Westonci.ca connects you with experts who provide insightful answers to your questions. Join us today and start learning! Connect with a community of experts ready to help you find solutions to your questions quickly and accurately. Explore comprehensive solutions to your questions from knowledgeable professionals across various fields on our platform.

The Browns wish toaccumulate at least $150,000 at the time of their last deposit in a college fund for their daughter by contributing an amount A into the account at the end of each year for eighteen years. What is the smallest annual payment A that will suffice if the college fund earns a level annual effective interest rate of 5%? If at the end of ten years, it is announced that the annual effective interest rate will drop to 4.5%, how much must the Browns increase their payments in order to reach their accumulation goal? Assume that the Browns wish to continue to make level payments except for a slightly reduced final payment.

Sagot :

Answer:The smallest initial annual payment ( A ) required at a 5% interest rate is approximately $5,330.88. If the interest rate drops to 4.5% after 10 years, the Browns need to increase their annual payments by approximately $3,304.28 to reach their goal of $150,000.

Step-by-step explanation: To solve this problem, we need to break it down into two parts:

1. Finding the initial annual payment (A) with a 5% interest rate for the entire 18 year.

2. Recalculating the required payment after the interest rate changes to 4.5% after 10 years.

Part 1: Initial Annual Payment with a 5% Interest Rate

The future value of an ordinary annuity can be calculated using the formula:

[ FV = A cdot frac{(1 + r)^n - 1}{r} ]

where:

- ( FV ) is the future value of the annuity,

- ( A ) is the annual payment,

- ( r ) is the annual effective interest rate,

- ( n ) is the number of payments.

Given:

- ( FV = 150,000 ) dollars,

- ( r = 0.05 ),

- ( n = 18 ).

Substituting these values into the formula, we get:

[ 150,000 = A cdot frac{(1 + 0.05)^{18} - 1}{0.05} ]

Let's solve for ( A ):

[ 150,000 = A cdot frac{(1.05)^{18} - 1}{0.05} ]

Calculating the value of ((1.05)^{18}):

[ (1.05)^{18} approx 2.4066 ]

So,

[ 150,000 = A cdot frac{2.4066 - 1}{0.05} ]

[ 150,000 = A cdot frac{1.4066}{0.05} ]

[ 150,000 = A cdot 28.132 ]

Finally,

[ A = frac{150,000}{28.132} approx 5,330.88 ]

Part 2: Adjusted Payments After Interest Rate Drops

After 10 years, the Browns have already accumulated some amount which we need to calculate. Then we will adjust the payment for the remaining 8 years at the new interest rate.

Step 1: Calculate the accumulated amount after 10 years at 5% interest rate

Using the future value formula for the first 10 years:

[ FV_{10} = A cdot frac{(1 + r)^{10} - 1}{r} ]

[ FV_{10} = 5,330.88 cdot frac{(1.05)^{10} - 1}{0.05} ]

Calculating ((1.05)^{10}):

[ (1.05)^{10} approx 1.6289 ]

So,[ FV_{10} = 5,330.88 cdot frac{1.6289 - 1}{0.05} ]

[ FV_{10} = 5,330.88 cdot frac{0.6289}{0.05} ]

[ FV_{10} = 5,330.88 cdot 12.578 approx 67,059.62 ]

Step 2: Calculate the new annual payment needed for the next 8 years at 4.5% interest rate

The remaining amount needed is:

[ 150,000 - 67,059.62 = 82,940.38 ]

Using the future value formula with the new interest rate ( r = 0.045 ) and ( n = 8 ):

[ 82,940.38 = A_{new} cdot frac{(1 + 0.045)^8 - 1}{0.045} ]

Calculating ((1.045)^8):

[ (1.045)^8 approx 1.4323 ]

So,

[ 82,940.38 = A_{new} cdot frac{1.4323 - 1}{0.045} ]

[ 82,940.38 = A_{new} cdot frac{0.4323}{0.045} ]

[ 82,940.38 = A_{new} cdot 9.6067 ]

Finally,

[ A_{new} = frac{82,940.38}{9.6067} approx 8,635.16 ]

Increase in Payments

The Browns must increase their annual payment from ( 5,330.88 ) to ( 8,635.16 ).

[ text{Increase in payment} = 8,635.16 - 5,330.88 approx 3,304.28]

Conclusion

The smallest initial annual payment ( A ) required at a 5% interest rate is approximately $5,330.88. If the interest rate drops to 4.5% after 10 years, the Browns need to increase their annual payments by approximately $3,304.28 to reach their goal of $150,000.