Finance
posted by Lyra .
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?

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=110001200/(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)^n1))
=13150.65*((.105(1.105^2)/(1.105^21)
=13150.65*0.580059
=$7628.16
Case K:
PV=130001200/1.1051300/1.105^21400/1.105^3
=16188.28
annual expense over 3 years
A=P*(A/P,10.5%,3)
=16188.28*((i*(1+i)^n)/((1+i)^n1))
=16188.28*((.105(1.105^3)/(1.105^31)
=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=13150.65213150.652/1.105^213150.652/1.105^4
=32741.43
Machine K:
PV=16188.2816188.28/1.105^3
=28186.42
Again, Machine K has a lower (negative) cash flow over 6 years. 
Many thanks for your help again, MathMate!