Question

Munson Performance Auto, Inc., modifies 375 autos per year. The manager, Adam Munson is interested in obtaining a measure of overall performance. He has asked you to provide him with a multifactor measure of last year’s performance as a benchmark for future comparison. You have assembled the following data. Resource inputs were labor, 10,000 hours; 500 suspension and engine modification kits; and energy, 100,000 kilowatt-hours. Average labor cost last year was $20 per hour, kits cost $1,000 each, and energy costs were $3 per kilowatt-hour. What do you tell Mr. Munson?

Answer 1

Multifactor productivity measure for last year = Total Output / Total Input Cost

Total output for last year = 375 autos

Total Input cost for last year = Total labor cost + Total kit cost + Total energy cost

Total labor cost = Labor hours x Per hour Labor cost = 10000 x 20 = $ 200,000

Total kit cost = Number of kits x Cost of each kit = 500 x 1000 = $ 500,000

Total energy cost = Total Killo-watt hours x Energy cost per kilowatt-hour = 100000 x 3 = $ 300,000

Therefore,

Total Input cost for last year = 200000 + 500000 + 300000 = $ 1,000,000

Hence,

Mutlifactor productivity measure for last year = 375 / 1000000 = 0.000375 autos per dollar