Transcript Document

Mutually Exclusive
Only one can be selected
Are compared against each other
Independent
More than one can be selected
Are compared only against do-nothing
For the alternatives shown below, which should be selected
if they are (a) Mutually exclusive, and (b) Independent
Project ID
A
B
C
D
Solution:
Present Worth
$30,000
$12,500
-$4,000
$2,000
(a) Select project A
(b) Select projects A, B, & D
Convert all cash flows to PW using MARR
Costs are preceded by minus sign; receipts plus
For mutually exclusive, select numerically largest
Alternative X has a first cost of $20,000, an operating cost of
$9,000 per year, and a $5,000 salvage value after 5 years.
Alternative Y will cost $35,000 with an operating cost of
$4,000 per year and a salvage value of $7,000 after 5 years.
At an MARR of 12% per year, which should be selected?
Solution:
PWX = -20,000 - 9000(P/A,12%,5) + 5000(P/F,12%,5)
=-$49,606
PWY = -35,000 - 4000(P/A,12%,5) + 7000(P/F,12%,5)
= -$45,447
Select alternative Y
Must compare alts for equal service(i.e. alts must end at the same time)
Two ways to compare for equal service:
(1) Least common multiple(LCM) of lives
(2) Specified planning period
(The LCM procedure is used unless otherwise specified)
Compare the machines shown below on the basis of
their (a) present worth, and (b) future worth. Use i =10%
Machine B
Machine A
30,000
First cost,$
20,000
Annual cost,$/yr
9000
7000
Salvage value,$
4000
6000
Life, yrs
3
6
Solution:
(a) PWA = -20,000 – 9000(P/A,10%,6) – 16,000(P/F,10%,3) + 4000(P/F,10%,6)
= -$68,961
PWB = -$30,000 – 7000(P/A,10%,6) + 6000(P/F,10%,6)
= -$57,100
(b) FWA = -20,000(F/P,10%,6) – 9000(F/A,10%,6) – 16,000(F/P,10%,3) + 4000
= -$122,168
FWB = -30,000(F/P,10%,6) –7000(F/A,10%,6) +6000
(both methods will always result in the same selection; in this case, machine B)
= -$101,157
Refers to the present worth of an infinite
Basic equation is :
series
A
P= i
Ex: Cap cost of $2,000 per yr forever at i=10% is $20,000
For finite life alternatives, convert all cash flow into
an A value over one life cycle and then divide by i
Compare the machines shown below on the basis of
their capitalized cost. Use i =10% per year.
Machine A
Machine B
First cost,$
20,000
300,000
Annual cost,$/yr
9000
7000
Salvage value,$
4000
----Life, yrs
3
∞
First convert machine A cash flow into AW and then divide by i:
AWA = -20,000(A/P,10%,3) – 9000 + 4000(A/F,10%,3)
= -$15,834
Cap CostA = -15,834/ 0.10 = -$158,340
Cap CostB = -300,000 – 7000/ 0.10 = -$370,000
(Select machine A)
Payback period refers to the time it takes(i.e. n) to recover
the initial investment cost (i.e. P) of an investment.
General equation is: 0 = -P  A(P/A,i,n)  F(P/F,i,n)
( frequently requires trial and error solution)
Business persons sometimes use simple payback (ignoring interest).
Such a procedure,while ‘simple’, obviously yields a lower n value
than the correct one.
Example: A racing team purchased a transporter for $175,000.They will be able
to sell the truck at any time within the next 5 years for $90,000, after which it will
sell for $70,000. If they expect to win an average of $25,000 more per year because
of the truck (i.e. being able to go to more races), how long will it take to recover
their investment at (a) i = 0%, and (b) i = 12% per year?
Solution:
(a) 0 = -175,000 + 25,000(n) + 90,000
n = 3.4 ,or 4 years
(b) 0 =- 175,000 +25,000(P/A,12%,n) + 90,000(P/F,12%,n)
for n≤5
0 =- 175,000 +25,000(P/A,12%,n) + 70,000(P/F,12%,n)
for n>5
By trial and error, n =12.6 , or 13 years
Bonds are IOU’s wherein entities get money now(V) and repay it later(n),
with interest paid(I) in between.
Important bond information:
V = bond face value
c = no. of interest pmts/yr
n = bond maturity date
b = bond interest rate/yr
I = bond interest amt/ period
The PW of a bond is represented by the following diagram:
V
I
1
PW=?
2
3
4
where I =
n
(V)(b)
c
A $10,000 bond with interest at 6% per year, payable semiannually is due in
20 years. The PW of the bond at 10% per year, comp’d semiannually is nearest:
(a) <$6000
(b) $6570
(c) $7120
(d) >$7500
Solution:
I=
(10,000)(0.06)
(2)
= $300 every six months
P = 300(P/A,5%,40) + 10,000(P/F,5%,40)
= $6568
Answer is (b)
Economic Analysis Problem
An economic analysis of four mutually exclusive revenue
projects yielded the results shown below. Which one(s) should
be selected?
A = - $10,000
B = -$7000
C = -$15000
D = -$8000
Answer: None
Capitalized Cost Problem
What is the capitalized cost of a machine with a first cost of
$200,000 and $10,000 annual cost at an interest rate of 10%?
CC = -200000 - 10000/.10 = $300,000
Answer: _Cap Cost = -$300,000 or $300,000