Question

A process control manager is considering two ro­bots to improve materials-handling capacity in the production of rigid shaft couplings that make dissimilar drive components. Robot X has a first cost of $90,000, an annual M&O cost of $31,000, and $43,000 salvage value, and it will improve revenues by $96,000 per year. Robot Y has a first cost of $146,000, an annual M&O cost of $28,000, and $47,000 salvage value, and it will increase revenues by $128,000 per year. The company’s MARR is 34% per year, and it uses a 3-year study period for economic evaluations. Calculate the incremental ROR, and identify the robot the manager should select.

The incremental ROR is %.

The manager should select robot .

 

 

Answer 1

 

Incremental initial cost (Y-X) = 146000 - 90000 = 56000

Incremental annual cost (Y-X) = 28000 - 31000 = -3000 (Annual savings)

Incremental annual revenue (Y-X) = 128000 - 96000 = 32000

Incremental salvage (Y-X) = 47000 - 43000 = 4000

Let Incremental IRR be i%, then

(32000-(-3000))*(P/A,i%,3) + 4000*(P/F,i%,3) = 56000

35000*(P/A,i%,3) + 4000*(P/F,i%,3) = 56000

Dividing by 1000

35*(P/A,i%,3) + 4*(P/F,i%,3) = 56

using trail and error method

When i = 41%, value of 35*(P/A,i%,3) + 4*(P/F,i%,3) = 35*1.568945 + 4*0.356732 = 56.340010

When i = 42%, value of 35*(P/A,i%,3) + 4*(P/F,i%,3) = 35*1.549408 + 4*0.349249 = 55.626259

using interpolation

i = 41%+ (56.340010-56)/(56.340010-55.626259)*(42%-41%)

i = 41% + 0.47% = 41.47%

As Incremental IRR > MARR, Robot Y should be selected