i=10.5%=.105
n=2 for machine J, =3 for machine K
We assume machine J lasts 2 years with no salvage value, same for machine K, but three years.
Case J:
PV=-11000-1200/(1.105)-1300/1.105^2=
=-13150.65
annual expense over 2 years
A=P*(A/P,10.5%,3)
=-13150.65*((i*(1+i)^n)/((1+i)^n-1))
=-13150.65*((.105(1.105^2)/(1.105^2-1)
=-13150.65*0.580059
=-$7628.16
Case K:
PV=-13000-1200/1.105-1300/1.105^2-1400/1.105^3
=-16188.28
annual expense over 3 years
A=P*(A/P,10.5%,3)
=-16188.28*((i*(1+i)^n)/((1+i)^n-1))
=-16188.28*((.105(1.105^3)/(1.105^3-1)
=-16188.28*0.40566
=-$6566.92
Since annual expense for machine K is lower, it is a better buy.
Another way to do this is compare the present values of buying 3 machine J with the PV of 2 machine K (total 6 years in each case).
It is less realistic because the machines may not cost the same 2,3 or 4 years from now.
Here it is anyway:
Machine J:
PV=-13150.652-13150.652/1.105^2-13150.652/1.105^4
=-32741.43
Machine K:
PV=-16188.28-16188.28/1.105^3
=-28186.42
Again, Machine K has a lower (negative) cash flow over 6 years.
The following are the investment and operating costs for two machines, J and K.
Year: 0, 1, 2, 3
J: 11000, 1200, 1300
K: 13000, 1200, 1300, 1400
The WACC is 10.5%. How do I use the WACC to determine which machine is a better buy?
2 answers
Many thanks for your help again, MathMate!
1 answer