Question

A firm is considering the purchase of one of two

new machines. The data on each are given.

Machine A Machine B

Initial cost 3400 6500

Service life 3 years 6 years

Salvage value 100 500

Net operating cost 2000/year 1800/year

If the MARR is 12%, which alternative should be

selected when using the following methods?

a. Annual equivalent cost approach ( method 2)

b. Present Worth comparison (method 2)

c. Incremental IRR comparison ( method 2)

Answer 1

Given

Project A

Initial cost A=$3400

Service life n=3 years

Salvage value S=100

Net operating cost P=2000/year

MARR r=10%

Project B

Initial cost A=$6500

Service life n=6 years

Salvage value S=500

Net operating cost C=1800/year

MARR r=10%

Question 1)

ECA of alternate A=- Equivalent annual cost of initial investment A for 3 years+Equivalent annual cost of Salvage value S for 3 years-Annual cost C

Equivalent annual cost of initial investment A for 3 years=P*r/(1-(1+r)^-n)=$3400*0.1/(1-(1+0.1)^-3)=$1367.19

Equivalent annual cost of initial investment S for 3 years=S*r/((1+r)^n-1)=100*0.1/((1+0.1)^3-1)=$30.21

ECA of alternate A=-1367.19+30.21-2000=-$3336.98 Eq 1

ECA of alternate B=- Equivalent annual cost of initial investment A for 6 years+Equivalent annual cost of Salvage value S for 6 years-Annual cost C

Equivalent annual cost of initial investment A for 6 years=P*r/(1-(1+r)^-n)=$6500*0.1/(1-(1+0.1)^-6)=$1492.45

Equivalent annual cost of initial investment S for 6 years=S*r/((1+r)^n-1)=500*0.1/((1+0.1)^6-1)=$64.80

ECA of alternate B=-1492.45+64.80-1800=-$3227.64 Eq 2

According to equation 1 and 2 it is clear that Project B will be selected because ECAb>ECAa.

Question 2)

Since project A has life cycle for 3 years and project B has life cycle for 6 years so we need to make the lifespan of project A needs to be doubled to equal the six-year lifespan of Machine B by reinvesting $3400 in third year and same cash flow for 4,5 and 6 years as cash flow in 1,2 and 3 year. So we get following cash flow for Project A and project B.

Present value of Project A=-$14533.41 eq 3

PV for annuity=PV=PMT*{1-(1+r)^-n}/r

PV =FV/(1+r)^n

Where PMT= payment per period

n= periods

r=MARR

Present value of Project B= -$14057.23 eq 4

According to equation 3 and 4 it is clear that Project B will be selected because PVb>PVa.

Question 3)

Since project A has life cycle for 3 years and project B has life cycle for 6 years so we need to make the lifespan of project A needs to be doubled to equal the six-year lifespan of Machine B.

So we get following cash flow for Project A and project B.

For incremental IRR we need to subtract lower (project A)cash flow value from higher one(project B) to get incremental cash flow.Following is the incremental cash flow and IRR.

Since Incremental IRR is greater than MARR(10%) we will choose project B.