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Value Engineering Unit 2 MG6863

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Engineering Economics
Final Year Mechanical Engineering

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Value Engineering Unit 2 MG6863

  1. 1. S.Balamurugan AP/Mech AAA College of Engg. & Tech.
  2. 2. Value Engineering VALUE - Reliable performance at Least cost FUNCTION - Performance COST - Expenditure of a resource(time, money, people, energy and material) S.Balamurugan AP/Mech AAA College of Engg. & Tech.
  3. 3. Make or Buy Decision S.Balamurugan AP/Mech AAA College of Engg. & Tech. • 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.
  4. 4. 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 S.Balamurugan AP/Mech AAA College of Engg. & Tech.
  5. 5. 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 S.Balamurugan AP/Mech AAA College of Engg. & Tech. 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
  6. 6. 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 S.Balamurugan AP/Mech AAA College of Engg. & Tech.
  7. 7. 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 S.Balamurugan AP/Mech AAA College of Engg. & Tech.
  8. 8. 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 S.Balamurugan AP/Mech AAA College of Engg. & Tech. Value = 𝑃𝑒𝑟𝑓𝑜𝑟𝑚𝑎𝑛𝑐𝑒(𝑈𝑡𝑖𝑙𝑖𝑡𝑦) 𝐶𝑜𝑠𝑡
  9. 9. 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 S.Balamurugan AP/Mech AAA College of Engg. & Tech.
  10. 10. 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? S.Balamurugan AP/Mech AAA College of Engg. & Tech.
  11. 11. 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
  12. 12. S.Balamurugan AP/Mech AAA College of Engg. & Tech. 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
  13. 13. 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 S.Balamurugan AP/Mech AAA College of Engg. & Tech. • 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
  14. 14. 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. S.Balamurugan AP/Mech AAA College of Engg. & Tech.
  15. 15. 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 S.Balamurugan AP/Mech AAA College of Engg. & Tech.
  16. 16. 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 S.Balamurugan AP/Mech AAA College of Engg. & Tech.
  17. 17. 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 S.Balamurugan AP/Mech AAA College of Engg. & Tech. i %
  18. 18. 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 S.Balamurugan AP/Mech AAA College of Engg. & Tech. i %
  19. 19. 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 S.Balamurugan AP/Mech AAA College of Engg. & Tech. i %
  20. 20. 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 S.Balamurugan AP/Mech AAA College of Engg. & Tech. i %
  21. 21. 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) S.Balamurugan AP/Mech AAA College of Engg. & Tech.
  22. 22. 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 S.Balamurugan AP/Mech AAA College of Engg. & Tech.
  23. 23. 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) S.Balamurugan AP/Mech AAA College of Engg. & Tech.
  24. 24. 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 S.Balamurugan AP/Mech AAA College of Engg. & Tech.
  25. 25. 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 S.Balamurugan AP/Mech AAA College of Engg. & Tech.

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