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With the high-low method, only the periods in which the lowest activity and the highest activity occurred are used to estimate the variable and fixed components ...
Typology: Study notes
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Horngren 13e
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The Dependent Variable: The cost that is
The Independent Variable: The cost driver being predicted
The cost driver
The slope of The Intercept: Fixed Costs
The slope of the line: variable cost per unitper unit
EXAMPLE: Kohlson Company has incurred the following shipping costs over the past eight months:
Units Sold
Shipping Cost January... 6,000 $66, FebruaryFebruary. 5 0005,000 $65 000$65, March ..... 7,000 $70, April ........ 9,000 $80, May......... 8,000 $76, JuneJune........ 10 00010,000 $85 000$85, July ......... 12,000 $100, August .... 11,000 $87,
With the high-low method, only the periods in which the lowest activity and the highest activity occurred are used to estimate the variable and fixed components of the mixed cost.
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Learning Objective 1: Explain the two assumptions frequently used in cost-behavior estimation... cost functions are linear and have a single cost driver Write a linear cost function equation for each of the following conditions. Use y for estimated costs and X for activity of the cost driver.
driver
a. Direct manufacturing labor is $10 per hour. b. Direct materials cost $9.20 per cubic yard. c. UtilitiUtilities have a minimum charge of $1,000, plus a charge of $0.05 per h i i h f $1 000 l h f $0 05 kilowatt-hour. d. Machine operating costs include $200,000 of machine depreciation per year plus $75 of utility costs for each day the machinery is inper year, plus $75 of utility costs for each day the machinery is in operation.
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Learning Objective 1: Explain the two assumptions frequently used in cost-behavior estimation... cost functions are linear and have a single cost driver Write a linear cost function equation for each of the following conditions. Use y for estimated costs and X for activity of the cost driver.
driver
a. Direct manufacturing labor is $10 per hour. b. Direct materials cost $9.20 per cubic yard. c. UtilitiUtilities have a minimum charge of $1,000, plus a charge of $0.05 per h i i h f $1 000 l h f $0 05 kilowatt-hour. d. Machine operating costs include $200,000 of machine depreciation per year plus $75 of utility costs for each day the machinery is inper year, plus $75 of utility costs for each day the machinery is in operation.
Answer:Answer:
a. y = $10X bb. y = $9 20Xy $9.20X c. y = $1,000 + $0.05X d. y = $200,000 + $75X
Learning Objective 4: Estimate a cost function using quantitative analysis... the end result is to evaluate the cost driver of the estimated costresult is to evaluate the cost driver of the estimated cost function
The Wildcat Company has provided the following information:The Wildcat Company has provided the following information:
Units of Output 30,000 Units 42,000 Units Direct materialsDirect materials $$ 180,000 180 000 $$ 252,000 252 000 Workers' wages 1,080,000 1,512, Supervisors' salaries 312,000 312, Equipment depreciationEquipment depreciation 151,200151,200 151,200151, Maintenance 81,600 110, Utilities 384,000 528, Total $2,188,800$ , , $2,865,600$ , ,
Using the high-low method and the information provided above, a. identify the linear cost function equation and b. estimate the total cost at 36,000 units of output.
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Learning Objective 4: Estimate a cost function using quantitative analysis... the end result is to evaluate the cost driver of the estimated costresult is to evaluate the cost driver of the estimated cost function
UsingUsing the highthe high-low method and the information provided above, low method and the information provided above a. identify the linear cost function equation and b. estimate the total cost at 36,000 units of output.
Answer: a. Variable cost = ($2,865,600 - $2,188,800) / (42,000 - 30,000) = $56. Fixed cost = $2 865 600Fixed cost = $2,865,600 - $56.40 × 42,000 = $496,800 $56 40 × 42 000 = $496 800
Cost function is y = $496,800 + $56.40X
b. Output level of 36,000 units = $496,800 + $56.40 × 36,000 = $2,527,200 total cost
Learning Objective 4: Estimate a cost function using quantitative analysis... the end result is to evaluate the cost driver of the estimated costresult is to evaluate the cost driver of the estimated cost function
Tessmer Manufacturing Company produces inventory in a highly automated assembly plant in OlatheTessmer Manufacturing Company produces inventory in a highly automated assembly plant in Olathe, Kansas. 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-hoursMachine hours Kilowatt-hoursKilowatt hours Total Overhead CostsTotal Overhead Costs January 3,800 4,520,000 $138, February 3,650 4,340,000 136, March 3,900 4,500,000 139, AprilApril 3 3003,300 4 290 0004,290,000 136 800136, May 3,250 4,200,000 126, June 3,100 4,120,000 120, Required: a.a. Use the high-low method to determine the estimating cost function with machine-hours as the costUse 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.c. For July, the company ran the machines for 3,000 hours and used 4,000,000 kilowatt-hours ofFor July, the company ran the machines for 3,000 hours and used 4,000,000 kilowatt hours of power. The overhead costs totaled $114,000. Which cost driver was the best predictor for July?
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Answer:
a. Machine-hours: Slope coefficient = ($139,200 - $120,000) / (3,900 - 3,100) = $24 00 per machine-hour= $24.00 per machine-hour Constant = $139,200 - ($24 × 3,900) = $45, 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$0.045 per kilowatt hour Constant = $138,000 - ($0.045 × 4,520,000) = -$65, Kilowatt-hour estimating equation = -$65,400 + $0.045KWH
c. July's estimated costs: with machine-hours = $45,600 + $24 × 3,000 = $117, with kilowatt-hours = -$65,400 + $0.045 × 4,000,000 = $114, The best estimator for July was the kilowatt-hour cost driver.
Least-Squares Regression & Statistical Analysis
Newton Company used least squares regression analysis to obtain the following output: Payroll Dept Cost Explained by # of Employees ConstantConstant $5 800$5, Standard error of Y estimate 630 R^2 0. Number of observations 20 X coefficient(s) $$1. Standard error of coefficient(s) 0. Required: aa. What is the total fixed cost?What is the total fixed cost? b. What is the variable cost per employee? c. Prepare the linear cost function. d. What is the coefficient of determination? Comment on the goodness of fit.
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Least-Squares Regression & Statistical Analysis
Newton Company used least squares regression analysis to obtain the following output: Payroll Dept Cost Explained by # of Employees ConstantConstant $5 800$5, Standard error of Y estimate 630 R^2 0. Number of observations 20 X coefficient(s) $$1. Standard error of coefficient(s) 0. Required: aa. What is the total fixed cost?What is the total fixed cost? b. What is the variable cost per employee? c. Prepare the linear cost function. d. What is the coefficient of determination? Comment on the goodness of fit.
Answer: a. The constant or intercept is the total fixed cost of $5,800. b. The variable cost per employee is the X coefficient of $1.902. cc. y = $5 800 + $1 902Xy = $5,800 + $1.902X d. The coefficient of determination is the R^2 of 0.8924. This represents a very high goodness of fit. The closer to 1.0, the better the cost driver explains the cost. Therefore, the conclusion can be drawn that there is a significant relationship between the cost of the payroll department and the number of employees employees.
Least-Squares Regression & Statistical Analysis
Schotte Manufacturing Company uses two different independent variables (machine-hours and number of packages) in two different equations to evaluate costs of the packaging department. The most recent results of the two regressions are as follows:g
Machine-hours: Variable Coefficient Standard Error t-Value ConstantConstant $748 30$748.30 $341 20$341.20 2 192. Independent Variable $52.90 $35.20 1.
R^2 = 0.
Number of packages: Variable Coefficient Standard Error t-Value ConstantConstant $242 90$242.90 $75 04$75.04 3 243. Independent Variable $5.60 $2.00 2. R^2 = 0.
Required: a. What are the estimating equations for each cost driver? b. Which cost driver is best and why?
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Least-Squares Regression & Statistical Analysis
Answer: a. Machine-hours y = $748.30 + $52.90X Number of packagesp g y = $242.90 + $5.60Xy $ $
b. Machine-hours has a low R^2 which implies that a small proportion of the variance is explained by machine-hours, thereby making it less attractive than number of packages as a cost predictor.t di t
Also, for the independent variable, number of packages, the t-value of 2.80 indicates that a relationship exists between the independent and dependent variables. For machine-hours,p p p , the t-value (1.50) is below 2.00, indicating that the coefficient is not significantly different from zero and that there may not be a relationship between the independent and dependent variables.
The t-values of the constant terms (g) for both drivers is greater than 2.00, therefore, there is no distinguishing characteristic between the constants.
Given the above findings, it appears that number of packages is the best predictor of costs of the packing department.
