# 12-Solved problem-Future value for uniform series deposits.

## The solved problem, How to find Future value for uniform series deposits?

### Solve For Future value for uniform series deposits in terms of n, I, A.

The solved problem is quoted from William G. Suliman’s book Engineering Economy 16th edition, Chapter 4. The solved problem can be written in the engineering economy in two ways; the first way deals with the lending and borrowing point of view. A is drawn downward as a deposit with a negative sign (-A).

The A direction is from the point of view of the man who deposits. Suppose eight annual deposits of \$187.45 each are placed in an account. How much money has accumulated immediately after the last deposit?

The money that the man will receive will be denoted by F as a future value and drawn upwards. This future value is an inflow. The interest is 10% annually. The same problem can be asked regarding the equivalence: What amount at the end of the eight years is equivalent to end-of-year payments of \$187.45 each?

The Timeline is drawn in years, with eight downward payments represented by symbol A and at the end of the eight years, there is an upward arrow for the future value. F is unknown? F is the amount to be received due to the investment of 8 annual deposits. The interest rate is 10%.

We will use the known formula to get F in terms of given A; the equation is to be written as F=A*(F/A,i,n). A will go with A and F as both sides are the same.

F=A*(F/A,10%,8). First, we will put A’s value in the equation equal to \$187.45.

We use the table of interest of i=10% to get the value of F/A after eight years, which is 11.4359; this value is bigger than one since F is greater than A. The final value of F will be equal to \$187.45*11.4359=\$2143.60 at the end of the eighth year.

### Solve For Future value for uniform series deposits in terms, of using the cash flow moments method.

If we use the concept of cash flow moment, we start by writing the given data as A=\$187.45&i=10%,n=8, while F is unknown.

Take a moment about a pivot point at n=8 and write the sum of all moments is equal to zero. F*(1+i)^0-A*(1+i)^7, the moment of first A is considered about a right point with time=8-1=7, A is minus, and the distance raised to the power with a positive sign, the second moment due to the second A will be -A*(1+i)^6.

The third moment due to the third  A will be-A*(1+i)^5-A*(1+i)^4-A*(1+i)^3-A*(1+i)^2-A*(1+i)^1-A*(1+i)^0=0.

We simplify the sum elements, for (1+i) is equal to 1.10, and for each aspect, A is a common factor. We will consider the corresponding value of the exponent.

The first exponent is 7, while the second exponent is 6, the third exponent is five and so with a decrease of 1 for the next term.

we can get F=A*((1.1)^7+(1.1)^6+(1.1)^5+(1.1)^5+(1.1)^4+(1.1)^3+(1.1)^2+(1.1)^1+(1.1)^0. The value of (1.1)^7 is equal to 1.9487, while (1.10)^6=1.7715 and (1.1)^5=1.6105. We will check the number of exponents as eight values.

Finally, we get (1.9487+1.7715+1.6105+1.4641+1.331+1.21+1.1+1). The sum of these values is 11.4358.

We will multiply the previous value by A’s value, which is \$182.45. Finally, we get F=\$2143.60, which is precisely the same value as we have estimated by using the formula.

### Solve For Future value for uniform series deposits, use the expression for P/A.

The expression of P/A  is used for a given A, and it is required to estimate the present value.P/A is equal to ((1+i)^n-1/(i)*(1+i)^n).

The reciprocal of P/A will be A/P and can get the A value for a given P-value or the present value. A/P=i*(1+i)^n/((1+i)^n-1)).

We have a given formula for the future value F in terms of P,i%,n, we have F=P(1+i)^n. We will substitute in the expression of P/A to get the value of F/A.

We remove P and substitute by F/(1+i)^n; the term of (1+i)^n is omitted due to substitution. Finally, F/A will be equal to ((1+i)^n-1/(i).

The reciprocal of F/A will be A/F and can get the A value for a given F value or the future value. A/F=(I)/((1+i)^n-1)). We can see Table 2-3, Giving the standard notations equation, to get F in terms of A, we use F/A. F=A*(F/A,i,n).

The function in Excel is FV(i,n,A). The F/A name is a uniform series compound amount. The term A/F name is sinking fund, Find A with a given F. This is the equation used to get the value of A/F. This is the standard notation A=F(A/F, i,n). the function in excel is PMT(i%,n, F).The pivot point B is at t=n, at a distance =-n from the force, (F). We will proceed to the next slide.

### Solve For Future value for uniform series deposits, use the interest table of 10%.

We need to estimate the value of F/A, by using a table for the interest rate of 10% for n of 8 years. Before starting using the table, first, we will use the expression for F/A which is equal to ((1+i)^n-1))/i. We have i=10% and n=8.  The value of F/A=((1.10)^8-1/(0.10)=11.4358 and can be approximated to 11.436.

If we wish to use the table, then select the table for the interest of 10%, and use the title uniform payment series. The Compound amount factor finds F, given A, title F/A. We go down until the intersection with the horizontal line at n=8. The intersection will give a value of 11.436.

This is the formula F=A(F/A,i=10%,n=8)=\$187.45*11.436=\$2143.67.

This is the formula F=A(F/A,i=10%,n=8)=\$187.45*11.436=\$2143.67.

### Solve For Future value for uniform series deposits, use the Excel function FV.

I can use a sheet and write the necessary elements arranged based on the generic formula of the FV function, which is the future value.

We write Fv, then open parentheses (rate, periods, payment). We can write Fv(10%,8,-182.45). The payments are outflows. That is why these payments are in negative signs. Then the result is a positive F value as inflows.

This serves as an alarm that you have deposited; use a negative sign to get output as positive, and then you will get inflow. Use an Excel sheet the first cell is rate =10%, for example, C19 in our case. Then we write n in years in the next cell, which is C20, which is n=8.

The A value is (+187.45) in cell C21.In the next cell, C22, we will write the future value as a built-in function, = Fv(C19, C20,-C21). The result is \$2143.66.

This is the Pdf file used for the illustration of this post. This is a good link -Engineering Economy. Applying Theory to Practice.

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