Question

A new bridge across the Cumberland River is being planned. The construction (first) cost of the bridge is $1,900,000 and annual upkeep is estimated to be $25,000. In addition, major maintenance work is anticipated every eight years at a cost of $350,000 per occurrence. The town government’s MARR is 8% per year.

1. For this problem, what analysis period (N) is, practically speaking, defined as forever?

2. If the bridge has an expected life of 50 years, what is the capitalized

worth (CW) of the bridge over a 100‐year study period?

(I need to solve it using the application of CW Concepts)

Answer 1

ANSWER

Part 1) The factor (A/P, i, N) is called equal payment series capital recovery factor. It is used when the objective is to find the annual equivalent amount (A) which is to be recovered at the end of every interest period for N interest periods for a loan (P) which is sanctioned now at an interest rate of i.

So, when N reaches 100, the factor attains the value of 0.08 which is equivalent to the interest rate of 8%. That is why, N = 100 years is, for practical purposes, forever in this example.

Part 2)

We have the following information

Initial Cost	$1,900,000
Annual Upkeep Cost	$25,000
Major Maintenance Work (every 8 years)	$350,000

Minimum acceptable rate of return (MARR) = 8% per annum or 0.08

Life = 100 years (given)

Present Worth (8%) = 1,900,000 + 25,000(P/A, 8%, 100) + 350,000(P/F, 8%, 8) + 350,000(P/F, 8%, 16) + 350,000(P/F, 8%, 24) + 350,000(P/F, 8%, 32) + 350,000(P/F, 8%, 40) + 350,000(P/F, 8%, 48) + 350,000(P/F, 8%, 56) + 350,000(P/F, 8%, 64) + 350,000(P/F, 8%, 72) + 350,000(P/F, 8%, 80) + 350,000(P/F, 8%, 88) + 350,000(P/F, 8%, 96)

Present Worth (8%) = 1,900,000 + (25,000 × 12.494) + (350,000 × 0.5403) + (350,000 × 0.2919) + (350,000 × 0.1577) + (350,000 × 0.0852) + (350,000 × 0.0460) + (350,000 × 0.0249) + (350,000 × 0.0134) + (350,000 × 0.0073) + (350,000 × 0.0039) + (350,000 × 0.0021) + (350,000 × 0.0011) + (350,000 × 0.0006)

Present Worth (8%) = $2,623,410.19

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