ANNUAL EQUIVALENCE ANALYSIS v Annual equivalent criterion v

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ANNUAL EQUIVALENCE ANALYSIS v Annual equivalent criterion v Applying annual worth analysis

ANNUAL EQUIVALENCE ANALYSIS v Annual equivalent criterion v Applying annual worth analysis

Annual Equivalent Worth Criterion v AE worth criterion provides a basis for measuring investment

Annual Equivalent Worth Criterion v AE worth criterion provides a basis for measuring investment worth by determining equal payments on annual basis. v Any lump-sum cash amount can be converted into a series of equal annual payments v 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) v We use this formula to evaluate the investment worth of projects. v Therefore, AE criterion provides basis for evaluating a project that is consistent with the PW criterion.

Annual Equivalent Worth Criterion v v Single Project Evaluation: The accept-reject decision rule for

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

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

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

Example 6. 1 Since energy savings are in two different geometric gradient series, we

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 Savings in electricity= $14, 000 (P/A 1, 4%, 10%, 12) P = $114, 301 Savings in coal usage PSavings in coal usage= $40, 950 (P/A 1, 5%, 10%, 12) P = $350, 356

Example 6. 1 Net present worth calculation: PW(10%) = $114, 301 + $350, 356

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.

Benefits of Annual Equivalent Analysis v In the real world situations, AE analysis is

Benefits of Annual Equivalent Analysis v In the real world situations, AE analysis is preferred, or demanded, over NPW analysis. v For example, corporations issue annual reports and develop yearly budgets. v 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.

v When only costs are involved, the AE method is called the annual equivalent

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

v Operating costs are incurred by operation of physical plants or equipment needed to

v 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. v Capital costs ( or ownership costs) are incurred by the purchasing of assets to be used in production and service. v Normally, Capital costs are nonrecurring (one time costs), where as operating costs recur as long as an asset is owned and being utilized. v Annual equivalent of a capital cost is given special name: Capital Recovery cost, designated CR ( i ). 9

Definition: The cost of owning a piece of equipment is associated with two different

Definition: The cost of owning a piece of equipment is associated with two different amounts - (1) initial cost ( I ) and (2) its salvage value ( S ) Capital costs: Taking these amounts into account, we calculate the capital costs as: Capital (Ownership) Costs S 0 N The first term (I – S) (A/P, i, N) implies that the balance (I – S ) will be paid in equal installments over the N-period at a rate of i, and the second term implies that simple interest in the amount i S is paid on until S is repaid. I 0 1 2 3 N CR ( i ) Figure 6 -2 Calculation of capital recovery cost

Table 6. 2 Will your car hold its value? SEGMENT BEST MODELS ASKING PRICE

Table 6. 2 Will your car hold its value? SEGMENT BEST MODELS ASKING PRICE AFTER 3 YEARS CR ( 6% ) Compact car Mini Cooper $19, 800 $12, 078 $3, 614 Midsize car Volkswagen Passat $28, 872 $15, 013 $6, 086 Sports car Porsche 911 $87, 500 $48, 125 $17, 618 Near luxury car BMW 3 Series $39, 257 $20, 806 $8, 151 Luxury car Mercedes CLK $51, 275 $30, 765 $9, 519 Minivan Honda Odyssey $26, 876 $15, 051 $5, 327 Subcompact SUV Honda CR-V $20, 540 $10, 681 $4, 329 Compact SUV Acura MDX $37, 500 $21, 375 $7, 315 Full size SUV Toyota Sequoia $37, 842 $18, 921 $8, 214 Compact truck Toyota Tacoma $21, 200 $10, 812 $4, 535 Full size truck Toyota Tundra $25, 653 $13, 083 $5, 488 CR(6%) = ($19, 800 - $12, 078)(A/P, 6%, 3) + (0. 06) $12, 078 = $3, 613. 55

Example 6. 2 Annual Equivalent Worth: Capital Recovery Cost Figure 6 -3 Justifying an

Example 6. 2 Annual Equivalent Worth: Capital Recovery Cost Figure 6 -3 Justifying an investment based on the AE method

Annual Worth Analysis

Annual Worth Analysis

Unit-Cost or Unit-Profit Calculation In many situations, we need to know the unit profit

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: 1. Determine the number of units to be produced (or serviced) each year over the life of the asset. 2. 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. 4. 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.

Example 6. 3 Unit Profit per Machine Hour when Annual Operating Hours Remain Constant

Example 6. 3 Unit Profit per Machine Hour when Annual Operating Hours Remain Constant PW (15%) = -$1, 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

MAKE – OR – BUY DECISION v Make or buy problems are among the

MAKE – OR – BUY DECISION v 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. v 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. v 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. v This unit cost comparison requires the use of annual worth analysis.

v The specific procedure is as follows: v v v v Determine the time

v The specific procedure is as follows: v v v v 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.

 AEC(12%) = ($35/unit) x (120, 000 units/year) = $4, 200, 000 / year

AEC(12%) = ($35/unit) x (120, 000 units/year) = $4, 200, 000 / year Example 6. 5 Figure 6 -5 Cash flows associated with make-or-buy option in manufacturing axial cams

EXAMPLE 6. 5 SOLUTION: BUY OPTION: The required annual production volume is 120, 000

EXAMPLE 6. 5 SOLUTION: BUY OPTION: The required annual production volume is 120, 000 units. AEC(12%) = ($35/unit) x (120, 000 units/year) = $4, 200, 000 / year MAKE OPTION: Capital Cost: CR(12%) = ($2, 200, 000 - $120, 000) (A/P, 12%, 5) + (0. 12) x ($120, 000) = $591, 412 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 BUY OPTION: Unit Cost = $35. 00 / unit 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 x 5 years = $2, 262, 000 saving for five years.

SOLVED PROBLEMS 3; 4; 7; 16; 24; 28; 31; 35;

SOLVED PROBLEMS 3; 4; 7; 16; 24; 28; 31; 35;

PROBLEM 6. 3 n An An Investment Revenue Consider the following cash flows and

PROBLEM 6. 3 n An An Investment Revenue Consider the following cash flows and 0 -$25, 000 compute the equivalent annual worth at i = 12% 1 -$10, 000 SOLUTION $14, 000 2 $13, 000 3 $13, 000 4 $13, 000 5 $8, 000 6 $5, 500

PROBLEM SOLUTION

PROBLEM SOLUTION

PROBLEM SOLUTION PW (9%) = -$1, 000 + $800 (P/F, 9%, 1) + $1,

PROBLEM SOLUTION PW (9%) = -$1, 000 + $800 (P/F, 9%, 1) + $1, 100 (P/F, 9%, 2) + $1, 600 (P/F, 9%, 3) + $800 (P/F, 9%, 4) + $800 (P/F, 9%, 5) PW (9%) = -$1, 000 + $734 + $926 + $1, 236 + $567 + $520 = $2, 983 AE (9%) = $2, 983 (A/P, 9%, 5) = $766. 92

PROBLEM SOLUTION

PROBLEM SOLUTION

PROBLEM SOLUTION

PROBLEM SOLUTION

PROBLEM SOLUTION

PROBLEM SOLUTION

PROBLEM SOLUTION

PROBLEM SOLUTION

SOLUTION

SOLUTION

6. 35) A large state university, currently facing a severe parking shortage on its

6. 35) A large state university, currently facing a severe parking shortage on its campus, is considering constructing parking decks off campus. A shuttle service composed of minibuses could pick up students at the off-campus parking deck and quickly transport them to various locations on campus. The university would charge a small fee for each shuttle ride, and the students could be quickly and economically transported to their classes. The funds raised by the shuttle would be used to pay for minibuses, which cost about $150, 000 each. Each minibus has a 12 -year service life, with an estimated salvage value of $3, 000. To operate each minibus, the following additional expenses must be considered: If student pay 10 cents for each ride, determine the annual ridership (i. e. , the number of shuttle rides per year) required to justify the shuttle project, assuming an interest rate of 6%.

SOLUTION ITEM ANNUAL EXPENSES Driver $40, 000 Maintenance $7, 000 Insurance $2, 000

SOLUTION ITEM ANNUAL EXPENSES Driver $40, 000 Maintenance $7, 000 Insurance $2, 000

HOMEWORK PROBLEMS 21, 27, 29, 30, 38

HOMEWORK PROBLEMS 21, 27, 29, 30, 38