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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.
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  • Chapter 1Q: Describe the concept of equivalence in a way that your brother in law can understand itSince 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 ProblemsSummer 2003Prepared by:Eng. Ahmed Taha

  • Chapter 1Q: 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

  • Chapter 1Starburst, 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 %

  • Chapter 1Q: 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 yearTotal interest = Pni = 1,000,000 (10)(0.06)= $600,000

  • Chapter 1Scenario b:F: the future value for the given yearP: the present value for the given year

    The difference between the a,b = 190,847.7 scenario b is better

    Sheet1

    YearPFInterest

    11000000106000060000

    21060000112360063600

    31123600119101667416

    411910161262476.9671460.96

    51262476.961338225.577675748.6176

    61338225.57761418519.11225680293.5346560001

    71418519.1122561503630.2589913685111.1467353601

    81503630.258991361593848.0745308490217.8155394816

    91593848.074530841689478.9590026995630.8844718507

    101689478.959002691790847.69654285101368.737540162

    Total790847.696542854

  • Chapter 1Q: 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:

  • Chapter 2How 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

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

    year

  • Chapter 2For 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=-800F= 4000-800(A/G,14%,4) = 4000-800(1.337) = $2930.4

    Year 1234Cash flow$4000320024001600

  • Chapter 2

    b: F= 4000(P/A,14%,4)-800(P/G,14%,4) = 4000(2.9137)-800(3.8957) = $853801423G=-800i = 3%year$4000$1600$2400$3200P=?

  • Chapter 2For 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

    Year 01234Cash flow0$200200+G200+2G200+3G

  • Chapter 2A 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 = 1179.252nd deposit = 1179.25/1.06 = 1112.5

  • Chapter 2You 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= 1.2i = 9.54%

  • Chapter 2You 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 %

  • Chapter 3Q: 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%

  • Chapter 3Q: what quarterly interest rate is equivalent to an effective annual rate of 6% per year compounded quarterly?i effective = (1+(r/m)m) 10.06 = (1+(r/4)4) 1(1+ 0.06 )1/4= 1+(r/4)R = 0.0587 per yearR = 1.468 % per quarter

  • Chapter 3Q: 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? SemiannuallyP = 50,000 (P/F, 6.5%,16) = 50,000 (0.3651) = 18,255Continuously i = er-1 i = e0.13-1= 13.88% per yearP = 50,000 (P/F,13.88%,16) = 50,000 (0.3535) = 17,675Difference 18,255 17,675 = $580

  • Chapter 3Q: 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 = $55,186Quarterly

  • Chapter 3Cont.b. i = 1%/Mo = (1+0.01)3-1= 3.03% /quarter750000 = A(P/A,3.03%,20)750000 = A(14.8363) A = $50,552

  • Chapter 3Q: 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

  • Chapter 4Q: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

    1

    Expense,$Income,$year

    900000

    200060001-5

    300060006-8

    500080009-14

    2

    3

  • Chapter 4Q: 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) + 3X0 = 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

  • Chapter 4Q: 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 AP=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) = $33,950A= 33950(A/P,10.25%,10)= 33950(0.1645) = $5,585

    1

    Disburcment,$Year

    3,5000

    3,5001

    3,5002

    3,5003

    5,0004

    5,0005

    5,0006

    5,0007

    5,0008

    5,0009

    15,00010

    2

    3

  • Chapter 4Q: 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

  • Chapter 4Q: 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)= $2,262.28F15= 2,262.28(F/A,12.68%,16) = 2,262.28(6.7538) = $15,279

  • Chapter 5Q: 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) = - $249,920PWy = -60000-13000(P/A,7.2%,10)+8000(P/F,7.2%,10) = -60000-13000(6.9591)+8000(0.4989) = -$146,480

    1

    Alternatives

    YX

    60,00040,000Initial Cost ,$

    05,000Monthly M&O Cost,$

    13,0000semiannual M&O Cost ,$

    8,00010,000Salvage Value, $

    55Life, years

    2

    3

  • Chapter 5Q: Compare the following machines on the basis of their present worth. Use i= 12% per year01 7$9,350$3,000$23,000Used Machine 14$9,350$3,000Year$23,00024691113$1,900$1,900$1,900$1,900$1,900$1,900PWnew= -44000-7210(P/A,12%,14)-2500(P/F,12%,5)-2500(P/F,12%,10) +4000(P/F,12%,14)= -$93,194PWused= -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

    1

    Alternatives

    Used MachineNew Machine

    23,00044,000Initial Cost ,$

    9,0007,000Annual Operating Cost,$

    350210Annual Repair Cost ,$

    1,9000Overhaul every 2 years,$

    02,500Overhaul every 5 years,$

    3,0004,000Salvage Value, $

    714Life, years

    2

    3

  • Chapter 5Q: 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) = $189,300PWR2= -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

    1

    Alternatives

    R2R1

    56,000147,000Initial Cost ,$

    30,000 in year 1: increasing by $1,000 per year11,000 in year 1: increasing by $500 per yearAnnual Cost,$

    2,0005,000Salvage Value, $

    36Life, years

    2

    3

  • Chapter 5Q: 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 = -$293,333i/yr = (1+0.16/12)12-1= 17.227%PW = -85,000-25,000/0.17227 PW = -$230,121

  • Chapter 5Q: 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= -$793,062Cap.cost Min = -900000-10000/0.113= -$988,496

    1

    Alternatives

    MINMAX

    900,000150,000Initial Cost ,$

    10,00050,000Annual Operating Cost,$

    1,000,0008,000Salvage Value, $

    Infinity5Life, years

    2

    3

  • Chapter 6Q: 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

  • Chapter 6Q: Compare the following machines on the basis of their annual worth . Using i= 12% per yearAWG = -62000 (A/P,12%,4)-15000+8000(A/F,12%,4)= -$33,738AWH = -77000 (A/P,12%,6)-21000+ 10000(A/F,12%,6)

    1

    Alternatives

    Machine HMachine G

    77,00062,000Initial Cost ,$

    21,00015,000Annual Operating Cost,$

    10,0008,000Salvage Value, $

    64Life, years

    2

    3

  • Chapter 6Q: Compare the following machines on the basis of their annual worth at i= 10% per yearAWP = -30000(A/P,10%,10)-15000+ [500(P/G,10%,6)(P/F,10%,4) (A/P,10%,10)]+7000(A/F,10%,10) = -$19,981AWQ = -42000(A/P,10%,12)-6000+ 11000(A/F,10%,12)= -$11,650

    1

    Alternatives

    Machine QMachine P

    42,00030,000Initial Cost ,$

    6,00015,000Annual Operating Cost year 1-4,$

    0500Annual Cost decrease year 5-n,$

    11,0007,000Salvage Value, $

    1210Life, years

    2

    3

  • Chapter 6Q: 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

  • Chapter 6Q: 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)= -$10,121AWH = -300000(0.08)-1000= -$25,000

    1

    Alternatives

    HG

    300,00040,000Initial Cost ,$

    1,0005,000-100(k-2)Annual Operating Cost,$

    50,0008,000Salvage Value, $

    Infinity10Life, years

    2

    3