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1. COURSE INSTRUCTOR : PROF. DR. SHAHAB KHUSHNOOD Chapter#10Techniques for making better engineering management decisions Lecture No. 08 Course: Engineering Management MED DEPARTMENT, U.E.T TAXILA

2. INTRODUCTION • During middle of 1940s, a significant growth of quantitative management techniques occurred. • The increase in the use of computers and the modern problem complexities have enhanced the importance of many of these techniques. For Example: • Linear Programming. • Non-Linear Programming.

3. Today, the mostly management techniques which are in practice: • Linear Programming • Decision Trees • Exponential Smoothing • Discounted Cash Flow Analysis

4. Optimization Techniques There are two widely known optimization techniques known as: • Lagrangian Multiplier Technique (Technique of undetermined Multipliers) • Linear Programming Technique (Advanced form of simplex method technique)

5. 1-LAGRANGIAN MULTIPLIER TECHNIQUE • This technique is demonstrated for a two variable function. • However on similar lines it can be extended fornvariables. • Assume that we have defined function f(y1,y2) Subject to the constraint function k(y1,y2)=0

6. 1-LAGRANGIAN MULTIPLIER TECHNIQUE • With the aid of these two functions the Lagrangian function , L(y1,y2,λ) formulated as follows:

7. 1-LAGRANGIAN MULTIPLIER TECHNIQUE • Subject to the following necessary conditions for estimating a relative maximum or minimum value: • Expressions for y1,y2 and λ can be obtained by solving the simultaneous equations.

8. EXAMPLE-1

9. Solution: The Lagrange function is formulated as follows Taking the partial derivative with respect to y1,y2 and λresults in

10. 2y1+2λ = 0 2y2- λ=0 2y1-y2=0 So Y1=8/5 Y2=-8/10 λ =-8/5 Thus the critical point of f, subjected to specified condition, is (8/5,-8/10) .

11. 2-LINEAR PROGRAMMING • In real life situations the objective and constraint functions will be rather complex. • However the simplest form of linear programming problem formulation may be expressed as follows

12. This can be subject to

13. Where

14. Example 2 For above equations following values for symbols are defined Write down the resulting equations by assuming that the objective function is to be maximized.

15. Example 2 (cont.…)

16. EXAMPLE-3

17. Solution:

18. (1) (2) (3) (4) (5) (6)

19. Plot Equation (1) to (6) as:

20. From Plot of equations 1-6: • Company profit is optimum at point F. (F point satisfies all the constraints) • F point shows maximum profit. The optimum values of Z1 and Z2 at point F are 6.5 and 2. substitute these values in equation.1.

21. Discounted Cash Flow Analysis • In various engineering investment decisions, the time value of money plays an important role. • Therefore it is necessary for the engineers to have some knowledge of engineering economics.

22. Discounted cash flow analysis The basics of engineering economics are • Simple Interest • This is the interest which is calculated on the original sum of money, called the original principal, for the period in which the lent or borrowed sum is being utilized. • The simple interest,St,is given by: (1) Where M is the principal amount lent or borrowed i is the interest rate per period ( this is normally a year) k is the interest periods (these are usually years)

23. Simple Interest (Cont.…) The total amount of money, Mt, after the specified lent or borrowed period is given by (2)

24. Example: 4

25. Solution: (1) (2)

26. 2. Compound Interest • In this case at the end of each equal specified period, the earned interest is added to the original principal or amount lent or borrowed at the beginning of that period • Thus, this new principal, or amount, acts as a principal for the next period and the process continues.

27. To calculate compound amount, Mck, the resulting formula is developed as follows:

28. Example:5

29. Solution:

30. 3.Present Worth • The present worth of a single payment is given by: • In simple terms, this formula is used to obtained the present worth, M, of money,Mck, after k periods, discounted at the periodic interest rate of i. • Sometime above equation can be written as

31. Example # 6

32. Solution:

33. Formula For Uniform Periodic Payments • For using formula it is assumed that at the end of each of the K period or years, the depositor adds D amount of money. The money is invested at the interest rate i, compounded annually or periodically. • Thus the total amount of money (1) where

34. It should be noted in this equation that the D amount of money is first time deposited at the end of first year or period.

35. Example # 07

36. Solution:

37. By substituting k=3, D=2000 and i= 0.15 in equation (1): From above equation the total amount of money after three year period is:

38. Example # 8 Given Data: (From Example 7)

39. Solution: Substitute the data in given equation:

40. PRESENT VALUE OF UNIFORM PERIODIC PAYMENTS • In present value of uniform periodic payment we wish to find the present value of uniform periodic payments after K periods or years instead of total amount. • Thus the present worth, PW, of uniform periodic payments is given by • Multiply both side with (1+i)-1,The resulting present worth formula

41. Example # 9

42. Solution: In this example following values are given: Substitute these values in equation:

43. Example 9 Cont. • Thus decision to purchase the machine will be profitable investment.

44. Depreciation Techniques • The term “Depreciation” means a decline in value. • In order to take into consideration the change in the value of the product, the depreciation charges are made during the useful life of the engineering products. • The three depreciation techniques are: 1-Declining-Balance Depreciation Method 2-Straight-line Depreciation Method 3-Sum Of Digits Depreciation Method

45. Depreciation Techniques (Cont..) 1. Declining-Balance Depreciation Method • This method dictates the accelerated write-off of the product worth in its early productive years and corresponding lower write-off near the end of useful life years. • The depreciation rate αd is given by where

46. The declining balance techniques requires a positive value of s. The product book value, Vbv(M), at the end of year M is given by:

47. Example # 10

48. Solution: From example statement Substitute in equation