Fundamentals of Machine Elements 3rd_Ch08

Chapter 4Time Value of Money:Valuing Cash Flow StreamsNote: All problems in this chapter are available in MyFinanceLab. An asterisk (*) indicates problemswith a higher level of difficulty. Editor’s Note: As we move forward to more complex financial analysis, the student will notice that someproblems may contain a large amount of data from different time periods that require more complicated and intensive analysis. Modern information technology has evolved in the form of financial calculators with built-in analysis functions and Time Value ofMoney functions that are built into computer-based electronic spreadsheet software such as Excel. The solutions for many data- and computationally intensive problems will be presented in formula form with a solution, as well as with the appropriate financialcalculator commands and Excel functions to produce the same correct answer. In this way, it is our expectation that students will develop proficiency in solving financial analysis problems by mathematical calculation as well as by using financial calculators and electronic spreadsheets.1. a.Year 1 2 3 4 5 CF1020304050()()()()23451020304050106.531.10 1.10 1.10 1.10 1.10PV =++++= b.()()()()23455040302010120.921.10 1.10 1.10 1.10 1.10PV =++++= Year 1 2 3 4 5 CF5040302010Chapter 4 Time Value of Money: Valuing Cash Flow Streams 27c. The present value is different because the timing of the cash flows is different. In the secondset, you get the larger cash flows earlier, so it is more valuable to you and shows up as a higher PV.2.Year 1 2 3 4 CF100-100200-200()()()23410010020020028.491.15 1.15 1.15 1.15PV =-+-=3.Year 1 2 3 4 5 CF10000-2000-3000-3500-3975()()()2342,0003,0003,5003,97510,000.121.085 1.085 1.085 1.085PV =+++=, so yes, the PV of your payments would just cover the loan amount (actually exceeding it by 12 cents). Thus, the maximum you could borrow would be 10,000.124. Month1 2 3 4 5 6 7 8 CF500 550 600 650 700 750 800 850()()()()()()()23456785005506006507007508008505,023.751.015 1.015 1.015 1.015 1.015 1.015 1.015 1.015PV =+++++++=So, the PV of your payments will exceed the loan balance. This means that your planned payments will be enough to payoff your credit card.5.From the bank’s perspective, the timeline is the same except all the signs are reversed.28 Berk/DeMarzo/Harford •Fundamentals of Corporate Finance, Third Edition, Global Edition6.0 1 2 3 4 48–300 –300 –300 –300 –300From the bank’s perspective, the timeline would be identical except with opposite signs.7. Plan: Draw the timeline and then compute the FV of these two cash flows.Execute:Timeline (because we are computing the future value of the account, we will treat the cashflows as positive—going into the account):800 600FV = ?FV = 800(1.05) + 600 = 1,440Evaluate:The timeline helps us organize our work so that we get the number of periods of compoundingcorrect. The first cash flow will have 1 year of compounding, but the second cash flow will bedeposited at the end of period 1, so it receives no compounding.8. Editor’s Note: In several previous problems, we used a financial calculator to solve a timevalue of money problem. Problems could be solved quickly and easily by manipulating the N,I/Y, PV, PMT, and FV keys. In each of these problems, there was a series of payments of equal amount over time, i.e., an annuity. All you had to do to input this series was enter the payment(PMT) and the number of payment (N).Many financial analysis problems involve a series of equal payments, but others involve aseries of unequal payments. A financial calculator can be used to evaluate an unequal series ofcash flows (using the cash flow (CF) key), but the process is cumbersome because each cashflow must be entered individually. I urge each student to study the Chapter 4 Appendix: “Usinga Financial Calculator,” as well as instructional materials that are prod uced by themanufacturers of the financial calculator.Here we will solve a problem with uneven cash flows mathematically and with a financialcalculator.Plan: It is wonderful that you will receive this windfall from your investment in your friend’sbusiness. Because the cash flow payments to you are of different amounts and paid over threeyears, there are different ways in which you can think about how much money you arereceiving.Chapter 4 Time Value of Money: Valuing Cash Flow Streams 29Execute:a. 2310,00020,00030,000PV 1.035 1.035 1.0359,66218,67027,05855,390=++=++=The Texas Instrument BA II PLUS calculator has a cash flow worksheet accessed with the CF key. To clear all previous values that might be stored in the calculator, press the CF, second, and CE/C buttons. The screen should show CF 0 = asking for the cash flow at time 0, which in this problem is 0. Press 0, then press enter, and then the down key button. Thescreen should show CO1 asking for the cash flow at time 1, which in this problem is 10,000. Input 10,000 followed by the enter key, followed by the down button. The screen should show FO1 = 1.0 asking for the frequency of this cash flow. Because it occurs only once, it is correct, and we press the down key. The screen now has CO2 asking for the time 2 cash flow, which is 20,000, which we input, followed by the enter key and the down key. The screen now has FO2 = 1.0, which is correct. Enter the down key, which asks for the third cash flow, which is 30,000. Input 30,000, followed by the enter and down keys. Now press the NPV key, and the calculator will display I = asking for the interest rate, which is 3.5. Input 3.5, press the enter key, and press the down key, and the screen will display NPV =. Then press the CPT button, and the screen should display 55,390.33, the Net Present Value.b. 3FV 55,390 1.03561,412=⨯=Evaluate: You may ask: “How much better off am I because of this windfall?” There are several answers to this question. The value today (i.e., the present value) of the cash you will receive over three years is $55,390. If you decide to reinvest the cash flows as you receive them, then in three years you will have $61,412 (i.e., future value) from your windfall.9. Plan: Use Eq. 4.3 to compute the PV of this stream of cash flows and then use Eq. 4.1 to compute the FV of that present value. To answer part (c), you need to track the new deposit made each year along with the interest on the deposits already in the bank.Execute: a. and b.233100100100257.71(1.08)(1.08)(1.08)257.71(1.08)324.64PV FV =++===30 Berk/DeMarzo/Harford • Fundamentals of Corporate Finance, Third Edition, Global Editionc. ()11Year 1: 100Year 2: 1001.08100208Year 3: 208(1.08)100324.64+=+=Evaluate:By using the PV and FV tools, we are able to keep track of our balance as well as quicklycalculate the balance at the end. Whether we compute it step by step as in part (c) or directly as in part (b), the answer is the same.10. Plan: First, create a timeline to understand when the cash flows are occurring.Second, calculate the present value of the cash flows.Once you know the present value of the cash flows, compute the future value (of this present value) at date 3.Execute: 231,0001,0001,000PV 1.05 1.05 1.0595********,723=++=++=NI/Y PV PMT FV Excel FormulaGiven: 35.00% -1,000 0Solve for PV:2,723.25=PV(0.05,3,-1,000,0)33FV 2,723 1.053,152=⨯=NI/Y PV PMT FV Excel FormulaGiven: 35.00% -2,723.43Solve for FV:3,152.71=FV(0.05,3,0,-2,723.43)Evaluate: Because of the bank’s offer, you now hav e two choices as to how you will repay this loan. Either you will pay the bank $1,000 per year for the next three years as originallypromised, or you can decide to skip the three annual payments of $1,000 and pay $3,152 in year three.You now have the information to make your decision.11. a. The FV of 100,000 invested for 35 years at 9% is 100,000 × (1.09)35 = 2,041,397b. The FV of 100,000 invested for 25 years at 9% is 100,000 × (1.09)25 = 862,308Chapter 4 Time Value of Money: Valuing Cash Flow Streams 31c. The difference is so large because of the effect of compounding, which is exponential. Inthe scenario where you invest earlier, those 10 years are critical to the compound growthyou achieve on your investment.12. Plan: This scholarship is a perpetuity. The cash flow is $5,000 and the discount rate is 6%.Execute:Timeline:5,000 5,000 …PV 5,000/0.06 = 83,333.33Evaluate:With a donation of $83,333.33 today and 6% interest, the university can withdraw the interest every year ($5,000) and leave the endowment intact to generate the next year’s $5,000. It can keep doing this forever.13. P lan: This is a deferred perpetuity. Here is the timeline:Do this in two steps:1. Calculate the value of the perpetuity in year 9, when it will start in only one year(we already did this in problem 12).2. Discount that value back to the present.Execute:The value in year 9 is 5,000/0.06 = 83,333.33.The value today is 83,333.33/1.069 = 49,324.87Evaluate:Because your endowment will have 10 years to earn interest before making its first payment, you can endow the scholarship for much less. The value of your endowment must reach$83,333.33 the year before it starts (in 9 years). If you donate $49,324.87 today, it will grow at 6% interest for 9 years, just reaching $83,333.33, one year before the first payment.32 Berk/DeMarzo/Harford • Fundamentals of Corporate Finance, Third Edition, Global Edition14. The timeline for this investment is:a. The value of the bond is equal to the present value of the cash flows. By the perpetuity formula, which assumes the first payment is at period 1, the value of the bond is:PV = 1,000 /0.08 = £12,500 b. The value of the bond is equal to the present value of the cash flows. The first payment will be received at time zero. The cash flows are the perpetuity plus the payment that will be received immediately.PV = 1,000 /0.08 + 1,000 = £13,50015. Plan: Draw the timeline of the cash flows for the investment opportunity. Compute the NPV ofthe investment opportunity at 7% interest per year to determine its value.Execute: The cash flows are a 100-year annuity, so by the annuity formula:1001,0001PV 10.07 1.0714,269.25⎛⎫=- ⎪⎝⎭=NI/Y PV PMT FV Excel FormulaGiven: 1007.00% 1,000 0Solve for PV:(14,269.25)=PV(0.07,100,1000,0)Evaluate: The PV of $1,000 to be paid every year for 100 years discount to the present at 7% is $14,269.25.16. Plan: Prepare a timeline of your grandmother’s deposits.The deposits are an 18-year annuity. Use Eq. 4.6 to calculate the future value of the deposits.Execute: 1811(1)11,000(1.03)123,414.430.03N FV C r r =⨯+-=-=⎡⎤⎡⎤⎣⎦⎣⎦Chapter 4 Time Value of Money: Valuing Cash Flow Streams 33NI/Y PV PMT FV Excel FormulaGiven: 183.00% 0 1,000Solve for FV:(23,414.43)=FV(0.03,18,1000,0)At age 18, you will have $23,414.43 in your account.Evaluate:The interest on the deposits and interest on that interest adds more than $5,414 to the account.17. a.First, we need to calculate the PV of $160,000 in 18 years.18160,000PV (1.08)40,039.84==NI/Y PV PMT FV Excel FormulaGiven: 188.00% 0 160,000Solve for PV:(40,039.84)=PV(0.08,18,0,160000)In order for the parents to have $160,000 in your college account by your 18th birthday, the 18-year annuity must have a PV of $40,039.84. Solving for the annuity payments:1840,039.841110.08 1.08$4,272.33C =⎛⎫- ⎪⎝⎭=which must be saved each year to reach the goal.NI/Y PV PMT FV Excel FormulaGiven: 188.00% 40,039.84Solve for PMT:(4,272)=PMT(0.08,18,40039.84,0)b. First, we need to calculate the PV of $200,000 in 18 years.18200,000PV (1.08)50,049.81==34 Berk/DeMarzo/Harford • Fundamentals of Corporate Finance, Third Edition, Global EditionNI/Y PV PMT FV Excel FormulaGiven: 188.00% 0 200,000Solve for PV:(50,049.81)=PV(0.08,18,0,200000)In order for the parents to have $200,000 in your college account by your 18th birthday, the 18-year annuity must have a PV of $50,049.81. Solving for the annuity payments:18$50,049.811110.08 1.08$5340.42C =⎛⎫- ⎪⎝⎭=which must be saved each year to reach the goal.N I/Y PV PMT FV Excel FormulaGiven: 18.00 0.08 50,049.810.00Solve for PMT:-5340.42=PMT(0.08,18,50049.81,0)*18. Plan:a. Draw the timeline of the cash flows for the loan.1 2 3 4 5To pay off the loan, you must repay the remaining balance. The remaining balance is equal to the present value of the remaining payments. The remaining payments are a four-year annuity, so:b.4 5Execute:a. 45,0001PV 10.06 1.0617,325.53⎛⎫=- ⎪⎝⎭=N I/Y PV PMT FV Excel FormulaGiven: 4 6.00% 5000 0Solve for PV:(17,325.53)=PV(0.06,4,5000,0)Chapter 4 Time Value of Money: Valuing Cash Flow Streams 35b. 5,000PV 1.064,716.98==Evaluate: To pay off the loan after owning the vehicle for one year will require $17,325.53. To pay off the loan after owning the vehicle for four years will require $4,716.98.19. Plan: This is a deferred annuity. The cash flow timeline is:Calculate the value of the annuity in year 17, one period before it starts using Eq. 4.5 and then discount that value back to the present using Eq. 4.2.The value of the annuity in year 17, one period before it is to start is:41100,000111331,212.68(1)0.08(1.08)n CF PV r r ⎡⎤⎡⎤=-=-=⎢⎥⎢⎥+⎣⎦⎣⎦To get its value today, we need to discount that lump sum amount back 17 years to the present:17331,212.68$89,516.50(1.08)=Evaluate:Even though the cash flows are a little unusual (an annuity starting well into the future), we can still value them by combining the PV of annuity and PV of a FV tool. If we invest $89,516.50 today at an interest rate of 8%, it will grow to be enough to fund an annuity of $100,000 per year by the time it is needed for college expenses.20. Plan: This is a deferred annuity. The cash flow timeline is:Calculate the value of the annuity in year 44, one period before it starts using Eq. 4.5 and then discount that value back to the present using Eq. 4.2.Execute:The value of the annuity in year 44, one period before it is to start is:16140,000111377,865.94(1)0.07(1.07)n CF PV r r r ⎡⎤⎡⎤=-=-=⎢⎥⎢⎥+⎣⎦⎣⎦To get its value today, we need to discount that lump sum amount back 44 years to the present:44377,865.94$19,250.92(1.07)=Evaluate:Even though the cash flows are a little unusual (an annuity starting well into the future), we can still value them by combining the PV of annuity and PV of a FV tool. The total value to you today of Social Security’s promise is less than $20,000.*21. Plan: Clearly, Mr. Rodriguez’s contract is complex, calling for payments over many years.Assume that an appropriate discount rate for A-Rod to apply to the contract payments is 7% per year.a. Calculate the true promised payments under this contract, including the deferred payments with interest.b. Draw a timeline of all of the payments.c. Calculate the present value of the contract.d. Compare the present value of the contract to the quoted value of $252 million. What explains the difference?Execute: Determine the PV of each of the promised payments discounted to the present at 7%.2001 2002 2003 2004 2005 2006 2007 2008 2009 2010 $18M 19M19M19M21M19M23M27M27M27M20112012 2013 2014 2015 2016 2017 2018 2019 2020 6.7196M 5.3757M 4.0317M 4.0317M 4.0317M 4.0317M 4.0317M 4.0317M 4.0317M 4.0317MThe PV of the promised cash flows is $165.77 million.Evaluate: The PV of the contract is much less than $252 million. The $252 million value does not discount the future cash flows or adjust deferred payments for accrued interest. *22. a.0 1 2343The amount in the retirement account in 43 years would be:43435,000FV (1.10)10.10$2,962,003.46=-⎡⎤⎣⎦=NI/Y PV PMT FV Excel FormulaGiven: 4310.00%0.00 -5,000Solve for FV:2,962,003.46=FV(0.1,43,-5000,0)b. To solve for the lump sum amount today, find the PV of the $2,962,003.46.432,962,003.46PV (1.10)$49,169.99==N I/Y PV PMT FV Excel FormulaGiven: 43 10.00%0 2,962,003Solve for PV:(49,169.99) =PV(0.1,43,0,2962003.46) c.Solve for the annuity cash flow that, after 20 years, exactly equals the starting value of the account.202,962,003.461110.10 1.10347,915.81C =⎛⎫-⎪⎝⎭=N I/Y PV PMT FV Excel FormulaGiven: 20.00 0.10 2,962,003.460.00Solve for PMT:-347,915.81=PMT(0.1,20,2962003.46,0)d.We want to solve for N , which is the length of time in which the PV of annual payments of $300,000 will equal $2,962,003.46. Setting up the PV of an annuity formula and solving for N :300,000112,962,003.460.10 1.1012,962,003.460.1010.98733451.10300,000110.98733450.01266551.101.1078.95456Log(78.95456)45.84Log(1.10)N N NN N ⎛⎫-= ⎪⎝⎭⨯⎛⎫-== ⎪⎝⎭=-==== e. If we can only invest $1,000 per year, then set up the PV formula using $1 million as the FV and $1,000 as the annuity payment. 431,000111,000,000(1)r r ⎛⎫-= ⎪+⎝⎭To solve for r , we can either guess or use the annuity calculator. You can check and see that r = 11.74291% solves this equation. So the required rate of return must be 11.74291%.N I/Y PV PMT FV Excel FormulaGiven: 43 0.00 -1,000 1,000,000Solve for Rate: 11.74291%=RATE(43,-1000,0,1000000)23. Plan: The bequest is a perpetuity growing at a constant rate. The bequest is identical to a firmthat pays a dividend that grows forever at a constant rate. We can use the constant dividend growth model to determine the value of the bequest.Execute: a.Using the formula for the PV of a growing perpetuity gives1,000PV 0.120.0825,000⎛⎫= ⎪-⎝⎭=which is the value today of the bequest.b.1 2 3 4Using the formula for the PV of a growing perpetuity gives:1,000(1.08)PV 0.120.0827,000=-=which is the value of the bequest after the first payment is made.Evaluate: The bequest is worth $25,000 today and will be worth $27,000 in 1 year’s time. *24. Plan: The machine will produce a series of savings that are growing at a constant rate. The rateof growth is negative, but the constant growth model can still be used.Execute:The timeline for the saving would look as follows:We must value a growing perpetuity with a negative growth rate of – 0.03:2,000PV 0.08(0.03)$18,181.82=--= Evaluate: The value of the savings produced by the machine is worth $18,181.82 today.25. Plan: Nobel’s bequest is a perpetu ity. The total amount is 5 ⨯ $45,000 = $225,000. With a cashflow of $225,000 and an interest rate of 7% per year, we can use Eq. 4.4 to solve for the total amount he would need to use to endow the prizes. In part (b), we will need to use the formula for a growing perpetuity (Eq. 4.7) to find the new value he would need to leave. Finally, in part (c), we can use the FV equation (Eq. 4.1) to solve for the future value his descendants would have had if they had kept the money and invested it at 7% per year.a. In order to endow a perpetuity of $225,000 per year with a 7% interest rate per year, he would need to leave $225,000/0.07 = $3,214,285.71.b. In order to endow a growing perpetuity with an interest rate of 7% and a growth rate of 4% and an initial cash flow of $225,000, he would have to leave:1225,0007,500,0000.070.04CF PV r g ===-- c. FV = PV (1 + r )n = 7,500,000(1.07)118 = $ 21,996,168,112Evaluate:The prizes that bear Nobel’s name were very expensive to endow—$3 million was an enormous sum in 1896. However, Nobel’s endowment has been able to generate enough interest each year to fund the prizes, which now have a cash award of approximately $1,500,000 each!26. Plan: The drug will produce 17 years of cash flows that will grow at 5% annually. The value ofthis stream of cash flows today must be determined. We can use the formula for a growing annuity (Eq. 4.8) or Excel to solve this. C = 2, r = 0.10, g = 0.05, N = 17Execute: 171 1.052121.860.100.05 1.10PV ⎛⎫⎛⎫⎛⎫=-= ⎪ ⎪ ⎪ ⎪-⎝⎭⎝⎭⎝⎭Because the cash flows from this investment will continue for 17 years, we decided to solve for the Net Present Value by using the NPV function in Excel. This is shown on the next page. The 17 cash flows are presented in columns C, D,…S. The initial cash flow of $2M is presented in cell C8, and each subsequent cash flow grows at 5% until $4.365749M is presented in year 17 in cell S8. (Note that columns G through Q are not presented.) The NPV of the project is calculated using the NPV formula = NPV(C3,C8:S8) in cell B10. The NPV of the future cash flows is $21.86M.ABCDEFR S 1 2 1+g 1.05 3 r 0.1 4 5 6 T 0 1 2 3 4 16 17 7 8 2 2.1 2.205 2.31525 4.1578564.3657499 10 NPV $21.86 11 12 EXCEL NPV FORMULA =NPV(C3,C8:S8) 13Evaluate: The value today of the cash flows produced by the drug over the next 17 years is $21.86 million. Because the cash flows are expected to grow at a constant rate, we can use the growing annuity formula as a shortcut.27. Plan: Your rich aunt is promising you a series of cash flows over the next 20 years. You mustdetermine the value of those cash flows today. This is a growing annuity and we can use Eq.4.8 to solve it, or we can also solve it in Excel. C= 5, r= 0.03, g= 0.05 and N= 20Execute:201 1.035179.820.050.03 1.05PV⎛⎫⎛⎫⎛⎫=-=⎪⎪ ⎪⎪-⎝⎭⎝⎭⎝⎭Because the cash flows from this investment will continue for 20 years, we decided to solve for the Net Present Value by using the NPV function in Excel. This is shown on the next page. The 20 cash flows are presented in columns C, D,…V. The initial cash flow of $5,000 is presented in cell C8, and each subsequent cash flow grows at 3% until $8,767.53 is presented in year 20 in cell V8. (Note that columns G through R are not presented.) The NPV of the project is calculated using the NPV formula =NPV(C3,C8:V8) in cell B10. The NPV of the future cash flows is $79,824.A B C D E …S T U V12 1 + g 1.033 r 0.05456 T 0 1 2 3 17 18 19 2078 5 5.15 5.3045 8.023532 8.264238 8.512165 8.76753910 NPV $79.821112 EXCEL NPV FORMULA =NPV(C3,C8:V8)13Evaluate: Because your aunt will be increasing what she gives each year at a constant rate, we can use the growing perpetuity formula as a shortcut to value the stream of cash flows. Her gift is quite generous: It is equivalent to giving you almost $80,000 today!28. Plan: This problem is asking us to solve for the rate of return (r). Because there are norecurring payments, we can use Eq. 4.1 to represent the problem and then just solvealgebraically for r. We have FV = 200, PV = 100, n = 10.Execute:FV = PV(1 +r)n200 = 100(1 +r)10, so r = (200/100) 1/10– 1 = 0.072 or 7.2%Evaluate:The implicit return we earned on the savings bond was 7.2%. Our money doubled in 10 years, which by the rule of 72 meant that we earned about 72/10 = 7.2% and our calculationconfirmed that.29. Plan: This problem is again asking us to solve for r. We will represent the investment with Eq.4.1 and solve for r. We have PV = 2,000, FV = 10,000, n = 10. The second part of the problemasks us to change the rate of return going forward and calculate the FV in another 10 years.Execute:a. FV= PV(1 + r)n10,000 = 2,000(1 +r)10, so r = (10,000/2,000) 1/10– 1 = 0.1746, or 17.46%b.FV = 10,000(1.12)10= $31,058.4830. Plan: Draw a timeline and determine the IRR of the investment.Execute:–IRR is the r that solves:15,000 = 20,000/(1 +r), so r = (20,000/15,000) – 1 = 33.33%Evaluate: You are making a 33.33% IRR on this investment.31. Plan: Draw a timeline to demonstrate when the cash flows will occur. Then solve the problemto determine the payments you will receive.Execute:–P = C/r,So, C = P × r= 500 × 0.08= $40Evaluate: You will receive $40 per year into perpetuity.32. Plan: Draw a timeline to determine when the cash flows occur. Solve the problem to determinethe annual payments. Timeline (from the perspective of the bank):Execute:–203,000,000$240,727.761110.05 1.05==⎛⎫- ⎪⎝⎭Cwhich is the annual payment.NI/YPVPMTFVExcel FormulaGiven: 205.00% -3000000Solve for PMT:$240727.76=PMT(0.05,20,-3000000,0)Evaluate: You will have to pay the bank $240,727.76 per year for 20 years in mortgage payments.*33. Plan: Draw a timeline to demonstrate when the cash flows will occur. Determine the annualpayments.Execute:0 2 4 6 20–This cash flow stream is an annuity. First, calculate the two-year interest rate: The one-year rate is 4%, and $1 today will be worth (1.04)2 = 1.0816 in two years, so the two-year interest rate is 8.16%. Using the equation for an annuity payment:1050,0001110.0816(1.0816)$7,505.34C =⎛⎫- ⎪⎝⎭=which is the payment you must make every two years.N I/Y PV PMT FV Excel Formula Given: 10 8.16% -50,000.00Solve for PMT: $7,505.34=PMT(0.0816,10,-50000,0)Evaluate: You must pay the art dealer $7505.34 every two years for 20 years.*34. Plan: Draw a timeline to determine when the cash flows occur. Timeline (where X is the balloon payment):0 1 2330–+ X Note that the PV of the loan payments must be equal to the amount borrowed.Execute:303023,5001300,00010.07 1.07(1.07)X⎛⎫=-+ ⎪⎝⎭ Solving for X :303023,5001300,0001(1.07)0.07 1.07$63,848X ⎡⎛⎫⎤=-- ⎪⎢⎥⎣⎝⎭⎦= NI/Y PV PMT FV Excel Formula Given: 307.00%-23,500Solve for PV:291,612.47=PV(0.07,30,-23500,0)The present value of the annuity is $291,612.47, which is $8,387.53 less than the $300,000.00. To make up for this shortfall with a balloon payment in year 30 would require a payment of $63,848.02.N I/YPVPMT FVExcel FormulaGiven: 30 7.00% 8,387.53 0Solve for FV:(63,848.02) =FV(0.07,30,0,8387.53)Evaluate: At the end of 30 years you would have to make a $63,848 single (balloon) payment to the bank.*35. Plan: Draw a timeline to demonstrate when the cash flows occur. We know that you intend tofund your retirement with a series of annuity payments and the future value of that annuity is $2 million.22 23 24 25 65 0 1 2343CC CCCExecute: FV = $2 million.The PV of the cash flows must equal the PV of $2 million in 43 years. The cash flows consist of a 43-year annuity, plus the contribution today, so the PV is:()431PV 10.05 1.05C C ⎛⎫=-+ ⎪⎝⎭The PV of $2 million in 43 years is:432,000,000$245,408.80(1.05)=N I/Y PV PMT FV Excel Formula Given: 43 5.00% 0 2,000,000Solve for PV: (245,408.80)=PV(0.05,43,0,2000000)Setting these equal gives434311245,408.800.05(1.05)245,408.80$13,232.5011110.05(1.05)C C C ⎛⎫-+= ⎪⎝⎭⇒==⎛⎫-+ ⎪⎝⎭We need $245,408.80 today to have $2,000,000 in 43 years. If we do not have $245,408.80 today, but wish to make 44 equal payments (the first payment is today, making the payments an annuity due) then the relevant Excel command is:=PMT(rate,nper,pv,(fv),type =PMT(.05,44,245,408.80,0,1) = 13,232.50Type is set equal to 1 for an annuity due as opposed to an ordinary annuity.Evaluate: You would have to put aside $13,232.50 annually to have the $2 million you wish to have in retirement.36. Plan: This problem is asking you to solve for n . You can do this mathematically using logs, orwith a financial calculator or Excel. Because the problem happens to be asking how long it will。

Fundamentals of Machine ToolsIn many cases products from the primary forming processes must undergo further refinements in size and surface finish to meet their design specifications.To meet such precise tolerance the removal of small amounts of material is ually machine tools are used for such operation.In the United States material removal is a big business—in excess of ＄36×109 per year,including material,labor,overhead,and machine-tools shipments,is spent.Since 60 percent of the machanical and industrial engineering and technology graduate have something connection with the machining industry either through sale,design,or operation of machine shops,or working in related industry,it is wise for an engineering student to devote some time in his curriculum to studying material removal and machine tools.A machine tool provide the means for cutting tools to shape a workpiece to required dimensions;the machine supports the tool and the workpiece in a controlled relationship through the functioning of its basic members,which are as follows:(a)Bed,Structure or Frame.This is the main member which provides a basis for,and a connection between,the spindles and slides;the distorion and vibration under load must be kept to a minimum.(b)Slides and Slideways.The translation of a machine element(e.g. the slide) is normally achieved by straight-line motion under the constraint of accurate guiding surface(the slideways).(c)Spindles and Bearings.Angular displacement take place about an axis of rotation;the position of this axis must be constant within extremely fine limits in machine tools,and is ensured by the provision of precision spindles and bearings.(d)Power Unit.The electric motor is the universally adopted power unit for machine tools.By suitably positioning individual motors,belt and gear transmissions and reduced to a minimum.(e)Transmission Linkage.Linkage is the general term used to denote the mechanical,hydraulic,pneumatic or electric mechanisms which connect angular andlinear displacements in defined relationship.There are two broad divisions of machining operations:(a)Roughing,for which the metal removal rate,and consequently the cutting force,is high,but the required dimensional accuracy relatively low.(b)Finishing,for which the metal removal rate,and consequently the cutting force,is low,but the required dimensional accuracy and surface finish relatively high.It follows that static loads and dynamic loads,such as result from an unbalanced grindingwheel,are rmore significant in finishing operations than in roughing operations.The degree of precision achieved in any machining process will usually be influenced by the magnitude of the deflections,which occur as a result of the force acting.Machine tool frames are generally made in cast iron,although some may be steel casting or mild-steel fabrications.Cast iron is chosen because of its cheapness,rigidity,compressive strength and capacity for damping the vibrations set-up in machine operations.To avoid massive sections in castings,carefully designed systems of ribbing are used to offer the maximum resistance to bending and torsional stresses.Two basic types of ribbing are box and diagonal.The box formation is convenient to produce,apertures in walls permitting the positioning and extraction of cores.Diagonal ribbing provides greater torsional stiffness and yet permits swarf to fall between the sections;it is frequently used for lathe beds.The slides and slideways of a machine tool locate and guide members which move relative to each other,usually changing the position of the tool relative to the workpiece.The movement genenally takes the forms of translation in a straight line,but is sometimes angulai rotation,e.g. tilting the wheel-head of a universal thread-grinding machine to an angle corresponding with the helix angle of the workpiece thread.The basic geometric elements of slides are flat,vee,dovetail and cylinder.These elements may be used separately or combined in various ways according to the applications.Features of slideways are as follows:(a)Accuracy of Movement.Where a slide is to be displaced in a straight line,this line must lie in two mutually perpendicular planes and there must be no sliderotation.The general tolerance for straightness of machine tool slideways is 0—0.02mm per 1000mm;on horizontal surfaces this tolerance may be disposed so that a convex surface results,thus countering the effect of “sag”of the slideway.(b)Means of Adjustment.To facilitate assembly,maintain accuracy and eliminate “play” between slideing members after wear has taken place,a strip is something inserted in the slides.This is called a ually,the grib is retained by socket-head screw passing through elongated slots;and is adjusted by grub-screws secured by lock nuts.(c)Lubrication.Slideways may be lubricated by either of the following systems:1)Intermittently through grease or oil nipples,a method suitable where movements are infrequent and speed low.2)Continuously,e.g. by pumping through a metering value and pipe-work to the point of application;the film of oil introduced between surfaces by these means must be extremely thin to avoid the slide “floating”.If sliding surfaces were optically flat oil would be squeezed out,resulting in the surfaces sticking.Hence in practice slide surfaces are either ground using the edge of a cup wheel,or scraped.Both processes produce minute surface depresssions,which retain “pocket” of oil,and complete separation of the parts may not occur at all points;positive location of the slides is thus retained.(d)Protection.To maintain slideways in good order,the following conditions must be met:1)Ingress of foreign matter,e.g. swarf,must be prevented.Where this is no possible,it is desirable to have a form of slideway,which does not retain swarf,e.g. the inverted vee.2)Lubricating oil must be retained.The adhensive property of oil for use on vertical or inclined slide surface is important;oils are available which have been specially developed for this purpose.The adhesiveness of oil also prevents it being washed away by cutting fluids.3)Accidental damage must be prevented by protective guards.A machine tool performs three major functions:1)it rigidly supports the workpiece or its holder and the cutting tool; 2)it provides relative motion between the workpieceand the cutting tool; 3)it provides a range of feeds and speeds.Machines used to remove metal in the form of chips are classified in four general groups:those using single-point tools,those using multipoint tools,those using random-point tools(abrasive),and those that considered special.Machines using basically the single-point cutting tools include:1)engine lathes,2)turret lathes , 3)tracing and duplicating lathes, 4)single-spindle automatic lathes,5)multi-single automatic lathes , 6)shapers and planers, 7)boring machines.Machines using multipoint cutting tools include:1)drilling machines, 2)milling machines, 3)broaching machines, 4)sawing machines, 5)gear-cutting machines.Machines using random-point cutting tools include:1)cylindrical grinder,2)centreless grinders, 3)surface grinders.Special metal removal methods include:1)chemical milling, 2)electrical discharge machining, 3)ultrasonic machining. The lathe removes material by rotating the workpiece against a cutter to produce external or internal cylindrical or conical surfaces.It is also commonly used for the production of flat surfaces by faing,in which the workpiece is rotated while the cutting tool is moved perpendicularly to the axis of rotation.The engine lathe is the basic turning machine from which other turning machines have been developed.The drive motor is located in the base and drives the spindle through a combination of belts and gears,which provides the spindle speeds from 25 to 1500 rpm.The spindle is a sturdy hollow shaft,mounted between heavy-duty bearings,with the forward end used for mounting a drive plate to impart positive motion to the workpiece.The drive plate may be fastened to the spindle by threads,by a cam lock mechanism,or by a thread collar and key.The lathe bed is cast iron and provides accurately ground sliding surfaces(way)on which the carriage rides.The lathe carriage is a H-shaped casting on which the cutting tool is mounted in a tool holder.The apron hangs from the front of the carriage and contains the driving gears that move the tool and carriage along or across the way to provide the desired tool motion.A compound rest,located above the carriage provides for rotation of the tool holder through any desired angle.A hand wheel and feed screw are provided with a hand wheeland feed screw for moving the compound rest perpendicular to the lathe way.A gear train in the apron provides power feed for the carriage both along and across the way.The feed box contains gears to impart motion to the carriage and control the rate at which the tool moves relative to the workpiece.On a typical lathe feeds range from 0.002 to 0.160 in. per revolution of the spindle,in about 50 steps.Since the transmission in the feed box is driven from the spindle gears,the feeds are directly related to the spindle speed.The feed box gearing is also used in thread cutting and provides from 4 to 224 threads per in.The connecting shaft between the feed box and the lathe apron are the feed rod and the lead screw.Many lathe manufacturers combine these two rods in one,a practice that reduces the cost of the machine at the expense of accuracy.The feed rod is used to provide tool motion essential for accurate workpiece and good surface finishes.The lead screw is used to provide the accurate lead necessary for the thread cutting.The feed rod is driven through a friction clutch that allows slippage in case the tool is overloaded.This safety device is not provided in the lead screw,since thread cutting cannot tolerate slippage.Since the full depth of the thread is seldom cut in one pass,a chasing dial is provided to realign the tool for subsequent passes.The lathe tailtock is fitted with an accurate spindle that has a tapered hole for mounting drills,drill chucks,reamers,and lathe centers.The tailstock can be moved along the lathe ways to accommodate various lengths of workpieces as well as to advance a tool into contact with the worpiece.The tailstock can be offset relative to the lathe ways to cut tapers or conical surfaces.The turret lathe is basically an engine lathe with certain additional features to provide for semiautomatic operation and to reduce the opportunity for human error.The carriage of the turret lathe is provided with T-slots for mounting a tool-holding device on both sides of the lathe ways with tools properly set for cutting when rotated into position.The carriage is also equipped with automatic stops that control the tool travel and provide good reproduction of cuts.The tailstock of the turret lathe is of hexagonal design,in which six tools can be mounted.Althogh a large amount of time is consumed in setting up the tools and stops for operation,the turret lathe,once set,can continue toduplicate operations with a minimum of operator skill until the tools become dulled and need replacing.Thus,the turret lathe is economically feasible only for production work,where the amount of time necessary to prepare the machine for operation is justifiable in terms of the number of part to be made.Tracing and duplicating lathes are equipped with a duplicating device to automatically control the longitudinal and cross feed motions of the single-point cutting tool and provide a finished part of required shape and size in one or two passes of the tools.The single-spindle automatic lathe uses a vertical turret as well as two cross slids.The work is fed through the machine spindle into the chuck,and the tools are operated automatically by cams.The multispindle automatic lathe is provided with four,five,six,or eight spindles,with one workpiece mounted in each spindle.The spindles index around a central shaft,with the main tools slide accessible to all spindles.Each spindle position is provided with a side tool-slide operated independently.Since all of the slides are operated by cams,the preparation of this machine may take several days,and a production run of at least 5000 parts is needed to justify its use.The principal advantage of this machine is that all tools work simultaneously,and one operator can handle several machines.For relatively simple parts,multispindle automatic lathes can turn out finished products at the rate of 1 every 5 sec.A shapers utilizes a single-point tool in a tool holder mounted on the end of the ram.Cutting is generally done on the forward stroke.The tool is lifted slightly by the clapper box to prevent excessive drag across the work,which is fed under the tool during the return stroke in preparation for the next cut.The column house the operating mechanisms of the shaper and also serves as a mounting unit for the work-supporting table.The table can be moved in two directions mutually perpendicular to the ram.The tool slide is used to control the depth of the cut and is manually fed.It can be rotated through 90 deg. On either side of its normal vertical position,which allows feeding the tool at an angle to the surface of the table.Two types of the driving mechanisms for shapers are a modified Whitworth quick-return mechanism and a hydraulic drive.For the Whitworth mechanism,the motor drives the bull gear,which drives a crank arm with an adjustable crank pin to control the length of the stroke.As the bull gear rotates,the rocker arm is forced to reciprocate,imparting this motion to the shaper ram.The motor on a hydraulic shaper is used only to drive the hydraulic pump.The remainder of the shaper motions are controlled by the direction of the flow of the hydraulic oil.The cutting stroke of the mechanically driven shaper uses 220 deg. of rotation of the bull gear,while the return stroke uses 140 deg..This gives a cutting stroke to return stroke ratio of 1.6 to 1.The velocity diagram shows that the velocity of the tool during the cutting stroke is never constant,while the velocity diagram for a hydraulic shaper shows that for most of the cutting stroke the cutting speed is constant.The hydraulic shaper has an added advantage of infinitely variable cutting speeds.The principal disadvantage of this type of machine is the lack of a definite limit at the end of the ram stroke,which may allow a few thousandths of an inch variation in stroke length.A duplicating device that makes possible the reproduction of contours from a sheet-metal template is available.The sheet metal template is used in conjunction with hydraulic control.Upright drilling machines or drill presses are available in a variety of sizes and types,and are equipped with a sufficient range of spindle speeds and automatic feeds to fit the needs of most industries.Speed ranges on a typical machine are from 76 to 2025 rpm.,with drill feed from 0.002 to 0.20 in. per revolution of the spindle.Radial drilling machines are used to drill workpieces that are too large or cumbersome to conveniently move.The spindle with the speed and feed changing mechanism is mounted on the radial arm;by combing the movement of the radial arm around column and the movement of the spindle assembly along the arm,it is possible to align the spindle and the drill to any position within reach of the machine.For work that is too large to conveniently support on the base,the spindle assembly can be swung out over the floor and the workpiece set on the floor beside the machine.Plain radial drilling machine provide only for vertical movement of thespindle;universal machines allow the spindle to swive about an axis normal to the radial arm and the radial arm to rotate about a horizontal axis,thus permitting drilling at any angle.A mutispindle drilling machine has one or more heads that drive the spindles through universal joints and telescoping splined shafts.All spindles are usually driven by the same motor and fed simultaneously to drill the desired number of holes.In most machines each spindle is held in an adjustable plate so that it can be moved relative to the others.The area covered by adjacent spindles overlap so that the machine can be set to drill holes at any location within its range.The milling operation involves metal removal with a rotating cutter.It includes removal of metal from the surface of a workspiece,enlarging holes,and form cutting,such as threads and gear teeth.Within an knee and column type of milling machine the column is the main supporting member for the other components,and includes the base cotaining the drive motor,the spindle,and the cutter.The cutter is mounted on an arbor held in the spindle,and supported on its outer extremity by a bearing in the overarm.The knee is held on the column in dovetail slots,the saddle is fastened to the knee in dovetail slots,and the table is attached to the saddle.Thus,the build-up of the knee and column machine provide three motions relative to the cutter.A four motion may be provided by swiveling the table around a vertical axis provided on the saddle.Fixed-bed milling machines are designed to provide more rigidity than the knee and column type.The table is mounted directly on the machine base,which provides the rigidity necessary for absorbing heavy cutting load,and allows only longitudinal motion to the table.Vertical motion is obtained by moving the entire cutting head.Tracer milling is characterized by coordinated or synchronized movements of either the paths of the cutter and tracing elements,or the paths of the workpiece and model.In a typical tracer mill the tracing finger follow the shape of the master pattern,and the cutter heads duplicate the tracer motion.。

材料科学与工程专业培养计划Undergraduate Program for Specialty inMaterials Science & Engineering一、业务培养目标1、Educational Objectives本专业培养具备材料科学、材料工程方面较宽的基础知识，能在各种材料的制备、加工成型、材料结构与性能、材料应用等领域从事科学研究与教学、技术开发、工艺和设备设计、技术改造及经营治理等方面工作，适应社会主义市场经济进展的高层次、高素养、全面进展的科学研究与工程技术人才。

Materials Science & Engineering (MSE) is a major that is concerned with the structure and properties of all engineering materials. After graduation, MSE students have wide basic theory in materials science and technology, and can become materials research workers or materials engineers who are responsible for designing, producing, examining and testing diverse materials. Students graduated with honor, can also do some research, education, development and management work related to materials designing and producing, materials processing, manufacturing and relative areas etc.二、业务培养要求2、Educational Requirement本专业学生要紧学习材料科学与工程的基础理论，把握材料的制备、组成、组织结构与性能之间关系的差不多规律，同意各种材料的制备、性能分析与检测技能的差不多训练，把握材料设计和制备工艺设计，使学生具有开发新材料、研究新工艺、提高和改善材料性能和提高产品质量的差不多能力。

生态系统论的基本观点简介生态系统论是现代生态学的核心理论之一，它主要研究生态系统内部的各种相互作用和调节机制，以及生态系统与外部环境的相互关系。

生态系统论通过研究物种间的相互依赖、能量流动和物质循环等基本过程，揭示了生态系统的结构和功能，为生态学研究提供了基本的理论基础。

生态系统论的基本观点生态系统论的基本观点包括以下几个方面：1. 生态系统是一个整体生态系统不仅仅包括各种生物（如植物、动物等），还包括非生物组成部分（如土壤、水、空气等）。

生物组成部分与非生物组成部分相互作用，共同构成一个复杂的整体。

生态系统中的各个组成部分之间存在着相互联系和相互作用。

2. 生态系统是动态的生态系统是一个处于不断变化和发展中的动态系统。

它受到内部和外部因素的影响，通过自身的调节和适应机制维持着相对稳定的状态。

生态系统的动态性表现在各种物质和能量的流动、循环和转化过程中。

3. 生态系统具有自我组织和自我调节能力生态系统对于变化有着自我组织和自我调节的能力。

在外界环境发生变化时，生态系统可以通过内部的调节机制，使系统重新达到一个相对稳定的状态。

这种自我调节能力是生态系统适应环境变化的关键。

4. 生态系统具有层次结构生态系统包括多个层次的组织结构，从个体、种群、群落到生物圈，呈现出明显的层次结构和组织关系。

不同层次的组织结构之间存在着复杂的相互作用和相互制约，构成了生态系统的整体特征。

5. 生态系统具有物质循环和能量流动生态系统中的物质循环和能量流动是维持生态系统正常运转的重要基础。

物质循环主要包括碳、氮、磷等元素在生态系统内的循环和转化过程；能量流动主要指以太阳能为基础的能量流动和转化过程。

物质循环和能量流动是生态系统内部各种生物和非生物组成部分之间相互联系的纽带。

6. 生态系统具有稳定性和脆弱性生态系统具有一定的稳定性，可以在一定范围内保持相对稳定的状态。

但是，生态系统也具有一定的脆弱性，当受到外界的强烈干扰或破坏时，可能导致系统无法恢复或进入到新的稳定状态。

热能与动力工程专业本科培养计划Undergraduate Program for Specialty inThermal Energy and Power Engineering一、业务培养目标1、Educational Objectives本专业培养具备热能与动力工程等方面。

基础知识和应用能力，能从事动力机械与动力工程的设计、制造、试验研究、开发、管理等方面工作，具有创新精神与实践能力的高级工程技术人才。

This program aims at cultivating the senior engineering and technical talents who have the basic acknowledge and application ability in thermal energy and power engineering, are capable of designing, manufacturing, experimental study, developing, managing etc in power machinery and engineering, and possess the spirit of innovation and practical ability.二、业务培养要求2、Educational Requirement本专业学生主要学习动力工程热物理的基础理论知识，学习能量转换及有效利用的理论和技术，受到现代动力工程师的基本训练；具有进行动力机械与热工设备设计、运行、实验研究的基本能力。

毕业生应获得以下几方面的知识和能力:1. 掌握本专业的技术理论知识，主要包括工程力学、机械设计基础、工程热物理、流体力学、电工与电子学、控制理论及企业管理等基础知识。

2.具有热能与动力工程的专业知识，获得相关的工程实践训练，具备开展动力机械与热工设备设计、运行、实验研究的基本能力；3.具有一定的计算机和外语应用能力；4.了解该学科的前沿知识及其发展趋势；5. 具有较强的自学能力，创新意识和较高的综合素质。