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ANNUAL EQUIVALENCE ANALYSIS

ANNUAL EQUIVALENCE ANALYSIS. CHAPTER 6. ANNUAL EQUIVALENCE ANALYSIS. Annual equivalent criterion Applying annual worth analysis. Annual Equivalent Worth Criterion. AE worth criterion provides a basis for measuring investment worth by determining equal payments on annual basis.

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ANNUAL EQUIVALENCE ANALYSIS

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  1. ANNUAL EQUIVALENCE ANALYSIS CHAPTER 6

  2. ANNUAL EQUIVALENCE ANALYSIS • Annual equivalent criterion • Applying annual worth analysis

  3. Annual Equivalent Worth Criterion • AE worth criterion provides a basis for measuring investment worth by determining equal payments on annual basis. • Knowing that any lump-sum cash amount can be converted into a series of equal annual payments • Find the net present worth of the original series and then multiply this amount by the capital-recovery factor: AE ( i) = PW ( i) x (A/P, i, N) • We use this formula to evaluate the investment worth of projects. • Therefore, AE criterion provides basis for evaluating a project that is consistent with the PW criterion.

  4. Annual Equivalent Worth Criterion continue ....... Single Project Evaluation: The accept-reject decision rule for a single revenue project is as follows: • If AE( i) > 0, accept the investment. • If AE( i) = 0, remain indifferent to the investment • If AE( i) < 0, reject the investment Comparing Mutually Exclusive Alternatives: • Service projects: select the alternative with the minimum annual equivalent cost • Revenue projects: select the alternative with the maximum AE(i ).

  5. EXAMPLE 6.1 Finding Annual Equivalent Worth by Conversion from Net Present Worth • A hospital uses three coal-fired boilers to supply steam for space heating, domestic hot water, and the hospital laundry. One boiler is operated at times of low load and at weekends, two are operated during the week, and the fourth boiler is normally off-line. The design efficiency on a steady load is generally about 78%. The boilers at the hospital were being run at between 70% and 73 % efficiency, due to inadequate instrumentation and controls. Engineers have proposed that the boiler controls be upgraded. The upgrade would consist of installing variable speed drives for the boiler fans and using the fans in conjunction with oxygen trim equipment for combustion control. • The cost of implementing the project is $159,000. The boilers have a remaining service life of 12 years. Any upgrade will have no salvage value at the end of 12 years.

  6. EXAMPLE 6.1 continue......... • The annual electricity use in the boiler house is expected to be reduced from 410,000 kWh to 180,000 kWh as a result of variable speed control of the boiler fan (because with the variable speed drives, the fan motor draw only the power actually required to supply air to the boilers). This is equivalent to $14,000 per year. This savings is expected to increase at annual rate of 4% as the cost of electricity increases over time. • Coal use will be 2% lower due to the projected improvement in boiler efficiency. This corresponds to a cost reduction of $40,950 per year. This savings is projected to increase as the coal price increases at an annual rate of 5%. If the hospital uses a 10% interest rate for any project justification, what would be the annual equivalent energy savings due to the improvement?

  7. Example 6.1 Figure 6-1 Computing equivalent annual worth

  8. PSavings in electricity=$14,000 (P/A1, 4%, 10%, 12) Example 6.1 Since energy savings are in two different geometric gradient series, we calculate the equivalent present worth in the following steps: Savings in electricity P = $114,301 Savings in coal usage P = $350,356 PSavings in coal usage=$40,950 (P/A1, 5%, 10%, 12)

  9. Example 6.1 Net present worth calculation: PW(10%) = $114,301 + $350,356 - $159,000 =$305,657 Since PW(10%) > 0 the project would be acceptable Now, spreading the NPW over the project life gives (0.1468) AE(10%) = $305,657 (A/P, 10%, 12) =$44,870 Net annual benefit Since AE(10%) > 0, the project is also worth undertaken. The positive value indicates that the project is expected to bring in a net annual benefit of $44,870 over the life of the project.

  10. Benefits of Annual Equivalent Analysis • In the real world situations, AE analysis is preferred, or demanded, over NPW analysis. • For example, corporations issue annual reports and develop yearly budgets. • For these purposes, a company may find it useful to present the annual cost or benefit of ongoing project rather than its overall cost or benefit.

  11. Annual Equivalent Cost • When only costs are involved, the AE method is called the annual equivalent cost method. • In this case, revenues must cover two kinds of costs: Operating costs and capital costs. Capital costs + Annual Equivalent Costs Operating costs

  12. Operating costs are incurred by operation of physical plants or equipment needed to provide service; examples include the costs of items such as labor and raw materials. • Capital costs (or ownership costs) are incurred by the purchasing of assets to be used in production and service. • Normally, Capital costs are nonrecurring (one time costs), whereas operating costs recur as long as an asset is owned and being utilized. • Annual equivalent of a capital cost is given special name: Capital Recovery cost designated CR ( i ).

  13. Capital (Ownership) Costs Definition: The cost of owning a piece of equipment is associated with two amounts – (1) initial cost( I ) and (2) its salvage value ( S ) Capital costs: Taking these amounts into account, we calculate the capital costs as: S 0 N I 0 1 2 3 N CR( i) Figure 6-2 Calculation of capital recovery cost

  14. Table 6.2 Will your car hold its value? CR(6%) = ($19,800 - $12,078) (A/P, 6%, 3) + (0.06) x $12,078 = $3,613.55

  15. Example 6.3 Annual Equivalent Worth: Capital Recovery Cost • Ferguson Company is considering an investment in computer-aided design equipment. The equipment will cost $110,000 and will have a five year useful economic life. It has a salvage value of $10,000. The expected annual operating costs for the equipment would be $20,000 for the first two years and $25,000 for the remaining three years. Assume that Ferguson’s desired return on its investment (MARR) is 15%, what is required annual savings to make investment worthwhile? • Given: I = $110,000, S = $10,000, N = 5 years, i=15% Find: AEC, and determine whether the firm should or should not purchase the machine.

  16. SOLUTION Capital Costs: CR( i) = ( I – S ) ( A / P, i, N ) + iS CR( i) = [($110,000 - $10,000) ( A/P, 15%, 5 ) + (0.15) x $10,000] = $31,332 Operating Cost: OC ( i) = [$20,000(P/A, 15%, 2) + $25,000(P/A, 15%, 3)(P/F, 15%, 2)] x (A/P, 15%, 5) OC ( i) = $22,573 Annual Equivalent Cost: AEC(15%) =CR(15%) + OC(15%) = $31,332 + $22,573 = $53,905 The required annual savings must be at least $53,905 to recover the investment made in the asset and cover the annual operating expenses.

  17. Annual Worth Analysis

  18. Unit-Cost or Unit-Profit Calculation • In many situations, we need to know the unit profit or unit cost of operating an asset. To obtain a unit profit (or cost), we may proceed as follows: • Determine the number of units to be produced (or serviced) each year over the life of the asset. • Identify the cash flow series associated with production or service over the life of the asset. • 3. Calculate the present worth of the project’s cash flow series at a given interest rate, and then determine the equivalent annual worth. • Divide the equivalent annual worth by the number of units to be produced or serviced during each year. When the number of units varies each year, you may need to convert the units into equivalent annual units.

  19. Example 6.4 Unit Profit per Machine Hour with constant or varying Annual Operating Hours Harrison Company experiences frequent industrial accidents involving workers who perform the spot-welding. The firm is looking into the possibility of investing in a specific robot for welding task. The required investment will cost Harrison $1 million upfront and this robot has a five year useful life and a salvage value of $100,000. The robot will reduce labor costs, worker insurance costs, and material usage cost and will eliminate accidents involving workers at the spot welding operations. The savings figure translates into a total of $800,000 a year. The additional operating and maintenance costs associated with the robot amount to $300,000 annually. Compute the equivalent savings per machine hour at i= 15% compounded annuallyfor the following situations: • Suppose that this robot will be operated for 2,000 hours per year. • Suppose that the robot will be operated according to varying hours: 1,500 hours in the first year, 2,500 hours in the second year, 2,500 hours in the third year, 2,000 hours in the fourth year, and 1,500 hours in the 5th year. The total number of operating hours is still 10,000 over 5 years.

  20. Solution Example 6.4 PW(15%) = -$1,000,000 + $500,000 (P/A, 15%, 5) + $100,000 (P/F, 15%. 5) =$725,795 AE (15%) = $725,795 (A/P, 15%, 5) = $216,516 Savings per machine hour = $216,516 / 2,000 =$108.26 / hour Figure 6-4 Computing equivalent savings per machine hour

  21. MAKE – OR – BUY DECISION • Make or buy problems are among the most common business decisions. At any given time, a firm may have the option of either buying an item or producing it. • If either the “make” or the “buy” alternative requires the acquisition of machinery or equipment besides the item itself, then the problem becomes an investment decision. • Since the cost of an outside service (the “buy” alternative) is usually quoted in terms of dollars per unit, it is easier to compare the two alternatives if the differential costs of the “make” alternative are also given in dollars per unit. • This unit cost comparison requires the use of annual worth analysis.

  22. The specific procedure is as follows: • Determine the time span (planning horizon) for which the product will be needed • Determine the annual quantity of the product • Obtain the unit cost of purchasing the product from outside firm. • Determine the cost of the equipment, manpower, and all other resources required to make the product. • Estimate the net cash flows associated with the “make” option over the planning horizon. • Compute the annual equivalent cost of producing the product. • Compute the unit cost of making the product by dividing the annual equivalent cost by the required annual quantity. • Choose the option with the smallest unit cost.

  23. EXAMPLE 6.6 UNIT COST: Make or Buy B & S Company manufactures several lines of pressure washers. (pressure washer is a high pressure mechanical sprayer that can be used to remove loose paint, mold, dust, mud, and dirt from surfaces and objects such as buildings, vehicles, concrete surfaces). One unique part, an axial cam, (A device used to transform the rotary motion of the motor/ engine into the reciprocating motion of a pump's pistons) requires specialized tools that need to be replaced. Management has decided that the only alternative to replacing these tools is to acquire the axial cam from an outside source. B & S’s average usage of the axial cam is 120,000 units each year over the next five years. Buy Option: A supplier is willing to provide the axial cam at a unit sales price of $35 if at least 100,000 units are ordered annually.

  24. Make Option: If the specialized tools are purchased, they will cost $2,200,000 and will have a salvage value of $120,000 after their expected economic life of five years. With these new tools, the direct labor and variable factory overhead will be reduced, resulting the following estimated unit production cost: Assuming that the firm’s interest rate is 12%, calculate the unit cost under each option and determine whether the company should replace the old tools or purchase the axial cam from an outside source.

  25. Example 6.6 BUY OPTION: The required annual production volume is 120,000 units. AEC(12%) = ($35/unit) x (120,000 units/year) = $4,200,000 / year  Figure 6-5 Cash flows associated with make-or-buy option in manufacturing axial cams

  26. MAKE OPTION:  Capital Cost: CR(12%) = ($2,200,000 - $120,000) (A/P, 12%, 5) + (0.12) x ($120,000) = $591,412 (unit capital cost = $591,412 /120000 = $4.93) Unit Cost = 26.3 + 4.93 = $31.23 Production Cost: OC(12%) = ($26.30/unit) x (120,000) = $3,156,000 / year Total Annual Equivalent cost is: AEC(12%) = $591,412 + $3,156,000 = $3,747,412 MAKE OPTION: Unit Cost = $3,747,412 / 120,000 = $31.23 / unit  $35.00 – $31.23 = $3.77 / unit saving by making axial cam in house. $3.77 x 120,000 =$452,400 savings per year.

  27. Comparing Mutually Exclusive Projects

  28. Analysis Period Differs from Project Lives • In present worth analysis, we must have a common analysis period when mutually exclusive • One of the approaches is the replacement chain method (or least common multiple), which assumes that each project can be repeated as many times as necessary to reach a common life span; the NPW over life span are then compared and the project higher NPW is chosen • Annual worth analysis also requires establishing common analysis periods, but AE offers some computational advantages over present-worth analysis, provided the following criteria are met: • The service of the selected alternative is required on a continuous basis • Each alternative will be replaced by an identical asset that has the same costs and performance • When these two criteria are met, we may solve for AE of each project on the basis of its initial life span rather than on the basis of the infinite streams of project cash flows.

  29. Analysis Period Differs from Project Lives

  30. Analysis Period Differs from Project Lives

  31. Analysis Period Differs from Project Lives

  32. Analysis Period Differs from Project Lives

  33. Analysis Period Differs from Project Lives

  34. PRACTICE PROBLEMS 17; 28; 30; 31; 33; 34; 43;

  35. 6.17 ) You invest in a piece of equipment costing $40.000. The equipment will be used for two years, at the end of which time the salvage value of the machine is expected to be $15,000. The machine will be used for 4,000 hours during the first year and 6,000 hours during the second year. The expected savings associated with the use of the piece of equipment will be $28,000 during the first year and $40,000 during the second year. If your interest rate is 10%. a) What is the capital recovery cost? b) What is the annual recovery cost? c) What would be the net savings generated per machine hour?

  36. 6.17 ) SOLUTION

  37. 6.28) You have purchased equipment costing $20,000. The equipment will be used for two years, and at the end of two years, its salvage value is expected to be $10,000. The equipment will be used 6,000 hours during the first year and 8,000 hours during the second year. The expected annual net savings will be $30,000 during the first year and $40,000 during the second year. If your interest rate is 10%, what would be the equivalent net savings per machine hour?

  38. 6.28) SOLUTION

  39. 6.30) Georgia Mills Company (GMC) purchase a milling machine, which it intends to use for the next five years, for $180,000. This machine is expected to save GMC $35,000 during the first operating year. Then, the annual savings are expected to decrease by 3% each subsequent year over the previous year due to increased maintenance costs. Assuming that GMC would operate the machine for an average of 5,000 hours per year and that the machine would have no appreciable salvage value at the end of the five year period, determine the equivalent dollar savings per operating hour at 15% interest compounded annually.

  40. CR (15%) = (180,000) (A/P, 15%, 5) = $53,694 PW (15%) = $35,000 (P/A1, -3%, 15%, 5) = $111,427.83 AE (15%) = $111,427.83 x (A/P, 15%, %5) = $33,238.67 annual savings AE = $33,238   and CR = $53,694 Differences will be CR – AE =$20,456 lost and $20,456 / 5000 hours = $4.09 lost per hour.  OR CR / 5000 =$ 10.718 cost per hour AE / 5000 =$ 6.647 savings per hour And differences will be  CR – AE = $4.07 lost/hour 6.30) SOLUTION

  41. 6.31) You invest in a piece of equipment costing $30,000. The equipment will be used for two years, at the end of which time the salvage value of the machine is expected to be $10,000. The machine will be used for 5,000 hours during the first year and 8,000 hours during the second year. The expected annual net savings in operating costs will be $25,000 during the first year and $40,000 during the second year. If your interest rate is 10%, what would be the equivalent net savings per machine hour?

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