George Kyparisis makes bowling balls in his Miami plant. With recent increases in his​ costs, he has a newfound interest in efficiency. George is interested in determining the productivity of his organization. He would like to know if his organization is maintaining the manufacturing average of a​ 3% increase in productivity. He has the following data representing a month from last year and an equivalent month this​ year: Last Year Now Cost Per Input UnitUnits Produced 1,000 1,000 Labor (hours) 280 255 $12Resin (pounds) 48 43 $5Capital Invested ($)9,000 10,000 1%Energy (BTU) 2,900 2,750 $0.50The percent change in productivity for one month last year versus one month this year on a multifactor basis with dollars as the common denominator = ___%?

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Answer:

7.313%

Explanation:

In the problem above, the total cost for last year production is equivalent to the addition of labor cost, energy cost, capital cost, and material cost. Thus,

Total cost (last year) = (300*12) + (3000*0.6) + (9000*0.02) + (48*6) = 3600 + 1800 + 180 + 288 = 5868.

Last year productivity = unit produced/total cost = 1000/5868 = 0.17042

Total cost (this year) = (275*12) + (2850*0.60) + (10000*0.02) + (43*6) = 3300 + 1710 + 200 + 258 = 5468

This year productivity = 1000/5468 = 0.18288

%change in productivity = [(0.18288 - 0.17042)/0.17042]*100% = 7.313%

The percent change in productivity for one month last year versus one month this year on a multifactor basis with dollars as the common denominator is 7.313%

  • The calculation is as follows;

 

Total cost (last year) is

= (300 × 12) + (3000 × 0.6) + (9000 × 0.02) + (48 × 6)

= 3600 + 1800 + 180 + 288

= 5868.

Now

Last year productivity = unit produced ÷ total cost

= 1000 ÷ 5868

= 0.17042

Now

Total cost (this year) = (275 ×12) + (2850 ×0.60) + (10000 × 0.02) + (43 × 6)

= 3300 + 1710 + 200 + 258

= 5468

Now  

This year productivity = 1000  ÷ 5468

= 0.18288

Now  

%change in productivity = [(0.18288 - 0.17042) ÷ 0.17042]

= 7.313%

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