Question

Hi, can you please solve this?

A machine that produces a certain piece must be turned off by the operator after each piece is completed. The machine "coasts" for 15 seconds after it is turned off, thus preventing the operator from removing the piece quickly before producing the next piece. An engineer has suggested installing a brake that would reduce the coasting time to 3 seconds.

The machine produces 50,000 pieces a year. The time to produce one piece is 1 minute 45 seconds, excluding coastint time. The operator earns $12 an hour and direct costs for operation are $2 an hour. The direct costs are incurred whenever the operator has to work. The brake will require servicing every 446 hours of operation. It will take the operator 30 minutes to perform the necessary maintenance and will require $54 in parts and material. The brake is expected to last 7,500 hours of operation (with proper maintenance) and will have no salvage value.

How much could be spent for the brake if the Minimum Attractive Rate of Return is 10% compounded annually?

Answer 1

Cost incurred without the brake = No. of pieces * (No. of minutes for producing one product / total number of minute in an hour) * cost per peice

No. of minutes for producing a product without the brake system = 105 seconds + 15 seconds of coastint time = 120 seconds or we can say 2 minutes

Cost incurred without break is = 50000*(2/60)*(12+2)$ = $ 23333.33

Now we will compute no. of minutes with installing a break = 105+3 seconds = 108 seconds = 1.8 minutes

Cost incurred with break = 50000*(1.8/60)*(12+2)$ = $ 21000.00

Now computation of maintainence cost, This will be parts & material cost and labor cost for operator

= (50000 * (1.8)/60)/446*(30/60*(12+54)) = $111

No. of years for which break will last

=7500 / 50000/(1.8/60) = 5 years

Max amount that can be spent on brake will be differential in cost incurred with brake and without brake * present value of annuity factor of 5 years @ 10%

= (23333-21111)*PVAF@10% for 5 years = $8423 (rounded off)