In: Economics
given
Project A
Initial cost P=3400
Service life n=3 years
Salvage value S=100
Operating cost C=2000
Project B
Initial cost P=6500
Service life n=6 years
Salvage value S=500
Operating cost C=1800
Question 1)
ECA of alternate A=- Equivalent annual cost of initial investment P for 3 years+Equivalent annual cost of Salvage value S for 3 years-Annual cost C=-P*r/(1-(1+r)^-n)+S*r/((1+r)^n-1)-C=-1367.19+30.21-2000=3336.97 Eq1
ECA of alternate B=- Equivalent annual cost of initial investment P for 6 years+Equivalent annual cost of Salvage value S for 6 years-Annual cost C=-P*r/(1-(1+r)^-n)+S*r/((1+r)^n-1)-C=-1492.45+64.80-1800=-3227.64 Eq2
From equation 1 and 2 it is clear that project B will be chosen because ECAb>ECAa.
Question 2) Since two projects have different life span so to compare the NPV So the lifespan of project A needs to be doubled to equal the six-year lifespan of Machine B by assuming that company reinvest same amount in third year of project A and cash flow for year 4, &6 will be same as year 1,2,&3.So new cashflow will be
PV of annuity =PMT*{1-(1+r)^-n}/r
PV=FV/(1+r)^n
where PMT =payment per period
n=period
r=MARR
NPV=PV -I
I=initial investment
NPV for project A=14533.41 Eq3
NPV for project B=14057.23 Eq4
From equation 3 and 4 it is clear that project B will be chosen because PVb>PVa.
Question 3)
For incremental cashflow we need to subtract cash flow of project lower initial investment amount (project A) from higher initial investment amount (project B) and calculate for IRR.
PMT*{1-(1+IRR)^-n}/IRR=I
Following table will be obtained for incremental cash flow and IRR.
From above table it is clear that incremental IRR for project B is greater that MARR(10%) so we will choose project B.