Question

The PW-based relation for the incremental cash flow series to find ?i* between the lower first-cost alternative X and alternative Y has been developed. 0 = -42,000 + 9000(P/A,?i*,10) + ( -5000(P/F,?i*,10)) Determine the highest MARR value for which Y is preferred over X. Any MARR value greater than % favors Y.

Answer 1

Sol:-

Here,Present Worth is given as:0=-42,000+9,000(P/A,i*,10)+(-5000(P/F,i*,10))

We can rewrite the PW equation as:

0=-42,000+9000{(1+i)^10-1)}/i(i+1)^10-5000/(1+i)^10

0=-42,000+9000(i+1)^10-9000/i(i+1)^10-5000/(i+1)^10

0=-42,000+9000(i+1)^10/i(i+1)^10-9000/i(i+1)^10-5000/(i+1)^10

0=(-42,000+9000i-9000-5000i)/i(1+i)^10

0=-42,000+9000i-9000-5000i

0=-51,000+4000i

51,000=4000i

51,000/4000=i

12.75 =i*

Here i represents the MARR in the calculation of Present Worth(PW)

Hence,for the Present Worth(PW) of the property to be zero the greatest value of MARR or the Minimum Acceptable Rate of Return has to be 12.75