Two roadway designs are under consideration for access to a permanent suspension bridge. Design 1A will cost $2.7 million to build and $175,000 per year to maintain. Design 1B will cost $3.8 million to build and $40,000 per year to maintain. Both designs are assumed to be permanent. Use an AW-based rate of return equation to determine (a) the breakeven ROR and (b) which design is preferred at an MARR of 10% per year. a) The breakeven ROR is %. b) At an MARR of 10% per year, design 1B Correctis preferred.

The answer is below

Explanation:

a)

The present cost of design 1A = 2700000 + 175000/r

The present cost of design 1B = 3800000 + 40000/r

Where r is the rate of return.

At breakeven rate of return, the present cost of both designs would be the same. Hence:

2700000 + 175000/r = 3800000 + 40000/r

3800000 - 2700000 = 175000/r - 40000/r

1100000 = 135000/r

r = 135000 / 1100000 = 0.1227

r = 12.27%

Therefore the breakeven rate of return is 12.27%

b) At an MARR of 10% per year, that is r = 0.1:

The present cost of design 1A = 2700000 + 175000/0.1 = $4.45 million

The present cost of design 1B = 3800000 + 40000/0.1 = $4.2 million

At an MARR of 10% per year, design 1B Correct is preferred because it has the lowest cost.

answer; the correct answer is (0.08);

the investment was $14,174.196

explanation:

a lump sum means that the father of the example made one deposit and leave it be.

we need to know the value of a capital that generated at 4.05% per year during 28 years the sum of $48,613.24

now clearing the formula to solve for capital we got:

and we reach the answer investment = $14,174.20