### Simple Time Value of Money Answers

1. Assuming end of year payments, your annual deposit is:

A = 1,000,000*(A/F, 0.1, 40) = 1,000,00*.002259 = \$2,259.

Without interest this corresponds to less than \$200 per month.

2. After 20 years deposited at 10% interest, the money will grow to:

F = 10000*(F/P, 0.1, 20) = 10000*6.7275 =\$67,274

3. First calculate the monthly payment. Note that since payments are monthly, interest is charged every month. This is the same as compounding once a month. Since the periods are expressed in months, the interest must be expressed per month. The interest per month is 0.09/12 or 0.0075, 0.75% per month.

A = 100000(A/P, 0.0075, 360) = 100000*0.008046 = 804.62

The sum of all payments is:

Total Payments 804.62*360 = 289,663

The principal paid is \$100,000. The remaining \$189,663 is interest. It costs a lot of money to borrow so much for so long.

4. A loan is money received now in return for payments in the future. Since the payments are a uniform series. The equivalent of a uniform series moved to one period before the series begins is found with the P/A factor. Again, we use a monthly interest rate since the payments are monthly.

P = 300(P/A, 0.005, 60) = 300*51.725 = \$15,517

This assumes that the first payment occurs one month after the loan is taken out.

5. Money that is to be received in the future has a smaller equivalent value at the present. We discount the future amount with the P/F factor.

P = 1,000,000(P/F, 0.1, 10) = 1,000,000*0.385543 = \$385,543

6. Here the interest rate is 3% every six months. Since the payments are every six months we identify a 6 month interval as one period. The payments start one period (six months) from now with a payment of \$1,100. The first \$1,000 is already gone. Two periods from now, the payment will increase by \$100, three periods by \$200, and so-on for the remaining 7 periods.

The cash flow is a uniform series of \$1,100, plus a uniform gradient of \$100. You must have in the bank the equivalent of the cash flow computed with an interest rate of 3% per period.

P = 1100(P/A, 0.03, 7) + 100(P/G, 0.03, 7)

= 1100*6.230283 + 100*17.9547

P = \$8,648.78

7. We must find the future equivalent of the monthly payment. Since the account is compounded monthly, the cash flow is expressed in month. The monthly interest rate is 0.5%, The savings will grow to:

F = 100(F/A, 0.005, 216) = 100*387.35 = \$38,735.

Since F/A factor for 216 is probably not in the tables, you would use the formula for the factor to compute the value.

8. The three different single payments can be considered individually and the results added. Note that the F/P factor moves money from one point in time to a later point in time. The N value in the factors is the number of periods moved.

F = 1000(F/P, 0.08, 10) + 1500(F/P, 0.08, 8) + 800(F/P, 0.08, 6) = 1000*2.158 + 1500*1.850 +800*1.568 = \$6,187

9.

Withdrawals are A11=\$1000, A12 = \$1000*(1.06), A13 = \$1000*(1.06)^2, A14 = \$1000*(1.06)^3, A15 = \$1000 *(1.06)^4. (The ^ indicates exponentiation.)

This is geometric series starting at \$1000 and growing at a 6% rate. Use the geometric series factor to find the quivalent worth at period 10, using i=8% and g = 6%.

P=1000(P/A, 0.06, 0.08, 5) = \$4,461.

Assuming that the uniform deposits are made at the end of the year, the deposits must grow to this value. Use the A/F factor to compute the deposit.

A = 4461 (A/F,0.08,10) = 4461*0.069 = \$307.9.

This is the annual deposit.