Question

A steel stamping corp is looking to buy a new high speed press. there are two offers: first offer: selling & shipping price$$1,050,000, annual operation $8,500 , maintenance $8,000 increasing $1,000 thereafter, salvage value $210,000, service life 15 years second offer: selling and shipping $1,225,000, annual oper. $7,500, maintenance $7,000 increasing $800 thereafter, salvage valu $300,000, service life 20 years. a) Which offer is better base on Annual Worth comparison b) Which offer is better using Future Worth comparison c) Do both always lead to the same result? d) for a 15-year-period, what salvage value for the press in offer 2 would make it better choice? e) state the assumption made to compare mutually exclusive alternatives of different lives?

Answer 1

ANSWER:

A) We need to find the annual worth of both the offers:

let i = 20% , n = 15 for first offer ,  

first offer:

selling and shipping price = $1,050,000

ao = $8,500

maintenace = $8,000

gradient in maintenance = $1,000

salvage value = $210,000

we will find the present worth and then the annual worth.

pw of 1st offer = selling and shipping price + ao(p/a,i,n) + maintenance(p/a,i,n) + gradient(p/g,i,n) + salvage value(p/f,i,n)

pw of 1st offer = -1,050,000 - 8,500(p/a,20%,15) -8,000(p/a,20%,15) -1,000(p/g,20%,15) + 210,000(p/f,20%,15)

pw of 1st offer = -1,050,000 - 8,500 * 4.675 - 8,000 * 4.675 - 1,000 * 18.509 + 210,000 * 0.0649

pw of 1st offer = -1,050,000 - 39,737.5 - 37,400 - 18,509 + 13,629

pw of 1st offer = -$1,132,017.5

aw of 1st offer = r * pw / (1- (1+r) ^ -n)

aw of 1st offer = 20% * -$1,132,017.5 / ( 1 - (1 + 20% ) ^ -15)

aw of 1st offer = -$226,403.5 / ( 1 - (1.2) ^ 15)

aw of 1st offer = -$226,403.5 / ( 1 - 0.0649)

aw of 1st offer = -$226,403.5 / 0.935

aw of 1st offer = -$242,118.3

second offer:

selling and shipping price = $1,225,000

ao = $7,500

maintenace = $7,000

gradient in maintenance = $800

salvage value = $300,000

we will find the present worth and then the annual worth.

pw of 2nd offer = selling and shipping price + ao(p/a,i,n) + maintenance(p/a,i,n) + gradient(p/g,i,n) + salvage value(p/f,i,n)

pw of 2nd offer = -1,225,000 - 7,500(p/a,20%,20) - 7,000(p/a,20%,20) - 800(p/g,20%,20) + 300,000(p/f,20%,20)

pw of 2nd offer = -1,225,000 - 7,500 * 4.87 - 7,000 * 4.87 - 800 * 21.739 + 300,000 * 0.0261

pw of 2nd offer = -1,225,000 - 36,525 - 34,090 - 17,391.2 + 7830

pw of 2nd offer = -$1,305,176.2

aw of 2nd offer = r * pw / (1- (1+r) ^ -n)

aw of 2nd offer = 20% * -$1,305,176.2 / ( 1 - (1 + 20% ) ^ -20)

aw of 2nd offer = -$261,035.24 / ( 1 - (1.2) ^ -20)

aw of 2nd offer = -$261,035.24 / ( 1 - 0.026)

aw of 2nd offer = -$261,035.24 / 0.9739

aw of 2nd offer = -$268,026.46

so at 20% interest rate the annual worth of ist offer is more then that of 2nd offer , so we choose the 1st offer.

B) now we will find the future worth of both the offers.

pw of 1st offer = -$1,132,017.5

fw = pw * (1+r) ^ n

r = 20% , n =15 years

fw = -$1,132,017.5 * (1+20%)^15

fw = -$1,132,017.5 * (1.2) ^ -15

fw = -$1,132,017.5 * 15.40

fw = -$17,441,018.5

pw of 1st offer = -$1,305,176.2

fw = pw * (1+r) ^ n

r = 20% , n =20 years

fw = -$1,305,176.2 * (1+20%)^20

fw = -$1,305,176.2 * (1.2) ^ 20

fw = -$1,305,176.2 * 38.33

fw = -$50,037,322.99

since the future worth of 1st offer is more then that of 2nd offer , therefore we choose the 1st offer.

c) yes both will lead to the same result as both annual and future worth are related to each other.

d) for finding this we will have to equate the pw of both the offers at 20% for 15 years as equal.

pw of 1st offer = -1,132,017.5

pw of 2nd offer = -1,225,000 -7,500(p/a,20%,15) - 7,000(p/a,20%,15) -800(p/g,20%,15) + salvage value(p/f,20%,15)

pw of 2nd offer = -1,225,000 - 7,500 * 4.675 - 7,000 * 4.675 - 800 * 18.509 + salvage value * 0.0649

pw of 2nd offer = -1,225,000 - 35,062.5 - 32,725 - 14,807.2 + salvage value * 0.0649

pw of 2nd offer = -1,225,000 -82,594.7 + salvage value * 0.0649

pw of 2nd offer = -1,307,595 + salvage value * 0.0649

now pw of 1st offer = pw of 2nd offer

-1,307,595 + salvage value * 0.0649 = -1,132,017.5

salvage value * 0.0649 = -1,132,017.5 + 1,307,595

salvage value * 0.0649 = 177,577.2

salvage value = 177,577.2 / 0.0649

salvage value = $2,705,330

so the salvage value of offer 2nd should be more then $2,705,330 to make it a better offer at a 20% rate and n = 15 years.

e) the assumption that was made was that interest rate was taken by myself and in this case as 20%.