Class:4th Branch: Industrial Engineering Examiner:-Msc.Rasha...

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



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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



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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



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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



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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)



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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



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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



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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



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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



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Lecture ten

Continuous interest

F= P e rn

P = F e -rn

ia = e r -1

SOLVE PROBLEMS



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Lecture eleven

4- A: Uniform series compound interest rate

SOLVE PROBLEMS



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Lecture one twelve

b- arithmetic gradient

SOLVE PROBLEMS



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Lecture thirteen

Present worth analysis I

PW =PB-PC

PW= present worth value

PB= present benefit value

pc= present cost value

SOLVE PROBLEMS



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Lecture fourteen

Present worth analysis II

PW =PB-PC

PW= present worth value


pc= present cost value

SOLVE EXTRA PROBLEMS



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Lecture fifteen

Equivalent uniform Annual analysis

a- Equivalent Uniform Annual Benefits ( EUAB)

( EUAB)= PB ( A/P ,i,n)


SOLVE PROBLEMS



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Lecture sixteen

Equivalent uniform Annual analysis

b- Equivalent Uniform Annual Costs( EUAC)

( EUAC)= PC ( A/P ,i,n)

PC= present costsvalue

SOLVE PROBLEMS



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Lecture seventeen

Equivalent uniform Annual worth

EUAW= EUAB - EUAC

SOLVE PROBLEMS



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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



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Lecture nineteen


Model (1)

If compound interest annually

i = (F/P)-n

-1

SOLVE PROBLEMS



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Lecture twenty


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



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Lecture twenty one


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



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Lecture twenty two


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



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Lecture twenty three


Model (5)

ZERO =PB-PC

Then try and error using and using linear interpolation equation

( )( ( )

( )

SOLVE PROBLEMS



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Lecture twenty four




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Lecture twenty five

Incremental analysis I

If ^ IRR<MARR

SOLVE PROBLEMS



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Lecture twenty six

Incremental analysis II

If ^ IRR>MARR

SOLVE PROBLEMS



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Lecture twenty seven

MAKING REVIEW

AND




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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



Class:4th



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.



Class:4th



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

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