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 Engineering Economy

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PostSubject: Engineering Economy   Engineering Economy Icon_minitimeSat Jan 02, 2010 2:05 am

Engineering Economy
The following variables will be used throughout the presentation


P = Principal (Present Sum)

S = Future Sum

N = Number of Payments

I = Interest Rate


Simple Interest


Equation: S = P + N I P = P (1 + N I)
Example: If $100.00 was deposited at 6% yearly interest, your account would have these balances after each year.


Year Balance
0 100.00
1 106.00
2 112.00
3 118.00
4 124.00


Compound Interest
"Interest on the Accrued Interest"
Equation: S = P (1 + I)^N

Example: If $100.00 was deposited at 6% interest and compounded annually, your account would have these balances after each year:

Year Balance Equation

0 100.00 P
1 106.00 P(1+I)^1
2 112.36 P(1+I)^2
3 119.10 P(1+I)^3
4 126.25 P(1+I)^4


Example: If you deposit $2000.00 today at 7% interest compounded annually, what will be the balance in 3 years?


Engineering Economy Line1


Compound Interest Time Line


"Backward Time"
Example:If $4000.00 is needed in 3 years, how much money should be deposited today, assuming 7% interest compounded yearly?

Engineering Economy Line2


Solution: P = Present Value = S (1 + I)^-N
= $4000.00 (1 + 0.07)^-3
= $3265.19



References:

Lecture written by: Dr. Larry Genalo

Authored for presentation by: David Kilzer

Revised by Mark Sobek and Lex Jacobson

HTML documentation by: Larry Genalo Jr.

Date last updated: 8/1/95
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PostSubject: Re: Engineering Economy   Engineering Economy Icon_minitimeSat Jan 02, 2010 10:38 pm

[size=12][font=Times New Roman]Tips for Doing Problems


When doing engineering economy problems, make sure:

"TIME FRAMES" MATCH:
If one period equals one quarter, then the interest rate should be quarterly and compounded quarterly. For example, if a yearly interest rate is given but interest is compounded monthly, divide the interest rate by 12.

ANSWER IS REASONABLE

ANSWER IS ROUNDED TO THE NEAREST PENNY


Sinking Fund

Engineering Economy Sink1


Definition: A sinking fund differs from compound interest in that we now have uniform payments over time in addition to the compound interest.




Sinking Fund Time Line


Example #1
Example: If you deposit $50 per month into an account that pays 6% interest, compounded monthly, for 2 years, how much is in the account immediately after the last deposit?

Engineering Economy Sink2

Please Note: Yearly interest in the equation is divided by 12 since payments are on a per-month basis.
Also Note: There are 24 payments, not 23 months.



Sinking Fund Time Line


Example #2

Example: Which is of more value to receive:
(a) $8000 today or
(b) 5 annual payments of $2000, beginning in 1 year?
(Assume 8% interest compounded annually.)

Engineering Economy Sink3

Sinking Fund Time Line
Example #3
Example: Which is of more value to receive:
(a) $8000 today or
(b) 5 annual payments of $2000, beginning in 1 year?
(c) 5 annual payments of $2000, beginning today?
(Assume 8% interest compounded annually.)

Engineering Economy Sink3Engineering Economy Sink6

Annuity
NOTE: Follow the algebraic steps used to find the present value of a sinking fund. This is the formula for an annuity.

http://www.eng.iastate.edu/efmd/Captures/annline1.gif


APPLICATION: The first application of the annuity formula is to find the present value of a sinking fund. The second is using the formula in situations like the following example.

EXAMPLE: You borrow $5000 to buy a car. If you repay the loan with 36 monthly installments at 18% interest, what is the amount of each payment?


Engineering Economy Annline2

Summary

Compound Interest
Engineering Economy Summ1

Equation: S = P (1 + I)^N

Engineering Economy Summ2

Sinking Fund
Engineering Economy Summ3
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