Engineering Economics Arithmetic

Engineering Economics
Name
Course
Date
1) The rate of return (ROR)
AWR20 = -61,000 (A/P, R%, 10) +8,800+ 600 (A/F, R%, 10)
61,000- (12,000+8,800) = 40,200
39800=61,000/ 40,200= 1.517413
(1-r 10)/r= 1.517413--≈ 0.65=65%
2) EUAC = Equivalent Uniform Annual Cost
The machine has an Annual operating cost 3,000 and increases by 4,200 per year
v= 1/(1+ cost of capital/ interest rate) 1/(1+10%)=0.9091
Discount factor= (1-0.909110)/0.1 6.1442
Discounted investment = initial cost/ 6.14428600/6.1442= $1,399.69
EUAC= 1,399.69+3,000+4200= 8,600
3) Economic worth
Interest rate – 10%
AW14=) -12,000 (A/P, 10%, 7) =-10,500+ 200(A/F, 10%, 7) =- 10,500+ 200(0.1054)= (10,479) -12,000 (A/P, 10%, 7)= (12,943.78)
AW27=-7,100 (A/P, 10%, 5)-11,700+ 500(A/F, 10%, 5) =-11,700+ 500(0.1638) = (11,618) -7,100 (A/P, 10%, 5)= (13,491.06)
AW40=-7,100 (A/P, 10%, 5)-6,200+400(A/F, 10%, 3) =6,200+ 400(0.3021)= (12,779)- -12,000 (A/P, 10%, 3)= (15,272.27)
Option 14, which costs 12,000 and has salvage value of 200 is the best option
Question 4) Discounted Payback Period
Discounted Payback Period
Year

CF

NCF

PV Factor

DCF

CCF

0

(58,000)

(58,000)

1.0000

(58,000.00)

(58,000.00)

1

(5,000)

(63,000)

0.9174

(4,587.16)

(62,587.16)

2

12,000

(51,000)

0.8417

10,100.16

(52,487.00)

3

40,000

(11,000)

0.7722

30,887.34

(21,599.66)

4

32,000

21,000

0.7084

22,669.61

1,069.95

5

(4,000)

17,000

0.6499

(2,599...