Compound Amount Factor and Present Worth Factor identity: what is the product of CAF(S/P) and PWF(S/P) for the same interest rate and period count?
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A1/2
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B1
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C1/3
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D1/4
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EIt depends on n
Answer
Correct Answer: 1
Explanation
Introduction / Context:Engineering-economy factors come in inverse pairs. Understanding these dualities prevents algebra mistakes and speeds problem solving. The Compound Amount Factor from Present (often denoted F/P or CAF(S/P)) and the Present Worth Factor from Future (P/F or PWF(S/P)) are classic inverses.
Given Data / Assumptions:
- Interest rate i and period count n are the same for both factors.
- CAF(S/P) ≡ (1 + i)^n.
- PWF(S/P) ≡ 1 / (1 + i)^n.
Concept / Approach:By definition, a present amount P grows to F = P * (1 + i)^n; conversely, a future amount F discounts to P = F / (1 + i)^n. Multiplying the growth and discount factors cancels the compounding effect, returning the identity value 1. This property holds for any positive i and integer n (and indeed for real n under continuous compounding analogs with appropriately defined factors).
Step-by-Step Solution:
Write CAF(S/P) = (1 + i)^n.Write PWF(S/P) = 1 / (1 + i)^n.Multiply: (1 + i)^n * [1 / (1 + i)^n] = 1.Verification / Alternative check:
Test numerically with i = 10%, n = 5: CAF = 1.61051; PWF = 0.62092; product ≈ 1.000.Why Other Options Are Wrong:
Values 1/2, 1/3, 1/4 incorrectly imply dependence on n; the true product is exactly 1 for matching i and n.Common Pitfalls:
Mixing i or n between factors; the identity only holds when both factors use the same i and n.Final Answer:
1