Transcript 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
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 coventure 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

7,500
i
 0.15
50,000

i = 15 %
profit
Investment
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
Chapter 1
Year
• Scenario b:
F: the future
value for the
given year
P: the present
value for the
given year
P
F
Interest
.
.
.
.
.
.
Total
• 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:
F=?
F=?
i = 8%
i = 8%
year
0
$350
1
2
3
$350
4
5
6
0
1
2
3
A= $125
year
4
5
6
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 interestbearing 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
F=?
year
i = 3%
0
1
$2500
12
F=?
year
i = 16%
0
1
$2500
12
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.
Year
1
Cash flow $4000
2
3200
• a: G=-800
F= 4000-800(A/G,14%,4) 0 1
= 4000-800(1.337)
A=?
= $2930.4
$4000
3
2400
4
1600
i = 3%
2 G=-800 3
$2400
$3200
4 year
$1600
Chapter 2
year
i = 3%
0
1
2
3
G=-800
P=?
$2400
$3200
$4000
• b:
F= 4000(P/A,14%,4)-800(P/G,14%,4)
= 4000(2.9137)-800(3.8957)
= $8538
4
$1600
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.
Year
0
Cash flow 0
1
2
3
4
$200
200+G 200+2G 200+3G
• i= 20%
i = 20%
800 = 200+G(A/G,20,4) 0 1
2 G=? 3
800 = 200+G(1.2742) A=800 $200
$200+G
 G = 470.88
$200+2G
year
4
$200+3G
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
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%
i = ?%
0 1
P=25
2 year
F=30
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)
i = ?%
Using trial and error
0
1
2
3
4 year
i  13 %
P=100000
$10000
$11000
$12000
$13000
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 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
i = 13%
P = 50,000 (P/F, 6.5%,16)
1
2 year
= 50,000 (0.3651) = 18,255
P=?
F=50000
• 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
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 = $55,186
0 1
P=750,000
Quarterly
A=?
i = 16%
year
5
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
P=750,000
Monthly
A=?
i = 12%
year
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 = $261.69
F=18,000
year
i = 6%
0
1
A=?
5
Chapter 4
Q:what is the present worth of the following series of income
and disbursement if the interest rate is a nominal 8% per
year compounded semi annually?
year
Income,$ Expense,$
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
8000
1
0
9000
6000
2000
6
9
3000
i = 8%
14
5000
year
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
3x
$800
i = 14%
1
2
0
$1000
3
$500
$1200
5
4
6
x
8
9
year
10
7
5000
2x
Chapter 4
Q: calculate the annual Worth (years 1 through 10) of the
following series of disbursement. Assume i= 10% per year
Year
Disburcment,$
compounded semiannually.
,
,
,
,
,
,
,
,
,
,
(1+0.05)2-1=
i/yr =
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
3
1
0
$3500
,
9 10
i = 10%
$5000
$15000
year
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
1
0
$600
8
9
12
i = 12%
G
2G
3G
4G
year
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)
0 1 2 3 4 5 6 7 8
= $2,262.28
F15= 2,262.28(F/A,12.68%,16)
$340
= 2,262.28(6.7538)
360
380
= $15,279
400
480
500
440
460
420
year
Chapter 5
Q: compare the alternatives below on the basis of their
present worth using an interest rate of 14% per year
compounded monthly.
Initial Cost ,$
Monthly M&O Cost,$
semiannual M&O Cost ,$
Salvage Value, $
Life, years
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
$40,000
Alternative X
$10,000
month
$5,000
60
Alternatives
X
Y
,
,
,
,
,
$8,000
month
Alternative Y
0 1
,
$13,000 semiannually
60
$60,000
Best Alternative
Chapter
5
Q: Compare the following machines on the basis of their
present worth. Use i= 12% per year
Alternatives
New Machine Used Machine
,
,
,
,
Initial Cost ,$
PWnew= -44000-7210(P/A,12%,14)Annual Operating Cost,$
2500(P/F,12%,5)-2500(P/F,12%,10) Annual Repair Cost ,$
Overhaul every years,$
+4000(P/F,12%,14)= -$93,194
Overhaul every years,$
,
Salvage Value, $
,
Life,
years
PWused= -23000-9350(P/A,12%,14)23000(P/F,12%,7)-1900[(P/F,12%,2)
New Machine
+(P/F,12%,4)+(P/F,12%,6)+(P/F,12%,
9)+(P/F,12%,11)+(P/F,12%,13)]
0 1
$7,210
+3000(P/F,12%,7)+3000(P/F,12%,14)
5
10
$44,000
= -$98,758
$2,500
$2,500
Used Machine
0 1
$23,000 2
4
$1,900 $1,900
$4,000
Year
14
$3,000
Year
6
$1,900
,
$3,000
7
$9,350
,
9
$23,000
$1,900
$9,350
11
$1,900
14
13
$1,900
Best Alternative
Chapter 5
Q: Compare the following machines on the basis of their
Alternatives
present worth. Use i= 16% per year
R
R
Initial Cost ,$
,
,
PWR1= -147,000-11000(P/A,16%,6),
in year
,
in year
: increasing
: increasing
Annual Cost,$
500(P/G,16%,6)+5000(P/F,16%,6)
by $
per
by $ ,
per
year
year
= $189,300
Salvage Value, $
,
,
Life, years
PWR2= -56000-30000(P/A,16%,3)1000(P/G,16%,3)-54000(P/F,16%,3)$2,000
[30000(P/A,16%,3)+1000(P/G,16%,3)] 0 1 2 3 year
R2
*(P/F,16%,3)+2000(P/F,16%,6)
30,000
$2,000
= -$203,644
32,000
$56,000
31,000
R1
0
1
2
3
4
$5,000
$11,500
6
year
30,000
4
5
6
year
11,000
$147,000
5
G= $500
$13,50
0
$56,000
31,000
32,000
Best Alternative
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.
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
Chapter 5
Q: Compare the following machines on the basis of their
capitalized cost . Use i= 11% per year compounded
Alternatives
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
MAX
0 1
$150,000
$50,000
Initial Cost ,$
Annual Operating Cost,$
Salvage Value, $
Life, years
$8,000
Year
5
MIN
0 1
$10,000
MAX
,
,
,
MIN
,
,
, ,
Infinity
$1,000,000
Year

$900,000
Best Alternative
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
$11,000
Year
0 1
$38,000
$2,000
4
Chapter 6
Q: Compare the following machines on the basis of their
Alternatives
annual worth . Using i= 12% per year
Machine G
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)
Machine G
0 1
$62,000
$15,000
Initial Cost ,$
Annual Operating Cost,$
Salvage Value, $
Life, years
$8,000
Year
4
Machine H
0 1
$77,000
$21,000
,
,
,
,
,
,
$10,000
Year
6
Best Alternative
Chapter 6
Q: Compare the following machines on the basis of their
Alternatives
annual worth at i= 10% per year
Machine P
Machine 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)
= -$19,981
AWQ = -42000(A/P,10%,12)-6000+
11000(A/F,10%,12)
= -$11,650
Machine P
Initial Cost ,$
Annual Operating
Cost year 1-4,$
Annual Cost
decrease year 5-n,$
Salvage Value, $
Life, years
$7,000
Machine Q
Year
0 1
$30,000
$50,000
4
10
30,000
42,000
15,000
6,000
500
0
7,000
10
11,000
12
$11,000
Year
0 1
$6,000
12
$42,000
Best Alternative
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
Chapter 6
Q: Compare the following machines on the basis of their
annual worth . Using i= 8% per year (the index k varies
Alternatives
from 1-10)
G
H
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
G
$40,000
$8,000
,
,
H
10
1
$5,100
Initial Cost ,$
Annual Operating Cost,$
Salvage Value, $
Life, years
Year
$4,200
G=-$100
0 1
$1,000
,
,
(k - )
,
,
Infinity
$50,000
Year

$300,000
Best Alternative