Four proposals are available for funding The investment cost

Four proposals are available for funding. The investment costs and income associated with each are shown. Consider operating cost as a disbenefit P2 P3 30000 19600 P4 45000 P1 15000 -- 2,000 15000 | 18000 | 11000-t 15000 12000 First cost,$ 9000 Operating cost, $peryear Salvage value, S poryear 9000 11000 30000 8000 4000 Expected Benefits, spor years! 15000-18000| 21000128000 ll proposals have a 10-year life and the MARR is 10% per year. 21000 Assuming that a Determine which one should be selected using: posals have a (a) B/C Method (b) Incremental B/C Method

Solution

a) B/C of P1= PW of Expected benefits/(First cost+PW of operating cost)

= 15000(P/A,10%,10)/ (15000+9000(P/A,10%,10)

= (15000*6.145)/ (15000+(9000*6.145))

= 1.311

B/C of P2= PW of Expected benefits/(First cost+PW of operating cost)

= 18000(P/A,10%,10)/ (20000+11000(P/A,10%,10)

=(18000*6.145)/ (20000+(11000*6.145))

= 1.35

B/C of P3= PW of Expected benefits/(First cost+PW of operating cost)

= 21000(P/A,10%,10)/ (30000+11000(P/A,10%,10)

=(21000*6.145)/ (30000+(11000*6.145))

= 1.32

B/C of P4= PW of Expected benefits/(First cost+PW of operating cost)

= 28000(P/A,10%,10)/ (45000+18000(P/A,10%,10)

=(28000*6.145)/ (45000+(18000*6.145))

= 1.105

Since B/C for P2 is highest we will select P2 proposal.

b)for Incremental B/C we will compare the other proposals with P4 since the expected benefits are highest

Incremental B/C for P1= 13000(P/A,10%,10)/ (30000+9000(P/A,10%,10)

=(13000*6.145)/ (30000+(9000*6.145))

= 0.93

Incremental B/C for P2= 10000(P/A,10%,10)/ (25000+7000(P/A,10%,10)

=(10000*6.145)/ (25000+(7000*6.145))

= 0.903

Incremental B/C for P3= 7000(P/A,10%,10)/ (15000+3000(P/A,10%,10)

=(7000*6.145)/ (15000+(3000*6.145))

= 1.28

Since the Incremental B/C for P3 is greater than 1 for P3 so we select P3 proposal.