In engineering economics, if S is the future capital accumulated after n years at an annual interest rate i (compounded annually), what is the present worth (P) of that amount?

Introduction / Context:
Present worth (P) converts a known future amount (S) to its equivalent value today using the time value of money. This fundamental concept enables fair comparison of alternatives that occur at different times—vital in project evaluation, equipment replacement, and lifecycle costing.

Given Data / Assumptions:

• S: future amount received at year n.
• i: annual interest rate (effective per year).
• Compounding: annual.

Concept / Approach:

The present worth factor is the reciprocal of the compound amount factor. Future amount is related by S = P * (1 + i)^n. Solving for P gives P = S / (1 + i)^n. This discounts S back n periods at rate i to reflect opportunity cost of capital.

Step-by-Step Solution:

Verification / Alternative check:

Check with numbers: If S = ₹121 at n = 2 and i = 10% (0.10), P = 121 / 1.1^2 = 121 / 1.21 = ₹100.

Why Other Options Are Wrong:

• S(1 + i)^n increases the future value further; not discounting.
• S(1 + i)^(1/n) does not represent discounting over n periods.
• S * i * n is simple interest, not present worth of a future lump sum.
• 'None of these' is incorrect because a correct expression exists.

Common Pitfalls:

• Confusing simple and compound interest.
• Forgetting to use effective annual rate when nominal rates compound intra-year.

Final Answer:

S / (1 + i)^n