IE 305 ENGINEERING ECONOMIC ANALYSIS

1
IE 305ENGINEERING ECONOMICANALYSIS
2
Topic 1 Economic Analysis for Decision Making

• Why is Engineering Economy so Important?
• Role of Engineering Economy in Decision Making.
• Performing an Engineering Economy Study.

3
Why is engineering economy so important?
Engineering Economy- about determining the
economic factors and the economic criteria
utilized when one or more alternatives are
considered for selection. Another way to
define
4
EC- a collection of mathematical techniques for
evaluating and comparing different methods of
accomplishing a given objective.
5
Example
The president of two small businesses play
racquetball each year. After several
conversations, they have decided that, due to
their frequent commercial-airline travel around
the region, they should evaluate the purchase of
a plane jointly owned by the two companies.
6
What are some of the typical economic-based
questions the two presidents should answer as
they evaluate the alternatives to (1) co-own a
plane or (2) continue as is?
7
Some questions (and what is needed to respond to
them) might be

• How much will it cost each year? (Estimates
  needed )
• How do we pay for it? (A financing plan needed)
• Are there tax advantages? (Tax law information
  needed)

8

• Which alternative is more cost-effective?
  (Selection criteria needed)
• What is the expected rate of return? (Equations
  needed)
• What happens if we use different amounts each
  year than we estimated? (Sensitivity analysis
  needed)

9
Role of engineering economy in decision making
People make decisions
Computers do not!
The techniques and models of Engineering Economy
assist people in making decisions.
10
Role of engineering economy in decision making

• A scientific decision-making involves solving
  problems.
• Common names for the procedure are
• Decision-making process
• OR
• Problem solving approach

11
Steps to Problem Solving

• Understand the problem and goal.
• Collect relevant information.
• Define the alternative solutions.
• Evaluate each alternative.
• Select the best alternative using some criteria.
• Implement the solution and monitor the results.

12
But what is a problem in the scientific
sense? The minimal necessary and sufficient
conditions for the existence of a decision-making
problem are

• An individual or individuals who have the
  problem
• THE DECISION MAKER(S) (d.m.)
• An outcome that is designed by the d.m.
  OBJECTIVE

13

• At least 2 unequally efficient courses of action
  which have a chance of yielding the desired
  objective ALTERNATIVES
• A state of doubt in the d.m. As to which
  alternative is best.
• An environment or context of the problem.

14
Performing an engineering economy study
Engineering Economy Study Approach is used to
perform an engineering economy analysis.
15
New plant design
Upgrade old plant
Alternative 1
Alternative2
Description
Description

• Income, cost estimations
• Financing strategies
• Tax laws

Cash flows over some time period
Cash flows over some time period
Analysis using an engineering economy model
Analysis using an engineering economy model

• Planning horizon
• Interest
• Measure of worth

• Calculated value of
• measure of worth

Evaluated alternative2
Evaluated alternative 1
Noneconomic issues-environmental considerations
I select alternative 1
Rate of return (Alt 1) gtRate of return
(Alt 2)
16

• To compare different methods of achieving a given
  objective, it is necessary to have evaluation
  criteria.
• dollars or TL are used as a basis for comparison
  and the evaluation criteria is to choose the
  alternative with the
• Lowest overall cost
• highest overall profit

17

• Intangible factors are only considered if the
  alternatives have approximately the same
  equivalent costs.
• Costs as determined by usual accounting systems
  are not sufficient for future decision-making, so
  the concept of the time value of money will be
  introduced.

18
Topic 2 Interest and Time Value of Money

• Time Value of Money
• Interest
• Simple Interest
• Compound Interest

19
Time value of money
Money makes money Money available now has more
value than the same amount of money available
later. This change in the amount of money over a
given time period is called the time value of
money.
20
Interest- a measure of the increase between the
original sum barrowed (or lent) and the final
amount owned (or accrued).So,

interest total amount accumulated-original invest
ment (if money is lent) interestpresent
amount to be paid-original loan (if money is
borrowed) Percent interest rateinterest accrued
per time unit /original amount100
21
Definitions
PPrincipal-original investment or
loan iinterest rate-when interest is expressed
as a percentage of the original amount per unit
time. Interest period-time unit used for the
quoted interest rate. (4 per one year)
22
Simple and compound interest
For more than one interest period, the terms
simple interest and compound interest become
important!
23
Simple Interest(Basit Faiz)
It ignores any interest accrued in preceding
interest periods. If P principal i interest
rate n number of interest periods
where the interest rate is expressed
in decimal form.
Simple interest P.i.n
24
Compound Interest(Bilesik Faiz)
Interest on top of interest.i.e. It reflects the
effect of the time value of money on the
interest also.
Compound interest (Pall accrued
interest).i for one period.
25
(No Transcript)
26
Example1,000 loaned for 3 periods at an
interest rate of 10 compounded each period.
27
Topic 3 Equivalence and interest of formulas

• Equivalence
• Notations and their meanings
• Cash Flows and Cash Flow Diagram
• Interest Formulas

28

• Nominal and Effective interest rate
• Continuous Compounding
• Payment periodgt Compounding period
• Payment periodlt Compounding period

29
Equivalence
Equivalence-different sums of money at different
times can be equal in value. e.g. If interest
rate is 50 / year, 100 TL today is equivalent
150 TL one year from now. That means, you are
indifferent to 100TL today or 150TL one year from
now. It has the same economic value to you.
30
Notations and their meanings
P value or amount of money at a time denoted as
the present, called the present worth or present
value. F value or amount of money at some
future time, called future worth or future
value. A series of consecutive, equal,
end-of-period amounts of money, called the
equivalent value per period or annual worth. n
number of interest period years, months. i
interest rate per interest period percent per
year, percent per month.
31
Cash Flows
Cash receipts and cash disbursements in a
(revenue, (expenses) income)
given time interval referred to as cash flow.
receipts - disbursements Net
cash flow Cash receipts - cash disbursements
End-of-period convention a simplifying
assumption all cash flows occur at the
end of interest period.
32
Cash Flow Diagram
A graphical representation of cash flows on a
time scale. (should represent the statement of
the problem and what is to be solved.) Cash flow
diagram time t0 is the present, and t1 is the
end of time period 1.
33
Example
Positive cash flows (cash receipts)

6 million TL
1 million TL
1 million TL
Cash flows TL
1
2
3
5
time
0
4
5 million TL
-
Negative cash flow (disbursement)
34
Interest Formulas

• Finding F when Given P
• Finding P when Given F
• Finding F when Given A
• Finding A when Given F
• Finding P when Given A
• Finding A when Given P
• Standard Factor Notation and Use of Interest
  Tables
• Finding P when Given G
• Finding F when Given G
• Finding A when Given G
• Escalating Series
• Calculate i
• Calculate n

35
Finding F When Given P
F P(F/P, i, n)
P(1i)n
F/P factor or Single-payment compound amount
36
Finding P When Given F
P F F(P/F, i, n)
1
(1i)n
P/F factor or single-payment present worth amount
37
Uniform Series Payments

• Finding F when Given A
• Finding A when Given F
• Finding P when Given A
• Finding A when Given P
• Perpetual Endowment
• Capitalized Cost

38
Finding F when Given A
F A A(F/A, i,n)
(1i)n-1
i
Uniform series compound amount factor
39
Finding A when Given F
A F F(A/F, i,n)
i
(1i)n-1
Sinking Fund factor
40
Finding P when Given A
P A A(P/A, i, n)
(1i)n-1
i (1i)n
Uniform series present worth factor
41
Example
How much would you be willing to pay me now for
my agreement to pay you 100 at the end of each
year for 6 years, if you desire a 6/year return
on your loan?
42
Finding A when Given P
A P P(A/P, i, n)
i(1i)n
(1i)n-1
Capital recovery factor
43
Example
If I lend you 10,000 now with the understanding
that you will repay this principal sum, including
interest at 4 on all unpaid balances in uniform
annual payments, how much should you pay me at
the end of each year for 5 years?
44
Perpetual Endowment
You give a corporation 1000 in return for which
the corporation agrees to pay you and your heirs
4 interest at the end of each year forever,
never returning the principal. How much should
you receive each year? P 1000 i
4/year A? n
infinity AP P
limn infinity APi (The principal is
never repaid.) APi
i(1i)n
1
(1i)n-1
(1/i)-(1/i(1i)n)
45
Capitalized Cost
A corporation promises to pay you and your heirs
100 a year forever. How much should you be
willing to pay this promise if you want a return
of 5 on your money? A 100
i5/year P ? P 2000
100
A
i
0.05
46
Standard Factor Notation and Use of Interest
Tables
Notation to represent the factors (X/Y, i,
n) See Appendix C.
Numbers of interest periods involved
What you want to find
What is given
interest rate/interest period
I10
1.6105
47
Gradient Series

• Uniform Gradient Series
• Finding P when Given G
• Finding F when Given G
• Finding A when Given G
• Geometric Gradient (Escalating)

48
Uniform Gradient Series
When a cash flow series increases or decreases
uniformly receipts or disbursements
change by the same amount in each period
uniform gradient The amount of increase or
decrease gradient G The initial sum over the
gradient occurs base payment
49
Cash Flow diagram for a Uniform Gradient
Increasing by G amount per period
(n-1)G
(n-2)G
(n-3)G
i interest rate per period
3G
2G
G
3
1
2
4
n
n-2
n-1
Notice that the first cash flow occurs at the end
of period two.
50
Finding P when Given G
PG -


n
(1i)n-1
1
i
i (1i)n
(1i)n
Gradient to present equivalent conversion factor
51
Example
If you promise to pay me 1000 at the end of this
year and increase your payments by 100/year each
year thereafter for 10 years, and if I want a
return of 6/year. How much will I be willing to
pay you now? Soln Let PG PW of the uniform
gradient PA PW of the uniform- series
payment PT PW of the total payments
52
e.g(continued)
PT ?
3
10
1
2
9
0
G100 A1000 PG 100(P/G, 6, 10)100(29.602)

2960.232 PA1000(P/A, 6, 10)1000(7.3601)
7360.1 PT PG PA
2960.2327360.1 10320.332 OR PT
1000(1/1.06)1100(1/1.06)2.....1900 (1/1.06)10
10320.49
53
Finding F when Given G
F -
n
G
(1i)n-1
i
i
54
Example
If I deposit 400 one year from now, 500 2 years
from now, 600 3 years from now, and 700 4 years
from now. What will be the amount accumulates at
the end of the 4th year, if interest is
compounded annually at 6/year? Soln Let FG FW
of the uniform gradient FA FW of the uniform
series payment FT FW of total payments
55
e.g(continued)
F?
0
1
2
3
4
G100 i6/yr. n4 years
400
500
600
700
(1.06)4 1
(1.06)4-1
FG 100/0.06 -4 , FA 400
FT FG FA2374.4325
0.06
0.06
56
Finding A when Given G
A G -

n
1
i
(1i)n -1
Gradient to uniform series conversion factor
57
Example
Given the following uniformly decreasing income,
what is the equivalent uniform annual worth at
10 interest? End of year
Amount of income 1
5000 2
4600 3
4200 4
3800 5
3400 6
3000 7
2600 8
2200
58
e.g(continued)
5000
5000
5000
5000
5000
Soln G 400 i 10/year n
8


2
3
7
8
1
400
800
2400
2800
A1 400(1/0.1)-8/((1.1)8-1) 1201.6 A 5000-
A1 5000- 1201.6 3798.4/year.
59
Escalating Series
When cash flow increase or decrease by a constant
percentage in consecutive periods
escalating series
3
1
2
n
0
D
D(1E)
D(1E)2
D base amount E escalation rate
D(1E)n-1
60
Present Worth of Escalating Series
PED
(1E)n
-1
(1i)n
When E i
(E-i)
n
When Ei, PED
1E
61
Example
A truck has a first cost of 80 million TL, and it
is expected to last 6 years, with a 13 million TL
salvage value. The operating cost is expected to
be 17 million TL for the first year, and increase
by 11/yr. thereafter. What is the equivalent
present cost of the truck if interest rate is
8/year?
62
P?
13 million TL
i 8/yr.
17
17(1.11)
17(1.11)2
17(1.11)3
80 million TL
17(1.11)4
17(1.11)5
63
PT 80,000,000PE -13,000,000(P/F,8,6)
80,000,00017,000,000 -
13,000,000(P/F,8,6) 80,000,00017,000,000(
5.9559)-8,192,600 173,058,500 TL.
(1.11)6 / (1.08)6 -1
0.11-0.08
64
Calculation of Unknown Interest Rates
Amount to be invested Cash flows
known
Rate of return, or the interest rate ? Use trial
error, and interpolation
65
Example
If I invest 3000 now in return for 5000 5 years
from now, what rate of return will I be
earning? If I can receive 9/yr. from an
alternative investment, which alternative should
I prefer? P 3000 F 5000 n 5 years i ?
66
Calculation of Unknown Years
How long will it take for 1000 to double if
interest rate is 5/year? P 1000, F 2000, i
5/yr. n? FP(1i)n a 20001000(1.05)n
(1.05)n 2
n ln 1.05 ln 2
n 14.2 years
67
Example
If a person deposits 2000 now, 500 3 years from
now, how many years will it take from now for his
total investment to amount to 10000, if
interest rate is 6/yr.?
68
Nominal and Effective interest rate
Relationship between nominal and effective
interest rate is similar to the relationship
between simple and compound interest
rates. Difference Nominal and effective rates
are used when compounding period (interest
period) lt 1 year.
69
Interest rates usually quoted as nominal. e.g. 6
interest means 6 compounded annually unless
otherwise stated. Interest at 6 per year
compounded semiannually means 3 interest is
charged every 6 months.
70
Nominal interest rate
If r nominal interest rate, r interest rate
per period (lt a year) ( of periods) e.g. 1.5
interest / month r 4.5 /quarter
18 /year 9 /semiannually NOTE
However these calculations ignore the time value
of money like simple interest. When time value of
money is included, effective interest rate is
obtained.
71
Effective interest rate

• To find the effective rate, given the nominal
  rate, you need
• Compounding period
• The frequency of payments, or receipts means
  payment period
• If compounding period payment period then
• Where i effective interest rate,
• m of compounding periods,
• r nominal interest rate

i (1 r/m)m -1
72
e.g. If 6 per year is compounded
semiannually(nominal), then your money will
receive 3 at the end of the first 6 months and
another 3 on the amount accumulated in the
second 6 months. i.e. effective interest rate of
6/year compounded semiannually is (1.03)2 -1
1.0609-1 0.0609
6.09 per year.
73
Interest can be expressed as I. Stated over
some time period without specifying the
compounding period. effective rates e.g.
1 per month means effective 1 per month
compounded monthly
74
II. The compounding period is identified
(compounding periodlt period over which interest
is stated) means nominal rate e.g. 8 per year,
compounded monthly means nominal 8 per year,
compounded monthly
75

• The word effective precedes as follows the
  specified interest rate. Compounding period is
  given.
• e.g. Effective 6 per month means effective 6
  per month compounded monthly

76
Examples
77
Effective rates can be calculated for periods
longer than actual compounding period. e.g.
2/month effective ?/ semiannually r
2/month 6 months /semiannual period
12/semiannual period m 6 months in a
semiannual period i (10.12/6)6 -1 12.62per
semiannual period.
78
Example
5/quarter What is the effective rate per
semiannual period? R5210/months i
(10.10/2)2 -1 10.25 per semiannual period
79
NOTE ONLY EFFECTIVE INTEREST RATES CAN BE USED
IN THE EQUIVALENCE FORMULAS DEVELOPED.
80
Continuous Compounding
Interest compounded continuously means m tends to
infinity. Then the effective interest rate
i er-1
81
e.g. If interest rate is 20/year compounded
continuously, the effective rate is i e0.20
-1 1.2214-1 22.14/year e.g. What is the
effective monthly interest rate of 20/yr.
compounded continuously? Nominal monthly interest
rate 20/12 1.667 Effective monthly rate e
0.01667-1 1.68
82
Calculation for Payment periodgt Compounding
period
Use the effective rate per period, and the
appropriate of interest periods.
83
Example
If I deposit 1000 now, 4000 4 years from now,
and 1500 6 years from now at an interest rate of
15/year compounded semiannually, how much money
will I have in my account 10 years from now?
84

• Two ways of solving
• i effective interest rate/year
• n of years
• i effective interest rate/6 months n of 6
  months

85

• i(10.15/2)2-115.563/yr effective
• F1000(1.155563)10 4000(1.155563)6
• 1500(1.155563)4
• F 16450.659

86
2) i 15/2 7.5/semiannuall period n20(the
of 6 months) F1000(1.075)20
4000(1.075)12 1500(1.075)8 F
16450.186
87
Payment periodlt Compounding period

• Two cases are possible
• No interest paid on money deposited (or
  withdrawn) between compounding periods
• Money deposited (or withdrawn) earns simple
  interest between compounding periods

88

• Any amount deposited between compounding periods
  assumed to have been deposited at the beginning
  of the next compounding period.
• 2.Any amount deposited between compounding
  periods earns simple interest of (M/N)i where
• M of periods prior to the end of the
  compounding period
• N of periods in a compounding period
• i interest rate per compounding period.

89
Example
How much money would be in my account if I make
deposits as shown below? Assume bank pays 6/year
compounded semiannually, and pays simple interest
on the interperiod deposits.
90
F?
month
2
3
1
4 5 6 7 8 9 10
11 12
0
750
700
800
850
900
1000
91
Let F6 the total value of deposits made in the
first 6 months, at the end of the 6th month. F6
10001000(5/6)0.03 900900(3/6)0.03800
2738.5 Let F12 amount accumulated at the
end of the second 6 months. F12
750750(5/6)0.03 850850(4/6)0.03
700700(1/6)0.03 2339.3 F
2738.5(1.03)2339.3 5160
92
Location of PW and FW
When uniform series of payments begin at a time
later than the end of interest period 1, to
calculate PW e.g.
P?
1 2 3 4 5 6 7 8
9 10 11 12 13
A A A A A A A A A A
93

• FIND THE PW OF EACH DISBURSEMENT AT YEAR ZERO,
  AND ADD THEM UP.
• i.e.

94
2. FIND THE FUTURE WORTH AT YEAR 13 OF EACH
DISBURSEMENT AND ADD THEM UP, THEN FIND THE PW OF
THE TOTAL FW AT YEAR ZERO. i.e.
95
3. FIND FA A(F/A,i,10), THEN FA(P/F,i,13). 4.
FIND PA A(F/A,i,10), THEN FIND P
PA(P/F,i,13).
96
SINCE (P/A) FORMULA WAS DERIVES BY ASSUMING THAT
THE FIRST PAYMENT STARTED AT THE END OF PERIOD 1,
P MUST ALWAYS BE LOCATED ONE INTEREST PERIOD
PRIOR TO THE FIRST UNIFORM PAYMENT WHEN USING
(P/A) FACTOR.
97
SINCE (F/A) IS DERIVED WITH BEING THE SAME YEAR
AS THE LAST UNIFORM PAYMENT FW IS ALWAYS LOCATED
IN THE SAME PERIOD AS THE LAST UNIFORM PAYMENT
WHEN USING (F/A) FACTOR.
98
EXAMPLE
A PERSON BUYS PROPERTY FOR 50 million DOWN
PAYMENT(PESINAT) AND DEFERRED(DELAYED) ANNUAL
PAYMENTS OF 5 million FOR 6 YEARS STARTING 3
YEARS FROM NOW. WHAT IS THE PW OF INVESTMENT IF
INTEREST IS 8/YEAR?
99
i8/yr.
P?
PA
1 2 3 4 5 6 7 8
A A A A A A5 million
50 million
100
(No Transcript)
101
EXAMPLE
FIND THE 8 YEAR EQUIVALENT UNIFORM ANNUAL WORTH
AT 16/YEAR. INTEREST OF THE UNIFORM
DISBURSEMENTS SHOWN BELOW.
F
PT
PA
A A A A A A800
102

• PW METHOD

103
2. FW METHOD
104
EXAMPLE
A couple decided to sell their mineral rights on
their land to a mining corporation with the
objective of financing their childrens
education. their children are 12 and 2 years old
now, they would be in college 6 to 16 years from
now. The couple made a proposal to the company so
that it would pay them 60 million/year for 20
years beginning 1 year from now, plus 30 million
six years from now, and 45 million 16 years from
now. If the corporation wanted to pay off its
lease immediately, how much would it have to pay
now, if the interest rate is 16/yr.?
105
EXAMPLE
CONVERT THE CASH FLOW IN THE PREVIOUS EXAMPLE TO
AN EQUIVALENT UNIFORM ANNUAL SERIES OVER 22 YEARS
WITH i 16/yr.
106
PW OF EQUIVALENT UNIFORM SERIES OF SHIFTED
GRADIENTS
THE PW OF A UNIFORM GRADIENT IS ALWAYS LOCATED 2
PERIODS BEFORE THE GRADIENT STARTS.
107
EXAMPLE
COMPUTE THE EQUIVALENT ANNUAL WORTH OF THE
FOLLOWING CASH FLOWS IF i10/yr.
PG
0 1 2 3 4 5 6
7
108
BASE AMOUNT50 THAT APPLIES FOR 7 YEARS. 1.PW
OF GRADIENT G20 IS PG 20(P/G,10,5)
20(6.862) 137.24 2. P0 PG (P/F,10,2)
137.24(0.8264)113.415
109
3.A P0 (A/P,10,7) 113.415(0.20541)
23.2966 4.AT 50A5023.2966 73.2966