Question

B.10. The city manager proposes building a new airport terminal building six years from now at an estimated cost of $2,000,000. He proposes to raise the necessary money by charging a fee for each passenger using the existing terminal facilities for the next six years, and depositing the proceeds into an interest bearing account. The fee would go into effect two months from now and would be deposited at monthly intervals, with the first deposit made three months from now. The account will earn a nominal 6 percent annual interest to be compounded monthly. The number of passengers passing through the existing terminal for the next six years is estimated as: During years Passengers/month I through 2 3 through 6 25,000 passengers/month 30,000 passengers/month (a) How many deposits will be made? (Ans. 70) (b) How much must be collected from each passenger in order to accumulate $2,000,000 by six years from now? The charge per passenger will be constant over the entire collection period. (Ans. $0.848)

Answer 1

Part-a:

In year -1, No. of Deposits that will made = 12-2 = 10 (Since first deposit will be made three months from now)

In year – 2 = 12

In year – 3 = 12

In year – 4 = 12

In year – 5 = 12

In year – 6 = 12

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No. of (monthly) deposits that will be made = (10+12+12+12+12+12) = 70

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Part-b:

Let the amount to be collected from each passenger in order to accumulate $2,000,000 by six years from now be $x.

.

.

Annual interest = 6%

Therefore, monthly interest = 6%/12 = 0.5%

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Value after two years from now of amount collected and deposited in year 1 & year 2 = 25,000x*FVIFA(0.5%,22)

= 25,000x * 23.1944

= 579,860x

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Value after six years from now of amount collected and deposited in year 1 & year 2 = 579,860x*(1.005^48)

= 579,860x * 1.270489

= 736,705.84x

.

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Value after six years from now of amount collected and deposited in year 3 through year 6 (i.e. year-3,4,5,6) = 30,000x* FVIFA(0.5%,48)

= 30,000x*54.0978

= 1,622,934x

.

.

Total Value after six years from now of amount collected and deposited in year through year 6 (i.e. year-1,2,3,4,5,6) = 736,705.84x + 1,622,934x

= 2,359,639.845x

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.

Therefore, this value must be equal to $2,000,000.

.

.

So,

2,359,639.845x must be equal to $2,000,000

2,359,639.845x = 2,000,000

Therefore, x = 2,000,000/2,359,639.845x

x = 0.848

.

.

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So amount to be collected from each passenger in order to accumulate $2,000,000 by six years from now is $0.848