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ruskin ruskin
wrote...
Posts: 664
6 years ago
Tessmer Manufacturing Company produces inventory in a highly automated assembly plant in Windsor, Ontario. The automated system is in its first year of operation and management is still unsure of the best way to estimate the overhead costs of operations for budgetary purposes. For the first six months of operations, the following data were collected:

   Machine-hours   Kilowatt-hours   Total Overhead Costs
January   3,800   4,520,000   $138,000
February   3,650   4,340,000   136,800
March   3,900   4,500,000   139,200
April   3,300   4,290,000   136,800
May   3,250   4,200,000   126,000
June   3,100   4,120,000   120,000

Required:
a.   Use the high-low method to determine the estimating cost function with machine-hours as the cost driver.
b.   Use the high-low method to determine the estimating cost function with kilowatt-hours as the cost driver.
c.   For July, the company ran the machines for 3,150 hours and used 4,180,000 kilowatt-hours of power. The overhead costs totaled $114,000. Which cost driver was the best predictor for July?
Textbook 
Cost Accounting: A Managerial Emphasis, Canadian Edition

Cost Accounting: A Managerial Emphasis, Canadian Edition


Edition: 7th
Authors:
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Replies
wrote...
6 years ago
a.   Machine-hours:
   Slope coefficient   = ($139,200 - $120,000)/(3,900 - 3,100)
   = $24.00 per machine-hour
   Constant = $139,200 - ($24 × 3,900) = $45,600
   Machine-hour estimating equation = $45,600 + $24X

b.   Kilowatt-hours:
   Slope coefficient   = ($138,000 - $120,000)/(4,520,000 - 4,120,000)
   = $0.045 per kilowatt-hour
   Constant = $138,000 - ($0.045 × 4,520,000) = -$65,400
   Kilowatt-hour estimating equation = -$65,400 + $0.045KWH

c.   July's estimated costs:
   with machine-hours = $45,600 + ($24 × 3,150) = $121,200
   with kilowatt-hours = -$65,400 + ($0.045 × 4,180,000) = $122,700
   The best estimator for July was the machine-hour cost driver.
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