=. The following table shows the variation of profit with the final stock price:Problem 12.5.What trading strategy creates a reverse calendar spread?A reverse calendar spread is created by buying a short-maturity option and selling a long-maturity option, both with the same strike price.Problem 12.6.What is the difference between a strangle and a straddle?Both a straddle and a strangle are created by combining a long position in a call with a long position in a put. In a straddle the two have the same strike price and expiration date. In a strangle they have different strike prices and the same expiration date.Problem 12.7.A call option with a strike price of $50 costs $2. A put option with a strike price of $45 costs $3. Explain how a strangle can be created from these two options. What is the pattern of profits from the strangle?A strangle is created by buying both options. The pattern of profits is as follows:Problem 12.8.Use put –call parity to relate the initial investment for a bull spread created using calls to the initial investment for a bull spread created using puts.A bull spread using calls provides a profit pattern with the same general shape as a bullspread using puts (see Figures 12.2 and 12.3 in the text). Define 1p and 1c as the prices of put and call with strike price 1K and 2p and 2c as the prices of a put and call with strike price 2K . From put-call parity111rT p S c K e -+=+222rT p S c K e -+=+Hence:121221()rT p p c c K K e --=---This shows that the initial investment when the spread is created from puts is less than the initial investment when it is created from calls by an amount 21()rT K K e --. In fact as mentioned in the text the initial investment when the bull spread is created from puts is negative, while the initial investment when it is created from calls is positive.The profit when calls are used to create the bull spread is higher than when puts are used by 21()(1)rT K K e ---. This reflects the fact that the call strategy involves an additional risk-freeinvestment of 21()rT K K e -- over the put strategy. This earns interest of2121()(1)()(1)rT rT rT K K e e K K e ----=--.Problem 12.9.Explain how an aggressive bear spread can be created using put options.An aggressive bull spread using call options is discussed in the text. Both of the options used have relatively high strike prices. Similarly, an aggressive bear spread can be created using put options. Both of the options should be out of the money (that is, they should have relatively low strike prices). The spread then costs very little to set up because both of the puts are worth close to zero. In most circumstances the spread will provide zero payoff. However, there is a small chance that the stock price will fall fast so that on expiration both options will be in the money. The spread then provides a payoff equal to the difference between the two strike prices, 21K K -.Problem 12.10.Suppose that put options on a stock with strike prices $30 and $35 cost $4 and $7,respectively. How can the options be used to create (a) a bull spread and (b) a bear spread? Construct a table that shows the profit and payoff for both spreads.A bull spread is created by buying the $30 put and selling the $35 put. This strategy gives rise to an initial cash inflow of $3. The outcome is as follows:A bear spread is created by selling the $30 put and buying the $35 put. This strategy costs $3 initially. The outcome is as follows:Problem 12.11.Use put –call parity to show that the cost of a butterfly spread created from European puts is identical to the cost of a butterfly spread created from European calls.Define 1c , 2c , and 3c as the prices of calls with strike prices 1K , 2K and 3K . Define 1p , 2p and 3p as the prices of puts with strike prices 1K , 2K and 3K . With the usual notation 111rT c K e p S -+=+ 222rT c K e p S -+=+ 333rT c K e p S -+=+ Hence 1321321322(2)2rT c c c K K K e p p p -+-++-=+- Because 2132K K K K -=-, it follows that 13220K K K +-= and13213222c c c p p p +-=+-The cost of a butterfly spread created using European calls is therefore exactly the same as the cost of a butterfly spread created using European puts.Problem 12.12.A call with a strike price of $60 costs $6. A put with the same strike price and expiration date costs $4. Construct a table that shows the profit from a straddle. For what range of stock prices would the straddle lead to a loss?A straddle is created by buying both the call and the put. This strategy costs $10. The profit/loss is shown in the following table:This shows that the straddle will lead to a loss if the final stock price is between $50 and $70.Problem 12.13.Construct a table showing the payoff from a bull spread when puts with strike prices 1K and 2K with K 2>K 1 are used.The bull spread is created by buying a put with strike price 1K and selling a put with strike price 2K . The payoff is calculated as follows:Problem 12.14.An investor believes that there will be a big jump in a stock price, but is uncertain as to thedirection. Identify six different strategies the investor can follow and explain the differences among them.Possible strategies are:Strangle Straddle Strip StrapReverse calendar spread Reverse butterfly spreadThe strategies all provide positive profits when there are large stock price moves. A strangle is less expensive than a straddle, but requires a bigger move in the stock price in order to provide a positive profit. Strips and straps are more expensive than straddles but provide bigger profits in certain circumstances. A strip will provide a bigger profit when there is a large downward stock price move. A strap will provide a bigger profit when there is a large upward stock price move. In the case of strangles, straddles, strips and straps, the profitincreases as the size of the stock price movement increases. By contrast in a reverse calendar spread and a reverse butterfly spread there is a maximum potential profit regardless of the size of the stock price movement.Problem 12.15.How can a forward contract on a stock with a particular delivery price and delivery date be created from options?Suppose that the delivery price is K and the delivery date is T . The forward contract is created by buying a European call and selling a European put when both options have strike price K and exercise date T . This portfolio provides a payoff of T S K - under allcircumstances where T S is the stock price at time T . Suppose that 0F is the forward price. If 0K F =, the forward contract that is created has zero value. This shows that the price of a call equals the price of a put when the strike price is 0F .Problem 12.16.“A box spread comprises four options. Two can be combined to create a long forwardposition an d two can be combined to create a short forward position.” Explain this statement.A box spread is a bull spread created using calls and a bear spread created using puts. With the notation in the text it consists of a) a long call with strike 1K , b) a short call with strike 2K , c) a long put with strike 2K , and d) a short put with strike 1K . a) and d) give a long forward contract with delivery price 1K ; b) and c) give a short forward contract with delivery price 2K . The two forward contracts taken together give the payoff of 21K K -.Problem 12.17.What is the result if the strike price of the put is higher than the strike price of the call in astrangle?The result is shown in Figure S12.1. The profit pattern from a long position in a call and a putwhen the put has a higher strike price than a call is much the same as when the call has a higher strike price than the put. Both the initial investment and the final payoff are muchhigher in the first caseFigure S12.1:Profit Pattern in Problem 12.17Problem 12.18.A foreign currency is currently worth $0.64. A one-year butterfly spread is set up using European call options with strike prices of $0.60, $0.65, and $0.70. The risk-free interest rates in the United States and the foreign country are 5% and 4% respectively, and the volatility of the exchange rate is 15%. Use the DerivaGem software to calculate the cost of setting up the butterfly spread position. Show that the cost is the same if European put options are used instead of European call options.To use DerivaGem select the first worksheet and choose Currency as the Underlying Type. Select Black--Scholes European as the Option Type. Input exchange rate as 0.64, volatility as 15%, risk-free rate as 5%, foreign risk-free interest rate as 4%, time to exercise as 1 year, and exercise price as 0.60. Select the button corresponding to call. Do not select the implied volatility button. Hit the Enter key and click on calculate. DerivaGem will show the price of the option as 0.0618. Change the exercise price to 0.65, hit Enter, and click on calculate again. DerivaGem will show the value of the option as 0.0352. Change the exercise price to 0.70, hit Enter, and click on calculate. DerivaGem will show the value of the option as 0.0181.Now select the button corresponding to put and repeat the procedure. DerivaGem shows the values of puts with strike prices 0.60, 0.65, and 0.70 to be 0.0176, 0.0386, and 0.0690, respectively.The cost of setting up the butterfly spread when calls are used is therefore.+.-⨯.=.006180018120035200095The cost of setting up the butterfly spread when puts are used is.+.-⨯.=.001760069020038600094Allowing for rounding errors these two are the same.Problem 12.19.An index provides a dividend yield of 1% and has a volatility of 20%. The risk-free interest rate is 4%. How long does a principal-protected note, created as in Example 12.1, have to last for it to be profitable to the bank? Use DerivaGem.Assume that the investment in the index is initially $100. (This is a scaling factor that makes no difference to the result.) DerivaGem can be used to value an option on the index with the index level equal to 100, the volatility equal to 20%, the risk-free rate equal to 4%, the dividend yield equal to 1%, and the exercise price equal to 100. For different times to maturity, T, we value a call option (using Black-Scholes European) and the amount available to buy the call option, which is 100-100e-0.04×T. Results are as follows:This table shows that the answer is between 10 and 11 years. Continuing the calculations we find that if the life of the principal-protected note is 10.35 year or more, it is profitable for the bank. (Excel’s Solver can be used in conjunction with the DerivaGem functions to facilitate calculations.)Further QuestionsProblem 12.20.A trader creates a bear spread by selling a six-month put option with a $25 strike pricefor $2.15 and buying a six-month put option with a $29 strike price for $4.75. What isthe initial investment? What is the total payoff when the stock price in six months is (a) $23, (b) $28, and (c) $33.The initial investment is $2.60. (a) $4, (b) $1, and (c) 0.Problem 12.21.A trader sells a strangle by selling a call option with a strike price of $50 for $3 andselling a put option with a strike price of $40 for $4. For what range of prices of the underlying asset does the trader make a profit?The trader makes a profit if the total payoff is less than $7. This happens when the price of the asset is between $33 and $57.Problem 12.22.Three put options on a stock have the same expiration date and strike prices of $55, $60, and $65. The market prices are $3, $5, and $8, respectively. Explain how a butterfly spread can be created. Construct a table showing the profit from the strategy. For what range of stock prices would the butterfly spread lead to a loss?A butterfly spread is created by buying the $55 put, buying the $65 put and selling two of the+-⨯=initially. The following table shows the profit/loss $60 puts. This costs 38251$from the strategy.The butterfly spread leads to a loss when the final stock price is greater than $64 or less than $56.Problem 12.23.A diagonal spread is created by buying a call with strike price 2K and exercise date 2T and selling a call with strike price 1K and exercise date 121()T T T >. Draw a diagram showing the profit at time T 1 when (a) 21K K > and (b) 21K K <.There are two alternative profit patterns for part (a). These are shown in Figures S12.2 and S12.3. In Figure S12.2 the long maturity (high strike price) option is worth more than the short maturity (low strike price) option. In Figure S12.3 the reverse is true. There is no ambiguity about the profit pattern for part (b). This is shown in Figure S12.4.K 1K 2ProfitS TFigure S12.2: Investo r’s Profit/Loss in Problem 12.23a when long maturity call is worthmore than short maturity callProfitS TK1K2Figure S12.3Investo r’s Profit/Loss in Problem 12.23b when short maturity call is worthmore than long maturity callProfitS TK2K1Figure S12.4Investo r’s Profit/Loss in Problem 12.23bProblem 12.24.Draw a diagram showing the variation of an investor’s profit and loss with the terminal stock price for a portfolio consisting ofa.One share and a short position in one call optionb.Two shares and a short position in one call optionc.One share and a short position in two call optionsd.One share and a short position in four call optionsIn each case, assume that the call option has an exercise price equal to the current stock price.The variation of an investor’s profit/loss with the termina l stock price for each of the four strategies is shown in Figure S12.5. In each case the dotted line shows the profits from thecomponents of the investor’s position and the solid line shows the total net profit.Figure S12.5 Answer to Problem 12.24Problem 12.25.Suppose that the price of a non-dividend-paying stock is $32, its volatility is 30%, and the risk-free rate for all maturities is 5% per annum. Use DerivaGem to calculate the cost of setting up the following positions. In each case provide a table showing the relationship between profit and final stock price. Ignore the impact of discounting.a. A bull spread using European call options with strike prices of $25 and $30 and amaturity of six months.b. A bear spread using European put options with strike prices of $25 and $30 and amaturity of six monthsc. A butterfly spread using European call options with strike prices of $25, $30, and$35 and a maturity of one year.d. A butterfly spread using European put options with strike prices of $25, $30, and$35 and a maturity of one year.e. A straddle using options with a strike price of $30 and a six-month maturity.f. A strangle using options with strike prices of $25 and $35 and a six-monthmaturity.In each case provide a table showing the relationship between profit and final stock price. Ignore the impact of discounting.(a)A call option with a strike price of 25 costs 7.90 and a call option with a strike price.-.=.. The of 30 costs 4.18. The cost of the bull spread is therefore 790418372profits ignoring the impact of discounting are(b) A put option with a strike price of 25 costs 0.28 and a put option with a strike price of 30.-.=.. The profits ignoring costs 1.44. The cost of the bear spread is therefore 144028116the impact of discounting are(c) Call options with maturities of one year and strike prices of 25, 30, and 35 cost 8.92, 5.60, and 3.28, respectively. The cost of the butterfly spread is therefore8923282560100.+.-⨯.=.. The profits ignoring the impact of discounting are(d) Put options with maturities of one year and strike prices of 25, 30, and 35 cost 0.70, 2.14,.+.-⨯.=..4.57, respectively. The cost of the butterfly spread is therefore 0704572214099 Allowing for rounding errors, this is the same as in (c). The profits are the same as in (c). (e) A call option with a strike price of 30 costs 4.18. A put option with a strike price of 30.+.=.. The profits ignoring the costs 1.44. The cost of the straddle is therefore 418144562impact of discounting are(f) A six-month call option with a strike price of 35 costs 1.85. A six-month put option with a.+.=.. The strike price of 25 costs 0.28. The cost of the strangle is therefore 185028213profits ignoring the impact of discounting areProblem 12.26.What trading position is created from a long strangle and a short straddle when both have the same time to maturity? Assume that the strike price in the straddle is halfway between the two strike prices of the strangle.A butterfly spread (together with a cash position) is created.Problem 12.27. (Excel file)Describe the trading position created in which a call option is bought with strike price K1 and a put option is sold with strike price K2 when both have the same time to maturity andK2 > K1. What does the position become when K1 = K2?The position is as shown in the diagram below (for K1 = 25 and K2 = 35). It is known as a range forward and is discussed further in Chapter 17. When K1 =K2, the position becomes aregular long forward.Figure S12.6:Trading position in Problem 12.27Problem 12.28.A bank decides to create a five-year principal-protected note on a non-dividend-paying stock by offering investors a zero-coupon bond plus a bull spread created from calls. The risk-free rate is 4% and the stock price volatility is 25%. The low strike price option in the bull spread is at the money. What is the maximum ratio of the higher strike price to the lower strike price in the bull spread? Use DerivaGem.Assume that the amount invested is 100. (This is a scaling factor.) The amount available tocreate the option is 100-100e-0.04×5=18.127. The cost of the at-the money option can be calculated from DerivaGem by setting the stock price equal to 100, the volatility equal to 25%, the risk-free interest rate equal to 4%, the time to exercise equal to 5 and the exercise price equal to 100. It is 30.313. We therefore require the option given up by the investor to be worth at least 30.313−18.127 = 12.186. Results obtained are as follows:Continuing in this way we find that the strike must be set below 163.1. The ratio of the high strike to the low strike must therefore be less than 1.631 for the bank to make a profit. (Excel’s Solver can b e used in conjunction with the DerivaGem functions to facilitate calculations.)。

ICP DAS I-7050D 数字输入输出数据采集模块快速入门指南说明书

ICP DAS I-7050D 数字输入输出数据采集模块快速入门指南说明书

I-7050D7 Ch Digital Input (Counter) and 8 Ch DigitalOutput Data Acquisition ModuleQuick Start GuideProduct Website:https:///i_7050_d/dcon_utility_pro.html1. IntroductionI-7050D is intelligently designed to provide signal conditioning system monitoring and safe value settings. I-7050D is a cost-effective solution for a wide range of valuable industrial control signals and systems. The DCON utility can help users to configure and test I-7050D Digital Input/Output modules. Plenty of library functions and demo programs are provided to let users develop programs easily under Windows, Linux and DOS operating systems. Users may mount the modules on a DIN rail, panel or wall. Modules have a screw-terminal block to connect to the signals.2. Terminal AssignmentI-7050D - 7 Ch DI and 8 Ch DO DAQ Module- QuickStart (Nov/2017)3. Block/ Wiring DiagramI-7050D - 7 Ch DI and 8 Ch DO DAQ Module- QuickStart (Nov/2017)4. Default SettingsDefault settings for the I-7050/D DIO modules are as follows:。

HullOFOD eSolutionsCh 第九版期权期货及其他衍生品课后答案

HullOFOD eSolutionsCh 第九版期权期货及其他衍生品课后答案

CHAPTER 2Mechanics of Futures MarketsPractice QuestionsProblem 2.1.Distinguish between the terms open interest and trading volume.The open interest of a futures contract at a particular time is the total number of long positions outstanding. (Equivalently, it is the total number of short positions outstanding.) The trading volume during a certain period of time is the number of contracts traded during this period. Problem 2.2.What is the difference between a local and a futures commission merchant?A futures commission merchant trades on behalf of a client and charges a commission. A local trades on his or her own behalf.Problem 2.3.Suppose that you enter into a short futures contract to sell July silver for $17.20 per ounce. The size of the contract is 5,000 ounces. The initial margin is $4,000, and the maintenance margin is $3,000. What change in the futures price will lead to a margin call? What happens if you do not meet the margin call?There will be a margin call when $1,000 has been lost from the margin account. This will occur when the price of silver increases by 1,000/5,000 = $0.20. The price of silver must therefore rise to $17.40 per ounce for there to be a margin call. If the margin call is not met, your broker closes out your position.Problem 2.4.Suppose that in September 2015 a company takes a long position in a contract on May 2016 crude oil futures. It closes out its position in March 2016. The futures price (per barrel) is $88.30 when it enters into the contract, $90.50 when it closes out its position, and $89.10 at the end of December 2015. One contract is for the delivery of 1,000 barrels. What is the company’s total profit? When is it realized? How is it taxed if it is (a) a hedger and (b) a speculator? Assume that the company has a December 31 year-end.The total profit is ($90.50 - $88.30) ⨯ 1,000 = $2,200. Of this ($89.10 - $88.30) ⨯ 1,000 or $800 is realized on a day-by-day basis between September 2015 and December 31, 2015. A further ($90.50 - $89.10) ⨯ 1,000 or $1,400 is realized on a day-by-day basis between January1, 2016, and March 2016. A hedger would be taxed on the whole profit of $2,200 in 2016. A speculator would be taxed on $800 in 2015 and $1,400 in 2016.Problem 2.5.What does a stop order to sell at $2 mean? When might it be used? What does a limit order to sell at $2 mean? When might it be used?A stop order to sell at $2 is an order to sell at the best available price once a price of $2 or less is reached. It could be used to limit the losses from an existing long position. A limit order to sell at $2 is an order to sell at a price of $2 or more. It could be used to instruct a broker that a short position should be taken, providing it can be done at a price more favorable than $2.Problem 2.6.What is the difference between the operation of the margin accounts administered by a clearing house and those administered by a broker?The margin account administered by the clearing house is marked to market daily, and the clearing house member is required to bring the account back up to the prescribed level daily. The margin account administered by the broker is also marked to market daily. However, the account does not have to be brought up to the initial margin level on a daily basis. It has to be brought up to the initial margin level when the balance in the account falls below the maintenance margin level. The maintenance margin is usually about 75% of the initial margin.Problem 2.7.What differences exist in the way prices are quoted in the foreign exchange futures market, the foreign exchange spot market, and the foreign exchange forward market?In futures markets, prices are quoted as the number of US dollars per unit of foreign currency. Spot and forward rates are quoted in this way for the British pound, euro, Australian dollar, and New Zealand dollar. For other major currencies, spot and forward rates are quoted as the number of units of foreign currency per US dollar.Problem 2.8.The party with a short position in a futures contract sometimes has options as to the precise asset that will be delivered, where delivery will take place, when delivery will take place, and so on. Do these options increase or decrease the futures price? Explain your reasoning.These options make the contract less attractive to the party with the long position and more attractive to the party with the short position. They therefore tend to reduce the futures price.Problem 2.9.What are the most important aspects of the design of a new futures contract?The most important aspects of the design of a new futures contract are the specification of the underlying asset, the size of the contract, the delivery arrangements, and the delivery months. Problem 2.10.Explain how margin accounts protect investors against the possibility of default.A margin is a sum of money deposited by an investor with his or her broker. It acts as a guarantee that the investor can cover any losses on the futures contract. The balance in the margin account is adjusted daily to reflect gains and losses on the futures contract. If losses are above a certain level, the investor is required to deposit a further margin. This system makes it unlikely that the investor will default. A similar system of margin accounts makes it unlikely that the investor’s broker will default on the contract it has with the clearing house member and unlikely that the clearing house member will default with the clearing house.Problem 2.11.A trader buys two July futures contracts on frozen orange juice. Each contract is for the delivery of 15,000 pounds. The current futures price is 160 cents per pound, the initial margin is $6,000 per contract, and the maintenance margin is $4,500 per contract. What price change would lead to a margin call? Under what circumstances could $2,000 be withdrawn from the margin account?There is a margin call if more than $1,500 is lost on one contract. This happens if the futures price of frozen orange juice falls by more than 10 cents to below 150 cents per pound. $2,000 can be withdrawn from the margin account if there is a gain on one contract of $1,000. This will happen if the futures price rises by 6.67 cents to 166.67 cents per pound.Problem 2.12.Show that, if the futures price of a commodity is greater than the spot price during the delivery period, then there is an arbitrage opportunity. Does an arbitrage opportunity exist if the futures price is less than the spot price? Explain your answer.If the futures price is greater than the spot price during the delivery period, an arbitrageur buys the asset, shorts a futures contract, and makes delivery for an immediate profit. If the futures price is less than the spot price during the delivery period, there is no similar perfect arbitrage strategy. An arbitrageur can take a long futures position but cannot force immediate delivery of the asset. The decision on when delivery will be made is made by the party with the short position. Nevertheless companies interested in acquiring the asset may find it attractive to enterinto a long futures contract and wait for delivery to be made.Problem 2.13.Explain the difference between a market-if-touched order and a stop order.A market-if-touched order is executed at the best available price after a trade occurs at a specified price or at a price more favorable than the specified price. A stop order is executed at the best available price after there is a bid or offer at the specified price or at a price less favorable than the specified price.Problem 2.14.Explain what a stop-limit order to sell at 20.30 with a limit of 20.10 means.A stop-limit order to sell at 20.30 with a limit of 20.10 means that as soon as there is a bid at20.30 the contract should be sold providing this can be done at 20.10 or a higher price. Problem 2.15.At the end of one day a clearing house member is long 100 contracts, and the settlement price is $50,000 per contract. The original margin is $2,000 per contract. On the following day the member becomes responsible for clearing an additional 20 long contracts, entered into at a price of $51,000 per contract. The settlement price at the end of this day is $50,200. How much does the member have to add to its margin account with the exchange clearing house?The clearing house member is required to provide 20$2000$40000⨯,=, as initial margin for the new contracts. There is a gain of (50,200 - 50,000)⨯ 100 = $20,000 on the existing contracts. There is also a loss of (5100050200)20$16000,-,⨯=, on the new contracts. The member must therefore add,-,+,=,400002000016000$36000to the margin account.Problem 2.16.Explain why collateral requirements will increase in the OTC market as a result of new regulations introduced since the 2008 credit crisis.Regulations require most standard OTC transactions entered into between derivatives dealers to be cleared by CCPs. These have initial and variation margin requirements similar to exchanges. There is also a requirement that initial and variation margin be provided for most bilaterally cleared OTC transactions.Problem 2.17.The forward price on the Swiss franc for delivery in 45 days is quoted as 1.1000. The futures price for a contract that will be delivered in 45 days is 0.9000. Explain these two quotes. Which is more favorable for an investor wanting to sell Swiss francs?The 1.1000 forward quote is the number of Swiss francs per dollar. The 0.9000 futures quote is the number of dollars per Swiss franc. When quoted in the same way as the futures price the /.=.. The Swiss franc is therefore more valuable in the forward forward price is11100009091market than in the futures market. The forward market is therefore more attractive for an investor wanting to sell Swiss francs.Problem 2.18.Suppose you call your broker and issue instructions to sell one July hogs contract. Describe what happens.Live hog futures are traded by the CME Group. The broker will request some initial margin. The order will be relayed by telephone to your broker’s trading desk on the floor of the exchange (or to the trading desk of another broker). It will then be sent by messenger to a commission broker who will execute the trade according to your instructions. Confirmation of the trade eventually reaches you. If there are adverse movements in the futures price your broker may contact you to request additional margin.Problem 2.19.“Speculation in futures markets is pure gambling. It is not in the public interest to allow speculators to trade on a futures exchange.” Discuss this viewpoint.Speculators are important market participants because they add liquidity to the market. However, contracts must be useful for hedging as well as speculation. This is because regulators generally only approve contracts when they are likely to be of interest to hedgers as well as speculators.Problem 2.20.Explain the difference between bilateral and central clearing for OTC derivatives.In bilateral clearing the two sides enter into an agreement governing the circumstances under which transactions can be closed out by one side, how transactions will be valued if there is a close out, how the collateral posted by each side is calculated, and so on. In central clearing a CCP stands between the two sides in the same way that an exchange clearing house stands between two sides for transactions entered into on an exchangeProblem 2.21.What do you think would happen if an exchange started trading a contract in which the qualityof the underlying asset was incompletely specified?The contract would not be a success. Parties with short positions would hold their contracts until delivery and then deliver the cheapest form of the asset. This might well be viewed by the party with the long position as garbage! Once news of the quality problem became widely known no one would be prepared to buy the contract. This shows that futures contracts are feasible only when there are rigorous standards within an industry for defining the quality of the asset. Many futures contracts have in practice failed because of the problem of defining quality.Problem 2.22.“When a futures contract is traded on the floor of the exchange, it may be the case that the open interest increases by one, stays the same, or decreases by one.” Explain this statement.If both sides of the transaction are entering into a new contract, the open interest increases by one. If both sides of the transaction are closing out existing positions, the open interest decreases by one. If one party is entering into a new contract while the other party is closing out an existing position, the open interest stays the same.Problem 2.23.Suppose that on October 24, 2015, a company sells one April 2016 live-cattle futures contracts. It closes out its position on January 21, 2016. The futures price (per pound) is 121.20 cents when it enters into the contract, 118.30 cents when it closes out its position, and 118.80 cents at the end of December 2015. One contract is for the delivery of 40,000 pounds of cattle. What is the total profit? How is it taxed if the company is (a) a hedger and (b) a speculator?Assume that the company has a December 31 year end.The total profit is40,000×(1.2120−1.1830) = $1,160If the company is a hedger this is all taxed in 2016. If it is a speculator40,000×(1.2120−1.1880) = $960is taxed in 2015 and40,000×(1.1880−1.1830) = $200is taxed in 2016.Problem 2.24.A cattle farmer expects to have 120,000 pounds of live cattle to sell in three months. The live-cattle futures contract traded by the CME Group is for the delivery of 40,000 pounds of cattle. How can the farmer use the contract for hedging? From the farmer’s viewpoint, what are thepros and cons of hedging?The farmer can short 3 contracts that have 3 months to maturity. If the price of cattle falls, the gain on the futures contract will offset the loss on the sale of the cattle. If the price of cattle rises, the gain on the sale of the cattle will be offset by the loss on the futures contract. Using futures contracts to hedge has the advantage that the farmer can greatly reduce the uncertainty about the price that will be received. Its disadvantage is that the farmer no longer gains from favorable movements in cattle prices.Problem 2.25.It is July 2014. A mining company has just discovered a small deposit of gold. It will take six months to construct the mine. The gold will then be extracted on a more or less continuous basis for one year. Futures contracts on gold are available with delivery months every two months from August 2014 to December 2015. Each contract is for the delivery of 100 ounces. Discuss how the mining company might use futures markets for hedging.The mining company can estimate its production on a month by month basis. It can then short futures contracts to lock in the price received for the gold. For example, if a total of 3,000 ounces are expected to be produced in September 2015 and October 2015, the price received for this production can be hedged by shorting 30 October 2015 contracts.Problem 2.26.Explain how CCPs work. What are the advantages to the financial system of requiring all standardized derivatives transactions to be cleared through CCPs?A CCP stands between the two parties in an OTC derivative transaction in much the same way that a clearing house does for exchange-traded contracts. It absorbs the credit risk but requires initial and variation margin from each side. In addition, CCP members are required to contribute to a default fund. The advantage to the financial system is that there is a lot more collateral (i.e., margin) available and it is therefore much less likely that a default by one major participant in the derivatives market will lead to losses by other market participants. There is also more transparency in that the trades of different financial institutions are more readily known. The disadvantage is that CCPs are replacing banks as the too-big-to-fail entities in the financial system. There clearly needs to be careful oversight of the management of CCPs.Further QuestionsProblem 2.27.Trader A enters into futures contracts to buy 1 million euros for 1.3 million dollars in three months. Trader B enters in a forward contract to do the same thing. The exchange rate (dollarsper euro) declines sharply during the first two months and then increases for the third month to close at 1.3300. Ignoring daily settlement, what is the total profit of each trader? When the impact of daily settlement is taken into account, which trader does better?The total profit of each trader in dollars is 0.03×1,000,000 = 30,000. Trader B’s profit is realized at the end of the three months. Trader A’s profit is realized day-by-day during the three months. Substantial losses are made during the first two months and profits are made during the final month. It is likely that Trader B has done better because Trader A had to finance its losses during the first two months.Problem 2.28.Explain what is meant by open interest. Why does the open interest usually decline during the month preceding the delivery month? On a particular day, there were 2,000 trades in a particular futures contract. This means that there were 2000 buyers (going long) and 2000 sellers (going short). Of the 2,000 buyers, 1,400 were closing out positions and 600 were entering into new positions. Of the 2,000 sellers, 1,200 were closing out positions and 800 were entering into new positions. What is the impact of the day's trading on open interest?Open interest is the number of contract outstanding. Many traders close out their positions just before the delivery month is reached. This is why the open interest declines during the month preceding the delivery month. The open interest went down by 600. We can see this in two ways. First, 1,400 shorts closed out and there were 800 new shorts. Second, 1,200 longs closed out and there were 600 new longs.Problem 2.29.One orange juice future contract is on 15,000 pounds of frozen concentrate. Suppose that in September 2014 a company sells a March 2016 orange juice futures contract for 120 cents per pound. In December 2014 the futures price is 140 cents; in December 2015 the futures price is 110 cents; and in February 2016 it is closed out at 125 cents. The company has a December year end. What is the company's profit or loss on the contract? How is it realized? What is the accounting and tax treatment of the transaction if the company is classified as a) a hedger and b) a speculator?The price goes up during the time the company holds the contract from 120 to 125 cents per pound. Overall the company therefore takes a loss of 15,000×0.05 = $750. If the company is classified as a hedger this loss is realized in 2016, If it is classified as a speculator it realizes a loss of 15,000×0.20 = $3000 in 2014, a gain of 15,000×0.30 = $4,500 in 2015, and a loss of 15,000×0.15 = $2,250 in 2016.Problem 2.30.A company enters into a short futures contract to sell 5,000 bushels of wheat for 750 cents per bushel. The initial margin is $3,000 and the maintenance margin is $2,000. What price changewould lead to a margin call? Under what circumstances could $1,500 be withdrawn from the margin account?There is a margin call if $1000 is lost on the contract. This will happen if the price of wheat futures rises by 20 cents from 750 cents to 770 cents per bushel. $1500 can be withdrawn if the futures price falls by 30 cents to 720 cents per bushel.Problem 2.31.Suppose that there are no storage costs for crude oil and the interest rate for borrowing or lending is 5% per annum. How could you make money if the June and December futures contracts for a particular year trade at $80 and $86?You could go long one June oil contract and short one December contract. In June you take delivery of the oil borrowing $80 per barrel at 5% to meet cash outflows. The interest accumulated in six months is about 80×0.05×1/2 or $2. In December the oil is sold for $86 per barrel which is more than the $82 that has to be repaid on the loan. The strategy therefore leads to a profit. Note that this profit is independent of the actual price of oil in June and December. It will be slightly affected by the daily settlement procedures.Problem 2.32.What position is equivalent to a long forward contract to buy an asset at K on a certain date and a put option to sell it for K on that date?The long forward contract provides a payoff of S T−K where S T is the asset price on the date and K is the delivery price. The put option provides a payoff of max (K−S T, 0). If S T> K the sum of the two payoffs is S T–K. If S T< K the sum of the two payoffs is 0. The combined payoff is therefore max (S T–K, 0). This is the payoff from a call option. The equivalent position is therefore a call option.Problem 2.33.A company has derivatives transactions with Banks A, B, and C which are worth +$20 million, −$15 million, and −$25 million, respectively to t he company. How much margin or collateral does the company have to provide in each of the following two situations?a) The transactions are cleared bilaterally and are subject to one-way collateral agreements where the company posts variation margin, but no initial margin. The banks do not have to post collateral.b) The transactions are cleared centrally through the same CCP and the CCP requires a total initial margin of $10 million.If the transactions are cleared bilaterally, the company has to provide collateral to Banks A, B, and C of (in millions of dollars) 0, 15, and 25, respectively. The total collateral required is $40 million. If the transactions are cleared centrally they are netted against each other and thecompany’s total variation margin (i n millions of dollars) is –20 + 15 + 25 or $20 million in total. The total margin required (including the initial margin) is therefore $30 million.Problem 2.34.A bank’s derivatives transactions with a counterparty are worth +$10 million to the bank and are cleared bilaterally. The counterparty has posted $10 million of cash collateral. What credit exposure does the bank have?The counterparty may stop posting collateral and some time will then elapse before the bank is able to close out the transaction s. During that time the transactions may move in the bank’s favor, increasing its exposure. Note that the bank is likely to have hedged the transactions and will incur a loss on the hedge if the transactions move in the bank’s favor. For example, if the transactions change in value from $10 to $13 million after the counterparty stops postingcollateral, the bank loses $3 million on the hedge and will not necessarily realize an offsetting gain on the transactions.Problem 2.35. (Excel file)The author’s website (www.rotman.utoronto.ca/~hull/data) contains daily closing prices for the crude oil futures contract and gold futures contract. (Both contracts are traded on NYMEX.) You are required to download the data for crude oil and answer the following:(a) Assuming that daily price changes are normally distributed with zero mean, estimate the daily price movement that will not be exceeded with 99% confidence.(b) Suppose that an exchange wants to set the maintenance margin for traders so that it is 99% certain that the margin will not be wiped out by a two-day price move. (It chooses two daysbecause the margin calls are made at the end of a day and the trader has until the end of the next day to decide whether to provide more margin.) How high does the margin have to be when the normal distribution assumption is made?(c) Suppose that the maintenance margin is as calculated in (b) and is 75% of the initial margin. How frequently would the margin have been wiped out by a two-day price movement in the period covered by the data for a trader with a long position? What do your results suggest about the appropriateness of the normal distribution assumption.(a) For crude oil, the standard deviation of daily changes is $1.5777 per barrel or $1577.7 percontract. We are therefore 99% confident that daily changes will not exceed 1577.7×2.326 or 3669.73.(b) The maintenance margin should be set at 3263.227.1577⨯⨯ or 5,191 when rounded.(c)The initial margin is set at 6,921 for on crude oil. (This is the maintenance margin of5,191 divided by 0.75.) As the spreadsheet shows, for a long investor in oil there are 157 margin calls and 9 times (out of 1039 days) where the trader is tempted to walk away. As9 is approximately 1% of 1039 the normal distribution assumption appears to workreasonably well in this case.。

ch04_Motivation and Values

ch04_Motivation and Values

4-5
Motivational Strength
• Biological vs. Learned Needs:
– Instinct: Innate patterns of behavior universal in a species – Tautology: Circular explanation (e.g. instinct is inferred from the behavior it is supposed to explain)
4-4
Ads Reinforce Desired ates
• This ad for exercise shows men a desired state (as dictated by contemporary Western culture), and suggests a solution (purchase of equipment) to attain it.
4 - 17
Criticisms of Maslow’s Hierarchy
• The application is too simplistic:
– It is possible for the same product or activity to satisfy every need.
• It is too culture-bound:
– The assumptions of the hierarchy may be restricted to Western culture
• It emphasizes individual needs over group needs
– Individuals in some cultures place more value on the welfare of the group (belongingness needs) than the needs of the individual (esteem needs)

零能耗建筑 ZERO ENERGY BUILDING

零能耗建筑 ZERO ENERGY BUILDING

Switching to the high-efficiency…
...Thermassiv®.
will reduce the usage of energy by 60%–80% !!!
It takes human energy to stop CO2 emission
Licensed Thermassiv®Producer, Building system Patent pending nr. : INT, (L) PCT/CH2009/000257 © Copyrights Stone Treuhand AG 2004 Subject to alterations ISO■STUC® is a registered trademark of Huizenkluiz b.v. Patent pending Nr: PCT/NL2004/000417 CISCA® is a registered trademark of Huizenkluiz b.v.
Licensed Thermassiv®Producer, Building system Patent pending nr. : INT, (L) PCT/CH2009/000257 © Copyrights Stone Treuhand AG 2004 Subject to alterations ISO■STUC® is a registered trademark of Huizenkluiz b.v. Patent pending Nr: PCT/NL2004/000417 CISCA® is a registered trademark of Huizenkluiz b.v.
It takes human energy to stop CO2 emission

Eluktronics MECH-15HS 用户手册说明书

Eluktronics MECH-15HS 用户手册说明书

NoticeThe information in this user’s manual is protected by copyright laws, all parts of this manual, including the products and software described in it, can not be reproduced, transmitted, transcribed, stored in a retrieval system, nor translated into any language.THE MANUFACTURER OR RESELLER SHALL NOT BE LIABLE FOR ERRORS OR OMISSIONS CONTAINED IN THIS MANUAL AND SHALL NOT BE LIABLE FOR ANY CONSEQUENTIAL DAMAGES, WHICH MAY RESULT FROM THE PERFORMANCE OR USE OF THIS MANUAL.The illustrations in this user’s manual are for reference only. Actual product specifications may vary with territories.The information in this user’s manual is subject to change without notice.TABLE OF CONTENTSNotice (2)Preface (5)1.1 Regulations Information (6)1.2 Safety Instructions (7)1.3 Conventions for this Manual (8)1.4 Release History (9)Getting to know the basics (10)2.1 Product Specification (11)2.2 Preparing your Computer (15)2.3 Product Overview (16)Getting started (24)3.1 AC Adapter (25)3.2 Knowing the Keyboard (26)3.2.1 For keyboard users (27)3.3 Using the touchpad / clickpad (30)3.3.1 Windows 10 Touchpad Usage (31)BIOS setup (32)4.1 About BIOS Setup (33)4.1.1 When to Use BIOS Setup ? (33)4.1.2 How to Run BIOS Setup ? (33)4.2 BIOS Setup Menu (34)4.2.1 Main Menu (36)4.2.2 Advanced Menu (38)4.2.3 Security Menu (40)4.2.4 Boot Menu (41)4.2.5 Exit Menu (42)Chapter 1 Preface1.1 Regulations Information¤ CE complianceThis device is classed as a technical information equipment (ITE) in class B and is intended for use in living room and office. The CE-mark approves the conformity by the EU-guidelines:- EMC Directive 2014/30/EU,- Low Voltage Directive 2014/35/EU(equals A2 : 2013) ,- R&TTE Directive 1999/5/ECT he unit can be operated at an ambient temperature of max. 35°C (95°F). Do not subject it to temperatures below 5°C (41°F) orabove 40°C (104 °F).C AUTION: R ISK OF EXPLOSION IF BATTERY IS REPLACEDBY AN INCORRECT TYPE DISPOSE OF USEDBATTERIES ACCORDING TO THE INSTRUCTIONS.P REVENTION OF HEARING LOSSC AUTION: L istening to music at high volume levels and forextended durations can damage one’s hearing. In orderto reduce the risk of damage to hearing, one should lowerthe volume to a safe, comfortable level, and reduce theamount of time listening at high levels. Headsets shouldcomply with EN 50332-2 requirements.1.2 Safety InstructionsDo not apply heavy pressure to the computer or subject it to any form ofstrong impact as this can damage the computer's components or otherwisecause it to malfunction.To keep your computer in prime operating condition, protect your work area from direct sunlight.Never cover or block the air vents including those located at the base of thecomputer. Never cover your computer or AC adapter with any object.Do NOT expose to or use near liquid, rain, or moisture.Do NOT use the modem during electrical storms.Do not use or expose this device around magnetic fields as magneticinterference may affect the performance of the device.152341.3 Conventions for this Manual1.4 Release History1.009.2017Initial releaseChapter 2Getting to know the basics2.1 Product SpecificationThis User’s Manual provides technical information of instructions and illustrations on how to operate this notebook for the customer. Please read this manual carefully before using this notebook.・CPUSupport ProcessorKBLH, 4+2・MemoryDDR4DDR4 2400MHz・StorageHDD2.5” SATA HDD support7/9.5mm・Physical CharacteristicDimension386.9 x 270 x 35.6mm; Weight: 3.0Kg・GPUSupport ProcessorNVDIA Geforce 1070, 256bits(N17E-G2)・I/O PortDC-in USBRJ45Card Reader HDMIx 1x 1x 1 (2 in 1 Card reader connector type) x 1 2.0 x 2 + 3.0 x 33.1 x 1 (Type C, no Audio and Video support)・AudioAudio CodecSpeaker/MICAzalia standard support, D3 mode supportAudio out & Mic in x 2Build-in 2 speakers Speaker : 2W/eachDigitial Microphone support w/ CameraDP port x 1 Quick launch buttonx 1 (FAN full speed ON/OFF)・InputKeyboard15.6”Pointing DevicePS2 Touch Pad with 2 buttons 92X52mmLCD15.6" FHD, 16:9, Glare or AG・DisplayLAN10/100/1000 Mb/SecWireless LANM.2 2230 w/ PCIE interfaceIEEE802.11b/g/n support; AC mode support ・Communication PortHD webcamHD webcam with D-MIC*1・Webcam・PowerAC AdapterAutomatics Voltage adjustment between 100 and 240VAC 50/60Hz, 230 Watts (19.5V/11.8A), 3 PinsLi-ion 18650 Battery, hard pack, Removable,6 cells (High power 3S2P 4300mAh ; or Normal 3S2P 4400mAh)BatteryPlease use the genuine adapter based on the rating of your PC's model and BIOS. If damage is determined to have been caused by utilization of an unauthorized power supply, the warranty will be void.CAUTION: MODEL IS DESIGNED TO USE WITH THE DC INPUT: 19.5V/11.8A2.2 Preparing your ComputerAConnect the AC adapter’s DC output plug to the DC IN jack.Top-Open View2.3 Product OverviewPlease become familiar with each component before you operatethe computer.CAUTION:When you are not using the computer, keep the LCD-screen closed to protect it from dust.13WebcamA device that allows you to record video or take photographs with yourcomputer.274Power ButtonPress this button to turn the computer's power on or off.56Camera Status LEDThe Camera Status LED shows the Camera status.MicrophoneBuilt-in microphone.LCD screenDisplays of your notebook computer.Power indicator*Power ON: ON*Suspend: Slow / Smooth, Blinking Red *FAN full speed: ON*Power Off: OFFCharging & Battery indicator*Charging: Red*Battery Low (<6%): Blinking Red*Charging finish : OFF*Power Mode: Blinking Red between Power & Charging Indicator WIFI CONTROL BUTTONSCREEN CONTROL BUTTONBottom Side ViewTouchpad/Click padTouch-sensitive pointing device which functions like the mouse.The left and right buttons function like the left and right buttons on astandard mouse.Left and right touchpad/Clickpad buttons8KeyboardThe keyboard provides keys with comfortable travel.91012345NOTE :The product’s thermal vent design may vary depending on the model and internal hardware in which you purchased.1Battery Lock 1 and 2 - SpringThe spring battery lock 1 and 2 are used to keep the battery pack secured.Lock 1Switch lock 1 by manual to lock or unlock.Lock 2When the battery pack is inserted, lock 2 will automatically lock.To remove the battery pack, lock 2 must be held in the unlocked position.2Battery Module34VentsHDD / RAM / SSD / WLAN Compartment CoverThe cover can be removed to install or remove the HDD / RAM/ SSD / WLAN. If damage occurs while attempting to modify orremove parts, it will not be protected under warranty.5HDD / RAM / SSD / WLAN Compartment Cover Screws You can remove the screws to install or remove the HDD / RAM / SSD / WLAN compartment cover. ( Please refer to the HDD /RAM / SSD / WLAN compartment view on the following page.)The thermal vents help the processor to avoid overheating.The thermal vents are designed to cool the internal components and avoid overheating.HDD / RAM / SSD / WLAN Compartment View (Optional)If you intend to modify or upgrade your SSD, RAM or WLAN parts, please*******************************************.Theuseofunauthorized parts will not void your warranty. However, any aftermarket or third party part must be removed before requesting warranty assistance.Right Side ViewLeft Side View3USB Ports Connects an USB device.(such as USB Zip drive, keyboard or mouse) into this jack.Audio Out JackConnects amplified speakers, headphones or microphone into this jack.12Mic In JackConnects amplified speakers, headphones or microphone into this jack.4 2 in 1 Card ReaderInsert Memory Card.1Stereo speakersProduce stereo sound. Back Side View1 2VentsPower ConnectorConnects the AC adapter into this connector.The thermal vents are designed to cool the internal components and avoid overheating.34Network JackHDMI portSupports high-definition digital video connections.56Display PortUSB PortConnected to the screen, or home theater system.7USB PortConnects an USB device (such as USB Zip drive, keyboard ormouse) into this jack.This jack allows you to connect to a Local Area Network (LAN) connection.Connects an USB device (such as USB Zip drive, keyboard ormouse) into this jack.-Intelligent charging 1.8A function for external USB device in power off mode.Shutdown state, continue to press the power button for about 6seconds, Power USB is enabled; under-powered USB shutdownstate, rst press the power button to cancel this function, press the power button is turned on.Chapter 3 Getting started3.1 AC AdapterDC IN123NOTE: The power adapter can become hot when in use. Please be sure the AC adapter is not covered with any materials keep it away from exposed parts of your body. The AC adapter appearance may vary depending on your region.CAUTION: The use of inferior extension cords may result in damageto your notebook. Your notebook comes with its own authorized ACadapter. Use of a different AC adapter or cable extension whichis not authorized for use will void warranty protection if damage tohardware is found in association to said adapter or extension cable. Please plug in the AC adapter for initial use of your notebook until your battery levels are verified to provide sufficient power. The AC adapter should be plugged back in as soon as your PC indicates the battery level is low. If the battery is continuously utilized past the point of indication of a low battery, it can affect the longevityof your battery. If you intend to remove the battery and use the PC under AC power only, the battery should be stored at room temperature at a charge level between 55-65% before removal. Note that the AC adapter included in the package is approved for your notebook; using other adapter model may damage either the notebook or other devices attached to it.3.2 Knowing the KeyboardThe following defines the colored hot keys on the Keyboard. The colored commands can only be accessed by first pressing and holding the function key while pressing a key with a colored command.3.2.1 For keyboard usersTo activate these functions, press the hot key associated with the desired function as below :Function KeysFunction KeysFunction Keys3.3 Using the touchpad / clickpad Press the left 1 and right 2 buttons located on the edge of the touchpad / clickpad to make selections and run functions. These two buttons are similar to the left and right buttons on a mouse. Tapping on the touchpad / clickpad produces similar results.The touchpad / clickpad is a rectangular electronic panel located just below your keyboard. You can use the static-sensitive panel of the touchpad / clickpad and slide it to move the cursor. You can use the buttons below the touchpad as left and right mouse buttons.Modern touchpad gesture:A modern touchpad should support the core touch gestures described in the following table.Touch Pad Gesture 3.3.1 Touchpad UsageChapter 4 BIOS setupNOTE : The drivers, BIOS and utilities bundled in the support DVD may vary by models and are subject to change without notice.4.1 About BIOS Setup4.1.1 When to Use BIOS Setup ?4.1.2 How to Run BIOS Setup ?You may need to run the BIOS Setup when:・ A n error message appears on the screen during the system booting up and is requested to run SETUP.・ Y ou want to change the default settings for customized features.・You want to reload the default BIOS settings.To run the BIOS Setup Utility, turn on the notebook and press the [Del] key during the POST procedure.If the message disappears before you respond and you still wish to enter Setup, either restart the system by turning it OFF and ON, or simultaneously pressing [Ctrl]+[Alt]+[Del] keys to restart.Be noted that the screen snaps and setting options in this chapter are for your references only.The actual setting screens and options on your Notebook may be different because of BIOS update.The setup function only can be invoked by pressing [Del] or [F2] key during POST that provide a approach to change some setting and configuration the user prefer, and the changed values will save in the NVRAM and will take effect after the system rebooted. The setup uses a menu interface to allow the user to configure their system and the features are briefly listed as follow.Press [F7] key for Boot Menu.4.2 BIOS Setup MenuOnce you enter the BIOS Setup Utility, the Main Menu will appear on the screen. Select the tags to enter the other menus.Main MenuShow system overview about memory size,setting of system time and date.Advanced MenuTo select the XD feature enable or disable XD feature only work with Intel platform + Windows.Security MenuInstall or clear the password settings for supervisor and user. Boot MenuConfigure Settings during System Boot.EXIT MenuSave or discard the changes before leaving the BIOS Setup Menu.4.2.1 Main Menu・System TimeThis item allows you to set the system time. There is a small internal (CMOS) battery which is designed to maintain your system clock. It is designed to maintain time even when thePC is powered down or in sleep mode. The time format is [hour:minute:second].Use [+] or [-] to configure system Time.・System DateThis item allows you to set the system date. The date format is [day:month:date:year].Day Day of the week, from Sun to Sat, whichis determined by BIOS (read-only).Month (Month)The month from 01 (January) to 12(December).Date (Date)The date from 01 to 31.Year (Year)The year can be adjusted by users.・Total MemoryThis allows you to see the total amount of memory.4.2.2 Advanced Menu• Wake on LANEnable/Disableintegrated LAN to wake the system.• Serial ATA HDDIt will show AHCI SATA HDD info.・NVMe DeviceEnable/Disableintegrated NVMe to wake the system.・Intel Virtualization TechnologyWhen enabled, a VMM can utilize the additional hardware capabilities provided by Vanderpool Technology.・RAID mode reminderPlease update strip size in BIOS to 128K in order for RAID to optimize performance.4.2.3 Security Menu・Change Administrator PasswordWhen this item is selected, a message box shall appear on the screen as below:Enter New PasswordType a maximum of 20-digit password and press [Enter].The password typed now will replace any previously set password from CMOS memory. You may also press [ESC] to abandon new password setting. It is extremely important to keep record for any BIOS password which is set as it can not be reset even with the removal of a CMOS battery for the security of your PC.Select Change User Password to give or to abandon password setting same as Change Administrator Password item above.Note that Administrator Password field allows users to enter and change the settings of the BIOS SETUP UTILITY, while User Password field only allows users to enter the BIOS SETUP UTILITY without having the authorization to make any change.The Password Check item is used to specify the type of BIOS password protection that is implemented.To clear a set Administrator Password/ User Password, just press [Enter] under Change Administrator Password/ Change User Password field when you are prompted to enter the password. A message box will pop up confirming password will be disabled. Once the password is disabled, the system will boot and user can enter setup without entering password.4.2.4 Boot Menu・Boot ConfigurationConfigure Settings during System Boot.・LAN Remote Boot [Disable]: Boot from LAN or not.・Set Boot Priority (1st/2nd/3rd/..... Boot)Specifies the boot sequence from the available devices.A device enclosed in parenthesis has been disabled in the corresponding type menu.・ Hard Disk Drive / USB HardDisk DriveSpecifies the Boot Device Priority sequence.4.2.5 Exit Menu・Save Changes and ResetExit system setup after saving the changes.F4 key can be used for this operation.・Discard Changes and ExitExit system setup without saving any changes.ESC key can be used for this operation.・Restore DefaultsRestore/Load Defaults values for all the setup options.F3 key can be used for this operation.。

计算机组成与设计 第五版答案_CH01_Solution

计算机组成与设计 第五版答案_CH01_Solution

Chapter 1 Solutions S-3 1.1 P ersonal computer (includes workstation and laptop): Personal computersemphasize delivery of good performance to single users at low cost and usuallyexecute third-party soft ware.Personal mobile device (PMD, includes tablets): PMDs are battery operatedwith wireless connectivity to the Internet and typically cost hundreds ofdollars, and, like PCs, users can download soft ware (“apps”) to run on them.Unlike PCs, they no longer have a keyboard and mouse, and are more likelyto rely on a touch-sensitive screen or even speech input.Server: Computer used to run large problems and usually accessed via a network.Warehouse scale computer: Th ousands of processors forming a large cluster.Supercomputer: Computer composed of hundreds to thousands of processorsand terabytes of memory.Embedded computer: Computer designed to run one application or one setof related applications and integrated into a single system.1.2a.Performance via Pipeliningb.Dependability via Redundancyc.Performance via Predictiond.Make the Common Case Faste.Hierarchy of Memoriesf.Performance via Parallelismg.Design for Moore’s Lawe Abstraction to Simplify Design1.3 Th e program is compiled into an assembly language program, which is thenassembled into a machine language program.1.4a.1280 ϫ 1024 pixels ϭ 1,310,720 pixels ϭϾ 1,310,720 ϫ 3 ϭ 3,932,160bytes/frame.b. 3,932,160 bytes ϫ (8 bits/byte) /100E6 bits/second ϭ 0.31 seconds1.5a. performance of P1 (instructions/sec) ϭ 3 ϫ 109/1.5 ϭ 2 ϫ 109performance of P2 (instructions/sec) ϭ 2.5 ϫ 109/1.0 ϭ 2.5 ϫ 109performance of P3 (instructions/sec) ϭ 4 ϫ 109/2.2 ϭ 1.8 ϫ 109S-4 Chapter1Solutionsb. cycles(P1) ϭ 10 ϫ 3 ϫ 109ϭ 30 ϫ 109 scycles(P2) ϭ 10 ϫ 2.5 ϫ 109ϭ 25 ϫ 109 scycles(P3) ϭ 10 ϫ 4 ϫ 109ϭ 40 ϫ 109 sc. No. instructions(P1) ϭ 30 ϫ 109/1.5 ϭ 20 ϫ 109No. instructions(P2) ϭ 25 ϫ 109/1 ϭ 25 ϫ 109No. instructions(P3) ϭ 40 ϫ 109/2.2 ϭ 18.18 ϫ 109CPInew ϭ CPIoldϫ 1.2, then CPI(P1) ϭ 1.8, CPI(P2) ϭ 1.2, CPI(P3) ϭ 2.6fϭ No. instr. ϫ CPI/time, thenf(P1) ϭ 20 ϫ 109ϫ1.8/7 ϭ 5.14 GHzf(P2) ϭ 25 ϫ 109ϫ 1.2/7 ϭ 4.28 GHzf(P1) ϭ 18.18 ϫ 109ϫ 2.6/7 ϭ 6.75 GHz1.6a. C lass A: 105 instr. Class B: 2 ϫ 105 instr. Class C: 5 ϫ 105 instr.Class D: 2 ϫ 105 instr.Time ϭ No. instr. ϫ CPI/clock rateTotal time P1 ϭ (105ϩ 2 ϫ 105ϫ 2 ϩ 5 ϫ 105ϫ 3 ϩ 2 ϫ 105ϫ 3)/(2.5 ϫ109) ϭ 10.4 ϫ 10Ϫ4 sTotal time P2 ϭ (105ϫ 2 ϩ 2 ϫ 105ϫ 2 ϩ 5 ϫ 105ϫ 2 ϩ 2 ϫ 105ϫ 2)/(3 ϫ 109) ϭ 6.66 ϫ 10Ϫ4 sCPI(P1) ϭ 10.4 ϫ 10Ϫ4ϫ 2.5 ϫ 109/106ϭ 2.6CPI(P2) ϭ 6.66 ϫ 10Ϫ4ϫ 3 ϫ 109/106ϭ 2.0b. c lock cycles(P1)ϭ 105ϫ 1ϩ 2 ϫ 105ϫ 2 ϩ 5 ϫ 105ϫ 3 ϩ 2 ϫ 105ϫ 3ϭ 26 ϫ 105clock cycles(P2) ϭ 105ϫ 2ϩ 2 ϫ 105ϫ 2 ϩ 5 ϫ 105ϫ 2 ϩ 2 ϫ 105ϫ 2 ϭ 20 ϫ 1051.7a. CPI ϭ Texecϫ f/No. instr.Compiler A CPI ϭ 1.1Compiler B CPI ϭ 1.25b.fB /fAϭ (No. instr.(B) ϫ CPI(B))/(No. instr.(A) ϫ CPI(A)) ϭ 1.37c.TA /Tnewϭ 1.67T B /Tnewϭ 2.27Chapter 1 Solutions S-51.81.8.1 C ϭ 2 ϫ DP/(V2*F)Pentium 4: C ϭ 3.2E–8FCore i5 Ivy Bridge: C ϭ 2.9E–8F1.8.2 Pentium 4: 10/100 ϭ 10%Core i5 Ivy Bridge: 30/70 ϭ 42.9%1.8.3 (Snew ϩ Dnew)/(Soldϩ Dold) ϭ 0.90Dnew ϭ C ϫ Vnew2 ϫ FS old ϭ Voldϫ IS new ϭ Vnewϫ ITh erefore:V new ϭ [Dnew/(C ϫ F)]1/2Dnew ϭ 0.90 ϫ (Soldϩ Dold) Ϫ SnewS new ϭ Vnewϫ (Sold/Vold)Pentium 4:S new ϭ Vnewϫ (10/1.25) ϭ Vnewϫ 8Dnew ϭ 0.90 ϫ 100 Ϫ Vnewϫ 8 ϭ 90 Ϫ Vnewϫ 8V new ϭ [(90 Ϫ Vnewϫ 8)/(3.2E8 ϫ 3.6E9)]1/2Vnewϭ 0.85 V Core i5:S new ϭ Vnewϫ (30/0.9) ϭ Vnewϫ 33.3Dnew ϭ 0.90 ϫ 70 Ϫ Vnewϫ 33.3 ϭ 63 Ϫ Vnewϫ 33.3V new ϭ [(63 Ϫ Vnewϫ 33.3)/(2.9E8 ϫ 3.4E9)]1/2Vnewϭ 0.64 V1.91.9.11 2.56E9 1.28E9 2.56E87.94E1039.712 1.83E99.14E8 2.56E8 5.67E1028.3 1.449.12E8 4.57E8 2.56E8 2.83E1014.2 2.88 4.57E8 2.29E8 2.56E8 1.42E107.10 5.6S-6 Chapter1Solutions1.9.2141.0229.3414.687.331.9.3 31.101.10.1 die area15cmϭ wafer area/dies per wafer ϭ pi*7.52 / 84 ϭ 2.10 cm2yield15cmϭ 1/(1ϩ(0.020*2.10/2))2ϭ 0.9593die area20cmϭ wafer area/dies per wafer ϭ pi*102/100 ϭ 3.14 cm2yield20cmϭ 1/(1ϩ(0.031*3.14/2))2ϭ 0.90931.10.2 cost/die15cmϭ 12/(84*0.9593) ϭ 0.1489cost/die20cmϭ 15/(100*0.9093) ϭ 0.16501.10.3 die area15cmϭ wafer area/dies per wafer ϭ pi*7.52/(84*1.1) ϭ 1.91 cm2yield15cmϭ 1/(1 ϩ (0.020*1.15*1.91/2))2ϭ 0.9575die area20cmϭ wafer area/dies per wafer ϭ pi*102/(100*1.1) ϭ 2.86 cm2yield20cmϭ 1/(1 ϩ (0.03*1.15*2.86/2))2ϭ 0.90821.10.4 d efects per area0.92ϭ (1–y^.5)/(y^.5*die_area/2) ϭ (1Ϫ0.92^.5)/(0.92^.5*2/2) ϭ 0.043 defects/cm2defects per area0.95ϭ (1–y^.5)/(y^.5*die_area/2) ϭ (1Ϫ0.95^.5)/(0.95^.5*2/2) ϭ 0.026 defects/cm21.111.11.1 CPI ϭ clock rate ϫ CPU time/instr. countclock rate ϭ 1/cycle time ϭ 3 GHzCPI(bzip2) ϭ 3 ϫ 109ϫ 750/(2389 ϫ 109)ϭ 0.941.11.2 SPEC ratio ϭ ref. time/execution timeSPEC ratio(bzip2) ϭ 9650/750 ϭ 12.861.11.3. CPU time ϭ No. instr. ϫ CPI/clock rateIf CPI and clock rate do not change, the CPU time increase is equal to the increase in the of number of instructions, that is 10%.Chapter 1 Solutions S-71.11.4 CPU time(before) ϭ No. instr. ϫ CPI/clock rateCPU time(aft er) ϭ 1.1 ϫ No. instr. ϫ 1.05 ϫ CPI/clock rateCPU time(aft er)/CPU time(before) ϭ 1.1 ϫ 1.05 ϭ1.155. Th us, CPU timeis increased by 15.5%.1.11.5 SPECratio ϭ reference time/CPU timeSPECratio(aft er)/SPECratio(before) ϭ CPU time(before)/CPU time(aft er) ϭ1/1.1555 ϭ 0.86. Th e SPECratio is decreased by 14%.1.11.6 CPI ϭ (CPU time ϫ clock rate)/No. instr.CPI ϭ 700 ϫ 4 ϫ 109/(0.85 ϫ 2389 ϫ 109) ϭ 1.371.11.7 Clock rate ratio ϭ 4 GHz/3 GHz ϭ 1.33CPI @ 4 GHz ϭ 1.37, CPI @ 3 GHz ϭ 0.94, ratio ϭ 1.45Th ey are diff erent because, although the number of instructions has beenreduced by 15%, the CPU time has been reduced by a lower percentage.1.11.8 700/750 ϭ 0.933. CPU time reduction: 6.7%1.11.9 No. instr. ϭ CPU time ϫ clock rate/CPINo. instr. ϭ 960 ϫ 0.9 ϫ 4 ϫ 109/1.61 ϭ 2146 ϫ 1091.11.10 Clock rate ϭ No. instr. ϫ CPI/CPU time.Clock ratenew ϭ No. instr. ϫ CPI/0.9 ϫ CPU time ϭ 1/0.9 clock rateoldϭ3.33 GHz1.11.11 Clock rate ϭ No. instr. ϫ CPI/CPU time.Clock ratenew ϭ No. instr. ϫ 0.85ϫ CPI/0.80 CPU time ϭ 0.85/0.80,clock rateoldϭ 3.18 GHz1.121.12.1 T(P1) ϭ 5 ϫ 109ϫ 0.9 / (4 ϫ 109) ϭ 1.125 sT(P2) ϭ 109ϫ 0.75 / (3 ϫ 109) ϭ 0.25 sclock rate (P1) Ͼ clock rate(P2), performance(P1) < performance(P2) 1.12.2 T(P1) ϭ No. instr. ϫ CPI/clock rateT(P1) ϭ 2.25 3 1021 sT(P2) 5 N ϫ 0.75/(3 ϫ 109), then N ϭ 9 ϫ 1081.12.3 MIPS ϭ Clock rate ϫ 10Ϫ6/CPIMIPS(P1) ϭ 4 ϫ 109ϫ 10Ϫ6/0.9 ϭ 4.44 ϫ 103S-8 Chapter1SolutionsMIPS(P2) ϭ 3 ϫ 109ϫ 10Ϫ6/0.75 ϭ 4.0 ϫ 103MIPS(P1) Ͼ MIPS(P2), performance(P1) Ͻ performance(P2) (from 11a)1.12.4 MFLOPS ϭ No. FP operations ϫ 10Ϫ6/TMFLOPS(P1) ϭ .4 ϫ 5E9 ϫ 1E-6/1.125 ϭ 1.78E3MFLOPS(P2) ϭ .4 ϫ 1E9 ϫ 1E-6/.25 ϭ 1.60E3MFLOPS(P1) Ͼ MFLOPS(P2), performance(P1) Ͻ performance(P2)(from 11a)1.131.13.1 Tfp ϭ 70 ϫ 0.8 ϭ 56 s. Tnewϭ 56ϩ85ϩ55ϩ40 ϭ 236 s. Reduction: 5.6%1.13.2 Tnew ϭ 250 ϫ 0.8 ϭ 200 s, TfpϩTl/sϩTbranchϭ 165 s, Tintϭ 35 s. Reductiontime INT: 58.8%1.13.3 Tnew ϭ 250 ϫ 0.8 ϭ 200 s, TfpϩTintϩTl/sϭ 210 s. NO1.141.14.1 C lock cyclesϭ CPIfp ϫ No. FP instr. ϩ CPIintϫ No. INT instr. ϩ CPIl/sϫNo. L/S instr. ϩ CPIbranchϫ No. branch instr.TCPUϭ clock cycles/clock rate ϭ clock cycles/2 ϫ 109clock cycles ϭ 512 ϫ 106; TCPUϭ 0.256 sTo have the number of clock cycles by improving the CPI of FP instructions:CPIimproved fp ϫ No. FP instr. ϩ CPIintϫ No. INT instr. ϩ CPIl/sϫ No. L/Sinstr. ϩ CPIbranchϫ No. branch instr. ϭ clock cycles/2CPIimproved fp ϭ (clock cycles/2 Ϫ (CPIintϫ No. INT instr. ϩ CPIl/sϫ No. L/Sinstr. ϩ CPIbranchϫ No. branch instr.)) / No. FP instr.CPIimproved fpϭ (256Ϫ462)/50 Ͻ0 ϭϭϾ not possible1.14.2 Using the clock cycle data from a.To have the number of clock cycles improving the CPI of L/S instructions:CPIfp ϫ No. FP instr. ϩ CPIintϫ No. INT instr. ϩ CPIimproved l/sϫ No. L/Sinstr. ϩ CPIbranchϫ No. branch instr. ϭ clock cycles/2CPIimproved l/s ϭ (clock cycles/2 Ϫ (CPIfpϫ No. FP instr. ϩ CPIintϫ No. INTinstr. ϩ CPIbranchϫ No. branch instr.)) / No. L/S instr.CPIimproved l/sϭ (256Ϫ198)/80 ϭ 0.7251.14.3 C lock cyclesϭ CPIfp ϫ No. FP instr. ϩ CPIintϫ No. INT instr. ϩ CPIl/sϫNo. L/S instr. ϩ CPIbranchϫ No. branch instr.Chapter 1 Solutions S-9TCPUϭ clock cycles/clock rate ϭ clock cycles/2 ϫ 109CPIint ϭ 0.6 ϫ 1 ϭ 0.6; CPIfpϭ 0.6 ϫ 1 ϭ 0.6; CPIl/sϭ 0.7 ϫ 4 ϭ 2.8;CPIbranchϭ 0.7 ϫ 2 ϭ 1.4T CPU (before improv.) ϭ 0.256 s; TCPU(aft er improv.)ϭ 0.171 s1.15110025054100/54 ϭ 1.85 1.85/2 ϭ .9342529100/29 ϭ 3.44 3.44/4 ϭ 0.86 812.516.5100/16.5 ϭ 6.06 6.06/8 ϭ 0.7516 6.2510.25100/10.25 ϭ 9.769.76/16 ϭ 0.61。

MHS-15合CURY 氧化物系统产品说明书

MHS-15合CURY 氧化物系统产品说明书

Minimum investment and low operating costs The MHS-15 Mercury/Hydride System is a manual accessory for high-sensitivity determination of mercury and hydride-forming elements, such as As, Se, Sb, Te,Bi and Sn, by flame atomic absorp-tion (AA) spectroscopy. Mercury/hydride atomic absorption is the preferred technique for trace deter-mination of mercury and the metal-lic hydride-forming elements,because it offers the best possible detection limits, down to the ng range, while minimizing capital investment and operating costs.The technique is simple, easy-to-handle, and reliable and has been proven in tens of thousands of laboratories worldwide.Simple operationThe MHS-15 system includes a reaction assembly and a quartz-cell assembly. The analyzer is free-standing and is placed adjacent to the AA spectrometer sample compartment. It includes a reaction flask, a reservoir for the reducing agent and all pneumatic compo-nents for carrier transportation of the metallic vapors to the quartz cell.The cell mounts on a standard10-cm PerkinElmer ®burner head,permitting the cell to be heated in the spectrometer flame. The quartz cell is positioned in the light beam of the spectrometer and alignment is provided by the instrument’s burner adjustment controls. ItsMHS-15 Mercury/Hydride SystemP R O D U C T N O T EMinimum investment Low operating costs Simple operationWorks with all PerkinElmeratomic absorption spectrometersKey BenefitsAT O M I C A B S O R P T I O Nquartz cell can be easily tilted out of the flame using a swivel mount.Due to the reaction chemistry in the mercury/hydride process, the ana-lyte is completely separated from its matrix during analysis, eliminating interferences.Operating procedure During analysis, the sample contain-ing the analyte in ionic form is placed in the reaction flask. The flask is then installed into the analyzer assembly.By manually actuating a pneumatic valve, an argon stream transports the reducing agent (NaBH 4) from a reser-voir into the sample. The special conical form of the immersion tube and reaction flask provides excellent mixing of the sample and NaBH 4,guaranteeing spontaneous reaction.No stirring device is required for this procedure.Hydrogen, liberated during the reac-tion, reduces the metal ions to gaseous hydrides. In the case of mercury determination, NaBH 4reduces mer-cury into its metallic state. Excess hydrogen and the argon carrier gas transport the hydride (or metallic mercury) vapor into the quartz cell where the absorption of the metal is measured. During mercury determi-nation, the cell remains at ambient temperatures. Mercury determina-tion can also be performed using stannous chloride (SnCl 2) as the reducing agent. For the determina-tion of hydride-forming elements,the quartz cell is heated by an air/acetylene flame to thermally decompose the metal hydrides.For a complete listing of our global offices, visit /lasoffices©2004 PerkinElmer, Inc. All rights reserved. The PerkinElmer logo and design are registered trademarks of PerkinElmer, Inc. AAnalyst is a trademark and PerkinElmer is a registered trademark of PerkinElmer, Inc. or its subsidiaries, in the United States and other countries. All other trademarks not owned by PerkinElmer, Inc. or its subsidiaries that are depicted herein are theproperty of their respective owners. PerkinElmer reserves the right to change this document at any time without notice and disclaims liability for editorial, pictorial or typographical errors.006385C_01ELEC110400Printed in USAPerkinElmer Life and Analytical Sciences 710 Bridgeport AvenueShelton, CT 06484-4794 USA Phone: (800) 762-4000 or (+1) 203-925-4602Technical dataAbsorption cell Quartz cell, 165 mm long, 12 mm diameter Cell heating Up to approx. 1000 °C with air/acetylene flameReaction flaskPolypropylene flask with special, conically formed inner side.Connection to the analyzer assembly with special joint with O-ring gasket.Sample volume Max. 50-mL volume Reducing agent NaBH 4or SnCl 2Inert gas supply Argon, inlet pressure 2.5-3.5 bar (36-51 psig) approx. consumption 1.1 L/min Power supply No electrical supply is required.Dimensions 70 mm wide x 160 mm deep x 375 mm high WeightApproximately 4.3 kgDetection limits using the MHS-15 with the AAnalyst™ 700(EDL lamps were used for all elements)。

CH9121网络和UART透明传输芯片说明书

CH9121网络和UART透明传输芯片说明书

Network and UART Transparent Transmission Chip CH9121ManualVersion: 2B1.OverviewCH9121 is a chip realizing transparent transmission between network and UART. It integrates TCP/IP protocol stack, which can realize bidirectional transparent transmission between network data packets and serial data. It has 4 working modes: TCP CLIENT, TCP SERVER, UDP CLIENT and UDP SERVER. The serial baud rate can be up to 921600bps. It can be easily configured by upper computer software or serial commands, which is convenient and quick.The figure below is a general application block diagram of CH9121.2.Features●Internal Ethernet MAC layer and PHY layer●Realize bidirectional transparent transmission between serial data and network data●Support 10/100M, full/half duplex self-adaption Ethernet interface, compatible with 802.3 protocol●Support automatic MDI/MDIX line conversion●Support DHCP automatic access to IP address, and DNS domain name access●Set the chip working mode, port, IP and other network parameters through upper computer software andserial commands●Support four working modes: TCP CLIENT, TCP SERVER, UDP CLIENT and UDP SERVER●Support up to two independent UARTs, independent transparent transmission●Serial baud rate supports 300bps ~ 921600bps●Serial TTL level, compatible with 3.3V and 5V●UART supports full-duplex and half-duplex serial communication, and RS485 transceiving automaticswitch●Support and provide virtual serial software●Support KEEPALIVE mechanism3.PackageChipPackageName DescriptionCH9121 LQFP64M LQFP package; 64 pins; package body 10 x10mm 4.PinsCH9121 Pin No.PinNamePin Type Pin description2, 12, 17, 21, 29, 40, 42, 45, 63 VCC33 Power3.3V positive power input, require an external 0.1uF powerdecoupling capacitor6, 19, 28, 43, 54, 64 VCC18 Power1.8V positive power input, require an external 0.1uF powerdecoupling capacitor3, 9, 13, 18, 20,37, 41, 44, 48GND Power Common ground14, 15, 16, 22,23, 24, 25, 26,27, 34, 35, 38,39, 47, 49, 50,61, 62NC - Reserved, suspended1 RSETE Input Connected with an external 18K resistor to ground4 RXP EthernetsignalEthernet RXP signal5 RXN Ethernet Ethernet RXN signalsignal7 TXP EthernetsignalEthernet TXP signal8 TXN EthernetsignalEthernet TXN signal10 XI Input Input of the crystal oscillator, requires an external 30MHzcrystal11 XO Output Inverted output of crystal oscillator, requires an external 30MHzcrystal30 TCPCS1 Output In TCP client mode, connection status indicator of port 1, activeat low level31 RUN Output CH9121 running status indicator pin, multiplexed as ISPupgrade pin32 CFGEN Input Network configuration enable pin, detects when power-up, configuration disabled when at low level33 TCPCS2 Output In TCP client mode, connection status indicator of port 2, activeat low level36 RSTI Input External reset input, active at low level, built-in pull-up resistor 46 DIR2 Output Used to control RS485 receiving/transmitting switch of UART251 DIR1 Output Used to control RS485 receiving/transmitting switch of UART152 ACT# Output Ethernet connection communication indicator LED drive pin53 LINK# Output PHY connection indicator LED, active low55 RXD2 Input Serial data input of UART2, built-in pull-up resistor(Off by default)56 TXD2 Output Serial data output of UART2 (Off by default)57 RXD1 Input Serial data input of UART1, built-in pull-up resistor(On by default)58 TXD1 Output Serial data output of UART1 (On by default)59 RESET InputRestore factory settings, chip power-on detection, active at low level60 CFG0 InputUART configuration mode set pin, built in pull-up. If low level is detected, enter UART configuration mode. Exit the mode if high level is detected5.Function5.1. Function DescriptionCH9121 is a chip for transparent transmission between network and UART, and realizes bidirectional transparent transmission of serial data and network data. It supports 4 working modes (TCP CLEINT/SERVER, UDP CLIENT/SERVER), and the serial baud rate supported ranges from 300bps to 921600bps. Before use, the network parameters and UART parameters of the chip shall be configured by upper computer software NetModuleConfig.exe or serial commands. After the configuration is completed, CH9121 will save the configuration parameters to the internal storage space. And after the chip is reset, CH9121 will work according to the saved configuration value.The basic parameters of CH9121 include: name, MAC address display, dynamic access IP address setting, manual IP address setting (including CH9121 IP address, subnet mask, default gateway), and UART negotiation configuration.The name is mainly for the convenience of CH9121 module management within the LAN, with the length of not more than 20 bytes. The MAC address field shows the MAC address of the currently selectedmodule. There are two ways for CH9121 to set network parameters. 1) DHCP, that is, it automatically obtains network parameters from the gateway device with DHCP SERVER function; 2) manual setting. UART negotiation configuration function is to enter serial configuration mode by handshaking through the UART, which is disabled by default.CH9121 port parameters include: network mode, local port, target IP/domain name, destination port, serial baud rate/data bit/stop bit/parity check bit, network cable disconnection processing, RX packet length, RX packet timeout interval, network connection operation.The network mode (TCP SERVER/CLIENT, UDP SERVER/CLIENT), destination IP address, and local/destination port are the basic parameters of network communication. The destination IP address can also be accessed by domain name. The serial baud rate ranges from 300bps to 921600bps (the baud rate error of the serial port transmitting signal is less than 0.3%, and the allowable baud rate error of the serial port receiving signal is not less than 2%). It supports 5/6/7/8 data bits, 1/2 stop bits, odd/even/no parity check, blank 0, and mark 1 check mode. The network cable disconnection processing means that when the network cable is disconnected, CH9121 actively closes the connection or does not take any action. The length range of RX packet is 1-1024, which means that when the length of CH9121 UART receiving data reaches the set length, CH9121 will immediately packet the serial data and send it out via network. The timeout time setting range is 0-200, and the timeout unit is about 5ms. For example, when the timeout is 1, the data length of the serial port receiving buffer does not reach the length of the RX packet, and the serial port does not receive a new one in more than 5ms, the serial port timeout will occur. After the serial port timeout occurs, CH9121 will send the data received by the serial port over the network. When the timeout time is set to 0, the internal hardware timeout mechanism will be enabled (no new data is received after 4 data times), which is suitable for occasions where real-time requirements are high and large quantities of data are sent and received. Clear serial port buffer setting refers to how the data received by the serial port is processed before the network connection is established, the data is cleared (discarded) or retained after TCP connection.5.2. Default ConfigurationWhen CH9121 leaves factory, port 2 is closed and port 1 works in TCP CLIENT mode by default. The default parameters related to the network are as follows:(1)Device IP: 192.168.1.200(2)Subnet mask: 255.255.255.0(3)Default gateway: 192.168.1.1(4)Module port: 2000(5)Destination IP: 192.168.1.100(6)Destination port: 1000(7)Number of reconnection: unlimitedUART related default parameters are:(1)Baud rate: 9600(2)Timeout: 0(3)Data bit: 8. Stop bit: 1. Parity check bit: No(4)Clear UART buffer: Never6.Parameters6.1. Absolute Maximum ValueCritical value or exceeding the absolute maximum value may cause the chip to work abnormally or even be damaged.Name Parameter description Min. Max. UnitTA Ambient temperature during operation VCC33=3.3VVCC18=1.8V-40 85 ℃TS Ambient temperature during storage -55 125 ℃VCC33 Supply voltage (VCC33 connects to power, GND to ground) -0.4 4.2 V VCC18 Supply voltage (VCC18 connects to power, GND to ground) -0.4 2.3 V VIO Voltage on input or output pins -0.4 VCC33+0.4 V VIO5 Voltage on input or output pins that support 5V withstand voltage -0.4 5.4 V6.2. Electrical ParametersTest Conditions: TA=25℃, VCC33=3.3V, VCC18=1.8VName Parameter description Min. Typ. Max. UnitVCCxx Supply voltage VCC33 2.7 3.3 3.6V VCC18 1.65 1.8 1.95ICC Total supply current duringoperationVCC33=3.3V 160 190 mAVIL Low level input voltage -0.4 0.7 V VIH High level input voltage 2.0 VCC33+0.4 V VOL Low level output voltage (4mA draw current) 0.4 V VOH High level output voltage (4mA output current) VCC33-0.4 VIUP Input current at the input terminal with built-inpull-up resistor20 40 100 uAIDN Input current at the input terminal with built-inpull-down resistor-20 -40 -100 uAVR Voltage threshold when power-on reset 1.4 1.5 2.5 V7.Applications7.1 Hardware Circuit DesignNote: Due to space limitations, the circuit of the power supply and decoupling capacitors near the 3.3V and 1.8V pins is omitted in the figure, which must be added when designing the circuit. The detailed circuit reference file: CH9121PCB (please download in our official website).U1 is the main control chip CH9121. TXD1 and RXD1 are compatible with 3.3V and 5V level, and the RS485 control pin DIR can be suspended if not used.P1 is RJ45 port, with built-in network transformer, used to connect network equipments such as switches and router. It contains two pairs of Ethernet differential signals.When actually making a PCB, R5-R8, C6, and C7 should be as close as possible to the 5th pin of P1. The 0.1 uF decoupling capacitors for 3.3V and 1.8V pins are omitted in the figure. TXOP (RXIP) and TXON. (RXIN) are differential signals. When wiring, they should be wired close to parallel, and try to provide ground wire or copper coating on both sides to reduce external interference. The length of relevant signals of crystals XI and XO should be shortened as much as possible. In order to reduce the interference of high-frequency clock for the outside, the baselines should be surrounded or copper should be clad around relevant components.。

KREATECH FISH探针系列产品说明书

KREATECH FISH探针系列产品说明书

A D V A N C ED S T A I N I N G RE A G E N T SFULLY AUTOMATED FISH PROBES FOR BONDK R E AT E C HF I S H P R O B E S F O R B O N DADVANCING MOLECULAR TESTING IN CANCER DIAGNOSTICSREALIZE THE BENEFITS OF FISH PROBES FOR BONDIN YOUR LABORATORYExpanding FISH Probes for BOND, a range of fully automated FISH probes for FFPE tissue.Combining the benefits of Kreatech REPEAT-FREE probe technology with the workflow efficiency and reliability provided by BOND fully automated IHC/ISH stainers, the expanded menu of FISH Probes for BOND represents an advancement in automatic FISH testing.FISH staining using Kreatech SS18 (18q11) Break– XL FISH ProbeEASYReduce complexity and increase standardization.EFFICIENTAchieve rapid results withminimal hands-on time.ACCURATEDeliver consistent, high-quality stainingthat supports accurate patient diagnosis.Manual FISH testing is complex and can introduce protocol variation which impairs staining quality. Fully automated FISH testing on BOND increasesstandardization and delivers reliable results with walk-away convenience.REDUCE COMPLEXITY AND INCREASE STANDARDIZATIONEASYMINIMIZE REPEATSStandardized protocols deliver consistent performance, minimizing repeat testing.IMPROVE WORKFLOWBOND systems automate all stepsfrom dewax to post-hybridization wash, reducing complexity and hands-on time.SIMPLIFY FISH TESTINGThe BOND instruments andintuitive software make operation simple and straightforward.FISH staining using Kreatech MDM2 (12q15) / SE 12 (D12Z3) – XL FISH Probe0:21:161:26:3012:00:0012:00:00MANUAL FISHAUTOMATED FISHHANDS-ON-TIMETotal reduction in hands on time of 81% (108 mins to 20 mins)WAITING TIMEDENATURATION/HYBRIDIZATION•F ISH Probes for BOND can achieve FISH results faster than a typical manual workflow, with minimal hands-on time.•P atient cases can be continuously processed on the BOND system, as there's no need to batch cases as with manual FISH staining.Traditional FISH can be time-consuming and labor-intensive, requiring experienced staff for many hours. BOND delivers increased productivity and reliable performance, offering you efficiency improvements without compromising quality.DELIVER RAPID RESULTS WITH MINIMAL HANDS ON TIMEEFFICIENTMAXIMIZE USE OF RESOURCESAutomated testing frees up valuable, highly-skilled staff to work on other tasks.IMPROVETURNAROUND TIMEBOND systems facilitate continuousslide processing, which means there is no need to batch slides.BE RESPONSIVEBOND systems perform IHC and ISH at the same time. This provides your laboratory with the flexibility to adapt to changing caseloads.FISH Probes for BOND combine the benefits of advanced probe technologywith the consistent quality of the BOND system.DELIVER CONSISTENT, HIGH-QUALITY STAINING THAT SUPPORTS ACCURATE PATIENT DIAGNOSISACCURATENO COMPROMISEBOND systems can perform simultaneous FISH, CISH and IHC without compromise to turnaround time.A D V A N C E D S T A I N I N G R E A G E N T SK R E A T E C H F I S H P R O B E SPRODUCT CODE PRODUCT NAME ROLE PACK SIZEDS9636BOND FISH Kit Enables FISH to be performed on the BONDand provides post-hybridization wash step.60 testsAR9037BOND Hybridization solution Used for the dilution of individual in situ hybridization(ISH) probes for use on automated BOND system.100 mLLK-095A DAPI 0.1 µg/mL Used as a counterstain for chromosomes. 1 mLLK-096A DAPI 1 µg/mL Used as a counterstain for chromosomes. 1 mLLK-097A Counterstain Diluent Used as an antifade solution, which helpspreserving the fluorescence signal.1 mLOP309700BOND Open Containers 30 mL BOND reagent container for probe; holds up to 30 mL.May be refilled (maximum total volume 40 mL).10 containersOP79193BOND Open Containers 7 mL BOND reagent container for probe; holds up to 7 mL.May be refilled (maximum total volume 40 mL).10 containersOPT9049BOND Titration Kit BOND reagent container for probe, withremovable inserts; holds up to 6 mL. Maybe refilled (maximum total volume 40 mL).Useful during protocol optimization.10 titration containers and 50 insertsOPT9719BOND Titration Container Inserts Used with titration containers shouldadditional removable inserts be required.50 inserts• Enables FISH on the BOND system• Provides post-hybridization wash step• Number of tests: 60• Product code: DS9636BOND FISH KITANCILLARIES AND CONSUMABLES FOR FISH ON BONDProducts in this flyer are subject to regulatory approval. Please consult your Leica Biosystems Sales Representative for availability in your region.For more information on regulatory status and intended use, see the IHC & ISH Product Catalog or product IFUs.A D V A N C E D S T A I N I N G R E A G E N T SKREATECHFISHPROBESCE-IVD XL FISH PROBES FOR BONDPRODUCT CODE PROBE NAMEPATHOLOGYKBI-XL001ALK (2p23) Break – XL for BOND LUNG PATHOLOGY KBI-XL002ROS1 (6q22) Break – XL for BOND LUNG PATHOLOGY KBI-XL003MET (7q31) / SE 7 (D7Z1) – XL for BOND LUNG PATHOLOGY KBI-XL004FGFR1 (8p11) / SE 8 (D8Z1) – XL for BOND LUNG PATHOLOGY KBI-XL005RET (10q11) Break – XL for BOND LUNG PATHOLOGY KBI-XL011EGFR (7p11) / SE7 (D7Z1) – XL for BOND NEW LUNG PATHOLOGY KBI-XL012NTRK1 (1q23) Break – XL for BOND NEWLUNG PATHOLOGY KBI-XL006MYC (8q24) Break – XL for BOND HEMATOPATHOLOGY KBI-XL007IGH (14q32) Break – XL for BOND HEMATOPATHOLOGY KBI-XL008BCL2 (18q21) Break – XL for BOND HEMATOPATHOLOGY KBI-XL009BCL6 (3q27) Break – XL for BOND HEMATOPATHOLOGY KBI-XL010TP53 (17p13) / SE 17 – XL for BOND HEMATOPATHOLOGY KBI-XL013CCND1 (11q13) Break – XL for BOND NEW HEMATOPATHOLOGY KBI-XL014MDM2 (12q15) / SE 12 (D12Z3) – XL for BOND NEW SOFT TISSUE PATHOLOGY KBI-XL015DDIT3 (12q13) Break – XL for BOND NEW SOFT TISSUE PATHOLOGY KBI-XL016FOXO1 (13q14) Break – XL for BOND NEW SOFT TISSUE PATHOLOGY KBI-XL017FUS (16p11) Break – XL for BOND NEW SOFT TISSUE PATHOLOGY KBI-XL018SS18 (18q11) Break – XL for BOND NEW SOFT TISSUE PATHOLOGY KBI-XL019EWSR1 (22q12) Break – XL for BONDNEW SOFT TISSUE PATHOLOGYDISCLAIMER: For In Vitro Diagnostic Use Only.11Leica Biosystems is a global leader in workflow solutions and automation. As the only company to own the workflow from biopsy to diagnosis, we are uniquely positioned to break down the barriers between each of these steps. Our mission of “Advancing Cancer Diagnostics, Improving Lives” is at the heart of our corporate culture. Our easy-to-use and consistently reliable offerings help improve Leica Biosystems is an international company with a strong network of worldwide customer services. For detailed contact information on your nearest sales office or distributor please visit our website: LEICA BIOSYSTEMSADVANCED STAINING SOLUTIONBOND ANCILLARIESANTIBODIESCISH PROBESFISH PROBESBOND DETECTIONRNA ISH PROBES。

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Solutions for End of Chapter QuestionsChapter 15Easy1 Car firms buy in raw materials (steel), intermediate goods (windscreens, tyres) and labour to make cars. Windscreen and tyre companies hire workers and also buy raw materials from other industries. What is the value added of the car industry (the three firms shown below)?1 The value added is the value of the final product less the value of the raw materialused. Assuming that “the car industry” includes all three firms, then: 1000 – 120 = 8802 GNP at market prices is £300 billion. Depreciation is £30 billion, indirect taxes £20 billion. (a) What is national income? (b) Why does depreciation cause a discrepancy between GNP and national income? (c) Why do indirect taxes enter the calculation?2 (a) Both depreciation and indirect taxes need to be subtracted, hence £250 bn. (b)National income is net national product at basic prices. Depreciation is part of the cost of producing output. (c) Indirect taxes do not add to the value of the national income. If they did nations would be able to become richer and richer simply by taxing more.3 GNP = 2000, C = 1700, G = 50 and NX = 40. (a) What is investment I? (b) If exportsare 350, what are imports? (c) If depreciation = 130, what is national income? (d) In this example net exports are positive. Could they be negative?3 (a) GNP = C +I + G + NX.: C + G + NX = 1700+ 50+ 40 = 1790, so I must be 210.(b) The net value of international trade NX is +40, so if the value of exports is 350the value of imports must be 310.(c) 2000 – 130 = 1870.(d) Yes, if the value of imports exceeds the value of exports. The economy wouldhave a balance of payments deficit.4 Given the data below: (a) What is 2010 GNP in 2009 prices? (b) What is the growth rate of real GNP from 2009 to 2010? (c) What is the inflation rate?4 (a) 2400 x (100/110) = 2182. (b) 182 / 2000 x 100 = 9.1%. (c) In this example, also 10per cent.5 Should these be in a comprehensive measure of GNP: (a) time spent by students in lectures; (b) the income of muggers; (c) the wage paid to traffic wardens;(d) dropping litter?5 This is a difficult question to answer, because the word “should” implies whatought to happen, a concept economists normally avoid, rather than what actually does.a) This is not included unless the students are in paid employment (eg apprenticeson day release.) The grants of full time students, where they exist, are transfer payments, not earned income. However, a case could be made for including the time even of full-time grantless students, since investment in human capital occurs.(b) No –this is simply a “transfer“ payment, though if the mugged person claimedon the insurance that would be counted.(c) Ye, they are earnings. (d) Environmentalists argue that pollution should ideally be subtracted from GNP.Paradoxically, however, the wages paid to those who pick it up are counted, so in this way pollution actually adds to GNP (as, for example, does the cost of preventing crime)6 Common fallacies. Why are these statements wrong? (a) Unemployment benefit props up national income in years when employment is low. (b) A high per capita real GNP is always a good thing. (c) In 2010 Crummy Movie earned £1 billion more at the box office than Gone With the Wind earned 50 years ago. Crummy Movie is already a bigger box office success.6 (a) Unemployment benefit itself is a transfer payment, since it does not add to realoutput. However, its expenditure adds to demand and may create the employment of others, so the statement is not completely wrong. (b) It depends on how the income is distributed. It might conceivably be undesirable if achieved by very unequal income distribution. (c) The £1 billion has to be adjusted for inflation.. In real terms, Gone With The Wind wins by a mile!7. Which is the correct answer? (a) Increasing the staff of the police force in response to higher crime raises national income because government spending is higher (b) Increasing the staff of the police force reduces national income becausesociety is having to waste resources tackling crime (c) There is no effect on national income because the benefit of more police is offset by the cost of more crime7 (a) is correct because government expenditure in wages (eg of teachers, soldiers,cleaners, and, as in this question, the police) adds to the national income. The arguments put forwards in b) and c) are really couched in terms of national well-being or welfare (which may be more important than national income.).Medium8 Suppose a country is unable to borrow from abroad and must always equate the value of its exports and imports. If the private sector is saving a lot more than it is investing, is the government in surplus or deficit? Why?8 Since total leakages equal total injections, and since exports equal imports, thismeans that, if the economy is in equilibrium, remaining injections (investment + government spending) must be equal to remaining leakages (saving + taxes).Hence, if saving exceeds investment, government spending must exceed taxes.The private sector surplus is offset by the government deficit.9 It is the year 2060. Nation states have been abolished and there is a world government whose spending is entirely financed by income tax. Redraw Figure 15.6 explaining how you have changed it.Figure 15.6 Summarizing the National Accounts[GRAPHIC ARTIST – PLEASE REDRAW THIRD COLUMN SO THAT THE DEEP ORANGE PART FOR NET PROPERTY INCOME FROM ABROAD IS THE SAMEIN COLUMNS TWO AND THREE, but different from depreciation in column 4]9 Reduce all column heights by eliminating net property income from abroad (atleast until we trade with other planets), and eliminate indirect taxes so that last two columns are same height as NNP at market prices10 The world government now publishes world economic accounts that estimate not merely depreciation of the stock of physical capital but also of the stock of environmental capital. (a) If pollution and climate change are causing adverse effects, how does your new diagram on world income and output differ from that in question 9? (b) Suppose environmental depreciation was initially $500bn and the world government spends $100bn on pollution control with the consequence that environment depreciation is now only $300bn. How is your diagram now affected?10 (a) Depreciation (environmental + capital) is now bigger, reducing NNP incolumn 4. The heights of columns 5 and 6 fall by similar amount. (b) The spending on G raises GNP in columns 1-3. by $100bn, and reduces depreciation in column 4. NNP rises for both reasons. Columns 5 and 6 increase by a similar amount.11 You are head of the Leisure Commission that has to recommend to the government how to include the value of leisure in GDP. How do you come up with an estimate?11 The place to start is that by taking leisure people reveal that this is at least asvaluable to them as working more, provided that have that choice in the first place. For all such people, you would have to decide how many hours a day are the fixed costs of being alive – for sleeping, eating, washing etc – and hence how many hours could potentially be allocated between work and leisure. Knowing something about national wage rates and hours of work, you could have a stab at valuing total leisure hours.In theory at least decisions between work and leisure are made, like so many economic decisions, at the margin. Thus in many industries, “overtime” is paid ata higher rate than normal in order to make work more attractive than leisure. Aperson working a 35 hour week would probably value an extra hour’s leisure less than someone working a 45 hour week.Some other complications: (a) all the people who don’t have the choice of working or not –the young, the sick, those caught in a poverty trap (b) the distinction between those working but choosing some leisure and those on benefit schemes you simultaneously get money and leisure (c) how to treat parents who stay at home to look after children. Even so, it would be possible to make a guess.12The price of a new television has remained roughly constant for the last 30 years. What does this show?12 Disentangle some things (a) the nominal price has been constant therefore thereal price has fallen a lot (b) quality has improved hugely over the period. To explain both of these simultaneously, there must have been a lot of technical progress. Deciding how to revalue goods because of their superior quality is just one of the headaches of a statistician working on the national accounts.13 Suppose the injections to the circular flow (investment I, government spending G, and exports X) do not depend on the current level of national output Y. In contrast, suppose leakages increase as output increases. Specifically, saving S = 0.1Y, imports Z = 0.4Y, and taxes T = 0.5Y. (a) If total injections equal 100, what is the value of national output? (b) If G = 40, is the government budget (G-T) in surplus or deficit?13 a) the multiplier is 1 (1 divided by the sum of the propensities to leak at themargin, 0.1+0.4+0.5). hence Y= 100b) T = 50 (half of the addition to Y) and the government deficit is therefore 10 if G =40Hard14 Essay question ‘Economists are preoccupied with what they can measure. GDP is so misleading an indicator of welfare that it is almost pointless to gather statistics about it, either for international comparison across countries or to assess how well particular governments are doing.’ How useful is GDP? Could we easily have a better indicator?The statement in inverted commas is wrong – it is far from pointless to try to measure GDP, provided the result is used with an awareness to its limitations. There are many suggestions for constructing a better measure, but none is easy.In planning your essay try to break your answer down into several sections. One suggestion would be:Acknowledged limitations of GDPSuggestions for improving GDPAttempts at constructing a better measure.Under these headings you should explain and discuss issues such as : Acknowledged limitations of GDP∙Measure income per head, not just total income∙Does not give information about income distribution/equality∙Ignores untraded goods and services (eg housework, voluntary work, own produce, subsistence farming in developing countries, exchanges of services.) ∙Unreported output (the informal or “back pocket” economy, formerly known as the “black” economy, but this term is no longer acceptable ))∙Use of a common international currency for international comparisons (purchasing power parity rate rather than market rate)Suggestions for improving GDP∙Subtracting for resource use∙Subtracting for “bads” such as pollution and congestionAttempts at constructing a better measure.The United Nations already uses a Human Development Index, HDI, which, in addition to GDP includes life expectancy at birth, infant mortality rate, percentage of children in school, and level of adult literacy. The HDI is particularly useful for comparing developing countries.Suggestions for an index of Human Welfare or Well-being, in addition to GDP, include such things as:∙Leisure and working hours∙Measure of equality (the Gini coefficient)∙Stability of marriages, partnerships∙Age of retirement∙Incidence of mental illness, suicide.∙Percentage of young people in higher education.∙Access to theatres, museums, libraries, etc.“Happiness” itself is difficult to measure, but some of the above are easily quantifiable.。

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