(20 Pts.) 3. A medium sized municipality wants to develop an intelligent softwar

Question

(20 Pts.) 3. A medium sized municipality wants to develop an intelligent software system to assist in project selection during the next 10 years. A life-cycle cost approach has been selected to categorize costs into development, programming, operating, and support costs for each alternative. There are three alternatives under consideration identified as A (tailored system), B (adapted system), and C (current system). The costs are summarized below. Using a present worth analysis and an interest rate of 10% per year, select the best alternative. Alternative Cost Component Cost Development Programming Operation Support Development Programming Operation Support Operation 5100,000 now, $150,000 year 1 $45,000 now, $35,000 year 1 50,000 in years 1 through 10 S 30,000 in years through $10,000 now $45,000 year 0, $30,000 year l 80,000 in years 1 through 10 40,000 in years I through 10 $150,000 in years I through 10

Explanation / Answer

ALTERNATIVE A

PW of development cost = -$100,000 - $150,000(P/F, 10%, 1)

PW of development cost = -$100,000 - ($150,000 * 0.9091)

PW of development cost = -$100,000 - $136,365 = -$236,365

PW of programming cost = -$45,000 - $35,000(P/F, 10%, 1)

PW of programming cost = -$45,000 - ($35,000 * 0.9091)

PW of programming cost = -$45,000 - $31,818.5

PW of programming cost = -$76,818.5

PW of operation cost = -$50,000(P/A, 10%, 10)

PW of operation cost = -$50,000 * 6.145

PW of operation cost = -$307,250

PW of support cost = -$30,000(P/A, 10%, 10)

PW of support cost = -$30,000 * 6.145

PW of support cost = -$184,350

PW of Alternative A = PW of development cost + PW of programming cost + PW of operation cost + PW of support cost

PW of Alternative A = -$236,365 -$76,818.5 -$307,250 -$184,350

PW of Alternative A = -$804,783.5

The Present Worth of Alternative A is -$804,783.5

ALTERNATIVE B

PW of development cost = -$10,000

PW of programming cost = -$45,000 - $30,000(P/F, 10%, 1)

PW of programming cost = -$45,000 - ($30,000 * 0.9091)

PW of programming cost = -$45,000 - $27,273

PW of programming cost = -$72,273

PW of operation cost = -$80,000(P/A, 10%, 10)

PW of operation cost = -$80,000 * 6.145

PW of operation cost = -$491,600

PW of support cost = -$40,000(P/A, 10%, 10)

PW of support cost = -$40,000 * 6.145

PW of support cost = -$245,800

PW of Alternative B = PW of development cost + PW of programming cost + PW of operation cost + PW of support cost

PW of Alternative B = -$10,000 -$72,273 -$491,600 -$245,800

PW of Alternative B = -$819,673

The Present Worth of Alternative B is -$819,673

ALTERNATIVE C

PW of operation cost = -$150,000(P/A, 10%, 10)

PW of operation cost = -$150,000 * 6.145 = -$921,750

The Present Worth of Alternative C is -$921,750

The Present Worth of Alternative A is numerically higher (in case of negative value, lowest value is considered highest).

So, according to Present Worth Ananlysis, the best alternative is Alternative A.

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