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 ProblemsSummer 2003

Prepared by:

Eng. Ahmed Taha

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

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 %

0.1550,000

7,500i

Investment

profiti

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 yearTotal interest = Pni = 1,000,000 (10)(0.06)

= $600,000

Chapter 1

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

Year P F Interest1 1000000 1060000 600002 1060000 1123600 636003 1123600 1191016 674164 1191016 1262477 71460.965 1262477 1338226 75748.626 1338226 1418519 80293.537 1418519 1503630 85111.158 1503630 1593848 90217.829 1593848 1689479 95630.8810 1689479 1790848 101368.7

Total 790847.7

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

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:

0 1 62 3 4 5

$350 $350

i = 8%

F=?

year0 1 62 3 4 5

A= $125

i = 8%

F=?

year

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

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

0 1 12$2500

i = 3%F=?

0 1 12$2500

i = 16%F=?

year

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

Year 1 2 3 4

Cash flow $4000 3200 2400 1600

0 1 42 3G=-800

i = 3%

year

$4000

$1600$2400

$3200

A=?

Chapter 2

• b:

F= 4000(P/A,14%,4)-800(P/G,14%,4)

= 4000(2.9137)-800(3.8957)

= $8538

0 1 42 3G=-800

i = 3% year

$4000

$1600$2400

$3200

P=?

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

Year 0 1 2 3 4

Cash flow 0 $200 200+G 200+2G 200+3G

0 1 42 3G=?

i = 20% year

$200

$200+3G$200+2G

$200+G

A=800

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 = 1179.252nd deposit = 1179.25/1.06 = 1112.5

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= 1.2i = 9.54%

0 1 2

i = ?%

F=30

year

P=25

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 % 0 1 42 3

i = ?%

year

$10000

$13000$12000$11000

P=100000

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%

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

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?

• 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

1 2

i = 13%

F=50000

year

P=?

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.

a. 750000 = A(P/A,4%,20)

750000 = A(13.5903)

A = $55,1860 1 5

A=?

i = 16%

P=750,000

year

Quarterly

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

0 1 5

A=?

i = 12%

P=750,000

year

Monthly

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

0 1 5A=?

i = 6%

F=18,000

year

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?

Expense,$Income,$year

900000

200060001-5

300060006-8

500080009-14

0

i = 8%

year 149

50009000

6

3000

12000

6000

8000

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

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) + 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

$800

0

i = 14%

year

1075000

4

$10001

$500

$1200

x

8

6

5

3

2 9

2x

3x

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 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

Disburcment,$Year 3,50003,50013,50023,50035,00045,00055,00065,00075,00085,000915,00010

0i = 10% year 109

$5000

31$3500

$15000

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

0i = 12%

year 1281 $600

G

4G3G2G

9

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)

= $2,262.28

F15= 2,262.28(F/A,12.68%,16)

= 2,262.28(6.7538)

= $15,279

500480

460440

420400

380360

$340

0 1 2 3 4 5 6 7 8 year

YX60,00040,000Initial Cost ,$

05,000Monthly M&O Cost,$13,0000semiannual M&O Cost ,$8,00010,000Salvage Value, $

55Life, years

Alternatives

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) = - $249,920

PWy = -60000-13000(P/A,7.2%,10)+8000(P/F,7.2%,10)

= -60000-13000(6.9591)+8000(0.4989) = -$146,480

0 1 60$5,000

$10,000month

$40,000

Alternative X

0 1 60$13,000 semiannually

$8,000month

$60,000

Alternative Y

Best Alternative

Best Alternative

Chapter 5Q: Compare the following machines on the basis of their

present worth. Use i= 12% per yearUsed 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

Alternatives

0 1 14$7,210

$4,000Year

$44,000

New Machine

$2,500 $2,5005 10

0 1 7$9,350

$3,000

$23,000

Used Machine

14$9,350

$3,000Year

$23,000

2 4 69 11 13$1,900 $1,900 $1,900

$1,900 $1,900 $1,900

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

R2R156,000147,000Initial Cost ,$


Alternatives

11,000 in year 1: increasing by $500 per

year

Annual Cost,$

30,000 in year 1: increasing by $1,000 per

year


present worth. Use i= 16% per year

$11,500 $13,500

G= $500

11,000

0 1 2 3 4 5 6 year

$147,000

$5,000R1

30,000

0 1 2 3year

$2,000

$56,000 31,000 32,000

30,000

4 5 6year

$2,000

$56,000 31,00032,000

R2

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

Best Alternative

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.

a. PW = - (60,000 + 25,000) - 25000/0.12

PW = -$293,333

b. i/yr = (1+0.16/12)12-1= 17.227%

PW = -85,000-25,000/0.17227

PW = -$230,121

MINMAX900,000150,000Initial Cost ,$10,00050,000Annual Operating Cost,$

1,000,0008,000Salvage Value, $Infinity5Life, years

Alternatives


capitalized cost . Use i= 11% per year compounded semiannually.

0 1 5$50,000

$8,000Year

$150,000

MAX

0 1 $10,000

$1,000,000Year

$900,000

MIN

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

Best Alternative

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.

0 1 4$2,000

$11,000Year

$38,000

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

Machine HMachine G77,00062,000Initial Cost ,$21,00015,000Annual Operating Cost,$10,0008,000Salvage Value, $

64Life, years

Alternatives


annual worth . Using i= 12% per year

0 1 4$15,000

$8,000Year

$62,000

Machine GYear

0 1 6$21,000

$10,000

$77,000

Machine H

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)

Best Alternative


annual worth at i= 10% per year

Year

$42,000

0 1 12$6,000

$11,000Machine Q

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)

= -$19,981AWQ = -42000(A/P,10%,12)-6000+

11000(A/F,10%,12)= -$11,650

Best Alternative

Machine QMachine P42,00030,000Initial Cost ,$


Alternatives

Annual Operating Cost year 1-4,$Annual Cost decrease year 5-n,$

15,000 6,000

500 0

0 1 104$50,000

$7,000

Year

$30,000

Machine P

G=$500

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

HG300,00040,000Initial Cost ,$1,0005,000-100(k -2)Annual Operating Cost,$

50,0008,000Salvage Value, $Infinity10Life, years

Alternatives


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

Best Alternative

1 10

G=-$100

$8,000

Year

$40,000

G

$5,100

$4,200 0 1 $1,000

$50,000Year

$300,000

H

Chapter 1

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