Question & Answer: George makes bowling balls in his plant. With recent increases in his costs, he has…..

MGT-4310 Pr Homework: Assignment Ch. 1 Score: 0 of 1 pt Problem 1.6 Hw Score: 16 67%, on George Kyparisis makes bowling balls in his Miani plant. With recent increases in his costs, he has a r newlound interest in efficiency George is interested in determining the productvity of hes organizalion e would lke to know it he man dacturng average of a 3% ngease r, produchty He has-folowing dta represening a month horn last year and an equivalent months year 275 Labor (hours) Resin (pounds) Captas invested 5) 10,000 12.500 1,000 2.750 The productivity change for each of the inputs (Labor, Resin, Capital, and Energy) is Labor Produdity Change·ㄈ % fenter your response as a percentage rounded t eo decinal places and include a minus sign f necessary) Enter your answer in the answrer box and then cick Check Anser Pama Type here to se as 6 8

George makes bowling balls in his 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 organization is maintaining the manufacturing average of a 3% increase in productivity. He has the following data representing a month last year and an equivalent month this year: The productivity change for each of the inputs (Labor, Resin, Capital, and Energy) is: Labor Productivity Change = % (enter your response as a percentage rounded to two decimal places and include a minus sign necessary)

Expert Answer

Answer

1. Labor productivity for last year = 1000 units produced/300 labor hours = 3.33

Labor productivity for this year = 1000 units produced/275 labor hours = 3.64

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Thus labor productivity change = (3.64-3.33)/3/33 = 9.09%

2. Resin productivity for last year = 1000 units produced/50 pounds = 20

Resin productivity for this year = 1000 units produced/46 pounds = 21.74

Thus resin productivity change = (21.74-20)/20 = 8.70%

3. Capital productivity for last year = 1,000 units/$10,000 = 0.10

Capital productivity for this year = 1,000/12,500 = 0.08

Thus capital productivity change = (0.08-0.10)/0.10 = -20.00%

4. Energy productivity for last year = 1000 units/3000 BTU = 0.33

Energy productivity for this year = 1000 units/2750 BTU = 0.36

Thus energy productivity change = (0.36-0.33)/0.33 = 9.09%

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