Chapter 1 Q: Describe the concept of equivalence in a way that your brother in law can understand...
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Transcript of Chapter 1 Q: Describe the concept of equivalence in a way that your brother in law can understand...
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.
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) = -$98,758
R2R156,000147,000Initial Cost ,$
2,0005,000Salvage Value, $36Life, years
Alternatives
11,000 in year 1: increasing by $500 per
year
Annual Cost,$
30,000 in year 1: increasing by $1,000 per
year
Chapter 5Q: Compare the following machines on the basis of their
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
Chapter 5Q: Compare the following machines on the basis of their
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
Chapter 6Q: Compare the following machines on the basis of their
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
Chapter 6Q: Compare the following machines on the basis of their
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)= -$11,650
Best Alternative
Machine QMachine P42,00030,000Initial Cost ,$
11,0007,000Salvage Value, $1210Life, years
Alternatives
Annual Operating Cost year 1-4,$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
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
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