Hinson's Homegrown Farms needs a new irrigation system. System one will cost $145,000, have annual maintenance costs of $10,000, and need an overhaul at the end of year six costing $30,000. System two will have first-year maintenance costs of $5000 with increases of $500 each year thereafter. System two would not require an overhaul. Both systems will have no salvage value after 12 years. If Hinson's cost of capital is 4%, using annual worth analysis determine the maximum Hinson's should be willing to pay for system two. Contributed by Ed Wheeler, University of Tennessee at Martin

System I:

Calculate the present worth as shown below:

Present Worth = - Cost - Annual Cost (P/A, I, n) - overhaul (P/F, I, n)

= - 145000 - 10000 (P/ A, 4%, 12) - 30,000 (P/ F, 4%, 12)

= - 145000 - 10000 (9385) - 30000 (0.6246)

= - 145000 – 93850 - 18738  

= - $257588

Calculate the annual worth as shown below:  

Annual Worth PW (A/P, I, n)

= - 257588 (A/ P, 4%, 12)

= - 257588 × 0.1066

= - 27458.88

System 2:

Calculate the present worth as shown below:

Plasma Worth = - Annual Cost (P/A, I, n) – Growth (P/G, I, n)  

= - 5000 (P/A, i, n) – 500 (P/G, 4%, I2)

= - 5000 (9385) – 500 (47248)

= - 46925 - 23624

= - 70549

Calculate the annual worth as shown below:  

Amoral Worth = PW (A/P, I, o)

= - 70549 (A/P, 4%, 12)

= - 70549 x 0.1066

= - 7520.52

Therefore, Hinson would be willing to pay 7520.52 far system two.