Data from problem 1.6 in textbook: Last Year ; now Units produced 1,000 ; 1,000
ID: 462996 • Letter: D
Question
Data from problem 1.6 in textbook:
Last Year ; now
Units produced 1,000 ; 1,000
Labor (hours) 300 ; 275
Resin (pounds) 50 ; 45
Capital invested ($) 10,000 ; 11,000
Energy (BTU) 3,000 ; 2,850
1.7 Eric Johnson (using data from Problem 1.6 from the textbook.... Pictured Above!) determines his costs to be as follows:
labor $10 per hours;
resin $5 per pound;
capital 1% per month of investment;
energy $.50 per BTU
What is the percent productivity change, for one month last year versus on month this year, on a multifactor basis with dollars as the common denominator?
Explanation / Answer
Last Year
Labor cost ( 300hrs, $10) = 300*10 = $3000
Resin cost (50 pounds , $5) = 50 * 5 = $250
Capital ($10000 , 1%) = 10,000 * 0.01 = $100
Energy (3,000 , $.50) = 3,000 * .5 = 1500
Total cost = Labor cost + Resin cost + Capital + Energy = $3000 + $250 + $100 + 1500 = $4850
Multi productiivity = No of units produced / Total cost = 1,000/4850 = 0.2062 units/dollar
Present year:
Labor cost ( 275hrs, $10) = 270*10 = $2750
Resin cost (45 pounds , $5) = 45 * 5 = $225
Capital ($11000 , 1%) = 11,000 * 0.01 = $110
Energy (2,850 , $.50) = 2850 * .5 = 1425
Total cost = Labor cost + Resin cost + Capital + Energy = $2750 + $225 + $110 + $1425 =$4510
Multi productiivity = No of units produced / Total cost = 1,000/$4510 = 0.2217 units/dollar
Percent productivity change = ((0.2217 - 0.2062)/0.2062)*100 = 0.075388*100 = 7.54 %
Related Questions
Navigate
Integrity-first tutoring: explanations and feedback only — we do not complete graded work. Learn more.