Module 7 Replacement Analysis PDF

Chapter 7. REPLACEMENT ANALYSIS ©2017 Batangas State University 264  Introduction Replacement analysis plays an important role in the economic running of any concern for years or decades. As a business firm, they have to face different types of replace...

Chapter 7. REPLACEMENT ANALYSIS ©2017 Batangas State University 264  Introduction Replacement analysis plays an important role in the economic running of any concern for years or decades. As a business firm, they have to face different types of replacement decisions such as, the replacement of capital equipment as it wear out or becomes obsolete, the capital equipment required for expansion and the displacement of old technology by the new one. Replacement decision is not an easy job, you need to consider all the possible factors affecting an assets, its productivity, maintenance and especially its economic aspects. 265 ©2017 Batangas State University  Learning Objectives Determine and analyze the economic life of a certain assets. Evaluate the performance of existing assets and its possible replacement. 266 ©2017 Batangas State University  Major Reasons for Replacement The Four Major Reasons for Replacement 1. Physical Impairment The existing asset is completely or partially worn out and will no longer function satisfactorily without extensive repairs. 267 ©2017 Batangas State University  Major Reasons for Replacement The Four Major Reasons for Replacement 2. Inadequacy The existing asset does not have sufficient capacity to meet the present demands that are placed on it. 268 ©2017 Batangas State University  Major Reasons for Replacement The Four Major Reasons for Replacement 3. Obsolescence This may be caused either by a lessening in the demand for the service rendered by the asset or the availability of more efficient assets which will operate with lower out-of-pocket costs. 269 ©2017 Batangas State University  Major Reasons for Replacement The Four Major Reasons for Replacement 4. Rental or lease possibilities It is possible to rent identical or comparable asset or property, thus freeing capital for other and more profitable use. 270 ©2017 Batangas State University  Sunk Cost Due to Unamortized Value Unamortized value of an equipment or property is the difference between its book value and its resale value when replaced. Unamortizedva should be considered a sunk cost or a loss. 271 ©2017 Batangas State University  Basic Patterns for Replacement Studies Replacement economy studies may be made by any of the basic procedures or patterns which have been discussed previously. However, in most cases either the rate of return method or the annual cost method is used. 272 ©2017 Batangas State University  Sample Problems An existing factory must be enlarged or replaced to accomodate new production machinery. The structure was built at a cost of P2.6 million. Its present book value, based on straight line depreciation is P700,000 but it has been appraised at P800,000. If the structure is altered, the cost will be P1.6 million and its service life will be extended 8 years with a salvage value of P600,000. A new factory could be purchased or built for P5.0 million. It would have a life of 20 years and a salvage value of P700,000. Annual maintenance of the new building would be P160,000 compared with P100,000 in the enlarged structure. However, the improved layout in the new building would reduce annual production cost by P240,000. All other expenses for the new structure are estimated as being equal. Using an investment rate of 8 percent, determine which is more attractive investment for this firm. 273 ©2017 Batangas State University  Sample Problems Solution: Enlarged Building Annual Costs: P800,000  P1,600,000 - P600,000 Depreciation  F/A, 8%, 8 P1,800,000   P169,227 10. 6366 Maintenance = P 100,000 Production (excess) = P 240,000 Total Annual Cost = P 509,227 274 ©2017 Batangas State University  Sample Problems Solution: New Building Annual Costs: P5,000,000 - P700,000 Depreciation  F/A, 8%, 20 P4,300,000   P93,964 45.7620 Maintenance = P 160,000 Total Annual Cost = P 253,964 275 ©2017 Batangas State University  Sample Problems Annual Savings = P509,227 - P225,263 Additional Investment = P5,000,000 -P800,000 - P1,600,000 = P2,600,000 P255,263 Rate of Return on additional investment = x 100 P2,600,000 = 9.82% Construct the New Building 276 ©2017 Batangas State University  Sample Problems A decision must be made whether to replace a certain engine wi a new one, or to rebore the cylinder of the old engine and thoroughly recondition it. The original cost of the old engine 10 years ago was P70,000; to rebore and recondition it now will cost P28,000, but would extend its useful life for 5 years. A new engine will have a first cost of P62,000 and will have an estimated life of 10 years. It is expected that the annual cost of fuel and lubricants with the reconditioned engine will be about P20,000 and that this cost will be 15% less with the new engine. It is also believed that repairs will be P2,500 a year less with the new engine that with the reconditioned one. Assume that neither engine has any net realizable value when retired. If money is worth 16%, what would you recommend? 277 ©2017 Batangas State University  Sample Problems Solution: Reconditioned engine Annual costs: P28,000 P28,000 Depreciation =   P4,071 F/A,16%,5 6.877 Fuel and Lubricants = P 20,000 Repairs (excess) = P 2,500 Interest on capital = P28,000(0.16) = P 4,480 Total Annual Cost P 31,051 278 ©2017 Batangas State University  Sample Problems Solution: New engine Annual costs: P62,000 P62,000 Depreciation =   P 2,908 F/A, 16%, 10 21.32 Fuel and Lubricants = P 17,000 Interest on capital = P62,000(0.16) = P 9,920 Total Annual Cost P 29,828 The old engine should be replaced 279 ©2017 Batangas State University  Sample Problems Four years ago an ore-crushing unit was installed at a mine which cost P81,000. Annual operating costs for this unit are P3,540. This unit was estimated to have a life of 10 years. The quantity of ore to be handled is to be doubled and is expected to continue at this higher rate for at least 10 years. A unit that will handle the same quantity of ore and have the same operating costs as the one now in service can be installed for P75,000. This unit will have a useful life of 6 years. A unit with double the capacity of the one now in use can be installed for P112,000. Its life is estimated at 6 years and its annual operating costs are estimated at P4,950. The present realizable value of the unit now in use is P26,000. All units under consideration will have an estimated salvage value at retirement age of 12% of the original cost. Interest rate is 20%. Annual taxes and insurance are 2.5% of the original cost. What would you recommend? 280 ©2017 Batangas State University  Sample Problems Solution Augmentation Annual Costs: Old Unit P26,000 - (P81,000)( 0.12) Depreciation  F/A, 20%, 6 P16,280   P1,639 Operation 9. 9299 = P3,540 Taxes and Insurance = (P81,000)(0.025) = P2,025 New small unit P75,000 - (P75,000)(0.12) Depreciation  F/A, 20%, 6 P66,000   P6,647 9.9299 Operation = P3,540 Taxes and Insurance = (P75,000)(0.025) = P1,875 Total annual cost P19,266 281 ©2017 Batangas State University  Sample Problems Solution Replacement Annual Costs: New Big Unit P112,000 - (P112,000)(0.12) Depreciation  F/A, 20%, 6 P98,560   P9,926 Operation 9.9299 = P4,950 Taxes and Insurance = (P112,000)(0.025) = P2,800 Total annual cost P17,676 Annul Savings = P19,266 - P17,676 = P1,590 Additional Investment = P112,000 - P26,000 = P11,000 ROR on additional investment = P1,590 x100  14.5% P11,000 Buy the new small unit to augment the old unit. 282 ©2017 Batangas State University  Sample Problems A car can be purchased for P600,000 when new. There follows a schedule of annual operating expenses for each year and trade-in-values at the end of each year. Assume that these amounts would be repeated for future replacements, and that the car will not be kept more than 3 years. If interest on invested capital is 15% before taxes, determine at which year's end the car should be replaced so that costs will be minimized. Year 1 Year 2 Year 3 Operating expenses for year P34,000 P38,000 P41,000 283 ©2017 Batangas State University  Sample Problems Solution Cost of Keeping each year 1 2 3 Operation P34,000 P38,000 P41,000 Depreciation 192,000 144,000 96,000 Interest on Capital (15%) 90,000 61,200 50,400 Total P316,000 P243,200 P187,400 1 year: 0 1 0 1 P316,00 EUAC 284 ©2017 Batangas State University 0 EUAC = P316,000  Sample Problems 2 years: 0 1 2 0 1 2 P243,00 P316,000 0 EUAC EUAC EUAC = [P316,000 +P243,000 (P/F, 15%, 2)] (A/P, 15%, 2) = P307,385 285 ©2017 Batangas State University  Sample Problems 3 years: 0 1 2 3 0 1 2 3 P187,00 P243,00 0 EUAC EUAC EUAC 0 P316,000 EUAC = [P316,000 +P243,000 (P/F, 15%, 2) + P187,400 (P/F, 15%, 3)] (A/P, 15%, 3) = P272,851 Thus it is cheaper to keep the car three years. 286 ©2017 Batangas State University