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 Present Worth A $30,000 B $12,500 C -$4,000 D $2,000

Solution:

(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:

PW X = -20,000 - 9000(P/A,12%,5) + 5000(P/F,12%,5) =-$49,606 PW Y = -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% First cost,$ Annual cost,$/yr Salvage value,$ Life, yrs Machine A 20,000 9000 4000 3 Machine B 30,000 7000 6000 6 Solution: (a) PW A = -20,000 – 9000(P/A,10%,6) – 16,000(P/F,10%,3) + 4000(P/F,10%,6) = -$68,961 PW B = -$30,000 – 7000(P/A,10%,6) + 6000(P/F,10%,6) = -$57,100 (b) FW A = -20,000(F/P,10%,6) – 9000(F/A,10%,6) – 16,000(F/P,10%,3) + 4000 = -$122,168 FW B = -30,000(F/P,10%,6) –7000(F/A,10%,6) +6000 = -$101,157 (both methods will always result in the same selection; in this case, machine B )

Refers to the present worth of an

infinite series

Basic equation is : P = A 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.

First cost,$ Annual cost,$/yr Salvage value,$ Life, yrs Machine A 20,000 9000 4000 3 Machine B 300,000 7000 ---- ∞ First convert machine A cash flow into AW and then divide by i: AW A = -20,000(A/P,10%,3) – 9000 + 4000(A/F,10%,3) = -$15,834 Cap Cost A = -15,834/ 0.10 = -$158,340 Cap Cost B = -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)  ( frequently requires trial and error solution) F(P/F,i,n)

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 b = bond interest rate/yr I = bond interest amt/ period c = no. of interest pmts/yr n = bond maturity date The PW of a bond is represented by the following diagram: V I PW=?

1 2 3 4 where I = (V)(b) c n

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)