Question

A Research& Development Company is deciding to select one of the two prototypes to fully implement for the market. The costs associated with each prototypes are shown below. The MARR is 16% per year.

B	A	Alternative
$240,500	$248,000	Initial R&D costs
23,500	36,000	Non-recurring investment costs (Years 1, 3, and 5)
$7,500 in year 4, Increasing by $250 until year 10	$8,250 in year 4, Increasing by $200 until year 10	Recurring costs
$1500	$2500	Annual maintenance costs
$1500	$1750	Disposal costs
14	14	Life, years

a. Classify the alternatives as investment or cost.

b. Calculate the future worth of the alternatives (stating standard notation for interest factors used in engineering economy).

c. Which model should be selected? Explain your decision.

Answer 1

a.

As only costs are given, so given alternatives are costs

b.

FW of Alternative A = -248000*(F/P,16%,14) - 36000*((F/P,16%,13) + (F/P,16%,11) + (F/P,16%,9)) - 8250*(F/A,16%,7)(F/P,16%,4) - 200*(F/G,16%,7)(F/P,16%,4) - 2500*(F/A,16%,14) - 1750

= -248000*7.987518 - 36000*(6.885791 + 5.117265 + 3.802961) - 8250* 11.413873*1.810639 - 200*27.586708*1.810639 - 2500*43.671987 - 1750

= -2841338.79 ~ -2841339

FW of Alternative B = -240500*(F/P,16%,14) - 23500*((F/P,16%,13) + (F/P,16%,11) + (F/P,16%,9)) - 7500*(F/A,16%,7)(F/P,16%,4) - 250*(F/G,16%,7)(F/P,16%,4) - 1500*(F/A,16%,14) - 1500

= -240500*7.987518 - 23500*(6.885791 + 5.117265 + 3.802961) - 7500* 11.413873*1.810639 - 250*27.586708*1.810639 - 1500*43.671987 - 1500

= -2526932.88 ~ -2526933

c.

As Future cost of Alternative B is less, it should be selected