1. Engineering Economics
4/14/2017 5:03 AM
Concepts from the example problems
( F / P, i, N)  Problem 1
( P / F, i, N)  Problem 2
( F / A, i, N)  Problem 3
( A / F, i, N)  Problem 4
( P / A, i, N)  Problem 5
( A / P, i, N)  Problem 6
( P / G, i, N)  Problem 7
( A / G, i, N)  Problem 8
Copyright (c) , K. D. Douglas & D. H. Jensen

2. Example 1
How much would you have to deposit today to have $2000 in 4 years if you can get a 12% interest rate compounded annually?
GIVEN:
F4 = $2 000
i = 12%
FIND P: 1 2 3
n=4 P?
$2 000
DIAGRAM:
P = F4(P/F,i,n)
= 2 000(P|F,12%,4)
= 2 000(0.6355) = $1 271

3. Different Ways of Looking at P/F
From previous example, if you can earn 12% compounded annually, you need to deposit $1271 to have $2000 in 4 years.
You are indifferent between $1271 today and $2000 in 4 years, assuming you can earn a return on your money of 12%.
The present worth of $2000 in 4 years is $1271 (i = 12% cpd ‡ annually).
‡ my abbreviation for compounded
If you could get 13% on your money, would you rather have the $1271 today, or $2000 in 4 years ?

4. Example 2
Tuition costs are expected to inflate at the rate of 8% per year. The first year’s tuition is due one year from now and will be $10,000. To cover tuition cost for 4 years, a fund is to be set up today in an account that will earn interest at the rate of 5% per year, compounded annually. How much must be deposited into the fund today in order to pay the 4 years of tuition expenses?

5. Present Given Gradient (Geometric)
Engineering Economics
4/14/2017 5:03 AM
Present Given Gradient (Geometric)
g is the geometric gradient over the time period
(time period: Time 0 to Time n, 1st flow at Time 1)
P is the present value of the flow at Time 0
(n periods in the past)
i is the effective interest rate for each period
Note: cash flow starts with A1 at Time 1, increases by constant g% per period n A1
P ? 1 2 3
P = A1(P/A,g,i,n)
g = %
Copyright (c) , K. D. Douglas & D. H. Jensen

6. Example 2
Tuition costs are expected to inflate at the rate of 8% per year. The first year’s tuition is due one year from now and will be $10,000. To cover tuition cost for 4 years, a fund is to be set up today in an account that will earn interest at the rate of 5% per year, compounded annually. How much must be deposited into the fund today in order to pay the 4 years of tuition expenses?

7. Example 2 - Concept
If your rich Aunt Edna wanted to put a sum of money in the bank today to pay for your next four years of tuition, that sum would be $39,759 assuming 5% return on investment and tuition that begins at $10,000 increasing by 8% per year. This problem assumes tuition is due at the end of the year.

8. Complex Cash Flows
Complex Cash Flows – Break apart (or separate) complex cash flows into component cash flows in order to use the standard formulas.
Remember: You can only combine cash flows if they occur at the same point in time.
(This is like building with LEGOs!)

9. Problem 3
A construction firm is considering the purchase of an air compressor.
The compressor has the following expected end of year maintenance costs:
Year $800
Year $800
Year $900
Year 4 $1000
Year 5 $1100
Year 6 $1200
Year 7 $1300
Year 8 $1400
What is the present equivalent maintenance cost if the interest rate is 12% per year compounded annually?

10. Problem 3 – Alt Soln 1 GIVEN: MAINT COST1-8 PER DIAGRAM
i = 12%/YR, CPD ANNUALLY
FIND P:
P = PA + PG + PF = A(P/A,i,n) + G(P/G,i,n) + F(P/F,i,n)
= $700(P/A,12%,8) + $100(P/G,12%,8) + $100(P/F,12%,1)
= $700(4.9676) + $100( ) + $100(0.8929) = $5014
DIAGRAM:
P ? 1 2 3
n=8
$700
$100 4
$200
$300
PA ? 1 2 3
n=8
$700 4
$100
PF ?
n=1
PG ? 1 2 3
n=8
$100
$700
$200
$300 4
NOTE: CAN BREAK INTO 3 CASH FLOWS: ANNUAL, LINEAR GRADIENT, AND FUTURE

11. Problem 3 – Alt Soln 2 GIVEN: MAINT COST1-8 PER DIAGRAM
i = 12%/YR, CPD ANNUALLY
FIND P:
P = PA + PG(PPG) = A(P/A,i,n) + G(P/G,i,n-1)(P/F,i,1)
= $800(P/A,12%,8) + $100(P/G,12%,7)(P/F,12%,1)
= $800(4.9676) + $100( )(0.8929) = $5014
DIAGRAM:
P ? 1 2 3
n=8
$800
$100
$600 4
$200
PA ? 1 2 3
n=8
$800 4
PG ?
PPG ? 1
PG ? 1 2
n=7 3
$100
$600
$200
NOTE: PG MUST BE OFFSET ONE YEAR – SO BRING THE OFFSET YEAR BACK TO TIME ZERO

12. Problem 4
A young couple has decided to make advance plans for financing their 3 year old daughter’s college education. Money can be deposited at 8% per year, compounded annually.
What annual deposit on each birthday, from the 4th to the 17th (inclusive), must be made to provide $7,000 on each birthday from the 18th to the 21st (inclusive)? 17
21 yrs
$7 000
A ?
DIAGRAM: 4 18
GIVEN:
WITHDRAWALS = $7 000
i = 8%/YR, CPD YEARLY
FIND A4-17:
P17 = A(F/A,i,n) = A(P/A,i,n)
= A(F/A,8%,14) = 7 000(P/A,8%,4)
= A( ) = 7 000(3.3121)
 A = $957
STRATEGY: CAN BREAK INTO 2 CASH FLOWS, SO PICK A CONVENIENT POINT IN TIME AND SET DEPOSITS EQUAL TO WITHDRAWALS…