Post a New Question

Finance

posted by .

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?

  • Finance -

    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.

  • Finance -

    Many thanks for your help again, MathMate!

Respond to this Question

First Name
School Subject
Your Answer

Similar Questions

More Related Questions

Post a New Question