13-Solved problems, Deposit value for uniform series.

How to find the deposit value for uniform series?

The first solved problem, find the deposit value for uniform series in terms of n, I, and F.

The title is College Savings Plan: find A, given F, N, and I. F is the future value, N is the number of years, number of deposits, number of time periods and i is the interest.

You want to set up a college savings plan for your daughter. She is currently 10 years old and will go to college at the age of 18.

She needs in the bank at least an amount of $100,000. To do this in action, you need to deposit in the bank.

The bank gives a rate of interest equal to 7%. The deposits are done at the end of each year with equal deposits. This is the case of uniform series or A. For the factor used (A/F), it is called the sinking fund factor.

You need $100,000, this sum is the future value, A is unknown, given n and interest rate i%.N value is 8 deposits in eight years and i=7%.

The first solved problem, find the deposit value for uniform series -using cash flow moment.

Suppose we start by using the method of cash flow moment at a pivot point. The pivot point is t=0. The moment caused by the future value F will be equivalent to the moment caused by the uniform series of deposits at the same point.

To get the moment due to the future value, it is equal to F as positive, and the point is at the left side of F; the moment arm to pivot point t=0 will be(1+i)^-8. The moment value is $100,000*(1.07)^-8.

While for the moment, caused by the uniform series of deposits. It will be as -A*(1+i)^-8+(-A)*(1+i)^-7 and so on till we complete the deposit expression. The sum of moments is zero. After adjustments the result can be found as A is multiplied by 5.97129.

The product of A*5.97129 will be equal to $100,000/(1.07)^8=$100,000/1.7181. We simplify the result as 100,000/10.2598=$974.78. The final value of A is $974.78 to be paid at the end of each year for 8 years to get the future value of $100,000.

The first solved problem, find the deposit value for a uniform series and use A/F.

The second method is by using the equation, A =$100,000*(A/F,7%,8)=$9746.71. The value is the same as estimated by the cash flow moment method.

A/F can be estimated by using the formula, A/F=(i)/((1+i)^n-1), i=7%,n=8. We substitute to get A/F as (0.07)/((1.07)^8-1))=0.09746. The value will be multiplied by $100,000, to get the A value. A value will be $100,000*0.09746=$9746.78.

The first solved problem, find the deposit value for uniform series, i=10%.

The next method is by using the interest rate table. We use the interest rate table of 7%; we have an n value equal to 8. We need to find the A/F value. Check the sinking fund factor A/F for finding A in terms of F. We go vertically and find the intersection with n=8; we get the value of 0.0975. This number matches the previous calculations.

The first solved problem, find the deposit value for uniform series, use Excel function PMT.

Another way to get the value of A/F is by using the Excel function P.M.T. The P.M.T. (rates, which is 0.07, Nper is 8 deposits, PV stands for present value,0, Fv stands for the future value which is $100,000, type is zero). The A value is by a minus sign since uniform series are outflows.

If we use an Excel sheet, use cells. Cell C19 has an interest rate of 7%. The N periods are 8. Cell C21 is for a

future value that is equal to 100,000. In the last cell, C23, we can get the A value as a minus. As we can see, I have taken a screenshot of the cell.

PMT(C19,C20,0,+C21). The A value can be obtained by using the Excel sheet.

The second solved problem, find the deposit value for uniform series -Use cash flows moment.

Let us check another problem 2.12, paying off an education Load: Find A, Given P,I, and N. This is another way to get A, but this time in terms of P,  where P is the present value. You borrow from the bank to finance your educational expenses. The loan should be paid back to the bank during the next five years.

The rate of interest given is 6% yearly. the payment should be in equal annual installments. The loan will be paid off over five years. As we know for the uniform series, paying is done starting after one year after getting the loan.

The first installment will be due a year later. It is required to estimate the amount of the annual installments A. Refer to figure 2.20. This solved problem requires estimating A in terms of P, while the previously solved problem requires estimating A in terms of F.

In this solved problem, I and n are given. First, we will use the method of the estimate of the moment of cash flow about the pivot point. Take the pivot point at time t=0. The P is pointing upwards; then it is +P.

The moment arm for P is (1+i)^0, whether from right or left, the P coincides with the point, and the arm length is equal to 1.

. The sum of moments is equal to zero. P will be as L.H.S, and the product of A by the arm will be set as R.H.S with a changing sign.
The negative exponents are considered fractions.

The equation is modified as we can see , P=A*(1/(1.06)+1/1.06^2)+1/(1.06)^3++1/(1.06)^4+1/(1.06)^5)). The P-value is $21061.82. we equate to the correct value (4.2123)*A. Finally, the A value is $5000.

This is another alternative, by using the pivot point at time t=1. The sum of moments is shown, based on the selected pivot point. Please refer to the right side of the slide picture for the sake of comparison between cash flow moments based on a pivot point at t=0, and at a pivot point at t=1. Finally, the A value is $5000.

The second solved problem, find the deposit value for uniform series, but use the expression A/P.

Here we use an equation, where A=P(A/P, i,n), P is given A is unknown. P=$21061.82. A/P value can be obtained from the known equation as equal to (i*(1+i)^n/(1+i)^n-1)). The A/P value is equal to 0.2374. A/P will be multiplied by P. The deposit value A =21061*0.2374=$5000. The same value for A matches with the previous calculation by the moment of cash flows.

This is the previous post-solved problem value for uniform series deposits.

This is a link to the PDF file used for the illustration of this post.

This is a good link -Engineering Economy. Applying Theory to Practice.