AREA 1 PROBLEM 58: FUTURE WORTH ANALYSIS
AN AGRICULTURAL COOPERATIVE IS EVALUATING WHETHER TO PURCHASE A SOLAR-POWERED IRRIGATION SYSTEM. THE COST OF THE SYSTEM IS ₱100,000 TODAY. IT WILL SAVE THE COOPERATIVE ₱25,000 ANNUALLY IN DIESEL FUEL COSTS FOR THE NEXT 5 YEARS. IF THE INTEREST RATE IS 12%, IS THE PROJECT ECONOMICALLY FEASIBLE BASED ON FUTURE WORTH ANALYSIS?
a. Yes, FW = ₱11,400.50 (Feasible)
b. Yes, FW = ₱23,450.20 (Feasible)
c. No, FW = -₱31,240.20 (Not Feasible)
d. No, FW = -₱17,450.15 (Not Feasible)
a. Yes, FW = ₱11,400.50 (Feasible)
b. Yes, FW = ₱23,450.20 (Feasible)
c. No, FW = -₱31,240.20 (Not Feasible)
d. No, FW = -₱17,450.15 (Not Feasible)
CORRECT ANSWER: D. NO, FW = -₱17,450.15 (NOT FEASIBLE)
Given Parameters:• Initial Cost (P) = ₱100,000
• Annual Savings (A) = ₱25,000
• Interest Rate (i) = 12% or 0.12
• Project Life (n) = 5 years
Step 1: Compound Initial Cost to Future Worth (FW_cost)
FW_cost = P × (1 + i)ⁿ
FW_cost = 100,000 × (1.12)⁵
FW_cost = 100,000 × 1.76234 = ₱176,234.17
Step 2: Convert Annual Savings to Future Worth (FW_savings)
FW_savings = A × [ ((1 + i)ⁿ - 1) / i ]
FW_savings = 25,000 × [ ((1.12)⁵ - 1) / 0.12 ]
FW_savings = 25,000 × [ 0.76234 / 0.12 ]
FW_savings = 25,000 × 6.35285 = ₱158,784.02
Step 3: Calculate Total Future Worth (FW_net)
FW_net = FW_savings - FW_cost
FW_net = 158,784.02 - 176,234.17 = -₱17,450.15
Conclusion:
Since the net Future Worth is negative, the return on the investment does not satisfy the 12% threshold rate, making the project not economically feasible.
A project is considered feasible in FW analysis only if the Net Future Worth is greater than or equal to zero.

0 Comments