Engineering Economics

Report
The Engineering
Economic Equations
Every Safety Professional
Should Know
Presented By:
Jarred O’Dell, ASP
Safety Director
Syracuse Utilities
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Sleeping during the
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Why Engineering Economics
For those of you pursuing your CSP:

7 – 15 questions on the ASP Exam

7 – 15 questions on the CSP Exam
For everyone else:

Ever buy a car or a house?

Who has a credit card?
Engineering Economics
0-4
Engineering Economics
You do not
need an
app for
that!!!
Engineering Economics
Uff Da!
Uff da is an expression of
Norwegian origin adopted by
Scandinavian-Americans.
This exclamation is an
announcement that, that
person is going into a state
of sensory overload.
Uff Da!
If you find yourself going
into sensory overload and
need to ask a question say:
Uff Da!
Uff
Da!
Question: 1
Engineering Economics: Q1
A wealthy relative died and left you her
estate. You can choose to either accept
$6,000,000 today or wait and receive
$10,000,000 in five years. Assume the
annual interest rate over this period is
10%. You decide to…
Engineering Economics: Q1
You decide to:
A)
Wait the 5 years
B)
B) Take the money and run
C)
Do nothing, its probably the same people
that brought you the Nigerian Lottery
Engineering Economics
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Engineering Economics
F = Future
P = Present
A = Amount of periodic receipt/payment
n = Number of years*
i = Annual Interest* expressed in decimal
form (e.g. 10% = .10)
Page 1
Engineering Economics
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Engineering Economics
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Engineering Economics
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Engineering Economics
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(1 + ) −1
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(1 + )
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(1 + ) −1
Engineering Economics: Q1
 = (1 + )

F = 10,000,000

P = 6,000,000

A = N/A

i = 10% or 0.10

n = 5 years
Page 2
 = (1 + )−
Engineering Economics: Q1
 = (1 + )
 = (1 + )−

F = 6,000,000 (1+ 0.10)5

P = 10,000,000(1+0.10)-5
Page 2
Question: 2
Engineering Economics: Q2
You decided to go back to school and eared
your Masters Degree in mathematics.
Having heard this,
your supervisor
throws this
scenario at you:
Engineering Economics: Q2
The chief financial officer of Widget Inc.
expects a 10% annual return on
investments for all capital projects. What
is the maximum cost that will be approved
from a project that is expected to save
$8,000 per year over 10 years? Assume
the project will be fully depreciated in the
10 years.
Question 2
Uff
Da!
Engineering Economics: Q2

F = N/A

P = ???

A = $8,000

i = 10% or 0.10

n = 10 years
Page 3
Engineering Economics
 = (1 + )
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(1 + ) −1
=

=
(1 + ) −1
=
(1 + )
(1 + )
=
(1 + ) −1

(1 + ) −1
Engineering Economics: Q3
(1 + ) −1
=
(1 + )
10
.10) −1
(1 +
 = 8000
.10(1 + .10)10
Page 3
Question: 3
Engineering Economics: Q3
You recently obtained 6σ Black Belt status.
Congratulations! Understandably, you are
very anxious to test out your new skills.
Soon you face this problem:
Engineering Economics: Q3
The financial policy of Acme requires that
capital investments must have an annual
return of 12%. An engineering solution to
a safety problem will cost $250,000 for the
initial installation, and it will cost $12,000
annually to maintain for 15 years. What is
the required annual savings from this
project in order for it to be approved?
Question 3
Engineering Economics: Q3

F = N/A

P = $250,000

A = ???

i = 12% or 0.12

n = 10 years
Page 4
Engineering Economics
 = (1 + )
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(1 + ) −1
=

=
(1 + ) −1
=
(1 + )
(1 + )
=
(1 + ) −1

(1 + ) −1
Engineering Economics: Q3
(1 + )
=
(1 + ) −1
15
.12)
.12(1 +
A = 250000 (1 + .12)15 −1
Page 4
Question: 4
Engineering Economics: Q4
Having recently been conferred as a Doctor
in Safety and Engineering Science,
your are now in
a position to poses
this scenario to your
employer:
Engineering Economics: Q4
Your company decided to hire an EHS/6σ,
executive. If a balloon payment of
$10,000,000 is due in 10 years, what
amount would management have to
deposit monthly into a savings account
(paying interest of 6% per year) to
accumulate adequate funds to pay the
note?
Question 4
Engineering Economics: Q4

F = $10,000,000

P = N/A

A = ???

i = 6% or 0.06

n = 10 years
Page 5
Engineering Economics
 = (1 + )
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(1 + ) −1
=

=
(1 + ) −1
=
(1 + )
(1 + )
=
(1 + ) −1

(1 + ) −1
Engineering Economics
=

1+ −1
A = 10,000,000
 = 10,000,000
.06
1+.06 10 −1
.005
1+.005 120 −1
Engineering Economics
F = Future
P = Present
A = Amount of periodic receipt/payment
n = Number of years*
i = Annual Interest* expressed in decimal
form (e.g. 10% = .10)
Page 1
Engineering Economics
=

1+ −1
A = 10,000,000
 = 10,000,000
.06
1+.06 10 −1
.005
1+.005 120 −1
Question: 5
Engineering Economics: Q4
One of your faceless
pawns is having
trouble figuring out
the following
scenario. He
humbly/fearfully
asks for your help:
Engineering Economics: Q5
Calculate the monetary value after ten
years of a behavior based safety program
that costs $40,000 per year at the start.
Assume an inflation rate of 4.3%
Question 5
Engineering Economics: Q4

F = ???

P = N/A

A = $40,000

i = 4.3% or 0.043

n = 10 years
Page 5
Engineering Economics
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(1 + ) −1
=

=
(1 + ) −1
=
(1 + )
(1 + )
=
(1 + ) −1

(1 + ) −1
Engineering Economics: Q5
(1 + ) −1
=

F = 40,000
(1+0.043)10 −1
0.043
Uff
Da!
Engineering Economics

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