Find the NAC of the wooden bridge.
NAC = 25000(A/P, .1, 20) + 3000 = 25000*.1175 + 3000 = 2937.5+ 3000 = 5937.5
Find the initial cost of the steel bridge to equal this annual cost.
NAC = P(A/P, .1, 35) + 7000(A/F, .1, 35) = 5937.5
NAC = P*0.1037 + 7000*0.0037 = 5937.5
P = (5937.- 25.90)/0.1037 = 5911.10/0.1037 = 57001.9286
The bid can be for $57,000 and still remain competitive.
NAC A = 750(A/P, .08, 12) + 100 + 50(A/G, 0.08, 12) =
= 750*0.13270 + 100 + 50*4.5957 = 99.52500+ 100 + 229.7850 = 429.31000
NAC B = 1200(A/P, 0.08, 8) + 200 + 400(A/F, 0.08, 8) = 446.35
NAC C = [(2000 + 600(P/F, 0.08, 4)](A/P, 0.08, 8) = 424.77
Choose C.
For a 24 year study period.
NPC A = 4520, NPC B = 4699, NPC C = 4472,
Cement: NAC = 5000(A/P, 0.2, 20) + 500(A/F, .2, 5) + 500(A/F,.2,20)
= 5000*.2054 + 500*.1344 + 1000*0.0054 = 1027.+67.20 + 5.40 = 1099.60
Lawn Service: NAC = 1000(F/P, .2, 1) = 1000(1.2) = 1200
Son: NAC = 1500(A/P, .2, 3) + 200 = 1500*.4747 + 200 = 712.05 + 200 = 912.05.
Considering costs positive.
NAC M = 1000(A/P, 0.12, 5) + 3000 = 3277.4
NAC A = (4000 - 1000)(A/P, 0.12, 3) + 1000(0.12) + 1000 = 2369.
Buy automatic equipment.
5. Note that the 300 is the same for all alternatives. It could be left out. The NPW would be 300 less in each case. The NPW could be expressed using costs negative. The signs will then change. The result would be the same.
Option A. NPW Costs = 300 + 20(P/G,.1,7)(P/F,.1,3) = 300 + 20*12.763*0.7513 = 300+ 192=492 | |
Option B: NPW Costs = 300+ 40(P/A,.1,5) = 300+ 40*3.791 = 300+ 152=452 | |
Option C. Use a 10 year study period NPW Costs = 300 + 300(P/F, i, 5) = 300 + 300*0.6209 = 300 + 186 = 486 |
Selection: Select Machine B Justification: B has the smallest NPW of Costs