Title | 5.2 Fixed, Increment, and Sunk Costs ME 005-ECE21S2 - Engineering Economy |
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Course | Engineering Economy |
Institution | Technological Institute of the Philippines |
Pages | 4 |
File Size | 130.5 KB |
File Type | |
Total Downloads | 83 |
Total Views | 314 |
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11/11/21, 2:57 PM
5.2: Fixed, Increment, and Sunk Costs: ME 005-ECE21S2 - Engineering Economy
5.2: Fixed, Increment, and Sunk Costs ▶ FIXED, INCREMENT, AND SUNK COSTS In this module, we will study the Fixed, Increment, and Sunk Costs. Types of Costs Fixed costs are those costs that remain constant, whether or not a given change in operations or policy is adopted. Variable costs are those costs that vary with output or any change in the activities of an enterprise. Increment costs are those that arise as the result of a change in operations or policy. Marginal cost is the additional cost of producing one or more units of a product. Sunk cost represents money which has been spent or capital which have been invested and that cannot be recovered due to certain reasons.
To illustrate the above principles please refer to the sample problem below. Sample Problem no. 37 A company having a capacity of 1,600 units per year currently is operating at a sales level of only 1,200 units, with a selling price of P720 per unit. The fixed cost of the plant is P365,000 per year, and the variable costs are P416 per unit. It has been estimated that a reduction of P50 per unit in the selling price would increase sales by 300 units per year. (a). Would this be a good program to follow? (b). An alternative being considered is to engage in a modernization plan that would increase the fixed costs by P58,000 per year but would reduce the variable costs by P56 per unit. Would this be a better procedure than the price reduction program? (c). Can you suggest any other program that might be superior to the foregoing? Solution (a) Present revenue = P720 (1,200)
= P864,000
Present costs: https://tip.instructure.com/courses/26120/pages/5-dot-2-fixed-increment-and-sunk-costs?module item id=1936844
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5.2: Fixed, Increment, and Sunk Costs: ME 005-ECE21S2 - Engineering Economy
Fixed Variable = P416 (1,200)
= P365,000 = P499,200
Total
= P864,200
Loss P200
(-)
New revenue = P670 (1,500)
= P1,005,000
Costs: Fixed Variable = P416(1,500)
= P365,000 = P624,000
Total
= P989,000
Profit P16,000 Therefore, reducing the price would be a profitable program. (b) Revenue
= P864,000
Costs: Fixed Variable = P360 (1,200)
= P423,000 = P432,000
Total
= P855,000
Profit P9,000 Modernization would be profitable, but it would not be as good a procedure as the price reduction. (c) Both programs should be combined. Sample Problem no. 38 An old machine was purchased 3 years ago at a cost of P50,000. It was estimated to have a useful life of 8 years, with a salvage value of P5,000. It is now going to be replaced by a new machine costing P70,000 and a P30,000 trade-in will be allowed for the old machine. Determine the sunk cost if depreciation has been computed by the (a) straight-line method (b) sinking fund method at 16%, and (c) SYD method.
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5.2: Fixed, Increment, and Sunk Costs: ME 005-ECE21S2 - Engineering Economy
Solution Co = P50,000 CL = P5,000 L=8 n=3 (a) Straight-line method D3 =
= P16,875
C3 = C o - D 3 = P50,000 - P16,875 = P33,125 Sunk cost = book value - resale value = P33,125 - P30,000 = P3,125 (b) Sinking fund method d=
=
= P3,160
D3 = d(F/A, 16%, 3) = P3,160
= P11,078
C3 = C o - D 3 = P50,000 - P11,078 = P38,922 Sunk cost = book value - resale value = P38,922 - P30,000 = P8,922 (c) SYD method Sum of digits = 1 + 2 + 3 +....7 + 8 = D3 =
(1 + 8) = 36
(P50,000 - P5,000) = P26,250
C3 = C o - D 3 = P50,000 - P26,250 = P23,750 Sunk cost = P23,750 - P30,000 = -P6,250 The negative sign means that the amount is not a loss but again.
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5.2: Fixed, Increment, and Sunk Costs: ME 005-ECE21S2 - Engineering Economy
After completing this module, please solve the exercise problem ($CANVAS_COURSE_REFERENCE$/modules/items/g870eac5d06a6f40b4b383e5db1b64a34) .
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