1. Mutually Exclusive Independent Are compared only against do-nothing More than one can be selected Are compared against each other Only one can be selected

2. 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

3. Convert all cash flows to PW using MARR Costs are preceded by minus sign; receipts plus For mutually exclusive, select numerically largest

4. 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

5. Must compare alts for equal service( i.e. alts must end at the same time ) Two ways to compare for equal service: (The LCM procedure is used unless otherwise specified) (1) Least common multiple(LCM) of lives (2) Specified planning period

6. Compare the machines shown below on the basis of their (a) present worth, and (b) future worth. Use i =10% Machine A Machine B First cost,$ Annual cost,$/yr Salvage value,$ Life, yrs 20,000 30,000 90007000 40006000 36 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)

7. Refers to the present worth of an infinite series Basic equation is : P = A i For finite life alternatives, convert all cash flow into an A value over one life cycle and then divide by i Ex: Cap cost of $2,000 per yr forever at i=10% is $20,000

8. Compare the machines shown below on the basis of their capitalized cost. Use i =10% per year. Machine A Machine B First cost,$ Annual cost,$/yr Salvage value,$ Life, yrs 20,000 300,000 90007000 4000 ----- 3∞ 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)

9. 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.

10. 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? (a) 0 = -175,000 + 25,000(n) + 90,000 Solution: 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

11. Bonds are IOU’s wherein entities get money now(V) and repay it later(n), with interest paid(I) in between. The PW of a bond is represented by the following diagram: where I = (V)(b) cPW=? I V 1234n Important bond information: b = bond interest rate/yr c = no. of interest pmts/yr n = bond maturity date I = bond interest amt/ period V = bond face value

12. Solution: I = (10,000)(0.06) (2) = $300 every six months P = 300(P/A,5%,40) + 10,000(P/F,5%,40) = $6568 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) $7500 Answer is (b)