Two processes can be used for producing a polymer that reduces friction loss in engines. Process K will have a first cost of $160,000, an operating cost of $7000 per month, and a salvage value of $40,000 after its 2-year life. Process L will have a first cost of $210,000, an operating cost of $5000 per month, and a $26,000 salvage value after its 4-year life. Which process should be selected on the basis of an annual worth analysis at an interest rate of 12% per year, compounded monthly?

process L selected.

Explanation:

Effective annual interest rate (r)

= (1 + 12% : 12) ^12 - 1

= 12.68%

K Process

Initial cost = $160000

Monthly operating cost = $7000

Salvage value after 2 year = $40000

Time = 2 years or 24 months

Annual interest rate = 12% annual compounding monthly

Monthly interest rate R = 12% : 12

= 1%

Present worth = $160,000 + $7,000 * (1 - 1 : (1 + R) ^24) : R - $40,000: (1 + R) ^24

= $160,000 + $7,000 * (1 - 1 : 1.01^24) : 0.01 - $40,000 : 1.01^24

= $277,201.06

Let, annual worth = AW

Present worth = AW * (1 - 1: (1+r) ^2) : r

= AW * (1-1 : 1.1268^2) /.1268

= AW*1.675

AW = $277,201.06 : 1.675

AW (for process K) = $165,493.17

For Process L

Initial cost = $210,000

Monthly operating cost = $5,000

Salvage value after 4 year = $26,000

Time = 4 years or 48 months

Annual interest rate = 12% annual compounding monthly

Monthly interest rate R = 12% : 12

= 1%

Present worth = 210000 + 5000 * (1 - 1 : (1 + R) ^48) : R - $26,000 : (1 + R) ^48

= $210,000 + $5,000 * (1 - 1: 1.01^48) : 0.01 - $26,000 : 1.01^48

= $383,743.03

If annual worth = AW

Present worth = $383,743.03 = AW * (1 - 1 : 1.1268^4) : 0.1268

= AW * 2.9943

AW = 383743.03 : 2.9943

AW (for process L) = $128157.84

Therefore, Process K is higher than the annual cost of process L or AW. So, process L selected.