Compound Interest Formulas III

5. Uniform Series Present-Worth Factor

The fifth group in Table 1-5 covers a set of problems that uniform series of equal investments, A, occurred at the end of each time period for n number of periods at the compound interest rate of i. In this case, the cumulated present value of all investments, P, needs to be calculated. In summary, P is unknown and A, i, and n are given parameters. And the problem can be noted as $P/A_{i,n}$ and displayed as:

P=?		A	A		A	A	0
	0	1	2	...	n-1	n

Figure 1-6: Uniform Series Present-Worth Factor, $P/A_{i,n}$

If we replace substitute F in Equation 1-3 from Equation 1-2, we will have the present value as:

$$\begin{array}{l}\text{Equation 1-3:} F=A\left[{\left(1+i\right)}^n-1\right]/i\\ \text{Equation 1-2:} F=P{\left(1+i\right)}^n\\ P{\left(1+i\right)}^n =A\left[{\left(1+i\right)}^n-1\right]/i\end{array}$$

then,

Equation 1-5 gives the cumulated present value, P, of all uniform series of equal investments, A, as $P=A\left[{\left(1+i\right)}^n-1\right]/[i{\left(1+i\right)}^n]$. And also can be noted as: $P=A*P/A_{i,n}$ .The factor $\left[{\left(1+i\right)}^n-1\right]/\left[i{\left(1+i\right)}^n\right]$is called the “uniform series present-worth factor” and is designated by $P/A_{i,n}$ . This factor is used to calculate the present sum, P that is equivalent to a uniform of equal end of period payments, A. Then $P/A_{i,n}=A\left[{\left(1+i\right)}^n-1\right]/\left[i{\left(1+i\right)}^n\right]$

Note that n is the number of time periods that equal series of payments occur.

Please review the following video, Uniform Series Present Worth Factor (Time 3:35).

Example 1-5:

Calculate the present value of 10 uniform investments of 2000 dollars to be invested at the end of each year for interest rate 12% per year compound annually.

P=?		A=$2000	A=$2000		A=$2000	A=$2000	0
	0	1	2	...	9	10

So,
A =$2000
n =10
i =12%
P=?

Using Equation 1-5, we will have:
$$\begin{array}{l}P=A*P/A_{i,n}=A\left[{\left(1+i\right)}^n-1\right]/\left[i{\left(1+i\right)}^n\right]\\ P=A*P/A_{12\%,10}=2000*\left[{\left(1+0.12\right)}^{10}-1\right]/\left[0.12{\left(1+0.12\right)}^{10}\right]\\ P=2000*5.650223=\$11,300.45\end{array}$$

Note that we use the factor $P/A_{i,n}$ when we have equal series of payments. i is the interest rate and n is the number of equal payments. There is an important assumption here, the first payment has to start from year 1. In that case $P/A_{i,n}$ will return the equivalent present value of the equal payments.

Now let's consider the case that we have equal series of payments and the first payment doesn't start from year 1. In that case the factor $P/A_{i,n}$ will give us the equivalent single value of equal series of payments in the year before the first payment. However, we want the present value of them (at year 0). So, we need to multiply that with the factor $P/F_{i,n}$ and discount it to the present time (year 0).
 

Example:

P=?			A=$2000		A=$2000	A=$2000	0
	0	1	2	...	10	11

Note that there are 10 equal series of $2,000 payments. But the first payment is not in year 1. The factor $P/A_{12\%,10}$ returns the equivalent value of these 10 payments to the year before the first payment, which is year 1.

P=?		$2000(P/A12%,10)					0
	0	1	2	...	10	11

However, we want the present value. So, we need to discount the value by one year to have the present value of 10 equal payments.

P=?	$2000(P/A12%,10)(P/F12%,1)						0
	0	1	2	...	10	11

$$\text{Present value}=2,000\left(P/A_{12\%,10}\right)\left(P/F_{12\%,1}\right)$$

Example: Now consider the the following case that the first payment starts at year 3:

P=?				A=$2000		A=$2000	A=$2000	0
	0	1	2	3	...	10	12

$$\text{Present value}=2,000\left(P/A_{12\%,10}\right)\left(P/F_{12\%,2}\right)$$

Table 1-10: Uniform Series Present-Worth Factor, P/A_i,n

Factor	Name	Formula	Requested variable	Given variables
$P/A_{i,n}$	Uniform Series Present-Worth Factor	$\left[{\left(1+i\right)}^n-1\right]/\left[i{\left(1+i\right)}^n\right]$	P: Present value of uniform series of equal investments	A: uniform series of equal investments n: number of time periods i: interest rate

6.Capital-Recovery Factor

The sixth group in Table 1-5 belongs to set of problems that A is unknown and P, i, and n are given parameters. In this category, uniform series of an equal sum, A, is invested at the end of each time period for n periods at the compound interest rate of i. In this case, the cumulated present value of all investments, P, is given and A needs to be calculated. It can be noted as $A/P_{i,n}$ .

P		A=?	A=?		A=?	A=?	0
	0	1	2	...	n-1	n

Figure 1-7: Capital-Recovery Factor,$A/P_{i,n}$

Equation 1-5 can be rewritten for A (as unknown) to solve these problems:

Equation 1-6 determines the uniform series of equal investments, A, from cumulated present value, P, as $A=P\left[i{\left(1+i\right)}^n\right]/\left[{\left(1+i\right)}^n-1\right]$. The factor $\left[i{\left(1+i\right)}^n\right]/\left[{\left(1+i\right)}^n-1\right]$ is called the “capital-recovery factor” and is designated by A/P_i,n. This factor is used to calculate a uniform series of end of period payment, A that are equivalent to present single sum of money P.

Note that n is the number of time periods that equal series of payments occur.

Please watch the following video, Capital Recovery Factor (Time 3:37).

Example 1-6:

Calculate uniform series of equal investment for 5 years from present at an interest rate of 4% per year compound annually which are equivalent to 25,000 dollars today. (Assume you want to buy a car today for 25000 dollars and you can finance the car for 5 years with 4% of interest rate per year compound annually, how much you have to pay each year?)

P=$25,000		A=?	A=?	A=?	A=?	A=?	0
	0	1	2	3	4	5

Using Equation 1-6, we will have:
$$\begin{array}{l}A=P*A/P_{i,n}=P\left[i{\left(1+i\right)}^n\right]/\left[{\left(1+i\right)}^n-1\right]\\ A=P*A/P_{4\%,5}=25,000*\left[0.04{\left(1+0.04\right)}^5/\left[{\left(1+0.04\right)}^5-1\right]\right]\\ A=25,000*0.224627=5615.68\end{array}$$

So, having $25,000 at the present time is equivalent to investing $5,615.68 each year (at the end of the year) for 5 years at annual compound interest rate of 4%.

Table 1-11: Capital-Recovery Factor, A/P_i,n

Factor	Name	Formula	Requested variable	Given variables
$A/P_{i,n}$	Capital-Recovery Factor	$\left[i{\left(1+i\right)}^n\right]/\left[{\left(1+i\right)}^n-1\right]$	A: uniform series of equal investments	P: Present value of uniform series of equal investments n: number of time periods i: interest rate

Note that

$$A/P_{i,n}=A/F_{i,n}*F/P_{i,n}=P\left[i{\left(1+i\right)}^n\right]/\left[{\left(1+i\right)}^n-1\right]$$

Using these six techniques, we can solve more complicated questions.

Example 1-7:

Assume a person invests 1000 dollars in the first year, 1500 dollars in the second year, 1800 dollars in the third year, 1200 dollars in the fourth year and 2000 dollars in the fifth year. At an interest rate of 8%:
1) Calculate time zero lump sum settlement “P”.
2) Calculate end of year five lump sum settlement “F”, that is equivalent to receiving the end of the period payments.
3) Calculate five uniform series of equal payments "A", starting at year one, that is equivalent to above values.

P=?		1000	1500	1800	1200	2000	F=?
	0	1	2	3	4	5

1) Time zero lump sum settlement “P” equals the summation of present values:

$$\begin{array}{l}P=1000*(P/F_{8\%,1})+1500*(P/F_{8\%,2})+1800*(P/F_{8\%,3})+1200*(P/F{8}_{\%,4})+2000*(P/F_{8\%,5})\\ P=1000*0.92593+1500*0.85734+1800*0.79383+1200*0.73503+2000*0.68058\\ P=5884.03\end{array}$$

2) End of year five lump sum settlement “F”, that is equivalent to receiving the end of the period payments equals the summation of future values:

$$\begin{array}{l}F=1000*\left(F/P_{8\%,(5-1)}\right)+1500*\left(F/P_{8\%,}{}_{(5-2)}\right)+1800*\left(F/P_{8\%,}{}_{(5-3)}\right)+1200*\left(F/P_{8\%,}{}_{(5-4)}\right)+2000\\ F=1000*\left(F/P_{8\%,4}\right)+1500*\left(F/P_{8\%,3}\right)+1800*\left(F/P_{8\%,2}\right)+1200*\left(F/P_{8\%,1}\right)+2000\\ F=1000*1.36049+1500*1.25971+1800*1.1664+1200*1.08+2000\\ F=8645.58\end{array}$$

Please note that in the factor subscript, n is the number of time period difference between F (the time that future value has to be calculated) and P(the time that the payment occurred). For example, 1800 payment occurs in year 3 but we need its future value in year 5 (2 year after) and time difference is 2 years. So, the proper factor would be:$\left(F/P_{8\%,}{}_{(5-3)}\right)\text{ or }\left(F/P_{8\%,2}\right)$ .

3) Uniform series of equal payments "A" can be calculated from either P or F :
$$A=5884.03*A/P_{8\%,5}=5884.03*0.25046=1473.7$$ or
$$A=8645.58*A/F_{8\%,5} =8800.71*0.17046=1473.7$$

Example 1-8: repeat your calculations for the following payments:

P=?	800	1000	1000	1600	1400		F=?
	0	1	2	3	4	5

1) Time zero lump sum settlement “P” equals the summation of present values: $$\begin{array}{l}P=800+1000*\left(P/F_{8\%,1}\right)+1000*\left(P/F_{8\%,2}\right)+1600*\left(P/F_{8\%,3}\right)+1400*\left(P/F_{8\%,4}\right)\\ P=800+1000*0.92593+1000*0.85734+1600*0.79383+1400*0.73503\\ P=4882.44\end{array}$$

2) End of year five lump sum settlement “F”, that is equivalent to receiving the end of the period payments equals the summation of future values: $$\begin{array}{l}F=800*\left(F/P_{8\%,5}\right)+1000*\left(F/P_{8\%,4}\right)+1000*\left(F/P_{8\%,3}\right)+1600*\left(F/P_{8\%,2}\right)+1400*\left(F/P_{8\%,1}\right)\\ F=800*1.46933+1000*1.36049+1000*1.25971+1600*1.1664+1400*1.08\\ F=7173.9\end{array}$$

3) Uniform series of equal payments "A" can be calculated from either P or F:
$$\begin{array}{l}A=4882.44*A/P_{8\%,5}=4882.44*0.25046=1222.84\\ \end{array}$$ or
$$A=7173.9*A/F_{8\%,5}=7173.9*0.17046=1222.84$$