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Ministry of Higher Education and Scientific Research University of Technology
Dep. of Production Engineering and Metallurgy (2014-2015)
Class:4th
Subject: Engineering Economic and Value Engineering
Branch: Industrial Engineering Examiner:-Msc.Rasha Jabbar
Lecture one
Making economic decisions:
Complexity of problem
1- Simple problems
2- Intermediate problems.
3- Complex problems
Engineering economic analysis
adet the results
choose the best alternative
predict each alternative's outcome
construct a mode
selact criterion to determine the best alternative
identify feasible alternatives
assemble data related
define the goal or objective
recognize the problem
Ministry of Higher Education and Scientific Research University of Technology
Dep. of Production Engineering and Metallurgy (2014-2015)
Class:4th
Subject: Engineering Economic and Value Engineering
Branch: Industrial Engineering Examiner:-Msc.Rasha Jabbar
Lecture two
Engineering costs and cost estimating
A- Engineering costs
fixed costs (FC)
marginal cost (MC)
variable cost (VC)
average cost (AVC)
total cost(TC)
opportunity costs
EX:- Electricity is sold for$0.12 per kilowatt-hour(KWH)for the first 10,000 units
each month and $0.09 KWH for all remaining units, if a firm uses 14,000
KWH/month, what is its average and marginal cost?
Solution:-
$0.12 for 10,000
$0.09 for 14,000-10,000=4000
Marginal cost (MC)= $0.09
Total cost = 0.12(10,000)+ 0.09(4000)+ (VC=0)
(TC)=$1,560
Average cost (AVC)= 1,560/14,000
=$ 0.111/KWH
Ministry of Higher Education and Scientific Research University of Technology
Dep. of Production Engineering and Metallurgy (2014-2015)
Class:4th
Subject: Engineering Economic and Value Engineering
Branch: Industrial Engineering Examiner:-Msc.Rasha Jabbar
Lecture three
Engineering costs and cost estimating
B- Cost estimating
Estimates that may use or need in economic analysis include:
purchase cost
annual revenue
yearly maintenance
interest rate for investments
annual labor & insurance costs
equipment salvage value
tax rates
Types of estimates:-
1- rough estimates
2- semi detailed estimates
3- detailed estimates
Ministry of Higher Education and Scientific Research University of Technology
Dep. of Production Engineering and Metallurgy (2014-2015)
Class:4th
Subject: Engineering Economic and Value Engineering
Branch: Industrial Engineering Examiner:-Msc.Rasha Jabbar
Lecture four
Estimating models
1- per unit model
EX:
Service/cost
Safety cost/employ
Gasoline cost/unit
Cost of defects/batch
Maintenance cost /window
2- cost index model
3- power sizing model
[ ( )
( ) ]
Where x is the power sizing-exponent, cost of A&B are at the same time
Ministry of Higher Education and Scientific Research University of Technology
Dep. of Production Engineering and Metallurgy (2014-2015)
Class:4th
Subject: Engineering Economic and Value Engineering
Branch: Industrial Engineering Examiner:-Msc.Rasha Jabbar
Lecture five
Interest and equivalence
Cash flow diagram
The costs and benefits of engineering projects occur over time & are summarized on a
cash flow diagram (CFD)
Ministry of Higher Education and Scientific Research University of Technology
Dep. of Production Engineering and Metallurgy (2014-2015)
Class:4th
Subject: Engineering Economic and Value Engineering
Branch: Industrial Engineering Examiner:-Msc.Rasha Jabbar
Lecture six
1- Simple interest
Is the interest that is computed only on the original summation , not on
occurred interest
Total interest earned = P*i*n =Pin
P = present sum of money
I= interest rate (%)
N= period of (n) years
At the end of year n the amount due is F,
F= P (1+i*n)
P= F (1+i*n) -1
SOLVE PROBLEMS
Ministry of Higher Education and Scientific Research University of Technology
Dep. of Production Engineering and Metallurgy (2014-2015)
Class:4th
Subject: Engineering Economic and Value Engineering
Branch: Industrial Engineering Examiner:-Msc.Rasha Jabbar
Lecture seven
2- Compound interest
Interest is computed by the compound interest formula
“Interest on top interest”
F= P (1+i) n
P= F (1+i) - n
SOLVE PROBLEMS
Ministry of Higher Education and Scientific Research University of Technology
Dep. of Production Engineering and Metallurgy (2014-2015)
Class:4th
Subject: Engineering Economic and Value Engineering
Branch: Industrial Engineering Examiner:-Msc.Rasha Jabbar
Lecture eight
3- Single payment compound interest formulas
F= P (1+i) n
F= P( F/P,i,n)
P= F (1+i) -n
P= F(P/F,i,n)
( F/P,i,n) and (P/F,i,n) is compound interest factor from compound
interest table
SOLVE PROBLEMS
Ministry of Higher Education and Scientific Research University of Technology
Dep. of Production Engineering and Metallurgy (2014-2015)
Class:4th
Subject: Engineering Economic and Value Engineering
Branch: Industrial Engineering Examiner:-Msc.Rasha Jabbar
Lecture nine
Nominal & effective interest rate
Nominal interest ( r)= m * i
M= no. of compounding sub period/ time period
i = interest rate
Effective interest rate (ia) = (1+ r/m)m
-1
Or = (1+ i)m
-1
SOLVE PROBLEMS
Ministry of Higher Education and Scientific Research University of Technology
Dep. of Production Engineering and Metallurgy (2014-2015)
Class:4th
Subject: Engineering Economic and Value Engineering
Branch: Industrial Engineering Examiner:-Msc.Rasha Jabbar
Lecture ten
Continuous interest
F= P e rn
P = F e -rn
ia = e r -1
SOLVE PROBLEMS
Ministry of Higher Education and Scientific Research University of Technology
Dep. of Production Engineering and Metallurgy (2014-2015)
Class:4th
Subject: Engineering Economic and Value Engineering
Branch: Industrial Engineering Examiner:-Msc.Rasha Jabbar
Lecture eleven
4- A: Uniform series compound interest rate
SOLVE PROBLEMS
Ministry of Higher Education and Scientific Research University of Technology
Dep. of Production Engineering and Metallurgy (2014-2015)
Class:4th
Subject: Engineering Economic and Value Engineering
Branch: Industrial Engineering Examiner:-Msc.Rasha Jabbar
Lecture one twelve
b- arithmetic gradient
SOLVE PROBLEMS
Ministry of Higher Education and Scientific Research University of Technology
Dep. of Production Engineering and Metallurgy (2014-2015)
Class:4th
Subject: Engineering Economic and Value Engineering
Branch: Industrial Engineering Examiner:-Msc.Rasha Jabbar
Lecture thirteen
Present worth analysis I
PW =PB-PC
PW= present worth value
PB= present benefit value
pc= present cost value
SOLVE PROBLEMS
Ministry of Higher Education and Scientific Research University of Technology
Dep. of Production Engineering and Metallurgy (2014-2015)
Class:4th
Subject: Engineering Economic and Value Engineering
Branch: Industrial Engineering Examiner:-Msc.Rasha Jabbar
Lecture fourteen
Present worth analysis II
PW =PB-PC
PW= present worth value
PB= present benefit value
pc= present cost value
SOLVE EXTRA PROBLEMS
Ministry of Higher Education and Scientific Research University of Technology
Dep. of Production Engineering and Metallurgy (2014-2015)
Class:4th
Subject: Engineering Economic and Value Engineering
Branch: Industrial Engineering Examiner:-Msc.Rasha Jabbar
Lecture fifteen
Equivalent uniform Annual analysis
a- Equivalent Uniform Annual Benefits ( EUAB)
( EUAB)= PB ( A/P ,i,n)
PB= present benefit value
SOLVE PROBLEMS
Ministry of Higher Education and Scientific Research University of Technology
Dep. of Production Engineering and Metallurgy (2014-2015)
Class:4th
Subject: Engineering Economic and Value Engineering
Branch: Industrial Engineering Examiner:-Msc.Rasha Jabbar
Lecture sixteen
Equivalent uniform Annual analysis
b- Equivalent Uniform Annual Costs( EUAC)
( EUAC)= PC ( A/P ,i,n)
PC= present costsvalue
SOLVE PROBLEMS
Ministry of Higher Education and Scientific Research University of Technology
Dep. of Production Engineering and Metallurgy (2014-2015)
Class:4th
Subject: Engineering Economic and Value Engineering
Branch: Industrial Engineering Examiner:-Msc.Rasha Jabbar
Lecture seventeen
Equivalent uniform Annual worth
EUAW= EUAB - EUAC
SOLVE PROBLEMS
Ministry of Higher Education and Scientific Research University of Technology
Dep. of Production Engineering and Metallurgy (2014-2015)
Class:4th
Subject: Engineering Economic and Value Engineering
Branch: Industrial Engineering Examiner:-Msc.Rasha Jabbar
Lecture eighteen
Rate of return analysis
By applying rate of return analysis (ROR), one can:
1- evaluate project CFS with the internal rate of return IRR
measures
2- use an incremental ROR analysis to evaluate competing
alternatives
3- plot a project PW against the interest rate
Ministry of Higher Education and Scientific Research University of Technology
Dep. of Production Engineering and Metallurgy (2014-2015)
Class:4th
Subject: Engineering Economic and Value Engineering
Branch: Industrial Engineering Examiner:-Msc.Rasha Jabbar
Lecture nineteen
Rate of return analysis
Model (1)
If compound interest annually
i = (F/P)-n
-1
SOLVE PROBLEMS
Ministry of Higher Education and Scientific Research University of Technology
Dep. of Production Engineering and Metallurgy (2014-2015)
Class:4th
Subject: Engineering Economic and Value Engineering
Branch: Industrial Engineering Examiner:-Msc.Rasha Jabbar
Lecture twenty
Rate of return analysis
Model (2)
If uniform series compound interest rate
P/A=(P/A,i,n)
X1 Y1
X2? Y2
X3 Y3
Using linear interpolation equation
( )( ( )
( )
SOLVE PROBLEMS
Ministry of Higher Education and Scientific Research University of Technology
Dep. of Production Engineering and Metallurgy (2014-2015)
Class:4th
Subject: Engineering Economic and Value Engineering
Branch: Industrial Engineering Examiner:-Msc.Rasha Jabbar
Lecture twenty one
Rate of return analysis
Model (3)
IF Nominal & effective interest rate
Compute the ROR from models (1&2) then apply:
Nominal interest ( r)= m * i
Effective interest rate (ia) = (1+ r/m)m
-1 Or = (1+ i)m
-1
SOLVE PROBLEMS
Ministry of Higher Education and Scientific Research University of Technology
Dep. of Production Engineering and Metallurgy (2014-2015)
Class:4th
Subject: Engineering Economic and Value Engineering
Branch: Industrial Engineering Examiner:-Msc.Rasha Jabbar
Lecture twenty two
Rate of return analysis
Model (4)
If compounded and uniform annually series
A(P/A,i,n)=F(P/F,i,n)
F/A= (F/A,i,n)
A/F= (A/F,i,n)
Using linear interpolation equation
( )( ( )
( )
SOLVE PROBLEMS
Ministry of Higher Education and Scientific Research University of Technology
Dep. of Production Engineering and Metallurgy (2014-2015)
Class:4th
Subject: Engineering Economic and Value Engineering
Branch: Industrial Engineering Examiner:-Msc.Rasha Jabbar
Lecture twenty three
Rate of return analysis
Model (5)
ZERO =PB-PC
Then try and error using and using linear interpolation equation
( )( ( )
( )
SOLVE PROBLEMS
Ministry of Higher Education and Scientific Research University of Technology
Dep. of Production Engineering and Metallurgy (2014-2015)
Class:4th
Subject: Engineering Economic and Value Engineering
Branch: Industrial Engineering Examiner:-Msc.Rasha Jabbar
Lecture twenty four
SOLVE EXTRA PROBLEMS
Ministry of Higher Education and Scientific Research University of Technology
Dep. of Production Engineering and Metallurgy (2014-2015)
Class:4th
Subject: Engineering Economic and Value Engineering
Branch: Industrial Engineering Examiner:-Msc.Rasha Jabbar
Lecture twenty five
Incremental analysis I
If ^ IRR<MARR
SOLVE PROBLEMS
Ministry of Higher Education and Scientific Research University of Technology
Dep. of Production Engineering and Metallurgy (2014-2015)
Class:4th
Subject: Engineering Economic and Value Engineering
Branch: Industrial Engineering Examiner:-Msc.Rasha Jabbar
Lecture twenty six
Incremental analysis II
If ^ IRR>MARR
SOLVE PROBLEMS
Ministry of Higher Education and Scientific Research University of Technology
Dep. of Production Engineering and Metallurgy (2014-2015)
Class:4th
Subject: Engineering Economic and Value Engineering
Branch: Industrial Engineering Examiner:-Msc.Rasha Jabbar
Lecture twenty seven
MAKING REVIEW
AND
SOLVE EXTRA PROBLEMS
Ministry of Higher Education and Scientific Research University of Technology
Dep. of Production Engineering and Metallurgy (2014-2015)
Class:4th
Subject: Engineering Economic and Value Engineering
Branch: Industrial Engineering Examiner:-Msc.Rasha Jabbar
Lecture twenty eight
Value engineerin
What is value?
A product or service is generally considered to have good value if that product or service
has appropriate performance and cost. Or, by reverse. A product is considered not to have
good value if it lacks either
appropriate performance or cost. It can almost truthfully be said that, by
this definition, value can be increased by either increasing the performance or decreasing
the cost. More precisely:
1. Value is always increased by decreasing costs (while, of course,
maintaining performance).
2. Value is increased by increasing performance if the customer needs, wants, and is
willing to pay for more performance.
It is perhaps worth taking a few lines to develop the word “value”. Four aspects of value
can be considered:
Cost Value – is the cost of manufacturing and selling an item
Exchange Value – is the price a customer is prepared to pay for the product, or service
Use Value – is the purpose the product fulfils
Esteem Value – is the prestige a customer attaches to the product
Ministry of Higher Education and Scientific Research University of Technology
Dep. of Production Engineering and Metallurgy (2014-2015)
Class:4th
Subject: Engineering Economic and Value Engineering
Branch: Industrial Engineering Examiner:-Msc.Rasha Jabbar
Lecture twenty nine
Value analysis
Value analysis
Value Analysis can be defined as a process of systematic review that is applied to
existing product designs in order to compare the function of the product required by a
customer to meet their requirements at the lowest cost consistent with the specified
performance and reliability needed.
Usually a company establishes a value analysis team who may adopt the following
processes:
Step 1 :-Selecting a product or service for study.
Step 2 Obtaining and recording information.
Step 3 Analysing the information and evaluating the product.
Step 4 Considering alternatives.
Step 5 Selecting of the least cost alternative.
Step 6 Recommendation.
Step 7 Implementation and follow-up.
The VA approach is almost universal and can be used to analyze existing products or
services offered by manufacturing companies and service providers alike. For new
products, the Value Engineering (VE) approach, which applies the same principles.
Ministry of Higher Education and Scientific Research University of Technology
Dep. of Production Engineering and Metallurgy (2014-2015)
Class:4th
Subject: Engineering Economic and Value Engineering
Branch: Industrial Engineering Examiner:-Msc.Rasha Jabbar
Lecture thirteen
Value Engineering
Value engineering (VE) is systematic method to improve the "value" of goods or
products and services by using an examination of function. The reasoning behind value
engineering is as follows: if marketers expect a product to become practically or
stylistically obsolete within a specific length of time, they can design it to only last for that
specific lifetime. The products could be built with higher-grade components, but with
value-engineering they are not because this would impose an unnecessary cost on the
manufacturer, and to a limited extend also an increased cost on the purchaser. Value
engineering will reduce these costs. A company will typically use the least expensive
components that satisfy the product's lifetime projections.
The Job Plan
Value engineering is often done by systematically following a multi-stage job plan..
Depending on the application, there may be four, five, six, or more stages. One modern
version has the following eight steps:
1. Preparation
2. Information
3. Analysis
4. Creation
5. Evaluation
6. Development
7. Presentation
8. Follow-up