Discounted Cash Flow Factors

  • Calculate present and future values of various payment methods such as single payment, uniform series, gradient series using formulae and vales extracted from table.

  • Estimate and compare rate of returns on different investment proposals.

Single payment compound amount factor: (F/P, i, N):


  • Moves a single payment to N periods later in time.

  • A single payment is made after n periods.

  • The interest earned at the end of each period is charged on the total amount owed (principal plus interest).

  • $1 now is worth (F/P,i,n) at time n if invested at i%

Single payment present worth factor:(P/F, i, N):


  • Moves a single payment to N periods earlier in time.

  • The reciprocal of the single-payment compound amount factor.

  • Discount rate: i

  • $1 n years in the future is worth (P/F, i, n) now.

Sinking Fund factor: (A/F, i, N):

Takes a single payment and spreads into a uniform series over N earlier periods. The last payment in the series occurs at the same time as F.

Uniform Series Compound Amount factor:(F/A, i, N):


  • Takes a uniform series and moves it to a single value at the time of the last payment in the series.

  • Equal payments, A, occur at the end of each period.

  • We will get back (F/A, i, n) at the end of period n if funds are invested at an interest rate i. \[F = A + A(1+i) + A(1+i)^2+...+ A(1+i)^{n-1}\]

Capital Recovery Factor:(A/P, i, N):


  • Takes a single payment and spreads it into a uniform series over N later periods. The first payment in the series occurs one period later than P.

  • An amount P is deposited now at an annual interest rate i. We will withdraw the principal plus the interest in a series of equal annual amounts A over the next n years.

  • The principal will be worth \(P(1+i)^n\) at the end of n years. This amount is to be recovered by receiving A every year the sinking-factor formula applies.

Uniform Series Present Worth Factor: (P/A, i, N):

Takes a uniform series and moves it to a single payment one period earlier than the first payment of the series.

Solved Examples

Solved Example:


An investor deposits \$1,000 in June in an initially empty account paying monthly 7%, and then makes withdrawals of \$400 and \$500 in July and August. What will be the compound amount of these cash flows in October using the compound amount formula?



Time value of \$1000 after 4 months = $1000 \times \left(1 + \dfrac{7}{100}\right)^4 = \$1310.80$

Time value of \$400 after 3 months $= 400 \times \left(1 + \dfrac{7}{100}\right)^3 = \$490.02$

Time value of \$500 after 2 months $= 500 \times \left(1 + \dfrac{7}{100}\right)^2 = \$572.45$

Net value after all transactions

\[= \$1310.80 - \$490.02 - \$572.45 = \$248.35\]

Correct Answer: A

Solved Example:


A company received a buy-out offer where it is offered \$37,000 now or \$45,000 after 3 years. If the interest rate is 6% per annum, which offer should it accept?

We will calculate the present value of \$45,000. \[F = P \left( 1 + \dfrac{r}{100} \right) ^t\] \begin{align*} P &= \dfrac{F}{\left( 1 + \dfrac{r}{100} \right) ^t}\\ &= \dfrac{45000}{\left( 1 + \dfrac{6}{100} \right) ^3}\\ &= \dfrac{45000}{(1.06)^3}\\ &= 37,782.87 \end{align*} So, the present value of \$45,000 now is \$37,782.87 which is higher than \$37,000. Hence, the company should accept \$45,000 after 3 years.
Note: Alternatively, you may calculate the future value of \$37,000 after 3 years and compare it with \$45,000.

Correct Answer: B

Solved Example:


A student borrows $15,000 education loan which he intends to pay off in 6 equal annual payments. The interest rate charged is 8%. How much will be his annual payment?

P = \$15,000 i = 8% N = 6 years A = ? \begin{align*} A &= P (A/P, 8\%, 6)\\ &= \$15000\ (A/P, 8\%, 6)\\ &= \$15000\ (0.2163)\\ &= \$3243 \end{align*}

Correct Answer: D

Solved Example:


If you can afford a $400 monthly car payment, how much car can you afford if interest rates are 7% on 36 month loans?

\[ i = \dfrac{0.07}{12} = 0.005833, \quad n = 36\] \begin{align*} (P/A,i\%,n) &= \dfrac{(1+i)^n - 1}{i((1+i)^n)}\\ &= \dfrac{1.2329 - 1}{0.005833 \times 1.2329}\\ &= 32.383 \end{align*} Amount which can be afforded = 400 $\times$ 32.383 = $12953.16

Correct Answer: D

Solved Example:


A person borrows \$85,000 for his 30-year mortgage. his monthly payments which include principal and interest only (no other factors such as insurance, etc. are included) are \$537. His interest rate will most likely be:

\begin{align*} 85,000 & = 537(P/A, i\%,n)\\ 85,000 & = 537(P/A, i\%,360)\\ 85,000 & = 537 * \dfrac{(1+i)^{360} -1}{i (1+i)^{360}} \end{align*} For 6.5% substitute i = 0.065/12 = 0.00541$\vec{6}$ and the right side will be \$84,959, which is very close to \$85,000.

Correct Answer: C