MS291: Engineering Economy

Chapter 2Factors: How Time andInterest Affect Money

Content of the Chapter

Single-Payment Compound Amount Factor (SPCAF) Single-Payment Present Worth Factor (SPPWF)

Uniform Series Present Worth Factor (USPWF) Capital Recovery Factor (CRF)

Uniform Series Compound Amount Factor Sinking Fund Factor (SFF)

Arithmetic Gradient Factor Geometric Gradient Series Factor (Optional Topic)

Simple InterestHereP=$100,000n= 3i= 10%Simple interest = P X n x iInterest = 100,000(3)(0.10)

= $30,000Total due = 100,000 + 30,000

= $130,000

Compound InterestInterest, year 1: I1 = 100,000(0.10) = $10,000 Total due, year 1: F1 = 100,000 + 10,000

=$110,000

Interest, year 2: I2 = 110,000(0.10) = $11,000 Total due, year 2: F2 = 110,000 + 11,000

= $121,000

Interest, year 3: I3 = 121,000(0.10) = $12,100 Total due, year 3: F3 = 121,000 + 12,100

= $133,100

Simple and Compound InterestExample: $100,000 lent for 3 years at interest rate i = 10%per year. What is repayment after 3 years ?

Simple: $130,000: Compounded: $133,100

Single Payment Compound AmountFactor (SPCAF)

If an amount “P” is invested at time “t=0”the amount accumulated after a year isgiven as

F1 = P + Pi= P(1 + i) ………. (1)

At the end of second year, the accumulatedamount F2 is given as;

F2 = F1 + F1 i= P(1+i) + P(1+i)i (from Eq. 1)= P + Pi + Pi+ Pi2= P(1+i)2…………(2)

Similarly;

F3 = F2 + F2 i= P(1+i)3………..(3)

to generalize the process for period “n” wecan write as;

F = P(1+i)n

The term “(1+i)n” is known as SinglePayment Compound Amount Factor(SPCAF)

It is also refer as F/P factor

This is a converting factor, whenmultiplied by “P” yields the futureamount “F” of initial amount “P”after “n” years at interest rate “i”

Simple InterestHereP=$100,000n= 3i= 10%Simple interest = P X n x iInterest = 100,000(3)(0.10)

= $30,000Total due = 100,000 + 30,000

= $130,000

Compound InterestNow we have F = P(1+i)n

P = $100,000n=3i=10%

So F = 100,000 (1+0.10)3

F= 100,000 (1.331)F = 133100

Simple and Compound InterestExample: $100,000 lent for 3 years at interest rate i = 10%per year. What is repayment after 3 years ?

Simple: $130,000: Compounded: $133,100

From SPCAF to SPPWF

Now we have the formula how to “convert” presentamounts into future amount at a given interest rate i.e.F = P(1+i)n

What if we are given a future amount (F) and we areasked to calculate present worth (P) ?

F = P(1+i)n

=> P = F [1/(1+i)n]or P = F(1+i)-n

Single Payment Present WorthFactor (SPPWF)

The term “(1+i)-n” is known as Single Payment PresentWorth Factor (SPPWF)

It is also refer as P/F factor

This is a converting factor, when multiplied by “F”yields the present amount “P” of initial amount “F”after “n” years at interest rate “i”

Example

Find the present value of $10,000 to bereceived 10 years from now at a discount rateof 10%F = $10,000r = 10%n = 10

P = F (1+i)-n

=> P = 10,000 (1+0.1)-10

= 10,000 x 0.385= $385

A Standard Notation Instead of writing the full formulas of SPCAF and SPPWF for simplicity

there is a standard notation

This notations includes two cash flows symbols, interest rate andnumber of periods

General form is: (X/Y, i, n) which means “X” represents what is sought, Yis given, i is interest rate and n is number of periods

Examples:

Name Equation withfactor formula

Notation Standard NotationEquation

Find/Given

Single-payment compoundamountSingle-payment presentworth

A = P(1+i)n(F/P, i, n) F = P(F/P, i, n) F/P

P = F(1+i)-n (P/F, i, n) P = F(P/F, i, n) P/F

Using Standard NotationExample

What will be the future value of Rs. 100,000compounded for 17 years at rate of interest 10% ?

F= (1+ i)n or F = P(1+0.1)n now writing that instandard notation we have F = P(F/P, i, n) F = 100,000(F/P, 10%, 17) F = 100,000 (5.054) F= 505400

That value youget from “Table”

(F/P, i, n)

Single Payments to Annuity

Normally, in real world we do not face Singlepayments mostly instead faces cash flows such ashome mortgage payments and monthly insurancepayments etc

An annuity is an equal annual series of cash flows.It may be equal annual deposits, equal annualwithdrawals, equal annual payments, or equalannual receipts. The key is equal, annual cashflows

Uniform Series Present Worth Factor(USPWF)

P = ?

t = given

A = given

n2

t=0

1 n-13

Capital Recovery Factor (CRF)

Name Equation with factorformula

Notation Standard Notation Equation

(P/A, i, n) P = A(F/P, i, n)

(A/P, i, n) A = P(A/P, i, n)

Uniform SeriesPresent Worth

Capital Recovery

P = givent = given

A = ? n2

t=0

1 n-13

Example 1: Uniform Series PresentWorth (P/A)

A chemical engineer believes that by modifying the structure of acertain water treatment polymer, his company would earn an extra$5000 per year. At an interest rate of 10% per year, how muchcould the company afford to spend now to just break even over a 5year project period?The cash flow diagram is as follows:

P = 5000(P/A,10%,5)= 5000(3.7908)= $18,954

0 1 2 3 4 5

A = $5000

P = ?i =10%

Solution:A = 5000 I = 10% n = 5

P = A(P/A, i, n)

Example 2: Uniform SeriesCapital Recovery (A/P)

A chemical product company is considering investment in costsaving equipment. If the new equipment will cost $220,000 topurchase and install, how much must the company save each yearfor 3 years in order to justify the investment, if the interest rate is10% per year?

0 1 2 3

A = ?

i =10%P = $220,000

The cash flow diagram is as follows:

A = 220,000(A/P,10%,3)= 220,000(0.40211)= $88,464

Solution:P = 220,000I = 10%n = 3

A = P(A/P, i, n)

Uniform Series Compound Amount Factor(USCAF)

F = ?t = given

A = given

n20 1 n-13

Sinking Fund Factor (SFF)

F = given

t = given

A = ?

n20 1 n-13

Example 3: Uniform Series Involving F/A

An industrial engineer made a modification to a chip manufacturingprocess that will save her company $10,000 per year. At an interest rateof 8% per year, how much will the savings amount to in 7 years?

The cash flow diagram is:

A = $10,000

F = ?

i = 8%

0 1 2 3 4 5 6 7

Solution:

F = 10,000(F/A,8%,7)

= 10,000(8.9228)= $89,228

F = A(F/A, i, n)

A =10,000i =8%n =7

PracticeExample 2.5

Uniform Series Compound Amount Factor(USCAF)

F = ?t = given

A = given

n20 1 n-13

Sinking Fund Factor (SFF)F = givent = given

A = ?

n20 1 n-13

Example 3: Uniform Series Involving F/A

An industrial engineer made a modification to a chip manufacturingprocess that will save her company $10,000 per year. At an interest rateof 8% per year, how much will the savings amount to in 7 years?

The cash flow diagram is:

A = $10,000F = ?

i = 8%

0 1 2 3 4 5 6 7

Solution:

F = 10,000(F/A,8%,7)

= 10,000(8.9228)= $89,228

F = A(F/A, i, n)

A =10,000i =8%n =7

PracticeExample 2.5

Arithmetic Gradient Factors(P/G, A/G)

Cash flows that increase or decrease by a constant amountare considered arithmetic gradient cash flows.

The amount of increase (or decrease) is called the gradient

CFn = base amount + (n-1)GCash Flow Formula

A A A A A

AA+G

A+2GA+3G

A+(n-1) G

0 1 2 3 4 n 0 1 2 3 4 n

0

G2G

3G

(n-1)G

0 1 2 3 4 n

=+

PT = PA + PG

Arithmetic Gradient Factors(P/G, A/G)

PA = A(P/A, i, n) or Uniform Series Present worth Factor

PG = G(P/G, i, n) or Arithmetic Gradient Present Worth Factor

Alternatively, PG can also be calculated by followingformula

PT = PA + PG

nn

n

G i

n

ii

i

i

GP

)1()1(

1)1(

Solving Arithmetic Gradient relatedproblems

Present value of the Arithmetic Gradient series can be calculated asfollows:

1. Find the gradient and base2. Cash flow diagram maybe helpful if u draw it3. Break the gradient series into a Uniform series and a

Gradient Series as shown on next slide4. The formula for calculating present value of the Arithmetic

Gradient series is as follows;

5. Calculate PA and PG and use the above formula to getthe present value of the Arithmetic Gradient

PT = PA + PG

Example (Problem 2.25)Profits from recycling paper, cardboard, aluminium,and glass at a liberal arts college have increased at aconstant rate of $1100 in each of the last 3 years.If this year’s profit (end of year 1) is expected to be$6000 and the profit trend continues through year5,(a) what will the profit be at the end of year 5 and(b) what is the present worth of the profit at an interest

rate of 8% per year?

G = $1100, Base = $6000

Example (Problem 2.25)(a) what will the profit be at the end of year 5 &(b) what is the present worth of the profit at an interest rate of 8% per year?

G = $1100 Base = $6000

$6000$7100

$8200

$9300

0 1 2 3 4 5

Find the cash flows as follows:CF = Base + G(n-1)GCF1 = 6000 + 1100(1-1)= 6000CF2 = 6000 + 1100(2-1)= 7100CF3 = 6000 + 1100(3-1)= 8200CF4 = 6000 + 1100(4-1)= 9300CF5 = 6000 + 1100(5-1)= 10400

$10400

=>0 1 2 3 4 5

$6000

+0 1 2 3 4 5

$1100$2200

$3300

$4400

P = A(P/A, i, n) G(P/G, i, n)+P = 6000(P/A, 8%, 5) + 1100(P/G, 8%, 5)

P = 6000(3.9927) + 1100(7.3724)

P = 326066

Arithmetic Gradient Future Worth

n

i

i

iGF

n

G

1)1(1

0

3G

2GG

(n-1)G

F ?

Converting Arithmetic Gradient (G) intoUniform Series (A) (to convert G into (A/G, i, n)

Formula for conversation from G to (A/G, i, n)

Also AT = AA + AG

Where AA = P(A/P, i, n) and AG = G(A/G, i, n)

1)1(

1nG i

n

iGA

The following formula is used for calculating future value of arithmeticgradient given the values of gradient (the term in bracket is calledArithmetic gradient future worth factor)

Geometric Gradient Factors(Pg /A)

A Geometric gradient is when the periodic payment isincreasing (decreasing) by a constant percentage:

It maybe noted that A1 is not consideredseparately in geometric gradients

0 1 2 3 ----- n

A1

where: A1 = cash flow in period 1 and g = rate of increaseNote: If g is negative, change signs in front of both g values

A1 (1+g)

A1 (1+g)2

A1 (1+g)n-1i = giveng = given

Pg

Pg is the sum of whole series including A1

A1 is the initial cash in Year 1

or1

11

1

i g fi

AnPg

ifor ggi

i

g

APg

n

Example: Geometric Gradient

Determine the present worth of a geometric gradientseries with a cash flow of $50,000 in year 1 andincreases of 6% each year through year 8. The interestrate is 10% per year.

250,321$

)425.6(000,50

04.0

257.0000,50

04.0

743.01000,50

06.010.010.016.01

150000

11

1

8

gi

i

g

APg

n

Thank You

Using Factor Tables

Using Factor Tables