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Engineering Economy Lec3

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Page 1: Engineering Economy Lec3
Page 2: Engineering Economy Lec3

Discount

Discount on a negotiable paper is the difference between the present worth (the amount received for the paper in cash) and the worth of the paper at some time in the future (the face value of the paper or principal).

Discount is the interest paid in advanced

Discount = Future Worth – Present Worth

The rate of discount is the discount on one unit of principal for one unit of time.

(1+ i)-1

P 1.00

d = 1 – (1+i)-1, i = d/1-d

Where: d = rate of discount for the period involved

I = rate of interest for the same period

Page 3: Engineering Economy Lec3

Example : Discount A man borrowed Php 5,000 from a bank and agreed to pay the

loan at the end of 9 months. The bank discounted the loan and gave him Php 4,000 in cash. (a) What is the rate of discount? (b) What was the rate of interest? (c) What was the rate of interest for 1 year?

4,000

5,000

9 mos.

0

0.80

1.00

9 mos.

0

Page 4: Engineering Economy Lec3

(a) d =

Another solution : d = 1 – 0.80 = 0.20 or 20%

(b) I =

Another solution : i =

(c) I =

d =

Page 5: Engineering Economy Lec3

Inflation Inflation is the increase in the prices for goods and

services from one year to another, thus decreasing the purchasing power of money

FC = PC (1+ f) n

where : PC = present cost of commodity FC = future cost of the same commodity f = annual inflation rate n = number of years

Page 6: Engineering Economy Lec3

Inflation Ex. 1. An item presently cost Php 1,000. If inflation

is the rate of 8% per year, what wuill be the cost of the item in two years?

Solution: FC = PC (1+ f)n = 1,000(1 + 0.08)2 = Php

1,116.40 In an inflationary economy, the buying power of

money decreases as cost increase. Thus, F =

where F is the future worth, measured in today’s pesos

Page 7: Engineering Economy Lec3

Inflation Ex. 2 An economy is experiencing inflation at an

annual rate of 8%. If this continues, what will Php 1,000 be worth two yaers from now, in terms of today’s pesos?

F = =

If interest is being compunded at the same time that inflation is occurring, the future worth will be :

Page 8: Engineering Economy Lec3

Annuities (Uniform Series)

• Uniform series are known as the equal annual payments made to an interest bearing account for a specified number of periods to obtain a future amount.

C

ash

Flo

w-

+

P

A1 A2 A3 A4 A5 A6

F

Time

Page 9: Engineering Economy Lec3

Types of Annuities• ordinary annuity - annuity where payments

are made at the end of each period

F = A

P = A

Where F = Future sum of money P = Present value of moneyn = number of interest periodsA = Series of n equal payments made at the end of each periodi = Effective interest rate per period

Derivation of this formula can be found in most engineering economics texts & study guides

P = A [ (1+i)n – 1 ] or i (1+i)n

Page 10: Engineering Economy Lec3

Types of Annuities• Deferred Annuity - A type of annuity contract that delays payments

of income, installments or a lump sum until the investor elects to receive them. This type of annuity has two main phases, the savings phase in which you invest money into the account, and the income phase in which the plan is converted into an annuity and payments are received

P = A

Page 11: Engineering Economy Lec3

Example:

What is the future value of a series payments of P10,000 each, for 5 years, if deposited into a savings account yielding 6% nominal interest compounded yearly?

Draw the cash flow diagram.

F = A [ (1+i)n – 1 ] = i

Check with Factor Values:

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

Page 12: Engineering Economy Lec3

Example:

What is the future value of a series payments of Php10,000 each, for 5 years, if deposited in a savings account yielding 6% nominal interest compounded yearly?

Draw the cash flow diagram.

F = A [ (1+i)n – 1 ] = 10,000 [ (1+0.06)5 – 1 ] = 10,000 [ 1.3382 – 1 ] i 0.06 0.06

= Php 56,370

Checking with Factor Values:

F = A (F/A,i,n) (obtained from tables) = 10,000 (F/A, 6%, 5) = 10,000 ( 5.6371 ) = Php 56,370

Page 13: Engineering Economy Lec3

Sinking Fund

• We can also get the corresponding value of an annuity (A) during (n) periods to an account that yields (i) interest to be able to get the future value (F) :

Solving for A: A = i F / [ (1+i)n – 1 ]

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

Example:How much money would you have to save annually in order to buy a car in 4 years which has a projected value of Php 800,000? The savings account offers 4.0% yearly interest.

Page 14: Engineering Economy Lec3

Sinking Fund

Example:How much money do we have to save annually to buy a car 4 years from now that has an estimated cost of $18,000? The savings account offers 4.0 % yearly interest. A = i F / [ (1+i)n – 1 ] A = (0.04 x 800,000) / [ (1.04)4 -1 ] = 32,000 / 0.16986 = Php 188,390

A = F (A/F,i,n) (from table value)A = (800,000)(A/F,4.0,4) = (800,000)(0.2355) = $188,400

Page 15: Engineering Economy Lec3

Present Worth of an Annuity (Uniform Series)

• Sometimes it is required to estimate the present value (P) of a series of equal payments (A) during (n) periods considering an interest rate (i)

From Eq. 1 and 5

P = A [ (1+i)n – 1 ] (7)

i (1+i)n

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

Example:What is the present value of a series of royalty payments of Php50,000 each for 8 years if nominal interest is 8%?

P =

Page 16: Engineering Economy Lec3

Present Worth of an Annuity (Uniform Series)

Example:What is the present value of a series of royalty payments of $50,000 each for 8 years if nominal interest is 8%?

P = A [ (1+i)n – 1 ] = 50,000 [ (1+0.08)8 – 1 ] = 50,000 [ 1.8509 – 1 ] i (1+i)n 0.08 (1.08)8 0.1481

= Php 287,300

P = A (P/A,i,n) = (50,000)(P/A,8,8) = (50,000)(5.7466) = Php 287,300

Page 17: Engineering Economy Lec3

Uniform Series Capital Recovery

• This is the corresponding scenario where it is required to estimate the value of a series of equal payments (A) that will be received in the future during (n) periods considering an interest rate (i) and are equivalent to the present value of an investment (P)

Solving Eq. 7 for A

A = i P (1+i)n (8)

(1+i)n -1

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

Example:If an investment opportunity is offered today for $5 Million, how much must it yield at the end of every year for 10 years to justify the investment if we want to get a 12% interest?

A =

Page 18: Engineering Economy Lec3

Uniform Series Capital Recovery

Example:If an investment opportunity is offered today for Php 5 Million, how much must it yield at the end of every year for 10 years to justify the investment if we want to get a 12% interest?

A = i P (1+i)n = 0.12 x 5 (1+0.12)10 = 0.6 [ 3.1058 ] (1+i)n -1 (1.12)10 - 1 2.1058

= 0.8849 Million Php 884,900 per year