1. Chapter 4 Simplex Method for Linear Programming Shi-Shang Jang Chemical Engineering Department National Tsing-Hua University

2. Example 1: Inspector Problem Assume that it is desired to hire some inspectors for monitoring a production line. A total amount of 1800 species of products are manufactured every day (8 working hours), while two grades of inspectors can be found. Maximum, 8 grade A inspector and 10 grade B inspector are available from the job market. Grade A inspectors can check 25 species/hour, with an accuracy of 98 percent. Grade B inspectors can check 15 species/hour, with an accuracy of 95 percent. Note that each error costs $2.00/piece. The wage of a grade A inspector is $4.00/hour, and the wage of a grade B inspector is $3.00/hour. What is the optimum policy for hiring the inspectors?

3. Problem Formulation Assume that the x 1 grade A inspectors x 2 grade B inspects are hired, then total cost to be minimized 4  8  x 1 +3  8  x 2 +25  8  0.02  2  x 1 +15  8  0.05  2  x 2 =40 x 1 +36 x 2 manufacturing constraint 25  8  x 1 +15  8  x 2  1,800  200 x 1 +120 x 2  1,800 no. of inspectors available: 0  x 1  8 0  x 2  10

4. The Graphical Solution

5. Theorem Property: If there exists an optimal solution to a LP, then at least one of the corner point of the feasible region will always qualify to be an optimal solution.

6. Special Cases Alternate Solutions (non-unique solutions) Max x 1 +2x 2 s.t. x 1 +2x 2  10 x 1 + x 2  1 x 2  4 x 1  0, x 2  0

7. Special Cases - continued Unbounded Optima : A system has a feasible region with open boundaries such that the optima may appear at the infinity. Example: For the previous example, in case the constraint x 1 +2x 2  10 is not given, then moving far away from the origin increases the objective function x 1 +2x 2, and the maxim Z would be + 

8. Unbounded Optima x 1 +x 2 =1 x 2 =4 x 1 +2x 2 =2 x 1 +2x 2 =6 x 1 +2x 2 =10 x 1 +2x 2 =15

9. Example 2: Student Fab The RIT student-run microelectronic fabrication facility is taking orders for four indigenously developed ASIC chips that can be used in (1) touch sensors ($6, s4hr, m1hr, v30), (2)LCD($10,s9hr, m1hr, v40), (3) pressure sensors($9, s7hr, m3hr, v20), and (4) controllers($20, s10hr, m8hr, v10). Constraints : Student hr ≦ 600, machine hr ≦ 420, space ≦ 800

10. The LP Problem

11. 4-2 The Basic Approach Standard Form of Linear Programming

12. Handling of in-equality Constraints Case 1: Slack Variable: x 1 +2x 2 +3 x 3 +4x 4  25 – Modified to x 1 +2 x 2 +3x 3 +4x 4 + x 5 =25 x5  0 is a slack variable. Case 2: Surplus Variable: 2 x 1 + x 2 -3 x 3  12 – Modified to 2 x 1 + x 2 -3 x 3 - x 4 =12 x 4  0 is a surplus variable.

13. Handling of Equality Constraints If s is unrestricted, i.e., s can be positive or negative, then we set s=s + -s - such that s +  0, s -  0.

14. Example Modify to

15. Definitions Definition: A feasible region, denoted by S is the set of all feasible solution. Mathematically,. Definition: An optimal solution is a vector x*  S, s.t. z 0 =c T x* is maximum or minimum in where Z is termed by the optimal value. Definition: Alternate optimal solution is a set X  S, s.t. all x  X has the same objective value z 0 and for all x  S, and z=c T x, z  z 0. Definition: If the solution set of LP contains only one element, it is termed the unique optimum. Definition: If the optimum value z approaches to infinity, then the LP is said to have unbounded optimum.

16. 4-3 The Simplex Method

17. Definitions Definition: A pivot operation is sequence of elementary row operations that reduce the coefficients of a specified variable to unity in one of the equation and zero elsewhere. Definition: In the above canonical form, x 1, ,x m are termed the basic variables or dependent variables, x m+1, ,x n are called nonbasic variables or the independent variables. Definition: The solution obtained from a canonical form is by setting the nonbasic variable or independent variable to zero is called a basic solution. Definition: A basic feasible solution is a basic solution in which the basic or dependent variables are non-negative.

18. Property  Remark: where  Definition: Property: A feasible basic solution is a simplex of the feasible region. Note: Given a canonical form and feasible basic solution, then the objective function: where x B is a basic variable.

19. Approach (Simplex Method): Start with an initial basic feasible solution in canonical form. Improve the solution by finding another basic feasible solution if possible. When a particular basic feasible solution is found, and cannot be improved by finding new basic feasible solution, the optimality is reached. Definition: An adjacent basic solution differs from a basic solution is exactly one basic variable. Question: If one wants to find an adjacent feasible basic solution from one feasible basic solution (i.e., switch to another simplex), which adjacent basic solution gives lowest objective function?

20. Derivation of Inner Product Rule Supposing, one wants to replace one of the original basic variable with nonbasic variable x s, we firstly, increase xs from zero to one, then for all i=1, ,m, and, for all j=m+1, ,n, j  s. x j =0

21. Theorem 1 (Inner Product Rule) Relative cost, (inner product rule) More: (1) In a minimization problem, a basic feasible solution is optimal if the relative costs of its all nonbasic variable are all positive or zero. (2) One should choose an adjacent basic solution from which the relative cost is the minimum. Corollary: The alternate optima exists if  z=0.

22. Theorem 2: (The Minimum Ratio Rule ) Given a nonbasic variable x s is change into the basic variable set, then one of the basic variable x r should leave from the basic variable set, such that: The above minimum happens at i=r. Corollary: The above rule fails if there exist unbounded optima.

23. Example: The LP Problem

24. The Standard Form

25. Table 1(s=4,r=6,rc=2) Basic={5 6 7};Nonbasic={1 2 3 4} 610920000 CBCB Basisx1x1 x2x2 x3x3 x4x4 x5x5 x6x6 x7x7 constraints 0x5x5 49710100600 (600/10=60) 0x6x6 1138010420(420/8=52.5) 0x7x7 30402010001800(800/10=80) ZZ 610920--- Z=0

26. Table 2(s=2,r=7,rc=3) Basic={5 4 7};Nonbasic={1 2 3 6} 610920000 CBCB Basisx1x1 x2x2 x3x3 x4x4 x5x5 x6x6 x7x7 constraints 0x5x5 2.757.753.2501-1.25075(75/7.75= 9.6774) 20x4x4 0.125 0.375100.125052.5(52.5/0.125=4 20) 0x7x7 28.7538.7516.2500-1.251275(275/38.75= 7.0798) ZZ 3.57.51.5---2.5- Z=1050

27. Table 3 Basic={5 4 2};Nonbasic={1 7 3 6} 610920000 CBCB Basisx1x1 x2x2 x3x3 x4x4 x5x5 x6x6 x7x7 constraints 0x5x5 -30001-0.220 x4x4 0.032300.3226100.129-0.003251.6129 10x2x2 0.741910.419400-0.03230.02587.0968 ZZ -2.0645--0.1935---1.6452-2.2581 Z=1103.226

28. 2. The Two Phase Simplex Method- Example

29. The Standard Form

30. Two Phase Approach-Phase I

31. Two Phase Approach-Phase I Table 1 0000011 CBBasisx1x2x3x4x5x6x7constraints 0x41-21100011(11/1=11) 1x6-4120103(3/2=1.5) 1x7-20100011(1/1=1) ZZ 6-3- 1-- Z=4

32. Two Phase Approach-Phase I Final 0000011 CBBasisx1x2x3x4x5x6x7constraints 0x43001-22-512 0x201001-21 0x3-2010001 ZZ ------- Z=0

33. Two Phase Approach-Phase II Table 1 -31100 CBBasisx1x2x3x4x5constraints 0x43001-212(12/3=4) 1x20100 1(1/0=  ) 1x3-201001(1/-2<0) ZZ ---5 Z=2

34. Two Phase Approach-Phase II Final -31100 CBBasisx1x2x3x4x5constraints -3x11000.3333-0.6674 1x201001 1x30010.6667-1.3339 ZZ ----- Z=-2

35. Example: Multi-Products Manufacturing A company manufactures three products: A, B, and C. Each unit of product A requires 1 hr of engineering service, 10 hr of direct labor, and 3lb of material. To produce one unit of product B requires 2hr of engineering, 4hr of direct labor, and 2lb of material. In case of product C, it requires 1hr of engineering, 5hr of direct labor, and 1lb of material. There are 100 hr of engineering, 700 hr of labor, and 400 lb of material available. Since the company offers discounts for bulk purchases, the profit figures are as shown in the next slide:

36. Example- Continued Formulate a linear program to determine the most profitable product mix. Product AProduct BProduct C Sales unitsUnit profitvariable s Sales unitsUnit profitvariablesSales unitsUnit profitvariables 0-4010X1X1 0-506X5X5 0-1005X8X8 40-1009X2X2 50-1004X6X6 Over 1004X9X9 100-1508X3X3 Over 1003X7X7 Over 1507X4X4

37. Problem Formulation Let’s denote the variables as shown in the table, then we have the following:

38. MATLAB PROGRAM f=[-10 -9 -8 -7 -6 -4 -3 -5 -4]'; A=[1 1 1 1 2 2 2 1 1; 10 10 10 10 4 4 4 5 5;3 3 3 3 2 2 2 1 1]; b=[100;700;400]; Aeq=[];beq=[]; LB=[0 0 0 0 0 0 0 0 0]; UB=[40 60 50 Inf 50 50 Inf 100 Inf]; [X,FVAL,EXITFLAG,OUTPUT,LAMBDA]=LINPROG( f,A,b,Aeq,beq,LB,UB)

39. Solution X’= 40.0000 22.5000 0.0000 0.0000 18.7500 0.0000 0.0000 0.0000 0.0000 FVAL =-715.0000 EXITFLAG =1 OUTPUT = iterations: 7 cgiterations: 0 algorithm: 'lipsol' LAMBDA = ineqlin: [3x1 double] eqlin: [0x1 double] upper: [9x1 double] lower: [9x1 double]

40. 3. Sensitivity Analysis Shadow Prices: To evaluate net impact in the maximum profit if additional units of certain resources can be obtained. Opportunity Costs: To measure the negative impact of producing some products that are zero at the optimum. The range on the objective function coefficients and the range on the RHS row.

41. Example A factory manufactures three products, which require three resources – labor, materials and administration. The unit profits on these products are $10, $6 and $4 respectively. There are 100 hr of labor, 600 lb of material, and 300hr of administration available per day. In order to determine the optimal product mix, the following LP model is formulated and solve:

42. Basic LP Problem

43. Optimal Solution and Sensitivity Analysis x1=33.33, x2=66.67,x3=0,Z=733.33 Shadow prices for row 1=3.33, row 2=0.67, row 3=0 Opportunity Costs for x3=2.67 Ranges on the objective function coefficients: 6 ≦ c 1 (10) ≦ 15, 4 ≦ c 2 (6) ≦ 10, -∞ ≦ c 3 (4) ≦ 6.67

44. Optimal Solution and Sensitivity Analysis- Continued 60 ≦ b 1 (100) ≦ 150, 400 ≦ b 2 (600) ≦ 1000, 200 ≦ b 3 (300) ≦ ∞

45. 100% Rules 100% rule for objective function coefficients 100% rule for RHS constants

46. Examples Unit profit on product 1 decrease by $1, but increases by $1 for products 2 and 3, will the optimum change?(δc 1 =-1, Δc 1 =- 4, δc 2 =1, Δc 2 =4, δc 3 =1, Δc 3 =2.67) Simultaneous variation of 10 hr decrease on labor 100 lb increase in material and 50hr decrease on administration