Value Engineering Unit 2 MG6863

S.Balamurugan AP/Mech AAA College of Engg. & Tech.

Value Engineering
VALUE - Reliable performance at Least cost
FUNCTION - Performance
COST - Expenditure of a resource(time, money,
people, energy and material)

Make or Buy Decision
• Any product can be made with in organization or bought from a sub contractor.
• Each decision involves its own cost. Select the best alternative which results in lower
cost.
• Make or buy decision should be reviewed periodically (1-3 years). Not static.

Criteria for Make or Buy
Making Decision
• Product can be made
cheaper by the industry
than by outside suppliers
• Unable to meet the
demand. Only few
company's manufacture the
product
• Important part & Requires
close quality control
• Company have Similar part
manufacturing experience
Buying Decision
• Requires High investments on
facility
• The company does not have
facility to make it & there are
more profitable opportunities
for investing company’s capital
• Existing facilities utilized
effectively.
• Skilled personnel not available.
• Patent or other legal barriers
• Demand is seasonal

Approaches for Make or Buy Decision
• Simple cost Analysis - Economics book Pg.No-206
• Break - Even Analysis- Economics book Pg.No-209
• Economic Analysis - Economics book Pg.No-208
Purchase Model Manufacturing Model
Q1 =
2 𝐶𝑜 𝐷
𝐶 𝑐
TC = D×P +
𝐷𝐶 𝑜
Q1
+
Q1× 𝐶 𝑐
2
Q2 =
2 𝐶𝑜 𝐷
𝐶 𝑐
1−
𝑟
𝑘
TC = D×P +
𝐷𝐶 𝑜
Q2
+ 𝐶 𝑐 (k-r)
Q2
2×𝑘
D – Demand / year, P-Purchase price / year, 𝐶 𝑐 – Carrying cost / unit /year,
𝐶 𝑜- Ordering cost / order or Setup cost / setup, k-production rate( No. of
units /year), r-demand/year, Q1-Economic order size, Q2- Economic
production size, TC- Total Cost per year

Value Analysis
• It is an organized approach to identify unnecessary costs associated with any
product (component, system or service) by analyzing the function and
eliminating such costs without affecting the quality, functional reliability or
capacity of the product to give service.
Symptoms for the need VA
– Product’s show decline in sales
– Price are higher than those of its competitors
– Raw material cost has grown disproportionate to the volume of production
– New designs are being introduced
– Cost of manufacture is rising disproportionate to the volume of production
– Rate of Return on investment has a falling trend.
– Inability of the firm to meet its delivery commitments
Rate of Return – It is the gain or loss on an investment over a specified time period

Value
• Based on person it differs
– Designer – Reliability
– Purchase person – Price
– Production person – Cost to manufacture
– Sales person – Customer willing to pay
Value Investigations, Economic Value
Cost Value - Cost required to produce the part or service (Material + Labour +
Overhead etc.)
Exchange Value – the price that a purchaser will offer for the product(Measure of
property, quality, features)
Use value (Fundamental)– Functional value (product performed its intended
functions (i) Buyer’s view- Price, Manufacturer’s view – cost involved to make, Without use
value, there is exchange or esteem value.
Esteem Value – Cost incurred by the manufacturer beyond the use value

Performance
• It is the measure of functional features and
properties that make it suitable for a specific
purpose.
• Appropriate performance requires
– Functional requirements
– Safety requirements
– Reliability requirements
– Maintainability requirements
– Appearance requirements
Value =
𝑃𝑒𝑟𝑓𝑜𝑟𝑚𝑎𝑛𝑐𝑒(𝑈𝑡𝑖𝑙𝑖𝑡𝑦)
𝐶𝑜𝑠𝑡

Function
• It is the purpose for which the
product is made.
• Primary functions – the basic
functions for which the product is
specially designed to achieve.
Essential function whose non-
performance would make the product
worthless.
• Secondary Functions – if not in built,
would not prevent the device from
performing its primary functions.
• Tertiary functions - Related to
esteem appearance

Aims of Value Engineering
• Simplify the product
• Use Cheaper & Better materials
• Modify & Improve product design
• Use efficient process
• Reduce the product cost
• Save money or Increase the profits
• Increase the utility of the product by Economical means
The value content of each piece of a product is assessed using the
following questions
Dose its use contribute to value?
Is its cost proportionate to its usefulness?
Does it need all its features?

ValueEngineering
Procedure
Blasting
1. Identify the Product
2. Collect relevant information
(Drawings, Production processes, Machine layout,
Time study, Manufacturing Capacity)
3. Define different Functions
Identify the types of functions, Specify the value
content of each function & Identify the High cost
areas.
Creating
4. Different Alternatives
Using the above information's, Generate
ideas & Create different alternatives
(Brain Storming)
5. Critically Evaluate the Alternatives
Based on financial & Technical requirements,
feasible ideas are found & involving lower
costs are further developed
Refining
.
6. Develop the Best Alternative
Development Plans comprise drawing the
sketches, building of models conducting
discussions with the purchase, Finance, Marketing
sections
7.Implement the Alternative
Best alternative Prototype

BasicPrinciplesofBrain
Storming
A quality idea comes from Quantity of ideas
Creative ideas emerge from unconventional thinking
Spontaneous evaluation of ideas curbs(control)
imaginative thinking and retards the flow of creative
ides
Hitch(Catch) -Hiking on the ideas often lead to better
ideas
Creativity is a regenerative process & the recording of
ideas as they emerge helps serve as a catalyst to
generate more ideas
When ideas cease to flow , short diversions enable the mind to
rebound with new ideas after recovery

Time Value of Money
• If an investor invests a sum of Rs.100
in a fixed deposit for five years with
interest rate 15% compounded
annually.
• Future Worth F = P × (1+i)n
Year
End
Interest
(Rs.)
Compound
amount (Rs.)
0
1 15.00 115.00
2 17025 132.25
3 19.84 152.09
4 22.81 174.90
5 26.24 201.14
• Interest rate is the Rental value of money.
Year
End
Present
Worth (Rs.)
Compound
amount after n
years (Rs.)
0 100
1 86.96 100
2 75.61 100
3 65.75 100
4 57.18 100
5 49.72 100
6 43.29 100
If we want Rs.100 at the end of the nth
year , interest rate 15%
Present Worth P =
𝐹
1+𝑖 𝑛

Time Value of Money
• A person won the prize during the festival contest, the prize will be
issued in any one of the following modes
1. Spot payment Rs.24.72
2. Rs.100 after 10 years from now (interest rate 15% compounded
annually)
Option 1- if he invests the prize amount in the business, he can get
24% interest rate worth amount. At the end of 10th year he will get
Rs.212.45.
This is the concept of Time Value of Money.

Single Payment Compound Amount
• Formula F = P × (1+i)n = P(F/P, i , n)
(?)F
0 1 2 3 4 n
P i %
Cash Flow Diagram
• A person deposits a sum of Rs.20,000 at the interest rate
of 10% compounded annually for 10 years. Find the
maturity value after 10 years.
P = Rs.20,000, i = 10%, n = 10 years
F = P × (1+i)n = P(F/P, i , n) = 20,000(F/P, 10% , 10)
= 20,000(2.594) = Rs.51,880

Single Payment Present Worth Amount
• Formula P =
𝐹
(1+i)n = F(P/F, i , n)
F
0 1 2 3 4 n
(?)P i %
Cash Flow Diagram
• A person wishes to have a future sum of Rs.1,00,000 for his son’s
education after 10 years from now. What is the single payment that
he should deposit now so that he gets the deposited amount after 10
years? The bank gives 15% interest rate compounded annually.
F = Rs.1,00,000, i = 15%, n = 10 years
• P =
𝐹
(1+i)n = F(P/F, i , n) = 1,00,000(P/F, 15%,10)
= 1,00,000(0.2472) = Rs.24,720

Equal Payment Series Compound Amount
• Formula F = A
1+𝑖 𝑛
−1
i
= A (F/A, i , n)
(?)F
0 1 2 3 4
n
A A A A A
Cash Flow Diagram
• A person who is now 35 years old is planning for his retired life. He plans to
invest an equal sum of Rs.10,000 at the end of every year for the next 25 years
starting from the end of the next year. The bank gives 10 % interest rate,
compounded annually. Find the maturity value of his account when he is
60years old.
A = Rs.10,000, i = 10%, n = 25 years
• F = A
1+𝑖 𝑛
−1
i
= A (F/A, i , n) =10,000(F/A, 10%, 25)
= 10,000(98.347) = Rs.9,83,470
i %

Equal Payment Series Sinking Fund
• Formula A = F
1
1+𝑖 𝑛
−1
= F (A/F, i , n)
F
0 1 2 3 4
n
A A A A (?) A
Cash Flow Diagram
A company has to replace a present facility after 15 years at an outlay of Rs.5,00,000. it
plans to deposit an equal amount at the end of every year for the next 15 years at an
interest rate of 9% compounded annually. Find the equivalent amount that must be
deposited at the end of every year for the next 15 years.
F = Rs.5,00,000, i = 9%, n = 15 years
• A = F
1
1+𝑖 𝑛
−1
= F (A/F, i , n) = 5,00,000(A/F, 9%,15)
= 5,00,000(0.0341) = Rs.17,050
i %

Equal Payment Series Present Worth Amount
• Formula P = A
1+𝑖 𝑛
−1
i 1+𝑖 𝑛 = A (P/A, i , n)
P(?) F
n
0 1 2 3 4
A A A A A
Cash Flow Diagram
A company wants to set up a reserve which will help the company to have an annual
equivalent amount of Rs.10,00,000 for the next 20 years towards its employees welfare
measures. The reserve is assumed to grow at the rate of 15% annually. Find the single-
payment that must be made now as reserve amount.
A = Rs.10,00,000, i = 15%, n = 20 years
• P = A
1+𝑖 𝑛
−1
i 1+𝑖 𝑛 = A (P/A, i , n) = 10,00,000(P/A,15%,20)
= 10,00,000(6.2593) = Rs.62,59,300
i %

Equal Payment Series Capital Recovery Amount
• Formula A = P
i 1+𝑖 𝑛
1+𝑖 𝑛
−1
= P (A/P, i , n)
P F
0 1 2 3 4 n
(?) A (?) A (?) A (?) A (?) A
Cash Flow Diagram
A bank gives a loan to a company to purchase an equipment worth
Rs.10,00,000 at an interest rate of 12% compounded annually. This amount
should be repaid in 15 yearly equal installments. Find the installment amount
that the company has to pay to the bank.
P = Rs.10,00,000, i = 12%, n = 15 years
• A = P
i 𝟏+𝒊 𝒏
𝟏+𝒊 𝒏
−𝟏
= P (A/P, i , n) = 10,00,000(A/P,12%,15)
= 10,00,000(0.1468) = Rs.1,46,800
i %

Uniform Gradient Series Annual Equivalent Amount
• Formula A = A1+G
1+𝑖 𝑛
−𝑖𝑛−1
𝑖 1+𝑖 𝑛
−𝑖
= A1+G(A/G, i , n)
Two cases
Annual Increase = A1+G, A = A1+ G
1+𝑖 𝑛
−𝑖𝑛−1
𝑖 1+𝑖 𝑛
−𝑖
= A1+G(A/G, i , n)
Annual Decrease = A1-G, A = A1 − G
1+𝑖 𝑛
−𝑖𝑛−1
𝑖 1+𝑖 𝑛
−𝑖
= A1+G(A/G, i , n)
Step1= Find the Annual Equivalent Amount
Step2= Find the Future Worth, F = A
1+𝑖 𝑛
−1
i
= A (F/A, i , n)

Uniform Gradient Series Annual Equivalent Amount
• A person is planning for his retired life. He has 10 more years of service. He would like
to deposit 20% of his Salary, which is Rs.4,000, at the end of the first year, and
thereafter he wishes to deposit the amount with an annual increase of Rs.500 for the
next 9 years with an interest rate of 15%. Find the total amount at the end of the 10th
year of the above series.
• A1=Rs.4,000, G=Rs.500, i=15%, n=10years, A=?, F=?
A = A1+ G
𝟏+𝒊 𝒏
−𝒊𝒏−𝟏
𝒊 𝟏+𝒊 𝒏
−𝒊
= A1+G(A/G, i , n)
A = 4,000 + 500 (A/G, 15%,10) = Rs.5691.60
Future Worth F = F = A
1+𝑖 𝑛
−1
i
= A (F/A, i , n)
F = 5691.60(F/A, 15%,10)
F = 5691.60(20.304) = Rs.1,15,562.25

Effective Interest Rate
• i be the nominal interest rate compounded
annually.
• The compounding may be monthly, quarterly or
semi-annually.
• Effective Interest Rate R = (1+
𝑖
𝐶
)C – 1
i – nominal interest rate
C – the number of interest periods in a year
• Monthly – 12 interest periods(C)
• Quarterly – 4 interest periods (C)
• Semi Annually – 2 interest periods (C)

Effective Interest Rate
• A person invests a sum of Rs.5,000 in a bank at a nominal
interest rate of 12% for 10 years. The compounding is
quarterly. Find the maturity amount of the deposit after 10
years.
• P = Rs.5,000, n = 10 years, i = 12%, F = ?
• Number of interest periods C = 4 (quarterly)
• Effective Interest Rate R = (1+
𝑖
𝐶
)C – 1
= (1 + (0.12/4))4 – 1
= 0.1255 = 12.55%
Future Worth F = P(1+R)n
= 5,000(1+.1255)10
= Rs.16,308.91

Bases for Comparison of Alternatives
• Worthiness of the project can be
determined by the following methods
– Present Worth method
– Future Worth method
– Annual Equivalent method
– Rate of Return method

Value Engineering Unit 2 MG6863

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Value Engineering Unit 2 MG6863