Carleton University
Department of Civil and Environmental Engineering
ECOR3800A
Engineering Economics
ASSIGNMENT 4 (SOLUTIONS)
Due Date: April 1, 2010
1.
These are mutually exclusive alternatives with unequal lives. The use of Annual Worth
Method (AW) is appropriate.
AW(Alt.1) = -$16,000(A/P,8%,4) - $3,000 + $11,000 + $0(A/F,8%,4) = $3,169.27
AW(Alt.2) = -$20,000(A/P,8%,8) - $5,000 + $11,500 + $4,000(A/F,8%,8) =
$3,395.76.
Alternative 2 is the choice.
2.
Let S be the Salvage of Alternative 1
Set 16,000(A/P,8%,4) + 8,500 + S(A/F,8%,4) = AW(2) found in question 1
Solve for S.
S = $1,021
Therefore S has to vary by $1,021 - $0 =$1,021 in order to reverse the choice in question
1.
3.
Profit = revenue
–
costs
Therefore, Z=[(3n/4)
–
(n
3
–
8n
2
+25n + 30)/25] = - [(4n
3
+ 32n
2
- 25n -120)/100]
(a) To find the break-even points, the profit equation is set equal to zero. The positive
roots of this equation yield break-even points of n = 3.26, and n = 6.23. This can be
shown graphically.
(b) To find the maximum profit, the profit equation is differentiated and set equal to zero.
dz/dn = - [(1/100)(12n
2
–
64n + 25)] = 0
The quadratic formula shows a minimum profit
at n = 0.425 and the desired maximum
profit at an output level of n = 4.91.