solution Chapter 4
财务决策Financial for Decision-makingCHAPTER 4
CHAPTER 4Problem 4.4: SolutionFirstly, we can determine the cost of supplies used as follows:£Opening balance 3,400Supplies purchased 36,00039,400Less: returns 1,020Supplies made available 38,380Supplies used ?Closing balance 4,200As we have determined that the cost of supplies made available is £38,380 and we know that we have ended the year with a stock of £4,200, we can conclude that £34,180 (£38,380 - £4,200) of supplies must have been used during the year.The year-end adjusting and closing entries necessary in this situation can be managed in three stages. Firstly, the balances on the “cleaning supplies purchases” and the “cleaning supplies returns” accounts can be trans ferred to the cleaning supplies inventory account. This results in the “cleaning supplies purchases” and the “cleaning supplies returns” accounts commencing the new accounting year with zero balances.Cleaning Cleaning CleaningAs an alternative to this approach, the “cleaning supplies purchases” and the “cleaning supplies returns” accounts could be closed directly to the profit and loss account.Secondly, the cleaning supplies expense account can be debited with the £34,180 cost of cleaning supplies used that was calculated above. The corresponding credit entry is to the cleaning supplies inventory account. These entries result in the recognition of an expense (the supplies expense account will be closed to the profit & loss statement). They also result in a £4,200 debit balance in the supplies inventory account, which reflects the result of the year-end stock-count. This inventory account balance will comprise part of the total assets recorded in the year-end balance sheet.Thirdly, as a closing entry, the cleaning supplies expense account must be closed to the profit and loss account.Problem 4.5: Solutiona)(1) Sales revenue Unearned revenue2,500 420,000 2,500(2) Salaries & wage expense Accrued salaries & wages 145,000 4,0004,000(3) Depreciation expense Accumulated depreciation 82,000 10,00010,000(4) Rent expense Prepaid rent3,000 900 600600(5) Interest revenue Interest receivable11,000 1,0001,000(6) Insurance expense Prepaid insurance18,000 12,000 6,0006,000b)TranquilStay HotelRevised Profit and Loss Statementfor the year-ended 30 June 20X1€€Sales Revenue 417,500 Less: Cost of Sales 80,000 Gross Profit 337,500 Add: Interest revenue 12,000349,500 Less: ExpensesSalaries and wages expense 149,000Depreciation expense 92,000Car park rental expense 3,600Insurance expense 24,000Sundry expense 14,000282,600€ 66,900Problem 4.6: Solutiona)Unearned service revenue Service revenue1,000 1,000b) Insurance expense Prepaid insurance600 600 c)Interest receivable Interest revenue400 400 d)Supplies expense Supplies7,200 7,200 Problem 4.8: SolutionContribution would be owner putting more money into the business. Dividends are funds paid to the owners.The question wants to test your understanding of the Balance sheet equation,Assets = Liabilities + Owners EquityAssets increased by $140,000 = Liabilities increased 50,000 + (Owner Equity increased by 100,000 - Dividends Paid 10,000) keeps the equation in balance.。
Chapter 04多组分系统热力学.ppt
§4.1 引言
多组分系统 两种或两种以上的物质（或称为组分）所
形成的系统称为多组分系统。
多组分系统可以是均相的，也可以是多相的。
混合物 多组分均匀系统中，各组分均可选用相同的方法处
理，有相同的标准态，遵守相同的经验定律，这种系统称为 混合物。
溶液(solution), 溶剂(solvent) 和溶质(solute)
如果组成溶液的物质有不同的状态，通常将液态物质称为 溶剂，气态或固态物质称为溶质。
如果都具有相同状态，则把含量多的一种称为溶剂，含量少的称为溶质。 溶质有电解质和非电解质之分，本章主要讨论非电介质所形成的溶液。
§4.2 多组分系统的组成表示法
在均相的混合物中，任一组分B的浓度表示法主要有如下几种：
化学势在判断相变和化学变化的方向和限度方面有重要作用。
保持热力学函数的特征变量和除B以外其他组分不变，
某热力学函数随物质的量 nB 的变化率称为化学势。
多组分系统的热力学基本公式应表示为：
dU TdS pdV BdnB dH TdS Vdp BdnB
B=1
B=1
dA SdT pdV BdnB dG SdT Vdp BdnB
律研究的对象是溶质，比例常数无确切的物理意义。
Herry定律使用注意事项
(1)式中 pB 为该气体的分压。对于混合气体，在总
压不大时，Henry定律分别适用于每一种气体。
(2)溶质在气相和在溶液中的分子状态必须相同。如 HCl，在气相为 HCl 分子，在液相为H+和Cl- ，则 Henry定律不适用。
“在一定温度和平衡状态下，气体在液体里的溶解度（用摩尔
分数 x 表示）与该气体的平衡分压 p 成正比。”
胶体化学专业英语
C. Properties of the solutions like colloids.
such as optical（光学的）, rheological（流变 的） properties.
(2) Lyophobic colloids： There is a interface between dispersed phase and continuous phase, the colloids are unstable thermodynamically.
From thermodynamics that spontaneous process occur in the direction of decreasing Gibbs free energy. Therefore， the separation of a two-phase dispersion system to form distinct（明显 不同的） layers（层） is a change in the direction of decreasing Gibbs free energy.
(2) Kinetic stability(动力稳定性)
Kinetic stability： The separation rate of two phases is slow enough that the thermodynamic instability is of very little. Two-phase dispersions will always spontaneously change into a smaller number of larger particles given sufficient time. (unstable)
Solutions_Chapter_4
P ( X = k ) = P (Y + Z = k ) = ∑ P (Y = m, Z = k − m )
m=0 k k
= ∑ P (Y = m )P ( Z = k − m )
m=0 k m m = ∑ Cn p (1 − p) m=0 k n−m k −m k −m × Cn (1 − p) −m p n − m −( k − m )
therefore,
P ({M = 2, Z = 0}) =
3 1 = 36 12 P ({S = 5, Z = 1}) = 0 P ({S = 8, Z = 1}) = 2 1 = 36 18
(c) Since P ( M = 2 ) =
3 1 25 = , P ( Z = 0) = , and that 36 12 36 1 1 25 ≠ × = P ( M = 2) P ( Z = 0) 12 12 36
F (2) = P ( X ≤ 2) = P ( X = 2)
For the event X = 2 , the sample space is
F (1) = P ( X ≤ 1) = P ( X = 1) = P ( Φ ) = 0
⎧ ⎪ ⎪ ⎪ ⎪ Ω=⎨ ⎪ ⎪ ⎪ ⎪ ⎩
(1, 1), ( 2, 1), (3, 1), ( 4, 1), ( 5, 1), ( 6, 1),
1 2
FX (π − 3) = P ( X ≤ π − 3) = P ( X = −1) + P ( X = 0 ) = FY (1) = P (Y ≤ 1) = P (Y = 0 ) + P (Y = 1) = 3⎞ 1 ⎛ 3⎞ ⎛ FY ⎜ ⎟ = P ⎜ Y ≤ ⎟ = P (Y = 0 ) = ⎝ 4⎠ ⎝ ⎠ 4 8 FY (π − 3) = P (Y ≤ π − 3) = P (Y = 0 ) = 1 8 1 2
《金融学》答案第四章 货币的时间价值与现金流贴现分析
CHAPTER 4THE TIME VALUE OF MONEY AND DISCOUNTED CASH FLOW ANALYSISObjectives∙To explain the concepts of compounding and discounting, future value and present value.∙To show how these concepts are applied to making financial decisions.Outline4.1Compounding4.2The Frequency of Compounding4.3Present Value and Discounting4.4Alternative Discounted Cash Flow Decision Rules4.5Multiple Cash Flows4.6Annuities4.7Perpetual Annuities4.8Loan Amortization4.9Exchange Rates and Time Value of Money4.10Inflation and Discounted Cash Flow Analysis4.11Taxes and Investment DecisionsSummary∙Compounding is the process of going from present value (PV) to future value (FV). The future value of $1 earning interest at rate i per period for n periods is (1+i)n.∙Discounting is finding the present value of some future amount. The present value of $1 discounted at rate i per period for n periods is 1/(1+i)n.∙One can make financial decisions by comparing the present values of streams of expected future cash flows resulting from alternative courses of action. The present value of cash inflows less the present value of cash outflows is called net present value (NPV). If a course of action has a positive NPV, it is worth undertaking.∙In any time value of money calculation, the cash flows and the interest rate must be denominated in the same currency.∙Never use a nominal interest rate when discounting real cash flows or a real interest rate when discounting nominal cash flows.How to Do TVM Calculations in MS ExcelAssume you have the following cash flows set up in a spreadsheet:A B1t CF20-1003150426053706NPV7IRRMove the cursor to cell B6 in the spreadsheet. Click the function wizard f x in the tool bar and when a menu appears, select financial and then NPV. Then follow the instructions for inputting the discount rate and cash flows. You can input the column of cash flows by selecting and moving it with your mouse. Ultimately cell B6should contain the following:=NPV(0.1,B3:B5)+B2The first variable in parenthesis is the discount rate. Make sure to input the discount rate as a decimal fraction (i.e., 10% is .1). Note that the NPV function in Excel treats the cash flows as occurring at the end of each period, and therefore the initial cash flow of 100 in cell B2 is added after the closing parenthesis. When you hit the ENTER key, the result should be $47.63.Now move the cursor to cell B7to compute IRR. This time select IRR from the list of financial functions appearing in the menu. Ultimately cell B7 should contain the following:=IRR(B2:B5)When you hit the ENTER key, the result should be 34%.Your spreadsheet should look like this when you have finished:A B1t CF20-1003150426053706NPV47.637IRR34%Solutions to Problems at End of Chapter1.If you invest $1000 today at an interest rate of 10% per year, how much will you have 20 years from now,assuming no withdrawals in the interim?2. a. If you invest $100 every year for the next 20 years, starting one year from today and you earninterest of 10% per year, how much will you have at the end of the 20 years?b.How much must you invest each year if you want to have $50,000 at the end of the 20 years?3.What is the present value of the following cash flows at an interest rate of 10% per year?a.$100 received five years from now.b.$100 received 60 years from now.c.$100 received each year beginning one year from now and ending 10 years from now.d.$100 received each year for 10 years beginning now.e.$100 each year beginning one year from now and continuing forever.e.PV = $100 = $1,000.104.You want to establish a “wasting” fund which will provide you with $1000 per year for four years, at which time the fund will be exhausted. How much must you put in the fund now if you can earn 10% interest per year?SOLUTION:5.You take a one-year installment loan of $1000 at an interest rate of 12% per year (1% per month) to be repaid in 12 equal monthly payments.a.What is the monthly payment?b.What is the total amount of interest paid over the 12-month term of the loan?SOLUTION:b. 12 x $88.85 - $1,000 = $66.206.You are taking out a $100,000 mortgage loan to be repaid over 25 years in 300 monthly payments.a.If the interest rate is 16% per year what is the amount of the monthly payment?b.If you can only afford to pay $1000 per month, how large a loan could you take?c.If you can afford to pay $1500 per month and need to borrow $100,000, how many months would it taketo pay off the mortgage?d.If you can pay $1500 per month, need to borrow $100,000, and want a 25 year mortgage, what is thehighest interest rate you can pay?SOLUTION:a.Note: Do not round off the interest rate when computing the monthly rate or you will not get the same answerreported here. Divide 16 by 12 and then press the i key.b.Note: You must input PMT and PV with opposite signs.c.Note: You must input PMT and PV with opposite signs.7.In 1626 Peter Minuit purchased Manhattan Island from the Native Americans for about $24 worth of trinkets. If the tribe had taken cash instead and invested it to earn 6% per year compounded annually, how much would the Indians have had in 1986, 360 years later?SOLUTION:8.You win a $1 million lottery which pays you $50,000 per year for 20 years, beginning one year from now. How much is your prize really worth assuming an interest rate of 8% per year?SOLUTION:9.Your great-aunt left you $20,000 when she died. You can invest the money to earn 12% per year. If you spend $3,540 per year out of this inheritance, how long will the money last?SOLUTION:10.You borrow $100,000 from a bank for 30 years at an APR of 10.5%. What is the monthly payment? If you must pay two points up front, meaning that you only get $98,000 from the bank, what is the true APR on the mortgage loan?SOLUTION:If you must pay 2 points up front, the bank is in effect lending you only $98,000. Keying in 98000 as PV and computing i, we get:11.Suppose that the mortgage loan described in question 10 is a one-year adjustable rate mortgage (ARM), which means that the 10.5% interest applies for only the first year. If the interest rate goes up to 12% in the second year of the loan, what will your new monthly payment be?SOLUTION:Step 2 is to compute the new monthly payment at an interest rate of 1% per month:12.You just received a gift of $500 from your grandmother and you are thinking about saving this money for graduation which is four years away. You have your choice between Bank A which is paying 7% for one-year deposits and Bank B which is paying 6% on one-year deposits. Each bank compounds interest annually. What is the future value of your savings one year from today if you save your money in Bank A? Bank B? Which is the better decision? What savings decision will most individuals make? What likely reaction will Bank B have? SOLUTION:$500 x (1.07) = $535Formula:$500 x (1.06) = $530a.You will decide to save your money in Bank A because you will have more money at the end of the year. Youmade an extra $5 because of your savings decision. That is an increase in value of 1%. Because interestcompounded only once per year and your money was left in the account for only one year, the increase in value is strictly due to the 1% difference in interest rates.b.Most individuals will make the same decision and eventually Bank B will have to raise its rates. However, it isalso possible that Bank A is paying a high rate just to attract depositors even though this rate is not profitable for the bank. Eventually Bank A will have to lower its rate to Bank B’s rate in order to make money.13.Sue Consultant has just been given a bonus of $2,500 by her employer. She is thinking about using the money to start saving for the future. She can invest to earn an annual rate of interest of 10%.a.According to the Rule of 72, approximately how long will it take for Sue to increase her wealth to $5,000?b.Exactly how long does it actually take?SOLUTION:a.According to the Rule of 72: n = 72/10 = 7.2 yearsIt will take approximately 7.2 years for Sue’s $2,500 to double to $5,000 at 10% interest.b.At 10% interestFormula:$2,500 x (1.10)n = $5,000Hence, (1.10)n = 2.0n log 1.10 = log 2.0n = .693147 = 7.27 Years.095310rry’s bank account has a “floating” interest rate on certain deposits. Every year the interest rate is adjusted. Larry deposited $20,000 three years ago, when interest rates were 7% (annual compounding). Last year the rate was only 6%, and this year the rate fell again to 5%. How much will be in his account at the end of this year?SOLUTION:$20,000 x 1.07 x 1.06 x 1.05 = $23,818.2015.You have your choice between investing in a bank savings account which pays 8% compounded annually (BankAnnual) and one which pays 7.5% compounded daily (BankDaily).a.Based on effective annual rates, which bank would you prefer?b.Suppose BankAnnual is only offering one-year Certificates of Deposit and if you withdraw your moneyearly you lose all interest. How would you evaluate this additional piece of information when making your decision?SOLUTION:a.Effective Annual Rate: BankAnnual = 8%.Effective Annual Rate BankDaily = [1 + .075]365 - 1 = .07788 = 7.788%365Based on effective annual rates, you would prefer BankAnnual (you will earn more money.)b.If BankAnnual’s 8% annual return is conditioned upon leaving the money in for one full year, I would need tobe sure that I did not need my money within the one year period. If I were unsure of when I might need the money, it might be safer to go for BankDaily. The option to withdraw my money whenever I might need it will cost me the potential difference in interest:FV (BankAnnual) = $1,000 x 1.08 = $1,080FV (BankDaily) = $1,000 x 1.07788 = $1,077.88Difference = $2.12.16.What are the effective annual rates of the following:a.12% APR compounded monthly?b.10% APR compounded annually?c.6% APR compounded daily?SOLUTION:Effective Annual Rate (EFF) = [1 + APR] m - 1ma.(1 + .12)12 - 1 = .1268 = 12.68%12b.(1 + .10)- 1 = .10 = 10%1c.(1 + .06)365 - 1 = .0618 = 6.18%36517.Harry promises that an investment in his firm will double in six years. Interest is assumed to be paid quarterly and reinvested. What effective annual yield does this represent?EAR=(1.029302)4-1=12.25%18.Suppose you know that you will need $2,500 two years from now in order to make a down payment on a car.a.BankOne is offering 4% interest (compounded annually) for two-year accounts, and BankTwo is offering4.5% (compounded annually) for two-year accounts. If you know you need $2,500 two years from today,how much will you need to invest in BankOne to reach your goal? Alternatively, how much will you need to invest in BankTwo? Which Bank account do you prefer?b.Now suppose you do not need the money for three years, how much will you need to deposit today inBankOne? BankTwo?SOLUTION:PV = $2,500= $2,311.39(1.04)2PV = $2,500= $2,289.32(1.045)2You would prefer BankTwo because you earn more; therefore, you can deposit fewer dollars today in order to reach your goal of $2,500 two years from today.b.PV = $2,500= $2,222.49(1.04)3PV = $2,500= $2,190.74(1.045)3Again, you would prefer BankTwo because you earn more; therefore, you can deposit fewer dollars today in order to reach your goal of $2,500 three years from today.19.Lucky Lynn has a choice between receiving $1,000 from her great-uncle one year from today or $900 from her great-aunt today. She believes she could invest the $900 at a one-year return of 12%.a.What is the future value of the gift from her great-uncle upon receipt? From her great-aunt?b.Which gift should she choose?c.How does your answer change if you believed she could invest the $900 from her great-aunt at only 10%?At what rate is she indifferent?SOLUTION:a. Future Value of gift from great-uncle is simply equal to what she will receive one year from today ($1000). Sheearns no interest as she doesn’t receive the money until next year.b. Future Value of gift from great-aunt: $900 x (1.12) = $1,008.c. She should choose the gift from her great-aunt because it has future value of $1008 one year from today. Thegift from her great-uncle has a future value of $1,000. This assumes that she will able to earn 12% interest on the $900 deposited at the bank today.d. If she could invest the money at only 10%, the future value of her investment from her great-aunt would only be$990: $900 x (1.10) = $990. Therefore she would choose the $1,000 one year from today. Lucky Lynn would be indifferent at an annual interest rate of 11.11%:$1000 = $900 or (1+i) = 1,000 = 1.1111(1+i)900i = .1111 = 11.11%20.As manager of short-term projects, you are trying to decide whether or not to invest in a short-term project that pays one cash flow of $1,000 one year from today. The total cost of the project is $950. Your alternative investment is to deposit the money in a one-year bank Certificate of Deposit which will pay 4% compounded annually.a.Assuming the cash flow of $1,000 is guaranteed (there is no risk you will not receive it) what would be alogical discount rate to use to determine the present value of the cash flows of the project?b.What is the present value of the project if you discount the cash flow at 4% per year? What is the netpresent value of that investment? Should you invest in the project?c.What would you do if the bank increases its quoted rate on one-year CDs to 5.5%?d.At what bank one-year CD rate would you be indifferent between the two investments?SOLUTION:a.Because alternative investments are earning 4%, a logical choice would be to discount the project’s cash flowsat 4%. This is because 4% can be considered as your opportunity cost for taking the project; hence, it is your cost of funds.b.Present Value of Project Cash Flows:PV = $1,000= $961.54(1.04)The net present value of the project = $961.54 - $950 (cost) = $11.54The net present value is positive so you should go ahead and invest in the project.c.If the bank increased its one-year CD rate to 5.5%, then the present value changes to:PV = $1,000= $947.87(1.055)Now the net present value is negative: $947.87 - $950 = - $2.13. Therefore you would not want to invest in the project.d.You would be indifferent between the two investments when the bank is paying the following one-year interestrate:$1,000 = $950 hence i = 5.26%(1+i)21.Calculate the net present value of the following cash flows: you invest $2,000 today and receive $200 one year from now, $800 two years from now, and $1,000 a year for 10 years starting four years from now. Assume that the interest rate is 8%.SOLUTION:Since there are a number of different cash flows, it is easiest to do this problem using cash flow keys on the calculator:22.Your cousin has asked for your advice on whether or not to buy a bond for $995 which will make one payment of $1,200 five years from today or invest in a local bank account.a.What is the internal rate of return on the bond’s cash flows? What additional information do you need tomake a choice?b.What advice would you give her if you learned the bank is paying 3.5% per year for five years(compounded annually?)c.How would your advice change if the bank were paying 5% annually for five years? If the price of thebond were $900 and the bank pays 5% annually?SOLUTION:a.$995 x (1+i)5 = $1,200.(1+i)5 = $1,200$995Take 5th root of both sides:(1+i) =1.0382i = .0382 = 3.82%In order to make a choice, you need to know what interest rate is being offered by the local bank.b.Upon learning that the bank is paying 3.5%, you would tell her to choose the bond because it is earning a higherrate of return of 3.82% .c.If the bank were paying 5% per year, you would tell her to deposit her money in the bank. She would earn ahigher rate of return.5.92% is higher than the rate the bank is paying (5%); hence, she should choose to buy the bond.23.You and your sister have just inherited $300 and a US savings bond from your great-grandfather who had left them in a safe deposit box. Because you are the oldest, you get to choose whether you want the cash or the bond. The bond has only four years left to maturity at which time it will pay the holder $500.a.If you took the $300 today and invested it at an interest rate 6% per year, how long (in years) would ittake for your $300 to grow to $500? (Hint: you want to solve for n or number of periods. Given these circumstances, which are you going to choose?b.Would your answer change if you could invest the $300 at 10% per year? At 15% per year? What otherDecision Rules could you use to analyze this decision?SOLUTION:a.$300 x (1.06)n = $500(1.06)n = 1.6667n log 1.06 = log 1.6667n = .510845 = 8.77 Years.0582689You would choose the bond because it will increase in value to $500 in 4 years. If you tookthe $300 today, it would take more than 8 years to grow to $500.b.You could also analyze this decision by computing the NPV of the bond investment at the different interest rates:In the calculations of the NPV, $300 can be considered your “cost” for acquiring the bond since you will give up $300 in cash by choosing the bond. Note that the first two interest rates give positive NPVs for the bond, i.e. you should go for the bond, while the last NPV is negative, hence choose the cash instead. These results confirm the previous method’s results.24.Suppose you have three personal loans outstanding to your friend Elizabeth. A payment of $1,000 is due today, a $500 payment is due one year from now and a $250 payment is due two years from now. You would like to consolidate the three loans into one, with 36 equal monthly payments, beginning one month from today. Assume the agreed interest rate is 8% (effective annual rate) per year.a.What is the annual percentage rate you will be paying?b.How large will the new monthly payment be?SOLUTION:a.To find the APR, you must first compute the monthly interest rate that corresponds to an effective annual rate of8% and then multiply it by 12:1.08 = (1+ i)12Take 12th root of both sides:1.006434 = 1+ ii = .006434 or .6434% per monthOr using the financial calculator:b.The method is to first compute the PV of the 3 loans and then compute a 36 month annuity payment with thesame PV. Most financial calculators have keys which allow you to enter several cash flows at once. This approach will give the user the PV of the 3 loans.Note: The APR used to discount the cash flows is the effective rate in this case, because this method is assuming annual compounding.25.As CEO of ToysRFun, you are offered the chance to participate, without initial charge, in a project that produces cash flows of $5,000 at the end of the first period, $4,000 at the end of the next period and a loss of $11,000 at the end of the third and final year.a.What is the net present value if the relevant discount rate (the company’s cost of capital) is 10%?b.Would you accept the offer?c.What is the internal rate of return? Can you explain why you would reject a project which has aninternal rate of return greater than its cost of capital?SOLUTION:At 10% discount rate:Net Present Value = - 0 + $5,000 + $4,000 - $11,000 = - 413.22(1.10)(1.10)2 (1.10)3c.This example is a project with cash flows that begin positive and then turn negative--it is like a loan. The 13.6% IRR is therefore like an interest rate on that loan. The opportunity to take a loan at 13.6% when the cost of capital is only 10% is not worthwhile.26.You must pay a creditor $6,000 one year from now, $5,000 two years from now, $4,000 three years from now, $2,000 four years from now, and a final $1,000 five years from now. You would like to restructure the loan into five equal annual payments due at the end of each year. If the agreed interest rate is 6% compounded annually, what is the payment?SOLUTION:Since there are a number of different cash flows, it is easiest to do the first step of this problem using cash flow keys on the calculator. To find the present value of the current loan payments:27.Find the future value of the following ordinary annuities (payments begin one year from today and all interest rates compound annually):a.$100 per year for 10 years at 9%.b.$500 per year for 8 years at 15%.c.$800 per year for 20 years at 7%.d.$1,000 per year for 5 years at 0%.e.Now find the present values of the annuities in a-d.f.What is the relationship between present values and future values?SOLUTION:Future Value of Annuity:e.f.The relationship between present value and future value is the following:FV = PV x (1+i)n28.Suppose you will need $50,000 ten years from now. You plan to make seven equal annual deposits beginning three years from today in an account that yields 11% compounded annually. How large should the annual deposit be?SOLUTION:You will be making 7 payments beginning 3 years from today. So, we need to find the value of an immediate annuity with 7 payments whose FV is $50,000:29.Suppose an investment offers $100 per year for five years at 5% beginning one year from today.a.What is the present value? How does the present value calculation change if one additional payment isadded today?b.What is the future value of this ordinary annuity? How does the future value change if one additionalpayment is added today?SOLUTION:$100 x [(1.05)5] - 1 = $552.56.05If you were to add one additional payment of $100 today, the future value would increase by:$100 x (1.05)5 = $127.63. Total future value = $552.56 + $127.63 = $680.19.Another way to do it would be to use the BGN mode for 5 payments of $100 at 5%, find the future value of that, and then add $100. The same $680.19 is obtained.30.You are buying a $20,000 car. The dealer offers you two alternatives: (1) pay the full $20,000 purchase price and finance it with a loan at 4.0% APR over 3 years or (2) receive $1,500 cash back and finance the rest at a bank rate of 9.5% APR. Both loans have monthly payments over three years. Which should you choose? SOLUTION:31.You are looking to buy a sports car costing $23,000. One dealer is offering a special reduced financing rate of 2.9% APR on new car purchases for three year loans, with monthly payments. A second dealer is offering a cash rebate. Any customer taking the cash rebate would of course be ineligible for the special loan rate and would have to borrow the balance of the purchase price from the local bank at the 9%annual rate. How large must the cash rebate be on this $23,000 car to entice a customer away from the dealer who is offering the special 2.9% financing?SOLUTION:of the 2.9% financing.32.Show proof that investing $475.48 today at 10% allows you to withdraw $150 at the end of each of the next 4 years and have nothing remaining.SOLUTION:You deposit $475.48 and earn 10% interest after one year. Then you withdraw $150. The table shows what happensAnother way to do it is simply to compute the PV of the $150 annual withdrawals at 10% : it turns out to be exactly $475.48, hence both amounts are equal.33.As a pension manager, you are considering investing in a preferred stock which pays $5,000,000 per year forever beginning one year from now. If your alternative investment choice is yielding 10% per year, what is the present value of this investment? What is the highest price you would be willing to pay for this investment? If you paid this price, what would be the dividend yield on this investment?SOLUTION:Present Value of Investment:PV = $5,000,000 = $50,000,000.10Highest price you would be willing to pay is $50,000,000.Dividend yield = $5,000,000 = 10%.$50,000,00034. A new lottery game offers a choice for the grand prize winner. You can receive either a lump sum of $1,000,000 immediately or a perpetuity of $100,000 per year forever, with the first payment today. (If you die, your estate will still continue to receive payments). If the relevant interest rate is 9.5% compounded annually, what is the difference in value between the two prizes?SOLUTION:The present value of the perpetuity assuming that payments begin at the end of the year is:$100,000/.095 = $1,052,631.58If the payments begin immediately, you need to add the first payment. $100,000 + 1,052,632 = $1,152,632.So the annuity has a PV which is greater than the lump sum by $152,632.35.Find the future value of a $1,000 lump sum investment under the following compounding assumptions:a.7% compounded annually for 10 yearsb.7% compounded semiannually for 10 yearsc.7% compounded monthly for 10 yearsd.7% compounded daily for 10 yearse.7% compounded continuously for 10 yearsa.$1,000 x (1.07)10 = $1,967.15b.$1,000 x (1.035)20 = $1,989.79c.$1,000 x (1.0058)120 = $2,009.66d.$1,000 x (1.0019178)3650 = $2,013.62e.$1,000 x e.07x10 = $2,013.7536.Sammy Jo charged $1,000 worth of merchandise one year ago on her MasterCard which has a stated interest rate of 18% APR compounded monthly. She made 12 regular monthly payments of $50, at the end of each month, and refrained from using the card for the past year. How much does she still owe? SOLUTION:Sammy Jo has taken a $1,000 loan at 1.5% per month and is paying it off in monthly installments of $50. We could work out the amortization schedule to find out how much she still owes after 12 payments, but a shortcut on the financial calculator is to solve for FV as follows:37.Suppose you are considering borrowing $120,000 to finance your dream house. The annual percentage rate is 9% and payments are made monthly,a.If the mortgage has a 30 year amortization schedule, what are the monthly payments?b.What effective annual rate would you be paying?c.How do your answers to parts a and b change if the loan amortizes over 15 years rather than 30?EFF = [1 + .09]1238.Suppose last year you took out the loan described in problem #37a. Now interest rates have declined to 8% per year. Assume there will be no refinancing fees.a.What is the remaining balance of your current mortgage after 12 payments?b.What would be your payment if you refinanced your mortgage at the lower rate for 29 years? SOLUTION:Exchange Rates and the Time Value of Money39.The exchange rate between the pound sterling and the dollar is currently $1.50 per pound, the dollar interest rate is 7% per year, and the pound interest rate is 9% per year. You have $100,000 in a one-year account that allows you to choose between either currency, and it pays the corresponding interest rate.a.If you expect the dollar/pound exchange rate to be $1.40 per pound a year from now and are indifferentto risk, which currency should you choose?b.What is the “break-even” value of the dollar/pound exchange rate one year from now?SOLUTION:a.You could invest $1 today in dollar-denominated bonds and have $1.07 one year from now. Or you couldconvert the dollar today into 2/3 (i.e., 1/1.5) of a pound and invest in pound-denominated bonds to have .726667(i.e., 2/3 x 1.09) pounds one year from now. At an exchange rate of $1.4 per pound, this would yield 0.726667(1.4) = $1.017 (this is lower than $1.07), so you would choose the dollar currency.b.For you to break-even the .726667 pounds would have to be worth $1.07 one year from now, so the break-evenexchange rate is $1.07/.726667 or $1.4725 per pound. So for exchange rates lower than $1.4725 per pound one year from now, the dollar currency will give a better return.。
电路基础英文版CHAPTER 4
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Linearity Property 线性定理 A linear circuit is one whose output is linearly related to its input.
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The superposition principle states that the voltage across (or current through) an element in a linear circuit is the algebraic sum of the voltages across (or currents through) that element due to each independent source acting alone.
Introduction to Management Science 5th Edition, 课后习题答案 Chapter 4
CHAPTER 4 THE ART OF MODELING WITH SPREADSHEETSSOLUTION TO SOLVED PROBLEMS4.S1Production and Inventory Planning ModelSurfs U p p roduces h igh-‐end s urfboards. A c hallenge f aced b y S urfs U p i s t hat t heir d emand i s highly s easonal. D emand e xceeds p roduction c apacity d uring t he w arm s ummer m onths, b ut is v ery l ow i n t he w inter m onths. T o m eet t he h igh d emand d uring t he s ummer, S urfs U ptypically p roduces m ore s urfboards t han a re n eeded i n t he w inter m onths a nd t hen c arries inventory i nto t he s ummer m onths. T heir p roduction f acility c an p roduce a t m ost 50 b oards per m onth u sing r egular l abor a t a c ost o f $125 e ach. U p t o 10 a dditional b oards c an b e produced b y u tilizing o vertime l abor a t a c ost o f $135 e ach. T he b oards a re s old f or $200. Because o f s torage c ost a nd t he o pportunity c ost o f c apital, e ach b oard h eld i n i nventory f rom one m onth t o t he n ext i ncurs a c ost o f $5 p er b oard. S ince d emand i s u ncertain, S urfs U p would l ike t o m aintain a n e nding i nventory (safety s tock) o f a t l east 10 b oards d uring t he warm m onths (May–September) a nd a t l east 5 b oards d uring t he o ther m onths (October–April). I t i s n ow t he s tart o f J anuary a nd S urfs U p h as 5 b oards i n i nventory. T he f orecast o f demand o ver t he n ext 12 m onths i s s hown i n t he t able b elow. F ormulate a nd s olve a l inear programming m odel i n a s preadsheet t o d etermine h ow m any s urfboards s hould b e p roduced each m onth t o m aximize t otal p rofit.Jan Feb Mar Apr May Jun July Aug Sep Oct Nov Dec10 14 15 20 45 65 85 85 40 30 15 15This i s a d ynamic p roblem w ith 12 t ime p eriods (months). T he a ctivities a re t he p roduction quantities i n e ach o f t he 12 m onths u sing r egular l abor a nd t he p roduction q uantities i n each o f t he 12 m onths u sing o vertime l abor.To g et s tarted, w e s ketch a s preadsheet m odel. E ach o f t he 12 m onths w ill b e a s eparate column i n t he s preadsheet. F or e ach m onth, t he r egular p roduction q uantity (a c hanging cell) m ust b e n o m ore t han t he m aximum r egular p roduction (50). S imilarly, f or e ach month t he o vertime p roduction q uantity (a c hanging c ell) m ust b e n o m ore t han t he maximum o vertime p roduction (10). E ach m onth w ill g enerate r evenue, i ncur r egular a nd overtime p roduction c osts, i nventory h olding c osts, a nd a chieve a r esulting p rofit. T he g oal will b e t o m aximize t he t otal p rofit o ver a ll 12 m onths. T his l eads t o t he f ollowing s ketch o f a s preadsheet m odel.The e nding i nventory e ach m onth w ill e qual t he s tarting i nventory (the g iven s tartinginventory f or J anuary, o r t he p revious m onth’s e nding i nventory f or f uture m onths) p lus a ll production (regular a nd o vertime) m inus t he f orecasted s ales. T he e nding i nventory a t t he end o f e ach m onth m ust b e a t l east t he m inimum s afety s tock l evel. T he r evenue w ill e qual the s elling p rice t imes f orecasted s ales. T he r egular (or o vertime) p roduction c ost w ill b e the r egular (or o vertime) p roduction q uantity t imes t he u nit r egular (or o vertime)production c ost. T he h olding c ost w ill e qual t he e nding i nventory t imes t he u nit h olding cost. T he m onthly p rofit w ill b e r evenue m inus b oth p roduction c osts m inus h olding c ost. Finally, t he t otal p rofit w ill b e t he s um o f t he m onthly p rofits. T he f inal s olved s preadsheet, formulas, a nd S olver i nformation a re s hown b elow.Unit Cost (Reg)Unit Cost (OT)Selling Price Holding Cost Starting Inventory<=Max Regular <=Max OTForecasted Sales Ending Inventory>=Safety StockThe v alues i n R egularProduction (C10:N10) a nd O TProduction (C14:N14) s how h ow m anysurf b oards S urfs U p s hould p roduce e ach m onth s o a s t o a chieve t he m aximum p rofit o f $31,150.Set Objective Cell: TotalProfit To: MaxBy Changing Variable Cells:RegularProduction, OTProduction Subject to the Constraints:RegularProduction <= MaxRegular OTProduction <= MaxOTEndingInventory >= SafetyStock Solver Options:Make Variables Nonnegative Solving Method: Simplex LP4.S2Aggregate Planning: Manpower Hiring/Firing/TrainingCool P ower p roduces a ir c onditioning u nits f or l arge c ommercial p roperties. D ue t o t he l owcost a nd e fficiency o f i ts p roducts, t he c ompany h as b een g rowing f rom y ear t o y ear. A lso, d ue to s easonality i n c onstruction a nd w eather c onditions, p roduction r equirements v ary f rommonth t o m onth. C ool P ower c urrently h as 10 f ully t rained e mployees w orking i nmanufacturing. E ach t rained e mployee c an w ork 160 h ours p er m onth a nd i s p aid a m onthly wage o f $4000. N ew t rainees c an b e h ired a t t he b eginning o f a ny m onth. D ue t o t heir l ack o f initial s kills a nd r equired t raining, a n ew t rainee o nly p rovides 100 h ours o f u seful l abor i n their f irst m onth, b ut a re s till p aid a f ull m onthly w age o f $4000. F urthermore, b ecause o f required i nterviewing a nd t raining, t here i s a $2500 h iring c ost f or e ach e mployee h ired. A fter one m onth, a t rainee i s c onsidered f ully t rained. A n e mployee c an b e f ired a t t he b eginning o f any m onth, b ut m ust b e p aid t wo w eeks o f s everance p ay ($2000). O ver t he n ext 12 m onths, Cool P ower f orecasts t he l abor r equirements s hown i n t he t able b elow. S ince m anagement anticipates h igher r equirements n ext y ear, C ool P ower w ould l ike t o e nd t he y ear w ith a t l east 12 f ully t rained e mployees. H ow m any t rainees s hould b e h ired a nd/or w orkers f ired i n e ach month t o m eet t he l abor r equirements a t t he m inimum p ossible c ost? F ormulate a nd s olve a linear p rogramming s preadsheet m odel.Jan Feb Mar Apr May Jun Jul Aug Sep Oct Nov Dec 1600 2000 2000 2000 2800 3200 3600 3200 1600 1200 800 800This i s a d ynamic p roblem w ith 12 t ime p eriods (months). T he a ctivities a re t he n umber o fworkers t o h ire a nd f ire i n e ach o f t he 12 m onths.To g et s tarted, w e s ketch a s preadsheet m odel. E ach o f t he 12 m onths w ill b e a s eparate column i n t he s preadsheet. F or e ach m onth, t here a re c hanging c ells f or b oth t he n umber o f workers h ired a nd f ired. B ased o n t he v alues o f t hese c hanging c ells, w e c an d etermine t he number o f t rainees a nd t rained e mployees. T he n umber o f l abor h ours g enerated b y t he employees m ust b e a t l east t he r equired l abor h ours e ach m onth. F inally, l abor c osts (for trainees a nd t he t rained w orkforce), h iring c ost, a nd s everance p ay l eads t o a t otal m onthly cost. T he g oal w ill b e t o m inimize t he t otal c ost o ver a ll 12 m onths. T his l eads t o t he following s ketch o f a s preadsheet m odel.Labor Monthly WageHiring Cost Severance PayLabor Hours/Trainee/MonthLabor Hours/Trained Worker/MonthStarting Trained WorkforceMinimum to Start the TraineesNext YearTrained Employees >=Labor Hours Available>=Required Labor HoursWhen a n e mployee i s f irst h ired, h e o r s he i s a t rainee f or o ne m onth b efore b ecoming afully-‐trained e mployee. T herefore, t he n umber o f t rainees (row 14) i s e qual t o t he n umber of w orkers h ired i n t hat m onth, w hile t he n umber o f t rained e mployees (row 15) i s t henumber o f t rained e mployees a nd t rainees f rom t he p revious m onth m inus a ny e mployee that i s f ired. T he l abor h ours a vailable i n e ach m onth e quals t he s umproduct o f t he l abor hours p rovided b y e ach t ype o f w orker (trained o r t rainees) w ith t he n umber o f e ach t ype of e mployee. T he l abor c osts i n e ach m onth a re t he m onthly w age m ultiplied b y t he number o f e mployees. T he h iring c ost i s t he u nit h iring c ost m ultiplied b y t he n umber o f workers h ired. T he s everance p ay i s t he u nit s everance c ost m ultiplied b y t he n umber o f workers f ired. T hen, t he t otal m onthly c ost i s t he s um o f t he l abor c osts, h iring c ost, a nd severance p ay. F inally, t he t otal c ost w ill b e t he s um o f t he m onthly c osts. F or a rbitrary values o f w orkers h ired a nd f ired e ach m onth, t his l eads t o t he f ollowing s preadsheet.The S olver i nformation i s s hown b elow, f ollowed b y t he s olved s preadsheet.Thus, W orkersHired (C11:N11) s hows t he n umber o f w orkers C ool P ower s hould h ire e achmonth a nd W orkersFired (C12:N12) s hows t he n umber o f w orkers C ool P ower s hould f ire each m onth s o a s t o a chieve t he m inimum T otalCost (O26) o f $787,500.Solver ParametersSet Objective Cell: TotalCost To: MinBy Changing Variable Cells: WorkersHired, WorkersFired Subject to the Constraints:N15 >= MinimumToStartNewYearLaborHoursAvailable >= RequiredLaborHours WorkersHired = integer WorkersFired = integer Solver Options:Make Variables Nonnegative Solving Method: Simplex LP。
Solutions
It is impossible to see the component particles of solutions or colloid or to separate them by passing the sopaper.
The terms unsaturated, saturated and supersaturated may be applied to solutions in which the solute has a finite solubility in the solvent. Unsaturated solutions contain less solute per 100 ml of solvent than the solubility. Saturated solutions contain the amount of solute equal to the solubility. Supersaturated solutions actually more solute per 100 ml of solvent than the solubility would seem to allow. One way to make a supersaturated solution is to evaporate solvent from a solution very slowly and carefully without stirring.
There are two general types of homogeneous mixtures, which are distinguished from each other by the size of their component particles. Solutions are homogeneous mixtures in which the particle sizes of the components (molecules or ions) range from about 1.0 to 10 nm. Colloids contain component particles from 10 to 100nm in size.
solution
solutionSolutionIntroductionIn today's fast-paced world, businesses face various challenges that require effective and innovative solutions. Whether it is improving efficiency, streamlining processes, reducing costs, or enhancing customer satisfaction, having a well-rounded solution is crucial. In this document, we will explore the concept of a solution, its importance, and how businesses can develop and implement effective solutions to address their issues.Definition of a SolutionA solution can be defined as a specific strategy or approach designed to address a problem or challenge effectively. It involves analyzing the root cause of the problem, identifying possible alternatives, and implementing the best course of action to solve the issue. A solution should be comprehensive, practical, and sustainable, delivering the desired results and providing long-term benefits.Importance of a SolutionHaving a well-thought-out solution is vital for businesses for several reasons. Firstly, it allows organizations to overcome obstacles and challenges that may hinder their growth and success. By identifying and resolving these issues, businesses can operate more efficiently and effectively, resulting in increased productivity and profitability.Secondly, a robust solution helps businesses stay ahead of the competition. It allows companies to differentiate themselves in the market by offering unique solutions and addressing customer pain points. Moreover, a well-implemented solution can enhance customer satisfaction, leading to improved brand loyalty and customer retention.Developing an Effective SolutionDeveloping an effective solution requires a systematic approach that involves thorough research, analysis, and collaboration. Here are some steps businesses can follow to develop a solution:1. Define the Problem: The first step is to clearly define the problem or challenge at hand. This involves identifying the symptoms, understanding the impact, and determining the root cause of the issue. A well-defined problem statementprovides a solid foundation for developing an effective solution.2. Research and Analysis: Once the problem is defined, thorough research and analysis are essential. This includes gathering relevant data, conducting market research, and analyzing internal processes and systems. The goal is to gain a comprehensive understanding of the problem and its underlying causes.3. Generate Alternatives: After conducting research, it is important to brainstorm and generate multiple alternative solutions. This can be done through group discussions, workshops, or individual ideation sessions. The more alternatives generated, the higher the chances of finding the most effective solution.4. Evaluate Alternatives: Once alternative solutions are generated, it is critical to evaluate each option against predetermined criteria. This evaluation can take into account factors such as feasibility, cost-effectiveness, impact, and sustainability. Narrowing down the list of alternatives provides a focused approach to selecting the best solution.5. Develop an Implementation Plan: Once the best solution is identified, a detailed implementation plan should be developed. This plan outlines the necessary steps, resources, timeline, and responsibilities required to execute the solution effectively. A well-structured implementation plan ensures a smooth transition from problem identification to solution implementation.6. Implement and Monitor: The next step is to implement the chosen solution and closely monitor its progress. Regular evaluation and monitoring help identify any bottlenecks or challenges encountered during implementation. Adjustments can be made accordingly to ensure the solution's effectiveness.7. Continuous Improvement: Solutions should not be considered final. Continuous improvement is essential to ensure effectiveness and adaptability in an ever-changing business environment. Regular evaluations, feedback loops, and revision of the solution help organizations stay agile and responsive.ConclusionIn conclusion, developing a solid solution is crucial for businesses in today's competitive landscape. It allowsorganizations to address challenges effectively, differentiate themselves from competitors, and improve customer satisfaction. By following a systematic approach to develop and implement solutions, businesses can enjoy increased efficiency, profitability, and long-term success. Remember, a well-implemented solution is not the end, but rather the beginning of continuous improvement and growth.。
通信网答案
SolutionChapter 114.The propagation delay is the time that is required for the energy of a signal to propagate from one point to another.a.Find the propagation delay for a signal traversing the following networks at the speed of light in cable (2.3 x 108meters/second): y a circuit board 10 cmi i b d10Solution:y To find the propagation delay, divide distance by thespeed of light in cable. Thus we have:y a circuit board t prop= 4.347 x 10-10seconds8y a room t= 4.3478 x 10-8secondsb. How many bits are in transit during the propagation delay in the above cases, if bits are entering the above networks at the following transmission speeds: 10,000 bits/second; 1 megabit/second; 100 megabits/second; 10 gigabits/second. Solution:y The number of bits in transit is obtained by multiplying thef15.In problem 14, how long does it take to send an L-bytefile and to receive a 1-byte acknowledgment back? Let L=109 bytes.Solution:S l i1.Message and ACK transmission time, which depends onthe message length and the transmission bit rate;2.Propagation delay, which depends solely on distance.y When the propagation delay is small, message and ACK transmission times determine the total delay. On the other transmission times determine the total delay On the otherMessage + ACK delay @10 kbps message +ACK delay@1 Mbpsmessage+ACK delay@100 Mbpsmessage+ACK delay@10 Gbps8.00E+118.00E+098.00E+078.00E+05 Table 1: Message length = 109bytesMessage + ACK delay @10 kbps message +ACK delay@1 Mbpsmessage+ACK delay@100 Mbpsmessage+ACK delay@10 Gbps8.00E+118.00E+098.00E+078.00E+05 Table 1: Message length = 1000 bytesSolutionChapter 219. Suppose an application layer entity wants to send an L-byte message to its peer process, using an existing TCP connection. The TCP segment consists of the message plus 20 bytes of header. The segment is encapsulated into an IP packet that has an additional 20 bytes of header. The IP packet in turn goes an additional20bytes of header The IP packet in turn goesSolution:y TCP/IP over Ethernet allows data frames with a payload size up to 1460 bytes. Therefore, L= 100, 500 and 1000 bytes are within this limit.y The message overhead includes:Th h d i l d20. Suppose that the TCP entity receives a 1.5 megabyte file from the application layer and that the IP layer is willing to carry blocks of maximum size 1500 bytes. Calculate the amount of overhead incurred from segmenting the file into packet-sized units.21.Suppose a TCP entity receives a digital voice stream from theapplication layer. The voice stream arrives at a rate of 8000bytes/second. Suppose that TCP arranges bytes into block sizes that result in a total TCP and IP header overhead of 50 percent.How much delay is incurred by the first byte in each block? 39.The internet below consists of three LANs interconnected by tworouters. Assume that the hosts and routers have the IP addresses as shown.a)Suppose that all traffic from network 3 that is destined to H1 is to berouted directly through router R2, and all other traffic from network 3 is to go to network 2.What routing table entries should be present in is to go to network2.What routing table entries should be present inb)Suppose that all traffic from network 1to network 3is tobe routed directly through R2. What routing table entriesshould be present in the network 1 hosts and in R2?R2R1H1H2p p p p Destination Next hop Destination Next hop Destination Next hop Destination Next hopSolutionChapter 360. Let g(x)=x3+x+1. Consider the information sequence1001.Solutions follow questions:a)Find the codeword corresponding to the precedinginformation sequence.information sequenceb)Suppose that the codeword has a transmission error inthe first bit. What does the receiver obtain when it doesits error checking?000162.Suppose a header consists of four 16-bit words: (11111111 11111111, 11111111 00000000, 11110000 11110000, 11000000 11000000). Find the internet checksum for this code.Solution 1 :Solution1:62.Suppose a header consists of four 16-bit words: (11111111 11111111, 11111111 00000000, 11110000 11110000, 11000000 11000000). Find the internet checksum for this code.Solution 2:S l ti2SolutionChapter 413.Suppose that a signal has twice the power as a noise signal that is added to it. Find the SNR in decibels. Repeat if the signal has 10 times the noise power? 2n times the noise power? 10k times the noise power?power?10times the noise power?32. Most digital transmission systems are “self-clocking” in that they derive the bit synchronization from the signal itself. To do this the systems use the transitions between positive and negative voltage levels. These transitions help define the boundaries of the bit intervals.y(b).In differential coding the sequence of 0s and 1s induces changes in the polarity of the signal; a binary 0results in no change in polarity, and a binary 1 results ina change in polarity. Repeat part (a). Does this schemehave a synchronization problem?have a synchronization problem?y The Manchester signaling method transmits a 0 as a +1 voltage for T/2 seconds followed by a −1 for T/2 seconds;a 1 is transmitted as a −1 voltage for T/2 secondsfollowed by a +1 for T/2 seconds. Repeat part (a) andexplain how the synchronization problem has beenexplain how the synchronization problem has beenSolutionChapter 560. Let g(x)=x3+x+1. Consider the information sequence1001.Solutions follow questions:a)Find the codeword corresponding to the precedinginformation sequence.information sequenceb)Suppose that the codeword has a transmission error inthe first bit. What does the receiver obtain when it doesits error checking?000162.Suppose a header consists of four 16-bit words: (11111111 11111111, 11111111 00000000, 11110000 11110000, 11000000 11000000). Find the internet checksum for this code.Solution:SolutionChapter 66.4 Suppose that the ALOHA protocol is used to share a 56k bps satellite channel. Suppose that packets are 1000 bits long. Find the maximum throughput of the system inppackets/second.6.526.531N N1N%80R R122R1×≤××+×⎟⎞⎜⎛+×8010101006×≤×106Repeater:R1N×80%RN6≤100×101080%10N80SolutionChapter 7Chapter77.20 / 7.26 / 7.32 / 7.33y20. A message of size m bits is to be transmitted over an L-hop path in a store-and-forward packet network as a series of N consecutive packets, each containing k data bits and h header bits. Assume that m >> k + h. The transmission rate of each link is R bits/second. Propagation and queueing delays are negligible.We know that:y message size = m bits = N packetsy one packet = k data bits + h header bitsy m >> k + h>>k+h(a) What is the total number of bits that must be transmitted?N (k + h) bits are transmitted.(b) What is the total delay experienced by the message (that is, the time between the first transmitted bit at the source and the last(c) What value of k minimizes the total delay?We determined D in the previous equation. We also know that N=[m/k] , where m is the ceiling. Therefore,Note that there are two components to the delay: 1. the first depends on the number of packets that need to be transmitted, m/k, as k increases this term decreases; 2. the second term increases as the size of the packets, k + h, is increased. The optimum value of k achieves a balance between these two tendencies.If we set dD/dk to zero,we obtain:It is interesting to note thatthe optimum payload sizeis the geometric mean ofth i d ththe message size and theheader, normalized to thenumber of hops (-1).7.26 Suppose a routing algorithm identifies paths that are “best” in the sense that: (1) minimum number of hops, (2) minimum delay, or (3) maximum available bandwidth. Identify conditions under which the paths produced by the different criteria are the same? are different?Solution:y The first criterion ignores the state of each link, but works well in Th fi t it i i th t t f h li k b t k ll iy The maximum available bandwidth criterion tries to route traffic along pipes with the highest “cross-section” to the destination.This approach tends to spread traffic across the various links in the network. This approach is inefficient relative to minimum hop routing in that it may use longer paths.y,yy At very low traffic loads, the delay across the network is the sum of the transmission times and the propagation delays. If all linksy7.32. Consider the network in Figure 7.30.3122154y The set of paths to destination 2 are shown below: y Now continue the algorithm after the link between node 2 and 4 goes down.The new set of paths are shown below:7.33. Consider the network in Figure 7.30.Solutions follow questions:(a) Use the Dijkstra algorithm to find the set of shortest paths from node 4 to other nodes.paths from node4to other nodes(b) Find the set of associated routing table entries.7.40. Assuming that the earth is a perfect sphere with radius 6400 km, how many bits of addressing is required to have a distinct address for every 1 cm x 1 cm square on the surface of the earth? y Solution:This is fewer bits than the 128 bits of addressing used in IP version 6 discussed in Chapter 8.SolutionChapter 8Chapter88.6 / 8.8 / 8.13 / 8.18 / 8.348.6. A host in an organization has an IP address 150.32.64.34 and a subnet mask 255.255.240.0. What is the address of this subnet? What is the range of IP addresses that a host can have on this subnet?this subnet?Solution:y8.8. A small organization has a Class C address for sevennetworks each with 24 hosts. What is an appropriate subnetmask?y Solution:y A Class C address requires 24 bits for its network ID,leaving 8 bits for the host ID and subnet ID to share.l i8bit f th h t ID d b t ID t hOne possible scheme would assign 4 bits to the hostand 4 to the subnet ID, as shown below. The number ofbits assigned to the host can be increased to 5 aswell.y8.12. Perform CIDR aggregation on the following /24IP addresses: 128.56.24.0/24; 128.56.25.0./24;128.56.26.0/24; 128.56.27.0/24.Solution:y Solution:y8.13 Perform CIDR aggregation on the following /24 IPaddresses:y200.96.86.0/24 ; 200.96.87.0/24 ;y200.96.88.0/24 ; 200.96.89.0/24 ;y8.18. Suppose a router receives an IP packet containing 600 data bytes and has to forward the packet to a network with maximum transmission unit of 200 bytes. Assume that the IP header is 20 bytes long. Show the fragments that the router creates and specify the relevant values in each fragment header (i.e., total length, fragment offset, and more bit).fragment offset and more bit)y The data packet must be divided into 4 frames, as shown by thefollowing calculations:y8.34. Consider the three-way handshake in TCP connection setup.y Solutions follow questions:y(a) Suppose that an old SYN segment from station A arrives atstation B, requesting a TCP connection. Explain how the three-way station B requesting a TCP connection Explain how the three wayy(b) Now suppose that an old SYN segment from station A arrives at station B, followed a bit later by an old ACK segment from A to a SYN segment from B. Is this connection request also rejected?。
Chapter-4-Exponential-and-Logarithmic-Functions
Example 1 – Converting from Exponential to Logarithmic Form
E xponentialF orm a. Since 5225 b. Since 3481 c. Since 1001
Logarithm icF orm
it follow s that it follow s that it follow s that
ln 2
Example 7 – Finding Half-Life
A 10-milligram sample of radioactive polonium 210 (which is denoted 210Po) decays according to the equation. Determine the half-life of 210Po. Solution:
Properties of Exponential Functions
2021/4/9
7
2007 Pearson Education Asia
Chapter 4: Exponential and Logarithmic Functions 4.1 Exponential Functions
Example 5 – Graph of a Function with a Constant Base
Chapter 4: Exponential and Logarithmic Functions
Chapter Outline
4.1) Exponential Functions 4.2) Logarithmic Functions 4.3) Properties of Logarithms 4.4) Logarithmic and Exponential Equations
solution计算器使用说明
Solution 2.0 SOLUTION TLB GROUP User’s ManualTable of Contents Introduction 3Quick Start4User’s Interface 6- Main Menu- Help- Functions Menu- Constants Menu- Custom Functions MenuPlot 12- General description- Navigation- More Functions MenuScientific Calculator 14- Supported functions- Complex computationsAnalysis 15- Limit- Symbolic Differentiation- Numeric Differentiation- IntegrationMatrix Operations 18Solvers 19- Non-linear Equation solver- Linear-System Solver- ODE SolverMore 22- Conversions- History- SettingsCredits 24End-User License Agreement 25Disclaimer 26IntroductionSolution is the most powerful mathematical application for mobile devices running Symbian OS Nokia Series 60 3rd Edition. Solution contains lots of different instruments and provides smart interface which makes using Solution quick and easy.This document describes all the Solution features and abilities and the right way to use them.In the 2nd chapter – Quick Start – you can find a detailed description of plotting exp(x) function. This simple example will introduce the basic interface features.In the 3rd chapter – User’s Interface – the interface utilities that are used in the whole program are widely discussed. The main idea of Solution interface is in fact this simply one – to enter any data press navigation centre key first and look at the appeared dialog if any.In the 4th – 8th chapters you can find full description and detailed how-to-use instructions for all the Solu-tion instruments such as Plot, Calculate, Analysis, Matrix and Solve.In the 9th chapter – More – you can find descriptions of some more utilities and settings that can be used in Solution.We hope that you would find Solution useful and worthwhile application.Good luck!Quick StartHow to plot exp(x) function?The interface of Solution is simple and provides a lot of opportunities at the same time. The best thing is just to try it on the simple example and get the main ideas. So let’s plot exp(x) function!In the Main Menu you see six icons that represent six main parts of Solutionprogram.Select Plot icon using navigation keys. Press navigation centre key to openPlot dialog.To enter new formula in F(x) box press navigation centre key and in the ap-peared Functions Menu dialog select exp using navigation keys.Press navigation centre key and see “exp()” added to F(x) box. Press naviga-tion centre key again. This time select “x” symbol and add it to F(x) box bypressing navigation centre key. Now formula in F(x) box is “exp(x)” and thatis precisely what we wanted. So press navigation centre key two more timesand get the first result – exp(x)-function’s plot appears on the screen.Press [*] to zoom in and press [#] to zoom out. Use navigation keys to move.Press back to return to Plot dialog.Let’s change the horizontal bounds Xmin and Xmax that specify the originalposition of function’s graph to -10 and 2 respectively. Select Xmin box us-ing navigation keys (press down three times). To enter minus press naviga-tion centre key, select minus sign “-” and press navigation centre key again.Use number keys to enter “10”. Then press navigation right key and changethe value of Xmax from default 5 to 2. Press navigation centre key two timesto view exp(x)-function’s plot again, but now on the different interval.To view help topic select the [?] icon in the upper-right corner and pressnavigation centre key.Now you are ready to start using Solution! To learn all the details concerning Solution instruments read the following chapters.User’s InterfaceMain MenuIn the Main Menu you see six icons that represent six main parts of Solutionprogram.The first item is Plot – using this utility you can plot any functions quicklyand easily. You can view several graphs on the same worksheet or varyfunctions using parametric dependencies.The second item is Calculate – you can evaluate any expression using thisscientific calculator that supports even complex computations. You maycreate your own functions and remember frequently used constants andthen refer to them by name.The third item is Analysis – it includes several powerful mathematical instru-ments concerning function analysis. Solution can evaluate limiting values,numerically integrate and differentiate functions and even find out symbolicderivatives.The next item is Matrix– this utility includes widely-used matrix opera-tions.The next item is Solve– this instrument will help you to solve non-linearequations, linear systems or ordinary differential equations quickly usingnumerical methods with high accuracy in calculations.The last item is More – there are several utilities and settings that could beused in Solution.HelpThere is a help topic for each instrument of Solution program.Help is available in any dialog. To view help topic for an instrument (plot, cal-culate etc) open the necessary utility and select [?] icon if it’s available andpress navigation centre key. In some menus (e.g. Functions Menu) there isno such icon. In this case use Options/Help to open the related help topic.Functions MenuFunctions Menu is the main tool that makes using Solution quick, smart,nice and easy. This menu is available nearly in every instrument dialog – tocall it just press navigation centre key. The main idea is simple – Func-tions Menu indicates what you are able to do in the particular situation, e.g.which function you can use and which one you can not. For mobile phonesit’s a well-known problem to enter formulas fast. The main purpose of thisutility is, of course, to make entering formulas as fast as possible using onlynavigation keys and numbers.The principle – “To enter any data press navigation centre key first” – issomething like Solution interface general rule.Functions Menu also provides the very important opportunity – creation ofuser-defined functions and quick access to previously saved ones. If ex-pressions are big and frequently used in calculations it takes a lot of timeto print them each time you need to. So this is time for creating a customfunction or save a constant.How to useStandard itemsTo enter standard function or constant value select the necessary one using navigation keys and press the navigation centre key. Any standard item that can’t be used in the particular case is marked grey.User-defined itemsYou can define your own constant/function using Constants and Custom Functions menus respectively. To call Constants menu press [*]. To call Custom Functions menu press [#]. Use these utilities to store any con-stants and functions that are frequently used. Give them unique names and then use them in any formula. To learn more about creation, changing and using constants and custom functions open the appropriate menu and call Options/Help there.DetailsHere are brief descriptions of every Functions Menu standard item.x – independent variablepi, e – constants.sin, cos, tan, ctg, asin, acos, atan, actg – trigonometric functionssh, ch, th, cth – hyperbolic functionsexp - exponentln – natural logarithmlog – logarithm of base 10sqrt – square root of a numberabs – absolute valuei – imaginary unityre, im – real and imaginary parts of a complex number+ - ^ * / – arithmetic operations! – factorial functionAnk = n!/(n-k)! – combinatorial functionCnk = n!/(k!*(n-k)!) – combinatorial function(), (, ) – brackets, – delimiter between arguments in multidimensional user-defined function A – numeric parameter in plotterE – scientific notation sign, e.g. 2E3=2*10^3=2000Ans – the last answery – desired function in Cauchy problem.Constants MenuUse this utility to store any constant values that are frequently used in cal-culations. Give them unique names and use these names in any formula youlike. This menu contains the full list of saved constants and provides quickaccess to them.How to useCreate newSelect “Add new…” item or use Options/Edit/Add new… Enter the nameand the value to be saved in the appeared dialogs. Some of the names (e.g.pi or e) are not allowed.Usage in formulaTo use saved constant in arbitrary formula set the cursor over constant’sname in the list (left column) using navigation keys. Press navigation centrekey or use Options/OK.Change existingTo change constant (name or value) set the cursor over the value of this constant in the list (right column) using navigation keys. Press navigation centre key or use Options/Edit/Change. In the appeared dialogs enter new name and then new value.DeleteTo delete constant select it in the list (any column) and press [C] or use Options/Edit/Delete.OrderConstants are listed in the same order as they were saved. To change current position of constant select it and call Options/Edit/Order.MoreThe last numerical result can be easily added to constants list. It appears as special button in Constants menu. There is also possibility to save the last result of any calculation directly to constants list. Set the cursor over the Answer box in parent dialog and press navigation centre key twice.Custom Functions MenuUse this utility to store any functions that are frequently used in calcula-tions. Give them unique names and use these names anywhere you need.This menu shows the full list of user-defined functions and provides quickaccess to them. User-defined functions can be multidimensional.How to useCreate newSelect “Add new…” item or use Options/Edit/Add new… Enter the uniquename and the formula to be saved in the appeared dialog. Some of thenames (e.g. sin or exp) are not allowed. To create multidimensional functionuse several independent variables – the names of them should be “x1”, “x2”etc. See the example below for details. The current function from parentdialog can be easily saved. It appears as special button in Custom Func-tions menu.Usage in formulaTo use saved function set the cursor over it’s name in the list (left column) using navigation keys. Press navi-gation centre key or use Options/OK. If function is multidimensional it is added to parent dialog with brack-ets and commas. You should specify all the necessary arguments. See the example below for details. Change existingTo change function (name or formula) set the cursor over the it’s formula (right column in the list) using navigation keys. Press navigation centre key or use Options/Edit/Change. In the appeared dialog enter new name and then new formula.DeleteTo delete function select it in the list (any column) and press [C] or use Options/Edit/Delete.OrderFunctions are listed in the same order as they were saved. To change current position of function select it and call Options/Edit/Order.Multidimensional function exampleSuppose you want to save the function fun(x1,x2) = sin(x1)+x2. Select “Add new…” item and enter the function’s name fun. Enter the formula (Functions Menu is available) sin(x1)+x2. Press OK or press centre key twice.To use new function select its name and press centre key. In the parent dialog fun(,) appears. The comma separates arguments. Enter the values for both arguments (that can be numbers, other functions, variables or expressions) and get the answer. fun(0,1) provides 1, fun(asin(1), -1) provides 0.PlotGeneral DescriptionUse this utility to create a linear plot of function F(x) in specified boundsXmin, Xmax, Ymin, Ymax. Parameter A can be used in F(x) formula, specifythis value if necessary. This utility also provides an opportunity to view sev-eral graphs of functions on the same worksheet.How to useFor entering F(x) formula quicker press navigation centre key and use theappeared Functions Menu. This menu provides quick access to standardand user-defined functions and constants. To learn more about FunctionsMenu read User’s Interface chapter above.When you finish entering data press navigation centre key two more times.On the appeared worksheet you will see plot of F(x) function. Press [*] tozoom in and press [#] to zoom out, use navigation keys to move throughthe worksheet.To view several functions on the same worksheet use “More functions…” menu (select this item and press navigation centre key). You can change existing functions and add new ones to the same worksheet. To learn more read the information below.More Functions MenuUse this utility to view several functions’ graphs on the same worksheet.How to useCreate newSelect “Add new…” item or use Options/Edit/Add new… Enter the functionin the appeared dialog (Functions menu is available).Change existingTo change the function (name or formula) select it from the list (left column)using navigation keys. Press navigation centre key or use Options/Edit/Change. In the appeared dialog make the necessary changes.To change the color of function’s graph set the cursor over the right columnin the list and press navigation centre key. In the appeared dialog make thenecessary changes.DeleteTo delete function from the list select it (any column) and press [C] or useOptions/Edit/Delete.OrderFunctions are listed in the same order as they were saved. To change cur-rent position of function select it and call Options/Edit/Order.PlotTo plot all the functions use Options/Plot.Scientific CalculatorThis is powerful scientific calculator with real and complex computationssupport.Supported functions1. Standard functions (with complex-variable support): +, – , /, *, ^, sqrt(x),sin(x), cos(x), tan(x), ctg(x), arcsin(x), arccos(x), arctan(x), arcctg(x) exp(x),log(x), ln(x), abs(x), re(x), im(x), sh(x), ch(x), th(x), cth(x).2. Combinatorial calculus: ! – factorial, A(n,k)= n!/(n-k)!, C(n,k) = n!/(k!*(n-k)!).3. User-defined functions and constants – Create your own functions (pos-sibly multidimensional) and remember constants that are frequently used incalculations.Complex computationsAll the standard functions support complex computations. To enter a complex number use Function Menu item “i” that represents imaginary unity.How to useFor entering F(x) formula quicker press navigation centre key and use the appeared Functions Menu. This menu provides quick access to standard and user-defined functions and constants. To learn more about Functions Menu call Options/Help there.When you finish entering data press navigation centre key two more times to get the answer that will appear in the Answer box below.By default pressing [#] causes entering “.” symbol. To switch to Alpha mode use More menu.By default everything is calculated in radians. To switch to degrees use More menu.AnalysisThis utility includes several function analysis tools such as: Limit, Symbolicand Numerical Differentiation and Numerical Integration. The detailed de-scription of each instrument is provided below.How to useFor entering F(x) formula in each of these instruments quicker press naviga-tion centre key and use the appeared Functions Menu. This menu providesquick access to standard and user-defined functions and constants. Tolearn more about Functions Menu read User’s Interface chapter above.When you finish entering data press navigation centre key two more timesto get the answer that will appear in the Answer box below.LimitCalculates the limiting value of F(x) function as x variable approaches speci -fied limiting point x0, possibly infinity , or –infinity .The limit is the real bidirectional limit, except in the case where the bidirec-tional limit does not exist. In this case the limit is directional, taken from theright and left, respectively.ExamplesF(x)sin(x)/x x00Answer1F(x)exp(x+1)/(x+1)x00Answer2.718281828F(x) (1+1/x)^x x0 +inf Answer 2.7182804690 F(x) ln(x)/(x+1)x0 0Answer Right: -inf; Left: Error;Symbolic DifferentiationComputes the symbolic derivative of the function F(x) with respect to vari -able x.ExamplesF(x)exp(2*x)Answer2*exp(2*x)F(x)sin(x)*exp(x)Answersin(x)*exp(x)+exp(x)*cos(x)F(x)x^(sin(x)+1)Answer(ln(x)*cos(x)+(1+sin(x))/x)*x^(1+sin(x))F(x) th(x)Answer 1-(th(x))^2Numeric DifferentiationNumerically computes the first, second, or higher–order derivative of the function F(x) with respect to variable x at specified point x0. Examples F(x) exp(2*x)x0 1 k 4Answer 118.224898 F(x) sin(x)*exp(x)x0 0 k 3Answer 2F(x) x^(sin(x)+1)x0 1 k 2Answer 2.63014901F(x) x^(2*pi*x)x0 1 k 1Answer 45.7616029Integration Numerically computes a definite integral of the function F(x) with respect tothe variable x over the straight line from a to b (finite numbers).The numerical answers are carried out with accuracy up to the five decimalplaces.ExamplesF(x) exp(2*x)a -1 b 1Answer 3.62687166380 F(x) sin(x)*exp(x)a 0 b 3Answer 11.8594882628F(x) x^(sin(x)+1)a -1 b 0Answer Can’t Do It! F(x) x^(sin(x)+1)a 0 b 1Answer 0.40824193777Matrix OperationsThis utility provides the most widely used matrix operations such as:A+B component-wise additionA-B component-wise subtractionAxB matrix multiplyingA-1computing the inverse matrix|A|computing the determinant of the given matrixHow to useTo fill A and B matrices set the cursor over [A] or [B] icon respectively. Pressnavigation centre key. In appeared dialog you can change the size and ele-ments of matrix you’ve chosen.To enter element set the cursor over its position and press navigation centrekey. Change the value and press navigation centre key once more or useOK. When you finish entering data press Save button. To call the matrix op-eration set the cursor over the operation’s icon and press navigation centrekey. The answer will appear in a new dialog.Examples(A): Width 3Height 31 2 31 12 32 4 5 637 8 9 AxB:Width 2Height 31 2122 28249 64376 100 (B) Width 2Height 31 21 1 22 3 43 5 6|A|:Determinant:SolversNon-linear Equation solverUse this utility to find numerical solution to a single equation f(x)=0 solvedfor a single unknown x over the straight line from a to b (finite numbers).The numerical answers are carried out with accuracy up to the five decimalplaces.How to useFor entering F(x) formula quicker press navigation centre key and use theappeared Functions Menu. This menu provides quick access to standardand user-defined functions and constants. To learn more about FunctionsMenu call Options/Help there.When you finish entering data press navigation centre key two more timesto get the answer that will appear in the Answer box below.ExamplesF(x) sin(x)*exp(x)a -1 b1 Answer 0F(x) ln(x)+tan(x)a 2b 4 Answer 2.418229103088F(x) x^2–2a 0b 2 Answer 1.41421222686 F(x) exp(2*x)a -1b 1 AnswerCan’t Do It!Linear-System SolverUse this utility to find numerical solution to a system of linear equations for several unknowns. The numeri-cal answers are carried out with accuracy up to the five decimal places.How to useThe system of linear equations is presented in Linear Algebra terms. Itmeans that there is a coefficients’ matrix that you should fill minding thesequence of unknown variables.Select “Coefficients” button and press navigation centre key. In a new dia-log set the number of equations in the system to be solved in Amount field.Then fill in the matrix displayed with all the system coefficients – select thenecessary element and press navigation centre key. Mind that the k–columnis for the k–unknown value only and the n–row – for the coefficients in then–equation. The last column represents the constant right part of equa-tions. See the example below for details.The answer is a vector, where the sequence of unknown variables is the sameas it was in the coefficients’ matrix. The answer appears in a new dialog. ExamplesIf you need to solve a system of the three linear equations like the following one you should fill the coef-ficients’ matrix as shown.Linear system to be solved: Eq12x – y + 3z = 9;Eq25x – y – z = 0;Eq3 –2y + z = 1.x is a1, y is a2 and z is a3.Then coefficients’ matrix is:a1 a2 a3 CEq1 2 –1 3 9Eq2 5 –1 –1 0Eq3 0 –2 1 1The elements of the first row are the coefficients of the first equation: 2 – x’s coefficient; (–1) – y’s coef-ficient; 3 – z’s coefficient. The last one – 9 – is the constant right part of the equation.In the third equation the first variable x has a zero–coefficient.The answer is (1 ; 2; 3), which means that x=1; y=2; z=3.ODE SolverUse this utility to solve initial value problems for ordinary differential equa-tions (ODEs). Specified equation dy/dx= f(x,y), where y(x) is scalar functiondepending on x, is numerically solved in the specified interval [x1, x2] usingEuler method.How to useFor entering F(x) formula quicker press navigation centre key and use theappeared Functions Menu. This menu provides quick access to standardand user-defined functions and constants. To learn more about FunctionsMenu call Options/Help there. Mind the special symbol “y” that representsthe desired unknown function and can be used in equation. When you finishentering data press navigation centre key two more times to get the answer.The numerically resolved answer is plotted in a new dialog.Examplesdy/dx cos(x)x1 0x2 6.3y(x1) 0sin(x) is plotteddy/dx 2*yx1 0x2 1y(x1) 1exp(2*x) is plotteddy/dx 2*abs(x)x1 -1X2 1y(x1) -1y=x^2 for x>0 and y=-x^2 else is plottedMoreConversionsYou can use this utility for converting values between various numerationsystems. Conversions from decimal numeration system return the posi-tive decimal numbers in a true representation; the whole negative decimalnumbers – in a complement representation. The fractional negative decimalnumbers are not represented in the complement form for not creating con-tradictions.Also there are functions for converting values from binary, octal or hexa-decimal numeration to decimal. These functions return the positive decimalnumbers.How to useEnter the number you want to convert to another notation (press naviga -tion centre key to enter letters) and then use special buttons that representsix available operations. In the first row there are “decimal to binary form”,“decimal to octal form”, “decimal to hexadecimal form”. In the second rowthere are “binary to decimal form”, “octal to decimal form” and “hexadeci -mal to decimal form”. Select the necessary operation and press navigationcentre key. Answer appears in the box below.ExamplesNumber 4.5 10->2Answer 100.1 Number 1000 [10->16]Answer3e8Number -1 [10->2]Answer 1111Number 11001[2->10]Answer 25Number abcdef[16->10]Answer 11259375HistoryThis utility provides quick access to the last 30 performed actions.How to useThe first line of each history item represents utility used and the second linerepresents object (function, matrix etc) upon which the action took place.Select the necessary item using navigation keys and press navigation cen-tre key. The proper already-filled dialog opens.Additional SettingsReset colors sets default values to all graph colors.Units of Angle changes the behavior of trigonometric functions. You can choose between radians (default) and degrees computations.Alpha Mode sets the F(x) editor mode. Digits mode “123” is set by default, letters mode “Abc” is recom-mended for users of QWERTY-keyboard.CreditsAlexander TaboriskiyNadezhda BaldinaSergey BasheleyshviliDmitriy Kuznetsovsupport@End-User License AgreementSolution is copyright (C) 2005 - 2008 SOLUTION TLB GROUP. All rights reserved.This license describes the conditions under which you may use version 2.0 of Solution («the program»). If you are unable or unwilling to accept these conditions in full, then, notwithstanding the conditions in the remainder of this license, you may not use the program at all.You are granted a non-exclusive license to use the program on one device at a time. The program may not be rented, leased or transferred.Any use of the program which is illegal under international or local law is forbidden by this license. Any such action is the sole respon-sibility of the person committing the action.The program is distributed «AS IS» and you assume full responsibility for determining the suitability of the program and for results obtained.SOLUTION TLB GROUP makes no warranty that all errors have been or can be eliminated from the program software and, with re-spect thereto, SOLUTION TLB GROUP shall not be responsible for losses, damages, costs, or expenses of any kind resulting from using or misusing the program including without limitation, any liability for business expenses, machine downtime, damages experi-enced by you or any third person as a result of any deficiency, defect, bug, error or malfunction. SOLUTION TLB GROUP shall not be liable for any indirect, special, incidental, or consequential damages relating to or arising out of the subject matter of this Agreement or actions taken hereunder.NO WARRANTY OF ANY KIND IS EXPRESSED OR IMPLIED. YOU USE THE PROGRAM AT YOUR OWN RISK. SOLUTION TLB GROUP DISCLAIMS ALL WARRANTIES, EITHER EXPRESS OR IMPLIED, INCLUDING THE WARRANTIES OF MERCHANTABILITY AND FITNESS FOR A PARTICULAR PURPOSE. NOBODY WILL BE LIABLE FOR DATA LOSS, DAMAGES, LOSS OF PROFITS OR ANY OTHER KIND OF LOSS WHILE USING OR MISUSING THIS SOFTWARE.You may not distribute, copy, emulate, clone, rent, lease, sell, modify, decompile, disassemble, otherwise reverse engineer, or transfer the program, or any subset of the program, except as provided for in this agreement. Any such unauthorized use shall result in im-mediate and automatic termination of this license and may result in criminal or civil prosecution.All rights not expressly granted here are reserved by SOLUTION TLB GROUP.SOLUTION TLB GROUP reserves the right to make exceptions to any of these conditions, or alter these conditions, at any time. How-ever, you may always use these conditions instead of any altered version if you prefer (note that this license explicitly applies only to one version of the program; therefore, if SOLUTION TLB GROUP make new conditions in connection with a future version, you do not then have the right to apply these conditions to that version instead).Installing or using the program signifies acceptance of these terms and conditions of the license.If you do not agree with the terms of this license you must remove the program files from your storage devices and cease to use the program.DisclaimerThis document describes the features and user interface of program Solution 2.0 by SOLUTION TLB GROUP. This program is intended for mobile phones running Symbian OS Nokia Series 60 3rd Edition. For complete list of supported devices please consult phone manufacturer’s website. However, SOLUTION TLB GROUP makes no warranty that the program will work properly at all devices listed there.The information contained in this document is for general information purposes only and should not be used or relied on for any other purpose whatsoever. While SOLUTION TLB GROUP has taken great care in the preparation of this document, it makes no warranty or guarantee about the suitability or accuracy of the information contained in this document. No part of this material may be reproduced without the express written permission of SOLUTION TLB GROUP.All registered trademarks mentioned in this document are property of their legal owners.The program Solution is protected by copyright law and international treaties. Unauthorized reproduction or distribution of this program, or any portion of it, may result in severe civil and criminal penalties, and will be prosecuted to the maximum possible under the law.。
Chapter 4 Recommended end_of_chapter problems.
Ch4 Recommended End-of-Chapter ProblemsE 4-2 Income statement format; singleLO1LO3LO5LO8 step and multiple stepp. 212The following is a partial trial balance for General Lighting Corporation as of December 31, 2011:300,000 shares of common stock were outstanding throughout 2011. Income tax expense has not yet been accrued. The income tax rate is 40%.Required:1. Prepare a single-step income statement for 2011, including EPS disclosures.2. Prepare a multiple-step income statement for 2011, including EPS disclosures.E 4-4 Income statement presentation;LO1LO5LO8 intraperiod tax allocationThe following incorrect income statement was prepared by the accountant of the Axel Corporation:Required:Prepare a multiple-step income statement for 2011 applying generally accepted accounting principles. The income tax rate is 40%. The gain from litigation settlement is considered an unusual and infrequent event.E 4-7 Discontinued operations; disposalLO4 in subsequent yearKandon Enterprises, Inc., has two operating divisions; one manufactures machinery and the other breeds and sells horses. Both divisions are considered separate components as defined by generally accepted accounting principles. The horse division has been unprofitable, and on November 15, 2011, Kandon adopted a formal plan to sell the division. The sale was completed on April 30, 2012. At December 31, 2011, the component was considered held for sale.On December 31, 2011, the company's fiscal year-end, the book value of the assets of the horse division was $250,000. On that date, the fair value of the assets, less costs to sell, was $200,000. The before-tax operating loss of the division for the year was $140,000. The company's effective tax rate is 40%. The after-tax income from continuing operations for 2011 was $400,000.Required:1. Prepare a partial income statement for 2011 beginning with income from continuing operations.Ignore EPS disclosures.2. Repeat requirement 1 assuming that the estimated net sales price of the horse division's assetswas $400,000, instead of $200,000.E 4-8 Discontinued operations; disposalLO4 in subsequent year; solving forunknownOn September 17, 2011, Ziltech, Inc. entered into an agreement to sell one of its divisions that qualifies as a component of the entity according to generally accepted accounting principles. By December 31, 2011, the company's fiscal year-end, the division had not yet been sold, but was being held for sale. The net fair value (fair value minus costs to sell) of the division's assets at the end of the year was $11 million. The pretax operating income of the division during 2011 was $4 million. Pretax income from continuing operations for the year totaled $14 million. The income tax rate is 40%. Ziltech reported net income for the year of $7.2 million.Required:Determine the book value of the division's assets on December 31, 2011.E 4-16 Statement of cash flowsLO11 preparationp. 215The following summary transactions occurred during 2011 for Bluebonnet Bakers:Required:Prepare a statement of cash flows for 2011 for Bluebonnet Bakers. Use the direct method for reporting operating activities.E 4-17 IFRS; statement of cash flows LO11LO12Refer to the situation described in Exercise 4-16.Required:How might your solution differ if Bluebonnet Bakers prepares the statement of cash flows according to International Financial Reporting Standards?P 4-1 Comparative income statements; multiple-step●LO1LO3through LO8formatSelected information about income statement accounts for the Reed Company arepresented below (the company's fiscal year ends on December 31):On July 1, 2011, the company adopted a plan to discontinue a division that qualifies as a component of an entity as defined by GAAP. The assets of the component were sold on September 30, 2011, for $50,000 less than their book value. Results of operations for the component (included in the above account balances) were as follows:p. 220In addition to the account balances above, several events occurred during 2011 that have not yet been reflected in the above accounts:1. A fire caused $50,000 in uninsured damages to the main office building. The fire was consideredto be an infrequent but not unusual event.2. An earthquake caused $100,000 in property damage to one of Reed's factories. The amount ofthe loss is material and the event is considered unusual and infrequent.3. Inventory that had cost $40,000 had become obsolete because a competitor introduced a betterproduct. The inventory was sold as scrap for $5,000.4. Income taxes have not yet been accrued.Required:Prepare a multiple-step income statement for the Reed Company for 2011, showing 2010 information in comparative format, including income taxes computed at 40% and EPS disclosures assuming 300,000 shares of common stock.P 4-2 Discontinued operations ●LO4 The following condensed income statements of the Jackson Holding Company are presented for the two years ended December 31, 2011 and 2010:On October 15, 2011, Jackson entered into a tentative agreement to sell the assets of one of its divisions. The division comprises operations and cash flows that can be clearly distinguished, operationally and for financial reporting purposes, from the rest of the company. The division was sold on December 31, 2011, for $5,000,000. Book value of the division's assets was $4,400,000. The division's contribution to Jackson's operating income before-tax for each year was as follows: 2011 $400,000 loss2010 $300,000 lossAssume an income tax rate of 40%.Required:1. Prepare revised income statements according to generally accepted accounting principles,beginning with income from continuing operations before income taxes. Ignore EPS disclosures.2. Assume that by December 31, 2011, the division had not yet been sold but was considered heldfor sale. The fair value of the division's assets on December 31 was $5,000,000. How would the presentation of discontinued operations be different from your answer to requirement 1?3. Assume that by December 31, 2011, the division had not yet been sold but was considered heldfor sale. The fair value of the division's assets on December 31 was $3,900,000. How would the presentation of discontinued operations be different from your answer to requirement 1?P 4-7 Income statement presentation; unusual items ●LO1LO3LO4LO7LO8LO9revenue, $15,300; cost of goods sold, $6,200; selling expenses, $1,300; general and administrative expenses, $800; interest revenue, $85; interest expense, $180. Incometaxes have not yet been accrued. The company's income tax rate is 40% on all items ofincome or loss. These revenue and expense items appear in the company's incomestatement every year. The company's controller, however, has asked for your help indetermining the appropriate treatment of the following nonrecurring transactions that alsooccurred during 2011 ($ in 000s). All transactions are material in amount.1. Investments were sold during the year at a loss of $220. Schembri also had unrealized gains of$320 for the year on investments accounted for as securities available for sale.2. One of the company's factories was closed during the year. Restructuring costs incurred were$1,200.3. An earthquake destroyed a warehouse causing $2,000 in damages. The event is considered tobe unusual and infrequent.4. During the year, Schembri completed the sale of one of its operating divisions that qualifies as acomponent of the entity according to GAAP. The division had incurred an operating loss of $560 in 2011 prior to the sale, and its assets were sold at a gain of $1,400.5. In 2011, the company's accountant discovered that depreciation expense in 2010 for the officebuilding was understated by $200.6. Foreign currency translation losses for the year totaled $240.Required:Prepare Schembri's combined statement of income and comprehensive income for 2011, including basic earnings per share disclosures. One million shares of common stock were outstanding at the beginning of the year and an additional 400,000 shares were issued on July 1, 2011.P 4-9 Statement of cash flows ●LO11 The Diversified Portfolio Corporation provides investment advice to customers. A condensed income statement for the year ended December 31, 2011, appears below:p. 223The following balance sheet information also is available:In addition, the following transactions took place during the year:1. Common stock was issued for $100,000 in cash.2. Long-term investments were sold for $50,000 in cash. The original cost of the investments alsowas $50,000.3. $80,000 in cash dividends was paid to shareholders.4. The company has no outstanding debt, other than those payables listed above.5. Operating expenses include $30,000 in depreciation expense.Required:1. Prepare a statement of cash flows for 2011 for the Diversified Portfolio Corporation. Use thedirect method for reporting operating activities.2. Prepare the cash flows from operating activities section of Diversified's 2011 statement of cashflows using the indirect method.The chief accountant for Grandview Corporation provides you with the company's 2011 statement of cash flows and income statement. The accountant has asked for your help with some missing figures in the company's comparative balance sheets. These financialstatements are shown next ($ in millions).p. 224Required:1. Calculate the missing amounts.2. Prepare the operating activities section of Grandview's 2011 statement of cash flows using theindirect method.Presented below are the 2011 income statement and comparative balance sheets forSantana Industries.p. 225Additional information for the 2011 fiscal year ($ in thousands):1. Cash dividends of $1,000 were declared and paid.2. Equipment costing $4,000 was purchased with cash.3. Equipment with a book value of $500 (cost of $1,500 less accumulated depreciation of $1,000)was sold for $500.4. Depreciation of $1,600 is included in operating expenses.Required:Prepare Santana Industries' 2011 statement of cash flows, using the indirect me thod to present cash flows from operating activities.The following questions are used in the Kaplan CPA Review Course to studythe income statement and statement of cash flows while preparing for theCPA examination. Determine the response that best completes thestatements or questions.1. Roco Company manufactures both industrial and consumer electronics. Due to a●LO4change in its strategic focus, the company decided to exit the consumer electronicsbusiness, and in 2011 sold the division to Sunny Corporation. The consumer electronicsdivision qualifies as a component of the entity according to GAAP. How should Rocoreport the sale in its 2011 income statement?a. Include in income from continuing operations as a nonoperating gain or loss.b. As an extraordinary item.c. As a discontinued operation, reported below income from continuing operations.d. None of the above.2. Bridge Company's results for the year ended December 31, 2011, include theLO3LO4LO7 following material items: ●Sales revenue $5,000,000Cost of goods sold 3,000,000Administrative expenses 1,000,000Gain on sale of equipment 200,000Loss on discontinued operations 400,000Loss from earthquake damage (unusual and infrequent event) 500,000Understatement of depreciation expense in 2010 caused by mathematical250,000 errorBridge Company's income from continuing operations before income taxes for 2011 is:a. $700,000b. $950,000c. $1,000,000d. $1,200,0003.In Baer Food Co.'s 2011 single-step income statement, the section titled●LO4LO5“Revenues” consisted of the following:Net sales revenue $187,00012,000Income on discontinued operations including gain on disposalof $21,000 and net taxes of $6,000Interest revenue 10,200Gain on sale of equipment 4,700Extraordinary gain net of $750 tax effect 1,500Total revenues $215,400In the revenues section of the 2011 income statement, Baer Food should have reported total revenues ofa. $201,900b. $203,700c. $215,400d. $216,3004. On November 30, 2011, Pearman Company committed to a plan to sell a division●LO4that qualified as a component of the entity according to GAAP, and was properlyclassified as held for sale on December 31, 2011, the end of the company's fiscalyear. The division was tested for impairment and a $400,000 loss was indicated.The division's loss from operations for 2011 was $1,000,000. The final sale wasexpected to occur on February 15, 2012. What before-tax amount(s) shouldPearman report as loss on discontinued operations in its 2011 income statement?a. $1,400,000 loss.b. $400,000 loss.c. None.d. $400,000 impairment loss included in continuing operations and a $1,000,000loss from discontinued operations.5. For 2010, Pac Co. estimated its two-year equipment warranty costs based on $100●LO7per unit sold. Experience during 2011 indicated that the estimate should have beenbased on $110 per unit. The effect of this $10 difference from the change inestimate is reportedAs an accounting change below 2011 income from continuing operations.p.219a.b. As an accounting change requiring the 2011 financial statements to berestated.c. As a correction of an error requiring 2011 financial statements to berestated.d. In 2011 income from continuing operations.6. Which of the following items is not considered an operating cash flow in the●LO11statement of cash flows?a. Dividends paid to stockholders.b. Cash received from customers.c. Interest paid to creditors.d. Cash paid for salaries.7. Which of the following items is not considered an investing cash flow in the●LO11statement of cash flows?a. Purchase of equipment.b. Purchase of securities.c. Issuing common stock for cash.d. Sale of land.The following questions dealing with the income statement are adapted fromquestions that previously appeared on Certified Management Accountant (CMA) examinations. The CMA designation sponsored by the Institute of ManagementAccountants () provides members with an objective measure ofknowledge and competence in the field of management accounting. Determine theresponse that best completes the statements or questions.1. Which one of the following items is included in the determination of income from●LO1continuing operations?a. Discontinued operations.b. Extraordinary loss.c. Cumulative effect of a change in an accounting principle.d. Unusual loss from a write-down of inventory.2. In a multiple-step income statement for a retail company, all of the following are●LO3included in the operating section excepta. Sales.b. Cost of goods sold.c. Dividend revenue.d. Administrative and selling expenses.3. When reporting extraordinary items,a. Each item (net of tax) is presented on the face of the income statement separately as acomponent of net income for the period.b. Each item is presented exclusive of any related income tax.c. Each item is presented as an unusual item within income from continuing operations.d. All extraordinary gains or losses that occur in a period are summarized as total gains andtotal losses, then offset to present the net extraordinary gain or loss.。
Chapter_9
Chapter 9, Solution 14. (a) 3 j14 14.318 77.91 0.7788169.71 0.7663 j0.13912 7 j17 18.385112.38 (62.116 j 231.82 138.56 j80)(60 j80) 24186 6944.9 1.922 j11.55 (67 j84)(16.96 j10.5983) 246.06 j 2134.7
z1 z 2 = [(10–53.13˚)(10–30˚)/(8–120˚)] = 12.536.87˚Ω = (10 + j7.5) Ω z3
Chapter 9, Solution 11. (a) V 21 15o V
(b) i(t ) 8sin(10t 70o 180o ) 8cos(10t 70o 180o 90o ) 8cos(10t 160o ) I = 8ס160° mA (c ) v(t ) 120sin(103 t 50o ) 120 cos(103 t 50o 90o ) V = 120–ס140° V (d) i(t ) 60 cos(30t 10o ) 60 cos(30t 10o 180o ) I = 60–ס170° mA
(a) 10cos(t – 60), (b) 9cos(8t + 90), (c) 20cos(t + 135)
Chapter 9, Solution 4. Design a problem to help other students to better understand sinusoids. Although there are many ways to work this problem, this is an example based on the same kind of problem asked in the third edition. Problem (a) Express v = 8 cos(7t + 15) in sine form. (b) Convert i = –10sin(3t - 85) to cosine form. Solution (a) (b) v = 8 cos(7t + 15) = 8 sin(7t + 15 + 90) = 8 sin(7t + 105) i = –10 sin(3t – 85) = 10 cos(3t – 85 + 90) = 10 cos(3t + 5)
博弈论 战略分析入门第四章课后题答案
Instructor’s Guide to Game Theory: A Nontechnical Introduction to theAnalysis of StrategyChapter 4. Nash Equilibrium1.Objectives and ConceptsThe principle objective of this chapter is to introduce the Nash equilibrium and to convey some notion of the range of possibilities and applications, including the possibilities that there may be no Nash equilibria in pure strategies and the possibility that there may be plural Nash equilibria. (Since mixed strategy equilibria are not introduced until Chapter 8, it is not possible to give a meaningful definition of pure strategies at this point, and is necessary to talk around it a bit.) Important subsidiary concepts are coordination games and Schelling points (or focal point equilibria), heuristic methods of finding the Nash equilibria, such as underlining, and refinement of Nash equilibrium.The chapter begins with an example that is based on Warren Nutter’s game-theoretic version of Bertrand competition, except that in this instance a kind of quality competition is considered. The solution to this game can be found by iterated elimination of dominated strategies (which will not be covered until Chapter 11) and reflects the intuition that it is best to be just one step ahead of the competition. Thus, while it does not have a dominant strategy equilibrium, it has some dominated strategies and a unique Nash equilibrium, and hopefully forms a natural bridge from the study of dominant strategy equilibrium.Games with plural equilibria are introduced with the game of Choosing Radio Formats. The idea that history (or other clues) can establish a Schelling point also comes in with this example. The Market Day game reinforces the idea that plural Nashequilibria can have explanatory value – explaining the persistence of what seem to be arbitrary conventions. Games without Nash equilibria (in pure strategies) are introduced with an escape-evasion game. This is an important category in itself, though the most important applications are in differential games and thus beyond the scope of the book.Accordingly, the concepts areNash EquilibriumUnique Nash EquilibriaFinding Nash EquilibriaPlural Nash EquilibriaThe difficulty of choosing among plural Nash equilibriaSchelling PointsCustom, convention and history as Schelling pointsSchelling points from the logic of the gameRefinementGames without Nash equilibria in pure strategies2. Common Study ProblemsStudents who have not yet grasped the best-response idea will find Nash equilibria even more difficult than dominant strategy equilibria. This is the crisis point for students who have not “got” best response. The best response tables (such as table 2 in the chapter) are designed to make this a little easier, so urge the student to rely on them and on underlining as intermediate steps in their analysis. I sometimes suggest to mystudents that they physically move their fingers along the column or row to pick out the biggest payoff. Making the solution as mechanical as possible will help students over that hump. Another (less troubling) problem is the relationship between Nash and dominant strategy equilibria. Taking dominant strategy equilibria first is a pedagogical convenience, since it is a little easier and will be familiar to students who have seen the Prisoner’s Dilemma in another class, but it can produce the impression that dominant strategy equilibria are not Nash equilibria. The Venn diagram (Figure 1) is meant to speak to that problem, and may need some stress in class.3. For Business StudentsThe key business concepts for this chapter are strategies of location and market niche, in the Location, Location, Location example, but also in the Radio Formats example and in the Hairstyle example in the exercises and discussion questions.4. Class AgendaFirst hour:1)Quiz on earlier material2)Introductory presentation: Nash Equilibria•Assignments3)Discussion: The Blonde Problem AgainSecond Hour:1)Discussion of quiz and assignments2)Play a coordination game in class, with random matching and without discussion.A handout description of the game is given on the next page.Another Random-Matching Two-Person GameOnce again, each person chooses between the strategies of collusion or defecting from the collusive arrangement.Put in your name and circle one of the two statements: either "my strategy is collude" or "my strategy is defect." Your instructor will tell you whether to follow directions A) or B) below.A)After you turn it in, your strategy choice will be matched with that of anotherclass member AT RANDOM, and your bonus points will be based on the payofftable above. There is to be no discussion of your strategy choices.B)You will be matched with your neighbor and may discuss your strategy choice ifyou wish.Payoffs are in GameBucks.TableArt's StrategyCollude DefectCollude (3,3)(0,2)Bob's StrategyDefect(2,0)(1,1)What will you do? Go for the big reward with a "collude" strategy or protect yourself with an "defect" strategy?Student name ____________________________My strategy is (circle one)ColludeDefect3)Discussion:a.Results of the in-class game.b.Give other examples of Schelling points in coordination games. Ideally,these should come from the students, but the following instances maystimulate the discussion if it comes slowly:i.Driving on the right or left-hand side of the road.ii.Speaking the same language.iii.Choosing a profession. Assumption: if both choose the sameprofession, it does not pay well because it is too crowded. Howmany business majors in the class? Engineering? Communications,etc?5. Answers to Exercises and Discussion Questions1.Solving the Game. Explain the advantages and disadvantages of NashEquilibrium as a solution concept for noncooperative games.Nash equilibrium is based on the idea that each player chooses the best response tothe strategy chosen by the other player. This is a clear concept of rationality wheneach person chooses in isolation from the other. Among the shortcomings are 1)Nash equilibrium may not be unique, posing the problem of determining which oftwo or more Nash equilibria may actually be chosen by rational agents, and 2) considering only the list of strategies for the game in normal form, that is, the“pure” strategies, there may not be a Nash equilibrium.2.Location, Location, Location (Again) Not all location problems have similarsolutions. Here is another one: Gacey's and Mimbel's are deciding where to puttheir stores in Metropolis, the town across the river from Gotham City. The three strategies for Metropolis are to locate downtown, in Old Town, or in the Garden District. The payoffs are shown in Table E1.Table E1 Payoffs in a New Location GameGacey'sDowntown Old Town Garden DistrictDowntown70,6060,12080,100Old Town110,7040,40120,110Mimbel'sGardenDistrict120,80110,12050,50Does this game have Nash equilibria? What strategies, if so? Which strategies would you predict that Gacey's and Mimbel's would choose? Compare and contrast this game with the location game in the chapter. What would you say about the relative importance of congestion in the location decisions of the firms in the two cases?A table modified to show the highest payouts for each player for each decision is as follows:Gacey's Downtown Old Town Garden District Downtown 70, 6060, 12080, 100Old Town110, 7040, 40120, 110M i m b e l 's Garden District 120, 80110, 12050, 50There are two Nash Equilibria. When Gacey’s locates in Old Town, Mimbels will locate in the Garden District, and vice versa. Which solution will actually be chosen is not definite.This problem is different from the one in the chapter since there are 2 NashEquilibriums instead of one, which requires a little guesswork as to which one will be the final solution. It is similar in that there is not a dominant strategy equilibrium.Congestion must be more of a problem in this scenario than in the chapterproblem. There is never a Nash equilibrium when both pick the same site. This could be explained by the congestion problem3. Drive on. Two cars meet, crossing, at the intersection of Pigtown Pike and Hiccup Lane. Each has two strategies: wait or go. The payoffs are shown in Table E2.Table E2. The Drive On Game Mercedeswait go wait0,01,5Buick go 5,1-100,-100Discuss this game, from the point of view of noncooperative solutions. Does it have a dominant strategy equilibrium? Does it have Nash equilibria? What strategies, if so? Would you predict which strategies rational drivers would choose in this game?Which? Why? Pigtown Borough has decided to put a stoplight at this intersection. How could that make a difference in the game?Here is a table modified to show the maximum payout for each driver:Mercedes Wait Go Wait0, 05, 1B u i c kGo 5, 1-100,-100Once again, there are 2 Nash Equilibria. They are for the Buick to wait and the Mercedes go, or vice versa.To determine which will happen requires guesswork. The personality of thedrivers might determine what happens. If I were in the Mercedes, I would probably not want to risk an expensive car getting damaged. Someone else, say in a CL600, mightfigure that his car is faster and that he can beat the other driver. Also, one of the drivers might just wave the other on rather than have both wait or both go.It is possible that both drivers might wait rather than run the risk of an accident, i.e. choose a risk dominant strategy.The stoplight would provide a Schelling Point to select for the equilibrium at which the driver with the green light chooses go.4. Rock, Paper, Scissors. Here is another common school-yard game called Rock, Paper, Scissors. Two children (we will call them Susan and Tess) simultaneously choose a symbol for rock, paper or scissors. The rules for winning and losing are:Paper covers rock (paper wins over rock)Rock breaks scissors (rock wins over scissors)Scissors cut paper (scissors win over paper)The payoff table is shown as Table E3.Table E3. Rock, Paper, ScissorsSusanpaper stone scissors.paper0,01,-1-1,1Tessstone-1,10,01,-1scissors1,-1-1,10,0Discuss this game, from the point of view of noncooperative solutions. Does it have a dominant strategy equilibrium? Does it have Nash equilibria? What strategies, if so? How do you think the little girls will try to play the game?Here is a table modified to show the best responses.Susanpaper stone scissors.paper0,01,-1-1,1Tessstone-1,10,01,-1scissors1,-1-1,10,0We see that there are no dominant strategies, nor are there Nash equilibriain terms of the strategies shown here. We have no basis (so far) to decide how the girls will play the game.NOTE TO INSTRUCTOR For the purist, it is not correct to say here that “thereare no Nash equilibria,” since this game has a mixed-strategy equilibrium. But, of course, we will not cover mixed strategy equilibria until a later chapter. Thecorrect statement is that there is no equilibrium in pure strategies.5. The Great Escape. Refer to Chapter 2, Question 2.Discuss this game, from the point of view of noncooperative solutions. Does it have a dominant strategy equilibrium? Does it have Nash equilibrium? What strategies, if so? How can these two opponents each rationally choose a strategy?WardenGuard walls Inspect cellsclimb No escape, success inpreventing escape Escape,failurePrisonerdig Escape,failure No escape, success inpreventing escapeThe numerical payoffs can be assigned in many different ways. Here is a simple version that interprets “no escape” as minus one for the prisoner, plus one for the warden, and “escape” as vice versa. As the underlines show, there is no Nash equilibrium. Thus far, we have no basis to say how a rational person would choose strategies in this case.WardenGuard walls Inspect cellsclimb-1,11,-1Prisonerdig1,-1-1,16. Sibling Rivalry. Refer to Chapter 2, Question 1.Discuss this game, from the point of view of noncooperative solutions. Does it have a dominant strategy equilibrium? Determine all the Nash equilibria in this game. Do some Nash Equilibria seem likelier to occur than others? Why?Irismath litmath 3.7, 3.8 4.0, 4.0Julialit 3.8, 4.0 3.7, 4.0If the siblings act independently, rationally and with self- interest (non-cooperatively), we can find two Nash equilibrium's strategies: (literature, math), (math, literature).We note that there is a Schelling point in this game: (Math, Lit) yields a certain 4.0 for both girls, which is a reason it might attract attention, and probably is more likely to be observed.7. Hairsyle.Shaggmopp, Inc. and Shear Delight are hair-cutting salons in the same strip mall, each groping for a market niche. Each can choose one of three styles: punker, contemporary sophisticate, or traditional. Those are their strategies. They already have somewhat different images, based on the personalities of the proprietors, as the names may suggest. The payoff table is shown as Table E4.Table E4. Payoffs for HaircuttersShearpunker sophisticate traditionalpunker35,2050,4060,30Shaggmoppsophisticate30,4025,2535,55traditional20,4040,4520,20Are there any dominant strategies in this game? Is there a dominant strategy equilibrium? Are there any Nash equilibria? How many? Which? How do you know?Once again, here is the modified table:ShearPunkerSophisticate Traditional Punker 35, 2050, 4060, 30Sophisticate 30, 4025, 2535, 55S h a g g m o p pTraditional20, 4040, 4520, 20Shaggmopp’s best strategy is to go punker regardless of what Shear does. This is his dominant strategy. Since Shear has no such dominant strategy, there is no dominant strategy equilibrium.The only Nash equilibrium is when Shear decides to go with the sophisticate look.Since Shear knows that Shaggmopp will probably go punk rather than sophisticate, it will choose sophisticate.6. Quiz questionPlaced on the next page for convenience in copying and printing.Student name ____________________________Quiz – Game TheoryFelix and Oscarina share their home with two cats. Felix, who has a sharp sense of smell, would like for the cat boxes to be cleaned twice a week. Oscarina, whose sense of smell is less acute, would be satisfied if they were cleaned once a week. Each would prefer not to be the one to clean the cat boxes. Their payoffs are shown on the following table.Oscarinadon't clean clean once clean twicedon't clean-5,-30,-17,-5Felixclean once-2,45,26,-4clean twice0,51,32,-3Find any and all Nash equilibria for the catbox game? Are there dominated strategies? Which? Is there a dominant strategy equilibrium? Explain.Answer:A payoff table with best responses underlined follows:Oscarinadon't clean clean once clean twicedon't clean-5,-30,-17,-5Felixclean once-2,45,26,-4clean twice0,51,32,-3The Nash equilibrium is where Felix cleans the cat box twice and Oscarina never cleans. “Clean twice” is a dominated strategy for Oscarina. Since the best response for each person depends on the strategy chosen by the other, there is no dominant strategy equilibrium.It seems that Felix, whose need is greater, will empty the catbox, if the two companions act noncooperatively. Now, it may seem odd that people who live together would act noncooperatively , but life is strange, and odd things do happen. However, a couple of years ago, Oscarina gave Felix a Christmas present – a year of catbox cleaning – and has renewed the gift, so love triumphs after all.。
弹性力学双语讲义(Chapter4)
Only the circumferential displacement takes place只有环向位移
• the angle of rotation of PA will be =(AA-PP)/ PA =[(u+u/rdr)-u ]/dr= u/r • the angle of rotation of PB will be =<pop =- PP/OP=-u/r • rr =+=u/r-u/r
• the angle of rotation of PA will be =0 • the angle of rotation of PB will be =(BBPP)/PB=[(ur+ur/d)-ur]/rd=ur/(r) • rr =+=ur/(r)
9
13
rxy=u/y+v/x中的一般规律
• 由rxy=u/y+v/x，总结出一般规律，即设有 两个正交坐标方向，一个坐标方向的位移（如 u）对另一个坐标方向（y）求导为该坐标方向 （y）线段的转角。 1. u/y--x方向的位移 u 对y坐标求导为y方向线 段的转角。 2. v/x--y方向的位移 v 对x坐标求导为x方向线 段的转角。 • 应用这一规律于极坐标，就能方便地解释 ur/(r) +u/r为rr中的项。
4.2 geometrical and physical equations in polar coordinates极坐标中的几何物理方程
• P57(E) Fig. 4.2.1;P60(中)图4-2
7
Only the radial displacement takes place只有径向位移
12
x=u/x y=v/y中的规律
讨论黑格尔哲学中的一个问题：苦恼意识及其出路
讨论黑格尔哲学中的一个问题：苦恼意识及其出路内容提要本文是讨论黑格尔哲学中的一个问题：苦恼意识及其出路。
这个问题不仅仅是黑格尔哲学中的一个问题，也是我们今天所同样要面临的问题，所以，本文的目的不仅仅在于阐明黑格尔哲学的一个观点，也要将黑格尔哲学中的这一问题与现实联系起来，揭示出其现实意义。
第一章并没有直接地提出苦恼意识这个问题，而是讨论了“精神”与“伦理生活”这两个概念，并探讨了两者的关系。
之所以这样做，是因为本文认为苦恼意识的问题所在及其解决都涉及到精神与伦理生活的内在联系。
首先讨论的是“精神”，指出精神的本质是绝对对立的统一，是真正的自我意识。
然后讨论了“伦理生活”，指出伦理生活的目的在于实现精神。
第二章提出本文所要讨论的问题。
在这一章，分析了什么是苦恼意识，并指出现代社会中存在着苦恼意识，它表现为功利意识和道德意识，前者以有限性为其本质，后者以无限性为其本质，但由于知性的思维方式将有限与无限分立起来，所以两者都沦为苦恼意识。
同时，还分析了这两种苦恼意识对伦理生活的态度，前者将伦理生活的目的遮蔽起来，把伦理生活等同于市民社会，后者把则是逃避伦理生活，把伦理生活沦为“游行”或宗教小团体。
最后，指出苦恼意识的苦恼是无法掩盖的，以奥斯维辛为例，说明人们最终还是不得不直面“苦恼”。
第三者讨论了希腊的伦理生活方式以及希腊的精神特征。
但通过本章的论证，可以看出希腊的伦理生活方式的前提是，“自我意识”或主体性尚未出现。
在这样的前提下，希腊的伦理生活确实与希腊人的个体生命相契，希腊因此而成为精神的诞生之地。
但是，苦恼意识是自我崛起之后出现的问题，我们既不可能恢复希腊的伦理生活方式，而希腊的精神也不能满足苦恼意识的需要。
第四章讨论了基督教所提出的伦理生活原则其及蕴含的精神。
自我意识出现后，人与人的对立达到了绝对的程度，伦理生活受到了极大的破坏，要重建伦理生活，不能通过相互批评指责的方式进行，而只能通过相互间的宽恕与真诚的忏悔。
材料科学与工程基础英文版第五版课后习题
Homework 11.1 What are materials? List eight commonly encountered engineering materials. Answer1.1: Materials are substances of which something is composed or made. Steels, aluminum alloys, concrete, wood, glass, plastics, ceramics and electronic materials.1.2 What are the main classes of engineering materials?Answer1.2: Metallic, polymeric, ceramic, composite, and electronic materials are the five main classes.1.3 What are some of the important properties of each of the five main classes of engineering materials?Answer1.3:Metallic Materials• many are relatively strong and ductile at room temperature• some have good strength at high temperature• most have relatively high electrical and thermal conductivitiesPolymeric Materials• generally are poor electrical and thermal conductors• most have low to medium strengths• most have low densities• most are relatively easy to process into final shape• some are transparentCeramic Materials• generally have high ha rdness and are mechanically brittle• some have useful high temperature strength• most have poor electrical and thermal conductivitiesComposite Materials• have a wide range of strength from low to very high• some have very high strength-to-weight ratios (e.g. carbon-fiber epoxy materials)• some have medium strength and are able to be cast or formed into a variety of sha (e.g. fiberglass-polyester materials)• some have useable strengths at very low cos t (e.g. wood and concrete)Electronic Materials• able to detect, amplify and transmit electrical signals in a complex manner• are light weight, compact and energy efficient1.8 What are nanomaterials? What are some proposed advantages of using nanomaterials over their conventional counterparts?Answer1.8: Are defined as materials with a characteristic length scale smaller than 100 nm. The length scale could be particle diameter, grain size in a material, layer thicknessin a sensor, etc. These materials have properties different than that at bulk scale or at themolecular scale. These materials have often enhanced properties and characteristics because of their nano-features in comparison to their micro-featured counterparts. The structural, chemical, electronic, and thermal properties (among other characteristics) are often enhanced at the nano-scale.Homework 2Chapter 3, Problem 4What are the three most common metal crystal structures? List five metals that have each of these crystal structures. Chapter 3, Solution 4The three most common crystal structures found in metals are: body-centered cubic (BCC), face-centered cubic (FCC), and hexagonal close-packed (HCP). Examples of metals having these structures include the following. BCC:iron,α-vanadium, tungsten, niobium, and chromium.FCC: copper, aluminum, lead, nickel, and silver. HCP: magnesium, titanium,α-zinc, beryllium, and cadmium.Chapter 3, Problem 5For a BCC unit cell, (a) how many atoms are there inside the unit cell, (b) what is the coordination number for the atoms, (c) what is the relationship between the length of the side a of the BCC unit cell and the radius of its atoms, and (d) APF = 0.68 or 68%Chapter 3, Solution 5(a) A BCC crystal structure has two atoms in each unit cell. (b) A BCC crystal structure has a coordination number of eight . (c) In a BCC unit cell, one complete atom and two atom eighths toucheach other along the cube diagonal. This geometry translates into the relationship 4.R =Chapter 3, Problem 6For an FCC unit cell, (a) how many atoms are there inside the unit cell, (b) What is the coordination number for the atoms, (c) 24R a =, and (d) what is the atomic packing factor?Chapter 3, Solution 6(a) Each unit cell of the FCC crystal structure contains four atoms. (b) The FCC crystal structure has acoordination number of twelve . (d) By definition, the atomic packing factor is given as:volume of atoms in FCC unit cellAtomic packing factor volume of the FCC unit cell=These volumes, associated with the four-atom FCC unit cell, are33416433atoms V R R ππ⎡⎤==⎢⎥⎣⎦and 3unit cellV a =where a represents the lattice constant. Substitutinga =33unit cellV a ==The atomic packing factor then becomes,3316APF (FCC unit cell)3632R R ππ⎛⎫⎛⎫== ⎪ ⎪ ⎪⎝⎭⎝⎭=0.74 Chapter 3, Problem 7For an HCP unit cell (consider the primitive cell), (a) how many atoms are there inside the unit cell, (b) What is the coordination number for the atoms, (c) what is the atomic packing factor, (d) what is the ideal c/a ratio for HCP metals, and (e) repeat a through c considering the “larger” cell.Chapter 3, Solution 7The primitive cell has (a) two atoms/unit cell; (b) The coordination number associated with the HCP crystal structure is twelve . (c)the APF is 0.74 or 74%; (d) The ideal c/a ratio for HCP metals is 1.633; (e) all answers remain the same except for (a) where the new answer is 6.Homework 3 Chapter 3, Problem 25Lithium at 20︒C is BCC and has a lattice constant of 0.35092 nm. Calculate a value for the atomic radius of a lithium atom in nanometers.Chapter 3, Solution 25For the lithium BCC structure, which has a lattice constant of a = 0.35092 nm, the atomic radius is,R ===0.152 nmPalladium is FCC and has an atomic radius of 0.137 nm. Calculate a value for its lattice constant a in nanometers.Chapter 3, Solution 27Letting a represent the FCC unit cell edge length and R the palladium atomic radius,Chapter 3, Problem 31 Draw the following directions in a BCC unit cell and list the position coordinates of the atoms whose centers are intersected by the direction vector: (a ) [100] (b ) [110] (c ) [111]Chapter 3, Solution 31Chapter 3, Solution 324 or R a R ====0.387 nm(1, 0, 0)yxzyxz[111]x = +1y = -1 z = -1 x = +1 y = -1 z = 0(a) (b)[110]x = -½ y = 1(c)x = – ⅓ y = – ⅓(d)A cubic plane has the following axial intercepts: . What are the Miller indicesof this plane?Chapter 3, Solution 46Given the axial intercepts of (⅓, -⅔, ½), the reciprocal intercepts are:Multiplying by 2 to clear the fraction, the Miller indices areChapter 3, Problem 50Determine the Miller indices of the cubic crystal plane that intersects the following positioncoordinates:Chapter 3, Solution 50First locate the three position coordinates as shown. Next, connect points a and b and extend the line to point d . Complete the plane by connecting point d to c and point c to b . Using (1, 0, 1) as the plane origin, x = -1, y = 1 and z = –1. The intercept reciprocals are thusThe Miller indices are121332, , a b c ==-=11313,, 2.2x y z ==-=.(634)1122(, 0, ); (0,0,1); (1,1,1).1111,1, 1.x y z =-==-.(111)a(½, 0, ½ )。
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Chapter 4:Q2:Microsoft Excel Sensitivity ReportAdjustable CellsFinal Reduced Objective Allowable AllowableIncrease DecreaseCell Name Value Cost Coefficient12 0 2 1E+30 0$C$4 Value:X10 0 4 0 1E+30$D$4 Value:X2ConstraintsFinal Shadow Constraint Allowable AllowableCell Name Value Price R.H. Side Increase Decrease$E$10 Used: 12 0 2 10 1E+30$E$8 Used: -12 0 8 1E+30 20$E$9 Used: 12 2 12 1E+30 10a. Constraint 2 is binding.b. There is an alternate optimal solution. Variable X2 is at its lower bound (of zero) but also has areduced cost of zero. This indicates that the value of X2 could be increased while having zero impact on the optimal objective function value.c. There is no way to answer this question directly from the sensitivity report. We know that the optimalsolution would change since the allowable decrease in the objective function value of X1 is zero, but we cannot tell what the new optimal solution would be. If we re-solve the revised model the solution is X1=2, X2=5.d. It can decrease by any amount without changing the solution.e. Constraint 2 is the only binding constraint. Therefore, we would want to increase its RHS value beforeany other.Q4. a. MIN: 260X13 + 220X14 + 290X15 + 230X23 + 240X24 + 310X25S.T.:X13 + X14 + X15 ≤ 20X23 + X24 + X25 ≤ 20X13 + X23≥ 10X14 + X24≥ 15X15 + X25≥ 10X ij≥ 0b.c. See belowMicrosoft Excel Sensitivity ReportAdjustable CellsFinal ReducedObjective Allowable AllowableCell Name Value Cost CoefficientIncrease Decrease$C$1Eustis Miami 0 50 260 1E+30 50$D$1Eustis Orlando 10 0 220 20 0$E$1Eustis Tallahassee 10 0 290 0 310$C$11Clermont Miami 10 0 230 50 230$D$11Clermont Orlando 5 0 240 0 20$E$1 1 ClermontTallahassee0 0 310 1E+30 0ConstraintsFinal Shadow ConstraintAllowable AllowableCell Name Value Price R.H. Side Increase Decrease $C$12Shipped Miami 10 230 10 5 10$D$12Shipped Orlando 15 240 15 5 5$E$1 2 ShippedTallahassee10 310 10 5 5$F$1Eustis Used 20 -20 20 5 5$F$11Clermont Used 15 0 20 1E+30 5d. The optimal solution is not degerate as none of the RHS values of the contraints have anallowable increase or allowable decrease of zero.e. No. From the sensitivity report, it can be seen that the allowable increase or allowabledecrease for some of the objective function coeffients equal zero. This indicates that the alternative optimal solutions exist.TotalCostf.available at Clermont.g. Reducing the capacity in Eustis would increase costs by $20 per unit decrease yielding anobjective function value of $8,600+20×5 = $8,700h. Every additional ton of concentrate unit shipped from Eustis to Miami would increase costsby $50.Q6:Microsoft Excel Sensitivity ReportAdjustable CellsFinal Reduced Objective Allowable Allowable Cell Name Value Cost Coefficient Increase Decrease$C$4 Value: X1 0 3 5 1E+30 3$D$4 Value: X2 0 1 3 1E+30 1$E$4 Value: X3 1 0 4 2 4ConstraintsFinal Shadow Constraint Allowable Allowable Cell Name Value Price R.H. Side Increase Decrease$F$8 Produced: 2 2 2 1E+30 1$F$9 Produced: 2 0 1 1 1E+30a. 0b. The new objective would be unbounded.c. Increasing the RHS of the first constraint reduces the feasible region. Therefore, the objectivefunction would be increased (made worse) at a rate of 2 per unit increase in the RHS value. Thus, the new objective function value would be 4 + 2 × 5 = 14.d. 4 - 2×1 = 2e. Yes, 1/3 + 1/4 = 7/12 < 100%Q11:Microsoft Excel Sensitivity ReportAdjustable CellsFinal Reduced Objective Allowable Allowable Cell Name Value Cost Coefficient Increase Decrease $B$6 - Make Model 1 3,000 0.00 50 4 57 $C$6 - Make Model 2 550 0.00 83 14 8 $D$6 - Make Model 3 900 0.00 130 8 137 $B$7 - Buy Model 1 0 4.00 61 1E+30 4 $C$7 - Buy Model 2 1,450 0.00 97 8 14 $D$7 - Buy Model 3 0 8.00 145 1E+30 8ConstraintsFinal Shadow Constraint Allowable Allowable Cell Name Value Price R.H. Side Increase Decrease $B$13 # Available Model 1 3,000 57.00 3000 380 2900 $C$13 # Available Model 2 2,000 97.00 2000 1E+30 1450 $D$13 # Available Model 3 900 137.00 900 211.1111 900 $E$17 - Wiring Used 9,525 0.00 10000 1E+30 475 $E$18 - Harnessing Used 5,000 (7.00) 5000 633.3333 1100a. No.b. The cost of making model 1 slip rings can increase by $4 without changing the solution.c. Yes. The allowable decrease on the objective coefficient for making model 2 slip rings is $8.d. No. There is presently a surplus of 475 hours in the wiring department. Overtime would only add tothis surplus.e. Yes. Harnessing represents a binding constraint with a shadow price of -$7. Each additional unit ofthis resource (up to 633.33) will reduce costs by $7. Since workers are paid an additional $6 per hourfor overtime, the company could save $633.33.f. See spider chart in the file. Costs are most sensitive to reductions in the harnessing requirements formodel 1 slip rings.。