As part of a study for the Department of Labor? Statistics, you are assigned the
ID: 419972 • Letter: A
Question
As part of a study for the Department of Labor? Statistics, you are assigned the task of evaluating the improvement in productivity of small businesses. Data for one of the small businesses you are to evaluate is shown below. The data are the monthly average of last year and the monthly average this year.
Labor ?$7 per? hour;
Capital 0.84?% per month of? investment;
Energy ?$0.70 per
?a) Determine the multifactor productivity with dollars as the common denominator for last year. The multifactor productivity with dollars as the common denominator for last year is nothing ?dozen/$. ?(Round your response to three decimal? places.)
?b) Determine the multifactor productivity with dollars as the common denominator for this year. The multifactor productivity with dollars as the common denominator for this year is nothing ?dozen/$. ?(Round your response to three decimal? places.)
?c) Determine the percent change in productivity for the monthly average last year versus the monthly average this year on a multifactor basis. The percent change in productivity for the monthly average is nothing?%. ?(Round your response to one decimal? place.)
more Info
LAST YEAR
THIS YEAR
Production? (dozen)
1, 300
1,300
Labor? (hours)
380
360
Capital investment? ($)
13,000
16,000
Energy? (BTU)
3,300
3,000
LAST YEAR
THIS YEAR
Production? (dozen)
1, 300
1,300
Labor? (hours)
380
360
Capital investment? ($)
13,000
16,000
Energy? (BTU)
3,300
3,000
Explanation / Answer
a)
Number of dozens produced = 1300
Total costs = Labor cost + Capital cost + Energy cost
= 380*7+(0.84/100)*13000+3300*0.7
=$ 5079.2
Multifactor productivity = 1300/5079.2 = 0.256 dozen/$
b)
Number of dozens produced = 1300
Total costs = Labor cost Capital cost+ Energy cost
= 360*7 + (0.84/100)*16000 + 3000*0.7
=$ 4754.4
Multifactor productivity = 1300/4754.4 = 0.273 dozen/$
c)
Percentage increase in productivity = ((0.273-0.256)/0.273)*100 = 6.23%
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