ch6 lecture notes

8/10/2019 Ch6 Lecture Notes

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Chapter 6

Annual Worth Analysis


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Advantages of Annual Worth

Popular Analysis Technique Easily understood results are

reported in $ / time period Eliminates the LCM problemassociated with the present worth

method Only have to evaluate one life cycle of a

project

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Annual Worth Calculations

Generally, A = P(A/P, i%, n)

A = F(A/F, i%, n)

Convert all cash flows to their end of periodequivalent amounts

For comparison, in Example 5.2 about office leaseoptions, a Panalysis was performed over 18 years, theLCM of 6 and 9 years.

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A project engineer is assigned to start up a new office in acity where a 6-year contract has been finalized. Two lease

options are available, each with a first cost, annual leasecost, and deposit-return estimates shown below.

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Determine which lease option should be selected onthe basis of a present worth comparison, if the MARRis 15% per year.

Present Worth Example: Unequal lives


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Comparison must be made over equal time periods;Compare over the least common multiple, LCM.

Since the leases have different terms (service lives),compare them over the LCM of 18 years.

For life cycles after the first, the first cost is repeated in year0 of each new cycle, which is the last year of the previouscycle. These are years 6 and 12 for location A and year 9for B.

The cash flow diagram is in Figure 52.

Calculate PW at 15% over 18 years.

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PA = -15,00014,000 (P/F, 0.15, 6)14,000 (P/F, 0.15, 12) + 1000 (P/F, 0.15, 18) -3,500(P/A, 0.15,18)

= -15,00014000 (0.4323)14000 (0.1869) + 1000 (0.0808) - 3,500 (6.1280)

= - 45036 + 80.80 = - $45,036PB = -18,000 - 16,000 (P/F, 0.15, 9) +2,000 (P/F, 0.15, 18) - 3,100(P/A, 0.15, 18)

= -18,000 - 16,000 (0.2843) + 2,000 (0.0808) - 3,100(6.1280)= -$41,384

Location B is selected, since it costs less in PW terms;


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Consider only location A with a 6-year life cycle.The cash flow diagram shows the cash flows for the

3 life cycles (first cost $15,000; annual costs $3500;deposit return $1000). Demonstrate the equivalence(at i = 15%) of P over 3 life cycles and A over one

cycle. In the previous example, present worth forlocation A was calculated as P = -$45,036.

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Equation [6.1] is applied to the PW value for 18 years.

A = - 45,036 (A/P,15%,18)

A = - 45,036 (0.16139) = - $7349.42

Calculate the equivalent uniform annual worth valuefor all cash flows in the firstlife cycle.

A = -15,000(A/P,15%,6) +1000(A/F,15%,6) 3500 = - $7349

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AW and Repeatability Assumption

If two or more alternatives possessunequal lives then evaluate the AW for

any given cycle The annual worth of one cycle is the

same as the annual worth of the other

cycles (by assumption)

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CAPITAL RECOVERY COST (CR)

CR = the equivalent annual worth of the assetgiven:

Capital Recovery (CR) is the annualized equivalent of the initial

investment, P0and the annualized amount of the future salvage

value Fn

.

P0

FN

0 1 2 3 N-1 N

S


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Lockheed Martin is increasing its booster thrust power to win more satellite launch

contracts from European companies interested in opening up new globalcommunications markets. An earth-based tracking equipment is expected to require aninvestment of $13 million with $8 million committednow and the remaining $5 millionexpended at the end of year 1of the project. Annual operating costs (AOC)for thesystem are expected to start the first year and continue at $0.9 million per year. Theuseful life of the tracker is 8 yearswith a salvage value of $0.5 million.

Calculate the AW value for the system, if the corporate MARR is currently 12%per year.

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The cash flows for the tracker system must be converted to an equivalent AW cash flowsequence over 8 years. The AOC is A = -$0.9 per year, and the capital recovery iscalculated by using Equation [6.3]. The present worth P in year 0 of the two separate

investment amounts of $8 and $5 is determined before multiplying by the A/P factor.

The correct interpretation of this result is very important to Lockheed Martin. Itmeans that each year for 8 years, the equivalent total revenue from the trackermust be at least $2,470,000 just to recover the initial present worth investment plus

the required return of 12% per year.This does not include the AOC of $0.9 million each year.

A = - 2.47 - 0.9 = $3.37 million per year

A = - 8 (A/P, 12%, 8) - 5 (P/F, 12%, 1) (A/P, 12%, 8) + 0.5 (A/F, 12%, 8) - 0.9

A = - 8 (0.2013) - 5 (0.8929) (0.2013) + 0.5 (0.0813) - 0.9 = -$3.36845 million

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The owner of PizzaRush.com plans to purchase and install 5portable,in-car systems to increase delivery speed and accuracy. The systemsprovide a link between the web order-placement software and the On-Star system for satellite-generated directions to any address in theLos Angeles area. The expected result is faster, friendlier service to

customers, and more income for PizzaRush.Each system costs $4600, has a 5-year useful life, and may besalvaged for an estimated $300. Total operating cost for all systems is$650for the first year, increasing by$50per year thereafter. The MARRis 10%.Perform an annual worth evaluation.

What annual incremental income is necessary to recover the investmentat the MARR of 10% per year?

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CR=- 5(4600)(A/P,10%, 5) + 5(300)(A/F,10%, 5) = -$5822

A=- 5(4600)(A/P,10%, 5) - 650 + 5(300)(A/F,10%, 5) - 50(A/G,10%, 5)

A =- 5(4600)(0.2638) - 650 +5(300)(0.1638) - 50(1.8101) = -$6562.21

A = - capital recovery + equivalent net income


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Annual Worth of a Permanent Investment

This section discusses the annual worth equivalent of the capitalizedcost.

Evaluation of public sector projects (flood control dams, irrigationcanals, bridges, or other large-scale projects), requires the comparison

of alternatives that have such long lives that they may be consideredinfinite in economic analysis terms.

For this type of analysis, the annual worth of the initial investment is theperpetual annual interest earned on the initial investment, that is,

CR = A =Pi. This is Equation [5.3]; however, theA value is also the capital recovery

amount.

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The U.S. Bureau of Reclamation is considering three proposals for increasing thecapacity of the main drainage canal in an agricultural region of Nebraska.

ProposalArequires dredging the canal to remove sediment and weeds that have

accumulated during previous years operation. The capacity of the canal will have tobe maintained in the future near its design peak flow because of increased waterdemand. The Bureau is planning to purchase the dredging equipment andaccessories for $650,000. The equipment is expected to have a 10-year life with a$17,000 salvage value. The annual operating costs are estimated to total $50,000. Tocontrol weeds in the canal itself and along the banks, environmentally safe herbicideswill be sprayed during the irrigation season. The yearly cost of the weed controlprogram is expected to be $120,000.

Proposal Bis to line the canal with concrete at an initial cost of $4 million. The lining isassumed to be permanent, but minor maintenance will be required every year at acost of $5000. In addition, lining repairs will have to be made every 5 years at a cost

of $30,000. Proposal Cis to construct a new pipeline along a different route. Estimates are: an

initial cost of $6 million, annual maintenance of $3000 for right-of-way, and a life of 50years.

Compare the alternatives on the basis of annual worth, using an interest rate of 5%per year.

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CR = A =Pi.

ch6 lecture notes

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