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Chapter 1

Chapter 1. Q: Describe the concept of equivalence in a way that your brother in law can understand it… Since money has time value, an amount today will be equal to a different amount tomorrow. These two different amounts can be related via interest rate. Practice Problems. Summer 2003.

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Chapter 1

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  1. Chapter 1 • Q: Describe the concept of equivalence in a way that your brother in law can understand it… • Since money has time value, an amount today will be equal to a different amount tomorrow. These two different amounts can be related via interest rate.

  2. Practice Problems Summer 2003 Prepared by: Eng. Ahmed Taha

  3. Chapter 1 • Q: which of the following has a better rate of return: $200 invested in 1 year with 6.25 paid in interest or $500 invested for 1 year with $18 paid in interest • Interest = present value (P) x interest rate (i) Scenario a: 6.25 = 200 i  i = 0.03125 = 3.125% Scenario b: 18 = 500 i  i = 0.036 = 3.6% • Scenario b is better because it has better rate of return

  4. Chapter 1 • Starburst, Inc. invested $50,000 in a foreign co-venture just one year ago and has reported a profit of $7,500. What is the annual rate that the investment is returning? • Rate of return   i = 15 %

  5. Chapter 1 • Q: Compute, graph, and compare the annual interest and total interest amount for 10 years on a million dollars under two different scenarios. First the $1 million is borrowed by a company at 6%-per-year simple interest. Second the $1 million is invested in a company at 6% per-year compounded annually. • Scenario a: Interest/y = 1,000,000 (0.06) = $60,000 per year Total interest = Pni = 1,000,000 (10)(0.06) = $600,000

  6. Chapter 1 • Scenario b: F: the future value for the given year P: the present value for the given year • The difference between the a,b = 190,847.7 •  scenario b is better

  7. F=? i = 8% year 0 1 2 3 4 5 6 A= $125 F=? i = 8% year 0 1 2 3 4 5 6 $350 $350 Chapter 1 • Q: Jaime wishes to invest at an 8%-per-year return so that 6 years from now. He can withdraw an amount of F in a lump sum. He has developed the following alternative plans. (a) Deposit $350 now and again 3 years from now. (b) Deposit $125 per year starting next year and ending in year 6 draw the cash-flow diagram for each plan if F is to be determined in year 6. • Scenario a: Scenario b:

  8. Chapter 2 • How much money would u have 12 years from now if u take your Christmas bonus of $2500 each year and (a) place it under your mattress. (b) put it in an interest- bearing checking account at 3% per year or (c) buy stock in a mutual fund which earns 16% per year? • Scenario a: i= 0% F= 2500 (12) = $30,000

  9. F=? i = 3% 0 1 $2500 12 F=? year i = 16% 0 1 12 $2500 Chapter 2 • Scenario b: i = 3% F= 2500(F/A,3%,12) = 2500(14.1920) = $35,480 • Scenario c: i = 16% F= 2500(F/A,16%,12) = 2500(30.85) = $77,125 year

  10. i = 3% G=-800 year 0 1 2 3 4 A=? $1600 $2400 $3200 $4000 Chapter 2 • For the cash flow shown below, calculate (a) the equivalent uniform annual worth in year 1 through year 4 and (b) the present worth in year 0. Assume i= 14% per year. • a: G=-800 F= 4000-800(A/G,14%,4) = 4000-800(1.337) = $2930.4

  11. Chapter 2 year i = 3% 0 1 2 3 4 G=-800 $1600 • b: F= 4000(P/A,14%,4)-800(P/G,14%,4) = 4000(2.9137)-800(3.8957) = $8538 P=? $2400 $3200 $4000

  12. i = 20% year 0 1 2 3 4 G=? A=800 $200 $200+G $200+2G $200+3G Chapter 2 • For the cash flow shown below determine the value of G that will make the equivalent annual worth = $800 at an interest rate of 20% per year. • i= 20% 800 = 200+G(A/G,20,4) 800 = 200+G(1.2742)  G = 470.88

  13. Chapter 2 • A company is planning to make s deposit such that each one 6% larger than the preceding one. How large must the second deposit be (at the end of year 2) if the deposit extended till year 15 and that fourth deposit is $1250? Use an interest rate of 10% per year. • 4th deposit= $1250 the third deposit will be less by 6% • 3rd deposit = 1250/1.06 = 1179.25 • 2nd deposit = 1179.25/1.06 = 1112.5

  14. i = ?% year 0 1 2 P=25 F=30 Chapter 2 • You have a just gotten a hot tip that you should buy stock in the GRQ company. The stock is selling for $25 per share. If u buy 500 shares and the stock increases to $30 per share in 2 years what rate of return would u realize on your investment. 25(F/P,i,2)= 30 (F/P,i,2) = 1.2 (1+i)2= 1.2 i = 9.54%

  15. i = ?% year 0 1 2 3 4 P=100000 $10000 $11000 $12000 $13000 Chapter 2 • You have just inherited $100,000 from your favorite uncle. His will stipulated that a certain bank will keep the money on deposit for you. His will also stipulated that you can withdraw $10,000 after 1 year, $11,000 after 2 years, and amounts increasing by $1000 per year until the amount is exhausted. If it takes 18 years for the inheritance to go to ZERO what interest rate was the money earning while on deposit? 100,000 = 10,000(P/A,i,18)+1000(P/G,i,18) Using trial and error i  13 %

  16. Chapter 3 Q: what is the nominal and the effective interest rates per year for an interest rate of 0.015% per day. a. Nominal i/year = 0.00015 (365) = 0.05475 = 5.48% b. Effective i= (1+i)n –1 i= (1+0.00015)365-1 i= 5.63%

  17. Chapter 3 • Q: what quarterly interest rate is equivalent to an effective annual rate of 6% per year compounded quarterly? • i effective = (1+(r/m)m) –1 • 0.06 = (1+(r/4)4) –1 • (1+ 0.06 )1/4= 1+(r/4) • R = 0.0587 per year • R = 1.468 % per quarter

  18. i = 13% year 1 2 P=? F=50000 Chapter 3 • Q: What is the difference in the present worth of $50,000 eight years from now if the interest rate is 13% per year compounded semiannually or continuously? • Semiannually • P = 50,000 (P/F, 6.5%,16) • = 50,000 (0.3651) = 18,255 • Continuously • i = er-1  i = e0.13-1= 13.88% per year • P = 50,000 (P/F,13.88%,16) • = 50,000 (0.3535) = 17,675 • Difference • 18,255 – 17,675 = $580

  19. year A=? i = 16% 0 1 5 P=750,000 Chapter 3 • Q: A jeans washing company is buying an ozone system for the washing machines and for the treatment of its dye wastewater. The initial cost of the ozone system is $750,000.how much money must the company save each quarter (in chemical costs) in order to justify the investment if the system will last 5 years and the interest rate is (a) 16% per year compounded quarterly . (b) 12% per year compounded monthly. • 750000 = A(P/A,4%,20) • 750000 = A(13.5903) •  A = $55,186 Quarterly

  20. Monthly year A=? i = 12% 0 1 5 P=750,000 Chapter 3 Cont. b. i = 1%/Mo = (1+0.01)3-1= 3.03% /quarter 750000 = A(P/A,3.03%,20) 750000 = A(14.8363)  A = $50,552

  21. F=18,000 year i = 6% A=? 0 1 5 Chapter 3 Q: a tool-and-die company expects to have one of its lathes replaced in 5 years at a cost of $18,000. How much would the company have to deposit every month in order to accumulate $18,000 in 5 years if the interest rate is 6% per year compounded semiannually? Assume no inter-period interest. 750000 = A(F/A,3%,10) 750000 = A(11.4639)  A = $1,570.15 per 6 month  A/month = 1,570.15/6 = $261.69

  22. 8000 6000 6 9 14 year 1 0 2000 3000 5000 9000 i = 8% Chapter 4 Q:what is the present worth of the following series of income and disbursement if the interest rate is a nominal 8% per year compounded semi annually? I/yr = (1+0.04)2-1 = 8.16% P = -9000+4000(P/A,8.16%,5) +3000(P/A,8.16%,9) (P/F,8.16%,5) Using the formulas = -9000+4000(3.9759)+ 3000(6.2055)(0.6756) = $19,481

  23. 3x $800 i = 14% $1200 1 2 $1000 5 8 9 year 0 10 3 4 6 7 5000 $500 x 2x Chapter 4 Q: find the value of X below such that the positive cash flows will be exactly equal to the negative cash flows if i= 14 % compounded semiannually i/yr = (1+0.07)2-1 = 14.49% Moving all cash flows to year 10 0 = 800 (F/P,14.49%,10)-500 (F/P,14.49%,8) +1000(F/A,14.49%,2)(F/P,14.49%,6)-X(F/P,14.49%,5)+ 1200(F/A,14.49%,2)(F/P,14.49%,3)- 2X(F/P,14.49%,2) + 3X 0 = 800(3.8697)-500(2.9522)+1000(2.1449)(2.2522)-X(1.9472) +1200(2.1449)(1.5007)-2X(1.3108)+3X 1.5888 X = 10313  X = $6,491

  24. 9 3 i = 10% 10 year 1 0 $3500 $5000 $15000 Chapter 4 Q: calculate the annual Worth (years 1 through 10) of the following series of disbursement. Assume i= 10% per year compounded semiannually. i/yr = (1+0.05)2-1= 10.25% Getting P then converting to A P=3500+3500(P/A,10.25%,3)+ 5000(P/A,10.25%,6)(P/F,10.25%,3)+ 15000(P/F,10.25%,10) = P= 3500+3500(2.4759)+5000(4.3235)(0.7462)+ 15000(0.3796) = $33,950 A= 33950(A/P,10.25%,10)= 33950(0.1645) = $5,585

  25. 9 12 $600 8 year 1 0 i = 12% G 2G 3G 4G Chapter 4 Q: find the value of G in the diagram below that would make the income stream equivalent to the disbursements stream, using an interest rate of 12% per year. 600(F/A,12%,8) = G(P/A,12%,3) + G(P/G,12%,4) 600(12.2997) = G(3.0373) + G(4.1273)  G = $1,030

  26. year 1 2 3 4 5 6 7 8 0 $340 360 380 400 420 440 460 480 500 Chapter 4 Q: for the diagram shown below, calculate the amount of money in year 15 that would be equivalent to the amounts shown, if the interest rate is 1% per month. i/yr = (1+0.01)12-1= 12.68% Getting P-1 then getting F15 P-1= 500(P/A,12.68%,9) -20 (P/G,12.68%,9) = 500(5.1933)-20(16.7183) = $2,262.28 F15= 2,262.28(F/A,12.68%,16) = 2,262.28(6.7538) = $15,279

  27. $10,000 Alternative X $8,000 month Alternative Y month 0 60 1 $5,000 0 60 1 $13,000 semiannually $40,000 $60,000 Best Alternative Chapter 5 Q: compare the alternatives below on the basis of their present worth using an interest rate of 14% per year compounded monthly. i/mo = 0.14/12= 0.01166 = 1.17% i/6mos = (1+ 0.01166)6-1= 7.2% PWx = -40000-5000(P/A,1.17%,60)+10000(P/F,1.17%,60) = -40000-5000(42.9787)+10000(0.4976) = - $249,920 PWy = -60000-13000(P/A,7.2%,10)+8000(P/F,7.2%,10) = -60000-13000(6.9591)+8000(0.4989) = -$146,480

  28. $4,000 New Machine Year 0 14 1 $7,210 5 10 $44,000 Best Alternative $2,500 $2,500 Chapter 5 Q: Compare the following machines on the basis of their present worth. Use i= 12% per year PWnew= -44000-7210(P/A,12%,14)-2500(P/F,12%,5)-2500(P/F,12%,10) +4000(P/F,12%,14)= -$93,194 PWused= -23000-9350(P/A,12%,14)-23000(P/F,12%,7)-1900[(P/F,12%,2) +(P/F,12%,4)+(P/F,12%,6)+(P/F,12%,9)+(P/F,12%,11)+(P/F,12%,13)] +3000(P/F,12%,7)+3000(P/F,12%,14) = -$98,758 $3,000 Used Machine $3,000 Year 0 7 1 $9,350 14 $9,350 2 4 6 $23,000 9 11 13 $1,900 $1,900 $1,900 $23,000 $1,900 $1,900 $1,900

  29. $5,000 R1 year 1 2 3 4 5 6 0 $2,000 $2,000 11,000 year year 4 1 5 2 6 3 0 R2 $147,000 G= $500 30,000 30,000 $13,500 $11,500 32,000 $56,000 $56,000 31,000 31,000 32,000 Best Alternative Chapter 5 Q: Compare the following machines on the basis of their present worth. Use i= 16% per year PWR1= -147,000-11000(P/A,16%,6)-500(P/G,16%,6)+5000(P/F,16%,6) = $189,300 PWR2= -56000-30000(P/A,16%,3)-1000(P/G,16%,3)-54000(P/F,16%,3)-[30000(P/A,16%,3)+1000(P/G,16%,3)] *(P/F,16%,3)+2000(P/F,16%,6) = -$203,644

  30. Chapter 5 • Q: Calculate the capitalized cost of $60,000 in year 0 and uniform beginning-of-year rent payments of $25,000 for an infinite time using an interest rate of (a) 12% per year (b) 16% per year compounded monthly. • PW = - (60,000 + 25,000) - 25000/0.12 •  PW = -$293,333 • i/yr = (1+0.16/12)12-1= 17.227% • PW = -85,000-25,000/0.17227 •  PW = -$230,121

  31. $8,000 $1,000,000 MAX MIN Year Year 0 5 0  1 $50,000 1 $10,000 $150,000 $900,000 Best Alternative Chapter 5 Q: Compare the following machines on the basis of their capitalized cost . Use i= 11% per year compounded semiannually. i/yr = (1+0.055)2-1= 11.3% Cap.cost max = [-150000(A/P,11.3%,5)-50000+8000(A/F.11.3%,5)]/0.113 = -$793,062 Cap.cost Min = -900000-10000/0.113 = -$988,496

  32. $11,000 Year 0 4 1 $2,000 $38,000 Chapter 6 Q: Find the annual worth amount (per month) of a truck which had a first cost of $38,000 an operating cost of $2000 per month and a salvage value of 11,000 after 4 years at an interest rate of 9% per year compounded monthly. i/mo = 0.09/12 = 0.75% AW = -38000 (A/P,0.75%,48)-2000+11000(A/F,0.75%,48) = -38000 (0.02489)-2000+11000(0.01739) = -$2,755

  33. $8,000 $10,000 Machine G Machine H Year Year 0 4 0 6 1 $15,000 1 $21,000 $62,000 $77,000 Best Alternative Chapter 6 Q: Compare the following machines on the basis of their annual worth . Using i= 12% per year AWG = -62000 (A/P,12%,4)-15000+8000(A/F,12%,4) = -$33,738 AWH = -77000 (A/P,12%,6)-21000+ 10000(A/F,12%,6)

  34. $7,000 $11,000 Machine P Machine Q Year Year 4 $6,000 0 10 0 12 1 1 G=$500 $50,000 $30,000 $42,000 Best Alternative Chapter 6 Q: Compare the following machines on the basis of their annual worth at i= 10% per year AWP = -30000(A/P,10%,10)-15000+ [500(P/G,10%,6)(P/F,10%,4) (A/P,10%,10)]+7000(A/F,10%,10) = -$19,981 AWQ = -42000(A/P,10%,12)-6000+ 11000(A/F,10%,12) = -$11,650

  35. Chapter 6 Q: What is the perpetual annual worth of $50,000 now and another $50,000 three years from now at an interest rate of 10% per year. AW = [50000+50000(P/F,10%,3)]*0.1 = [50000+50000(0.7513)]*0.1 = -$8,756.5

  36. $50,000 H $8,000 G Year 10 1 Year 0  1 $1,000 $4,200 $300,000 G=-$100 $5,100 $40,000 Best Alternative Chapter 6 Q: Compare the following machines on the basis of their annual worth . Using i= 8% per year (the index k varies from 1-10) AWG = -40000(A/P,8%,10)-5100+100(A/G,8%,10)+ 8000(A/F,8%,10) = -$10,121 AWH = -300000(0.08)-1000 = -$25,000

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