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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.
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Practice ProblemsSummer 2003Prepared by:Eng. Ahmed Taha
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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
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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 %
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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
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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
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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:
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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
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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
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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
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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=?
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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
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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
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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%
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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 %
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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%
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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
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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
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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
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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
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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
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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
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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
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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
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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
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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
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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
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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
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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
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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
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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
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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
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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
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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
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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
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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