Compound Interest 1m6h2n

Compound Interest

:Compound interest factors

Based on the concept of time value of money and bound to the mechanism of compound interest, it is often required to interrelate some of the following quantities: P is the value or sum of money at a time denoted as the present. F is the value or sum of money at a future time. A is the series of consecutive, equal, end-of-period amounts of money. n is the number of interest periods. i is the interest rate. The transformation multipliers (or time-value conversions) are called the “Compound Interest Factors" or the “Engineering Economy Factors".

n1 ((1 1 ii)nn)

Compound Interest Factors

Discrete Cash Flow, Discrete Compounding To Find

Given

F

P

P

F

Name of Factor Compound Amount Factor (single payment) Present Worth Factor (single payment)

F

A

Compound Amount Factor (uniform series)

A

F

Sinking Fund Factor

Factor

3

n i( (1 )[i)nn 1   1 n (21 n]i)n

Compound Interest Factors

Discrete Cash Flow, Discrete Compounding To Find

Given

Name of Factor

A

P

Capital Recovery Factor

P

A

A

G

P

G

Present Worth Factor (uniform series) Arithmetic Gradient Conversion Factor (to uniform series) Arithmetic Gradient Conversion Factor (to present value)

Factor

4

,,i% n PA///FPFPAA ,,ii% % n n %

Table 1. Standard factor notations Factor name Single-payment present worth (SPPWF) Single-payment compound amount (SPCAF) Uniform series present worth (USPWF) Capital recovery (CRF) Sinking fund (SFF) Uniform series compound amount (USCAF)

Standard notation

% n P  F P / F , i % n AP///PPFAF,,,iii%  P , %  nnP /A  A A i % n ,F P n A/FA,i%

Table 4.2: Computations using standard notation To find

given

P

F

F

P

P

A

A

P

A

F

F

A

factor

formula

F  P F//P1,inn1in

SINGLE-PAYMENT COMPOUND AMOUNT FACTOR (SPCAF) (S P C A F )

P = g ive n

0

1

2

3

4

n-2

n-1

n

F = ??

.This factor was implicitly derived previously

Example 1 If you have $2000 now and invest it at 10%, how much will it be ?worth in 8 years F = ??

i = 10 %

0

1

2

3

4 ye ars

2000

5

6

7

8

Solution

Given: P = $2000, i = 10% per year, and n = 8 years Find: F Using a calculator: Use a calculator to evaluate the (1+i) n term F = $2000 (1 + 0.10)8 $4287.18 = Using a spreadsheet package as Excel, the future worth calculation could be :determined from FV Function FV returns the future value of an investment based on periodic, constant .payments and a constant interest rate Syntax FV (rate, nper, pmt, pv, type)

Example 2

The following sums are to be invested at 5 % interest rate: 6x105 m.u. now, 3x105 m.u. two years from now and 4x105 m.u. five years from now. How much will be accumulated ten ? years from now

Solution

F = ??

i= 5 % 0

y e a rs

1

2

3

4

5

6

7

5

3 x1 0 m .u . 5

6 x 1 0 m .u .

4 x10 m .u .

5

F = P1 (F/P, 5, 10) + P2 (F/P, 5, 8) + P3 (F/P, 5, 5) = 6 × 105 × (1.05)10 + 3×105 × (1.05)8 + 4×105 × (1.05)5 = 1.9311×106 m.u.

8

9

10

P  F / P//F,1nin1in

SINGLE-PAYMENT PRESENT WORTH FACTOR (SPPWF) (SPPW F)

P= ??

0

1

2

3

4

n-2

n- 1

n

F = given

Example 3

Suppose that $1000 is to be received after 5 years. At an annual interest rate of 12%, What is the present worth of this amount?

Solution 1 0 0 0 L .E . i = 12 %

0

1

2

3

4

5

y e a rs

P = ??

:This example could be solved by one of three ways Using a calculator (Mathematical Formulas)-1 P = 1000 × (1 + 0.12)-5 = 1000 × (0.5674) = $567.40 In order To have L.E.1000 in your savings at the end of 5 years, .you must deposit $567.40 now

( 0 . 5 6 7 4 )      P = 1,0(P /F ,12% ,5)= L .E 567.40

Using interest tables. 2

%Interest Table For Discrete Compounding; i = 12 Periods

Compound Amount Factor

Present Worth Factor

Compound Amount Factor

Present Worth Factor

n

F/P

P/F

F/A

P/A

4

1.5735

0.6355

4.7793

3.0373

5

1.7623

0.5674

6.3528

3.6048

6

1.9738

0.5066

8.1152

4.1114

Note: this method might be obsolete as it requires lots of data tables. These tables are formed using the simple formulae developed here in this lecture

UNIFORM-SERIES COMPOUNDAMOUNT FACTOR (USCAF) (U S C A F )

0

1

2

3

4

F= ??

n-2

n-1

n

A = g iv e n

A series of equal installments paid at the ends of consecutive interest periods .will accumulate to a certain future value after n periods



nnn  123 M  F  A 1  iiii  1   ui1ltipyFng1bAyin 1FAi iiiii nnnF1A,in11 1211 1in1 1

The future compound amount is the summation of the right-hand column. Thus,

Subtractin

g

Example 5

Suppose you make 15 equal annual deposits of $1000 each into a bank paying 5% interest per year. The first deposit will be made one year from today. How much money can be withdrawn from this bank immediately after the 15th deposit?

Solution:

The value of A is $1000, n = 15 years, and i = 5% per year. Immediately after the 15th payment, the future equivalent amount is. F = l000 (F/A, 5%, 15) = 1000 (1.0515 – 1) / 0.05 = 1000 × (21.5786) = $21578.60



i A  F FAiF,i1nn1in1

SINKING FUND FACTOR (SFF)

(S S F )

0

1

2

3

4

A = ??

n-2

F = g iv e n

n-1

n

Example

An enterprising student is planning to have personal savings totaling $1000000 when she retires at age 65. She is now 20 years old. If the annual interest rate will average 7% over the next 45 years on her savings , what equal end-of-year amount must she save to accomplish her goal?

Solution

The future amount, F, is $1000000. The equal annual amount this student must place in a sinking fund that grows to $1000000 in 45 years at 7% annual interest A = 1000000 (A/ F, 7%, 45) A= 1000000 (0.07/((1+0.07)45 –1)) A= 1000000(0.0035) = $3500

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