Question

2. Old Southwest Canning Co. has determined that any one of four machines can be used in its chili- canning operation. The costs of the machines are estimated below, and all machines have a 5-year life. If the minimum attractive rate of return is 60% per year, determine which machine should be selected on the basis of a rate of return analysis. Machine First Cost, $ AOC, $ -46,000 -26,000 -45,000 -27,000 -42,000 -29,000 -43,000 -28,000

Answer 1

MARR = 60%

For incremental analysis first we arrange machines in increasing order of initial cost

Machine 3 < Machine 4 < Machine 2 < Machine 1

Incremental analysis between (machine 4 - machine 3)

incremental inital cost = -43000 - (-42000) = -1000

incremental annual cost = -28000 - (-29000) = 1000

Let ROR be i%, then

1000 * (P/A,i%,5) = 1000

(P/A,i%,5) = 1

using trail and error method

when i = 96%, value of  (P/A,i%,5) = 1.005655

when i = 97%, value of  (P/A,i%,5) = 0.996182

using interpolation

i = 96% + (1.005655 - 1) / (1.005655 - 0.996182)*(97%-96%)

i = 96% + 0.5969%

i = 96.60%

As incremental ROR is greater than MARR, machine 4 should be seelcted

Incremental analysis between (machine 2 - machine 4)

incremental inital cost = -45000 - (-43000) = -2000

incremental annual cost = -27000 - (-28000) = 1000

Let ROR be i%, then

1000 * (P/A,i%,5) = 2000

(P/A,i%,5) = 2

using trail and error method

when i = 41%, value of  (P/A,i%,5) = 2.001381

when i = 42%, value of  (P/A,i%,5) = 1.968562

using interpolation

i = 41% + (2.001381 - 2) / (2.001381 - 1.968562)*(42%-41%)

i = 41% + 0.04%

i = 41.04%

As incremental ROR is less than MARR, machine 4 should be selected

Incremental analysis between (machine 1 - machine 4)

incremental inital cost = -46000 - (-43000) = -3000

incremental annual cost = -26000 - (-28000) = 2000

Let ROR be i%, then

2000 * (P/A,i%,5) = 3000

(P/A,i%,5) = 1.5

using trail and error method

when i = 60%, value of  (P/A,i%,5) = 1.507721

when i = 61%, value of  (P/A,i%,5) = 1.487800

using interpolation

i = 60% + (1.507721 - 1.5) / (1.507721 - 1.487800)*(61%-60%)

i = 60% + 0.39%

i = 60.39%

As incremental ROR is greater than MARR, machine 1 should be selected

Machine 1 should be selected as per ROR analysis