Question

Bricol Confectionary manufactures candied treats for hotels in a highly automated assembly plant in Edmonton, Alberta.  Their 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:

                        Observation         Machine-hours    Kilowatt-hours   Total Overhead Costs

                       January                         1,880             2,260,000                $116,000

                      February                         1,825             2,170,000                114,000

                           March                         1,900             2,250,000                115,000

                              April                         1,650             2,145,000                114,000

                              May                         1,625             2,100,000                105,000

                             June                         1,500             2,060,000                100,000

Required (5 marks)

a) Using the high-low method, determine the cost estimating equation with machine-hours as the predictor variable. (2 marks)

b) Using the high-low method, determine the cost estimating equation with kilowatt-hours as the predictor variable. (2 marks)

c) For July the company ran the machines for 1,600 hours and used 2,075,000 kilowatt hours of power. The overhead costs totaled $95,000. Which driver was the best predictor for July? Explain why you chose the predictor you did. (2 marks)

Answer 1

Bricol Confectionary Requirement a Machine-hours: Slope coefficient = ($115,000 - $100,000)/(1,900 – 1,500) = $37.5 per machine-hour Constant = $115,000 - ($37.5 x 1,900) = $43,750 Machine-hour estimating equation = $43,750 + $37.5X Requirement b Kilowatt-hours: Slope coefficient = ($116,000 - $100,000)/(2,260,000 -2,060,000) = $0.08 per kilowatt-hour Constant = $116,000 - ($0.08 x 2,260,000) = $64,800 Kilowatt-hour estimating equation =-$64,800 + $0.08KWH Requirement c July's estimated costs: with machine-hours = $43,750 + $37.5 x 1600 = $103750 with kilowatt-hours = $64,800 + $0.08 x 2075000 = $101,200 The best estimator for July was the kilowatt-hour cost driver.