1. Simplex Linear Programming I. Concept II. Model Template III. Class Example IV. Procedure V. Interpretation MAXIMIZATION METHOD Applied Management Science for Decision Making, 2e © 2014 Pearson Learning Solutions Philip A. Vaccaro, PhD Resourc e Planning a nd Allocation Management MGMT E-5050

2. Simplex Linear Programming * * DECISION VARIABLES MIGHT BE PRODUCTS BEING CONSIDERED FOR PRODUCTION Most real-life LP problems have more than two decision variables*, and thus are too large for the simple graphical solution procedure. In simplex LP problems the optimal solution will lie at a corner point of a multi-sided, multi- dimensional figure called an n - dimensional polyhedron that represents the feasible region. THE FEASIBLE REGION WOULD BE SIMILAR TO THE SURFACE OF A DIAMOND

3. Simplex Linear Programming  The simplex method examines the corner points in a systematic fashion, using basic algebraic concepts.  The same set of procedures is repeated time after time until an optimal solution is reached.  Each repetition, or iteration, increases the value of the objective function so that we are always moving closer to the optimal solution.

4. Simplex Linear Programming   The simplex method yields valuable economic information in addition to the optimal solution.  The manual computations must be mastered in order to successfully employ the software pro- grams, and to interpret the computer printouts. WHY LEARN THE MANUAL COMPUTATIONS?

5. Problem Statement Black and White Color A firm produces two different types of television sets: Three resources are required to produce those televisions: Chassis24 units Labor160 hours Color Tubes10 units

6. Problem Statement TELEVISION RESOURCE REQUIREMENTS AND PROFIT MARGINS TelevisionChassis Labor HoursColor Tubes Unit Profit Black + White 150$6.00 Color 1101$15.00

7. Simplex Linear Programming Let X 1 = BLACK AND WHITE TELEVISIONS Let X 2 = COLOR TELEVISIONS Objective Function: Maximize Z = 6X 1 + 15X 2 subject to: 1X 1 + 1X 2 =< 24 chassis 5X 1 + 10X 2 =< 160 labor hours 0X 1 + 1X 2 =< 10 color tubes X 1, X 2 => 0 THE MODEL

8. Simplex Linear Programming CONVERSION TO LINEAR EQUALITIES S 1 1X 1 + 1X 2 + 1S 1 = 24 chassis S 2 5X 1 + 10X 2 + 1S 2 = 160 labor hours S 3 0X 1 + 1X 2 + 1S 3 = 10 color tubes : S1, S2 =, S3 = COLOR TUBES SLACK VARIABLES: S1 = CHASSIS, S2 = LABOR HOURS, S3 = COLOR TUBES ADDING A SLACK VARIABLE TO EACH CONSTRAINT AND SETTING IT EQUAL TO THE RIGHT – HAND SIDE

9. Simplex Linear Programming CONVERSION TO LINEAR EQUALITIES SUITABLE FOR INCLUSION IN THE SIMPLEX MATRIX 00 1X 1 + 1X 2 + 1S 1 + 0S 2 + 0S 3 = 24 chassis 00 5X 1 + 10X 2 + 0S 1 + 1S 2 + 0S 3 = 160 labor hours 000 0X 1 + 1X 2 + 0S 1 + 0S 2 + 1S 3 = 10 color tubes PRESENCE REWRITE EACH CONSTRAINT TO REFLECT THE PRESENCE OR ABSENCE ABSENCE OF ALL VARIABLES IN THE PROBLEM

10. Simplex Linear Programming 1 st FEASIBLE SOLUTION TRADITIONALLY THE FIRST FEASIBLE SOLUTION PRODUCES NO PRODUCT AND HAS ALL RESOURCES INTACT X1X1X1X1 BLACK AND WHITE TVs 0 X2X2X2X2 COLOR TVs 0 S1S1S1S1 CHASSIS 24 S2S2S2S2 LABOR HOURS 160 S3S3S3S3 COLOR TUBES 10

11. The 1 st Feasible Solution Cj BasisQty X1X1X1X1 X2X2X2X2 S1S1S1S1 S2S2 S3S3S3S3 S1S1S1S1 S2S2S2S2 S3S3S3S3 Zj Cj-Zj ----- THE BASIS OR MIX COLUMN SHOWS ALL VARIABLES GREATER THAN ZERO IN THE CURRENT SOLUTION

12. Toward 1 st Feasible Solution Cj BasisQty X1X1X1X1 X2X2X2X2 S1S1S1S1 S2S2 S3S3S3S3 S1S1S1S12411100 S2S2S2S2160510010 S3S3S3S31001001 Zj Cj-Zj INSERTING LINEAR EQUALITIES INTO THE MATRIX RIGHT-HAND SIDES ARE ENTERED INTO THE QUANTITYCOLUMN

13. Toward 1 st Feasible Solution Cj BasisQty X1X1X1X1 X2X2X2X2 S1S1S1S1 S2S2 S3S3S3S3 S12411100 Zj Cj-Zj INSERTING LINEAR EQUALITIES INTO THE MATRIX 1X 1 + 1X 2 + 1S 1 + 0S 2 + 0S 3 = 24

14. Toward 1 st Feasible Solution Cj BasisQty X1X1X1X1 X2X2X2X2 S1S1S1S1 S2S2 S3S3S3S3 S2160510010 Zj Cj-Zj INSERTING LINEAR EQUALITIES INTO THE MATRIX 5X 1 + 10X 2 + 0S 1 + 1S 2 + 0S 3 = 160

15. Toward 1 st Feasible Solution Cj BasisQty X1X1X1X1 X2X2X2X2 S1S1S1S1 S2S2 S3S3S3S3 S31001001 Zj Cj-Zj INSERTING LINEAR EQUALITIES INTO THE MATRIX 0X 1 + 1X 2 + 0S 1 + 0S 2 + 1S 3 = 10

16. Toward 1 st Feasible Solution Cj BasisQty X1X1X1X1 X2X2X2X2 S1S1S1S1 S2S2 S3S3S3S3 S1S1S1S12411100 S2S2S2S2160510010 S3S3S3S31001001 Zj Cj-Zj INSERTING LINEAR EQUALITIES INTO THE MATRIX

17. The 1 st Feasible Solution Cj$6.$15.$0.$0.$0. BasisQty X1X1X1X1 X2X2X2X2 S1S1S1S1 S2S2 S3S3S3S3 $0. S1S1S1S12411100 $0. S2S2S2S2160510010 $0. S3S3S3S31001001 Zj Cj-Zj ----- THE “Cj” or CONTRIBUTION MARGIN THE GROSS PROFIT PER UNIT FOR ALL VARIABLES IN THE PROBLEM. SURPLUS AND SLACK VARIABLES HAVE $0.00 Cj’s BY DEFINITION.

18. Toward 1 st Feasible Solution Cj$6.$15.$0.$0.$0. BasisQty X1X1X1X1 X2X2X2X2 S1S1S1S1 S2S2 S3S3S3S3 $0. S1S1S1S12411100 $0. S2S2S2S2160510010 $0. S3S3S3S31001001 Zj Cj-Zj

19. Toward 1 st Feasible Solution Cj$6.$15.$0.$0.$0. BasisQty X1X1X1X1 X2X2X2X2 S1S1S1S1 S2S2 S3S3S3S3 $0. S1S1S1S12411100 $0. S2S2S2S2160510010 $0. S3S3S3S31001001 Zj$0. Cj-Zj COMPUTING THE Zj ROW $0. $0. $0. MULTIPLY EACH SLACK VARIABLE’S Cj BY ITS QUANTITY AND ADD THE RESULTS

20. Toward 1 st Feasible Solution Cj$6.$15.$0.$0.$0. BasisQty X1X1X1X1 X2X2X2X2 S1S1S1S1 S2S2 S3S3S3S3 $0. S1S1S1S12411100 $0. S2S2S2S2160510010 $0. S3S3S3S31001001 Zj$0.$0. Cj-Zj COMPUTING THE Zj ROW $0. $0. $0. THEN MULTIPLY EACH SLACK VARIABLE’S Cj BY ITS VARIABLE COEFFICIENTS AND ADD THE RESULTS

21. Toward 1 st Feasible Solution Cj$6.$15.$0.$0.$0. BasisQty X1X1X1X1 X2X2X2X2 S1S1S1S1 S2S2 S3S3S3S3 $0. S1S1S1S12411100 $0. S2S2S2S2160510010 $0. S3S3S3S31001001 Zj$0.$0.$0. Cj-Zj COMPUTING THE Zj ROW $0. $0. $0. THEN MULTIPLY EACH SLACK VARIABLE’S Cj BY ITS VARIABLE COEFFICIENTS AND ADD THE RESULTS

22. Toward 1 st Feasible Solution Cj$6.$15.$0.$0.$0. BasisQty X1X1X1X1X2 S1S1S1S1 S2S2 S3S3S3S3 $0. S1S1S1S12411100 $0. S2S2S2S2160510010 $0. S3S3S3S31001001 Zj$0.$0.$0.$0. Cj-Zj COMPUTING THE Zj ROW $0. $0. $0. THEN MULTIPLY EACH SLACK VARIABLE’S Cj BY ITS VARIABLE COEFFICIENTS AND ADD THE RESULTS

23. Toward 1 st Feasible Solution Cj$6.$15.$0.$0.$0. BasisQty X1X1X1X1 X2X2X2X2 S1S1S1S1 S2S2S2S2 S3S3S3S3 $0. S1S1S1S12411100 $0. S2S2S2S2160510010 $0. S3S3S3S31001001 Zj$0.$0.$0.$0.$0. Cj-Zj COMPUTING THE Zj ROW $0. $0. $0. THEN MULTIPLY EACH SLACK VARIABLE’S Cj BY ITS VARIABLE COEFFICIENTS AND ADD THE RESULTS

24. Toward 1 st Feasible Solution Cj$6.$15.$0.$0.$0. BasisQty X1X1X1X1 X2X2X2X2 S1S1S1S1 S2S2 S3S3S3S3 $0. S1S1S1S12411100 $0. S2S2S2S2160510010 $0. S3S3S3S31001001 Zj$0.$0.$0.$0. $0. Cj-Zj COMPUTING THE Zj ROW $0. $0. $0. THEN MULTIPLY EACH SLACK VARIABLE’S Cj BY ITS VARIABLE COEFFICIENTS AND ADD THE RESULTS

25. Toward 1 st Feasible Solution Cj$6.$15.$0.$0.$0. BasisQty X1X1X1X1 X2X2X2X2 S1S1S1S1 S2S2 S3S3S3S3 $0. S1S1S1S12411100 $0. S2S2S2S2160510010 $0. S3S3S3S31001001 Zj$0.$0.$0.$0.$0.$0. Cj-Zj THE COMPUTED Zj ROW

26. Toward 1 st Feasible Solution Cj$6.$15.$0.$0.$0. BasisQty X1X1X1X1 X2X2X2X2 S1S1S1S1 S2S2 S3S3S3S3 $0. S1S1S1S12411100 $0. S2S2S2S2160510010 $0. S3S3S3S31001001 Zj$0.$0.$0.$0.$0.$0. Cj-Zj

27. Toward 1 st Feasible Solution Cj$6.$15.$0.$0.$0. BasisQty X1X1X1X1 X2X2X2X2 S1S1S1S1 S2S2 S3S3S3S3 $0. S1S1S1S12411100 $0. S2S2S2S2160510010 $0. S3S3S3S31001001 Zj$0.$0.$0.$0.$0.$0. Cj-Zj ----- $6. COMPUTING THE Cj – Zj ROW ON A COLUMN BY COLUMN BASIS, SUBTRACT Zj FROM Cj

28. Toward 1 st Feasible Solution Cj$6.$15.$0.$0.$0. BasisQty X1X1X1X1 X2X2X2X2 S1S1S1S1 S2S2 S3S3S3S3 $0. S1S1S1S12411100 $0. S2S2S2S2160510010 $0. S3S3S3S31001001 Zj$0.$0.$0.$0.$0.$0. Cj-Zj ----- $6.$15. COMPUTING THE Cj – Zj ROW ON A COLUMN BY COLUMN BASIS, SUBTRACT Zj FROM Cj

29. Toward 1 st Feasible Solution Cj$6.$15.$0.$0.$0. BasisQty X1X1X1X1 X2X2X2X2 S1S1S1S1 S2S2 S3S3S3S3 $0. S1S1S1S12411100 $0. S2S2S2S2160510010 $0. S3S3S3S31001001 Zj$0.$0.$0.$0.$0.$0. Cj-Zj ----- $6.$15.$0. COMPUTING THE Cj – Zj ROW

30. Toward 1 st Feasible Solution Cj$6.$15.$0.$0.$0. BasisQty X1X1X1X1 X2X2X2X2 S1S1S1S1 S2S2 S3S3S3S3 $0. S1S1S1S12411100 $0. S2S2S2S2160510010 $0. S3S3S3S31001001 Zj$0.$0.$0.$0.$0.$0. Cj-Zj ----- $6.$15.$0.$0. COMPUTING THE Cj – Zj ROW

31. Toward 1 st Feasible Solution Cj$6.$15.$0.$0.$0. BasisQty X1X1X1X1 X2X2X2X2 S1S1S1S1 S2S2 S3S3S3S3 $0. S1S1S1S12411100 $0. S2S2S2S2160510010 $0. S3S3S3S31001001 Zj$0.$0.$0.$0.$0.$0. Cj-Zj ----- $6.$15.$0.$0.$0. COMPUTING THE Cj – Zj ROW

32. The 1 st Feasible Solution Cj$6.$15.$0.$0.$0. BasisQty X1X1X1X1 X2X2X2X2 S1S1S1S1 S2S2 S3S3S3S3 $0. S1S1S1S12411100 $0. S2S2S2S2160510010 $0. S3S3S3S31001001 Zj$0.$0.$0.$0.$0.$0. Cj-Zj ----- $6.$15.$0.$0.$0. THE COMPLETED Cj - Zj ROW

33. The 1 st Feasible Solution Cj$6.$15.$0.$0.$0. BasisQty X1X1X1X1 X2X2X2X2 S1S1S1S1 S2S2 S3S3S3S3 $0. S1S1S1S12411100 $0. S2S2S2S2160510010 $0. S3S3S3S31001001 Zj$0.$0.$0.$0.$0.$0. Cj-Zj ----- $6.$15.$0.$0.$0.

34. Simplex Linear Programming 1 st FEASIBLE SOLUTION TRADITIONALLY THE FIRST FEASIBLE SOLUTION PRODUCES NO PRODUCT AND HAS ALL RESOURCES INTACT X1X1X1X1 BLACK AND WHITE TVs0 X2X2X2X2 COLOR TVs 0 S1S1S1S1 CHASSIS 24 S2S2S2S2 LABOR HOURS 160 S3S3S3S3 COLOR TUBES 10

35. Simplex Linear Programming X 1 ( B+W TVs ) = 0 NOT IN THE BASIS X 2 ( Color TVs ) = 0 NOT IN THE BASIS Z ( Profit ) = $0.00 SINCE X 1 AND X 2 = 0 1 st FEASIBLE SOLUTION

36. Toward 2nd Feasible Solution Cj$6.$15.$0.$0.$0. BasisQty X1X1X1X1 X2X2X2X2 S1S1S1S1 S2S2 S3S3S3S3 $0. S1S1S1S12411100 $0. S2S2S2S2160510010 $0. S3S3S3S31001001 Zj$0.$0.$0.$0.$0.$0. Cj-Zj ----- $6.$15.$0.$0.$0. WE BRING THE MOST PROFITABLE TELEVISION ( X2 ) INTO THE SOLUTION X2 BECOMES THE PIVOT COLUMN

37. Toward 2nd Feasible Solution Cj$6.$15.$0.$0.$0. BasisQty X1X1X1X1 X2X2X2X2 S1S1S1S1 S2S2 S3S3S3S3 $0. S1S1S1S12411100 $0. S2S2S2S2160510010 $0. S3S3S3S31001001 Zj$0.$0.$0.$0.$0.$0. Cj-Zj ----- $6.$15.$0.$0.$0. WE COMPUTE EACH ROW’S RATIO OF QUANTITY DIVIDED BY ITS PIVOT COLUMN COEFFICIENT PIVOT COLUMN 24 16 10

38. Toward 2nd Feasible Solution Cj$6.$15.$0.$0.$0. BasisQty X1X1X1X1 X2X2X2X2 S1S1S1S1 S2S2 S3S3S3S3 $0. S1S1S1S12411100 $0. S2S2S2S2160510010 $0. S3S3S3S31001001 Zj$0.$0.$0.$0.$0.$0. Cj-Zj ----- $6.$15.$0.$0.$0. THE LOWEST POSITIVE RATIO DENOTES THE PIVOT ROW PIVOT COLUMN 24 16 10 PIVOTROW MEANS WE CAN ONLY PRODUCE TEN COLOR TVs THIS IS THE LIMITED RESOURCE

39. Toward 2nd Feasible Solution Cj$6.$15.$0.$0.$0. BasisQty X1X1X1X1 X2X2X2X2 S1S1S1S1 S2S2 S3S3S3S3 $0. S1S1S1S12411100 $0. S2S2S2S2160510010 $15. X2X2X2X21001001 Zj$0.$0.$0.$0.$0.$0. Cj-Zj ----- $6.$15.$0.$0.$0. REAL VARIABLE X2 REPLACES SLACK VARIABLE S3 IN THE BASIS PIVOT COLUMN PIVOTROW TO PRODUCE TEN COLOR TVs MEANS THAT ALL COLOR TUBES MUST BE CONSUMED, MAKING S3 = 0, AND FORCING IT TO LEAVE THE BASIS

40. Toward 2nd Feasible Solution Cj$6.$15.$0.$0.$0. BasisQty X1X1X1X1 X2X2X2X2 S1S1S1S1 S2S2 S3S3S3S3 $0. S1S1S1S12411100 $0. S2S2S2S2160510010 $15. X2X2X2X21001001 Zj$0.$0.$0.$0.$0.$0. Cj-Zj ----- $6.$15.$0.$0.$0. THE PIVOT NUMBER IS LOCATED AT THE INTERSECTION OF THE PIVOT ROW AND PIVOT COLUMN PIVOT COLUMN PIVOTROW

41. Toward 2nd Feasible Solution Cj$6.$15.$0.$0.$0. BasisQty X1X1X1X1 X2X2X2X2 S1S1S1S1 S2S2 S3S3S3S3 $0. S1S1S1S12411100 $0. S2S2S2S2160510010 $15. X2X2X2X21001001 Zj$0.$0.$0.$0.$0.$0. Cj-Zj ----- $6.$15.$0.$0.$0. PIVOT COLUMN PIVOTROW THE PIVOT NUMBER MUST ALWAYS BE “1”

42. When the Pivot Number is Not “1” Cj$6.$15.$0.$0.$0. BasisQty X1X1X1X1 X2X2X2X2S1S2S3 $0.1 $0.10 $15. X2X2X2X26010500-10 Zj$0.$0.$0.$0.$0.$0. Cj-Zj ----- $15. PIVOT COLUMN PIVOTROW EXAMPLE THE ENTIRE PIVOT ROW MUST BE DIVIDED BY WHATEVER NUMBER NEEDED T0 FORCE A VALUE OF “1” FOR THE PIVOT NUMBER

43. When the Pivot Number is Not “1” Cj$6.$15.$0.$0.$0. BasisQty X1X1X1X1 X2X2X2X2 S1S1S1S1 S2S2 S3S3S3S3 $0.1 $0.10 $15. X2X2X2X2122100 - 2 Zj$0.$0.$0.$0.$0.$0. Cj-Zj ----- $15. PIVOT COLUMN PIVOTROW EXAMPLE

44. Toward 2 nd Feasible Solution ROW TRANSFORMATION Old 1 st Row S1S1 2411100 Pivot Row S3S3 1001001 New 1 st Row S1S1 141010 We subtract the pivot row from all other rows in the matrix ( except “Zj” and “Cj - Zj” ) in such a way as to force a “0” coefficient in the column above or below the pivot number. SUBTRACT THE PIVOT ROW ( S 3 ) FROM THE FIRST ROW ( S 1 ) Pivot Number

45. Toward 2 nd Feasible Solution ROW TRANSFORMATION Old 2 nd Row S2S2 160510010 Pivot Row S3S3 1001001 New 2 nd Row S2S2 0 SUBTRACT THE PIVOT ROW ( S 3 ) FROM THE SECOND ROW ( S 2 ) HERE, THE PIVOT ROW MUST BE MULTIPLIED BY “10” IN ORDER TO FORCE THE REQUIRED ZERO COEFFICIENT IN THE NEW ROW Pivot Number

46. Toward 2 nd Feasible Solution ROW TRANSFORMATION Old 2 nd Row S2S2 160510010 Pivot Row S3S3 10001000 New 2 nd Row S2S2 605001-10 SUBTRACT THE PIVOT ROW ( S 3 ) FROM THE SECOND ROW ( S 2 ) HERE, THE PIVOT ROW WAS MULTIPLIED BY “10” IN ORDER TO FORCE THE REQUIRED ZERO COEFFICIENT IN THE NEW ROW Pivot Number

47. Toward 2 nd Feasible Solution Cj$6.$15.$0.$0.$0. BasisQty X1X1X1X1 X2X2X2X2 S1S1S1S1 S2S2 S3S3S3S3 $0. S1S1S1S1141010 $0. S2S2S2S2605001-10 $15. X2X2X2X21001001 Zj Cj-Zj INSERTING THE TRANSFORMED ROWS INTO THE MATRIX

48. Toward 2 nd Feasible Solution Cj$6.$15.$0.$0.$0. BasisQty X1X1X1X1 X2X2X2X2 S1S1S1S1 S2S2 S3S3S3S3 $0. S1S1S1S1141010 $0. S2S2S2S2605001-10 $15. X2X2X2X21001001 Zj$150. Cj-Zj COMPUTING THE Zj ROW $0. $0. $150. MULTIPLY EACH BASIS VARIABLE’S “Cj” BY ITS QUANTITY AND ADD THE RESULTS

49. Toward 2 nd Feasible Solution Cj$6.$15.$0.$0.$0. BasisQty X1X1X1X1 X2X2X2X2 S1S1S1S1 S2S2 S3S3S3S3 $0. S1S1S1S1141010 $0. S2S2S2S2605001-10 $15. X2X2X2X21001001 Zj$150.$0. Cj-Zj COMPUTING THE Zj ROW $0. $0. $0. MULTIPLY EACH BASIS VARIABLE’S “Cj” BY ITS VARIABLE COEFFICIENTS AND ADD THE RESULTS

50. Toward 2 nd Feasible Solution Cj$6.$15.$0.$0.$0. BasisQty X1X1X1X1 X2X2X2X2 S1S1S1S1 S2S2 S3S3S3S3 $0. S1S1S1S1141010 $0. S2S2S2S2605001-10 $15. X2X2X2X21001001 Zj$150.$0.$15. Cj-Zj COMPUTING THE Zj ROW $0. $0. $15. MULTIPLY EACH BASIS VARIABLE’S “Cj” BY ITS VARIABLE COEFFICIENTS AND ADD THE RESULTS

51. Toward 2 nd Feasible Solution Cj$6.$15.$0.$0.$0. BasisQty X1X1X1X1 X2X2X2X2 S1S1S1S1 S2S2 S3S3S3S3 $0. S1S1S1S1141010 $0.S2605001-10 $15. X2X2X2X21001001 Zj$150.$0.$15.$0.$0.$15. Cj-Zj COMPUTING THE Zj ROW $0. $0. MULTIPLY EACH BASIS VARIABLE’S “Cj” BY ITS VARIABLE COEFFICIENTS AND ADD THE RESULTS $0.

52. Toward 2 nd Feasible Solution Cj$6.$15.$0.$0.$0. BasisQty X1X1X1X1 X2X2X2X2 S1S1S1S1 S2S2S2S2 S3S3S3S3 $0. S1S1S1S1141010 $0. S2S2S2S2605001-10 $15. X2X2X2X21001001 Zj$150.$0.$15.$0.$0.$15. Cj-Zj COMPUTING THE Zj ROW $0. $0. $0. MULTIPLY EACH BASIS VARIABLE’S “Cj” BY ITS VARIABLE COEFFICIENTS AND ADD THE RESULTS

53. Toward 2 nd Feasible Solution Cj$6.$15.$0.$0.$0. BasisQty X1X1X1X1 X2X2X2X2 S1S1S1S1 S2S2 S3S3S3S3 $0. S1S1S1S1141010 $0. S2S2S2S2605001-10 $15. X2X2X2X21001001 Zj$150.$0.$15.$0.$0.$15. Cj-Zj COMPUTING THE Zj ROW $0. $0. $15. MULTIPLY EACH BASIS VARIABLE’S “Cj” BY ITS VARIABLE COEFFICIENTS AND ADD THE RESULTS

54. Toward 2 nd Feasible Solution Cj$6.$15.$0.$0.$0. BasisQty X1X1X1X1 X2X2X2X2 S1S1S1S1 S2S2 S3S3S3S3 $0. S1S1S1S1141010 $0. S2S2S2S2605001-10 $15. X2X2X2X21001001 Zj$150.$0.$15.$0.$0.$15. Cj-Zj THE COMPLETED Zj ROW

55. Toward 2 nd Feasible Solution Cj$6.$15.$0.$0.$0. BasisQty X1X1X1X1 X2X2X2X2 S1S1S1S1 S2S2 S3S3S3S3 $0. S1S1S1S1141010 $0. S2S2S2S2605001-10 $15. X2X2X2X21001001 Zj$150.$0.$15.$0.$0.$15. Cj-Zj

56. Toward 2 nd Feasible Solution Cj$6.$15.$0.$0.$0. BasisQty X1X1X1X1 X2X2X2X2 S1S1S1S1 S2S2 S3S3S3S3 $0. S1S1S1S1141010 $0. S2S2S2S2605001-10 $15. X2X2X2X21001001 Zj$150.$0.$15.$0.$0.$15. Cj-Zj---$6. COMPUTING THE Cj - Zj ROW ON A COLUMN BY COLUMN BASIS, SUBTRACT Zj FROM Cj

57. Toward 2 nd Feasible Solution Cj$6.$15.$0.$0.$0. BasisQty X1X1X1X1 X2X2X2X2 S1S1S1S1 S2S2 S3S3S3S3 $0. S1S1S1S1141010 $0. S2S2S2S2605001-10 $15. X2X2X2X21001001 Zj$150.$0.$15.$0.$0.$15. Cj-Zj---$6.$0. COMPUTING THE Cj - Zj ROW ON A COLUMN BY COLUMN BASIS, SUBTRACT Zj FROM Cj

58. Toward 2 nd Feasible Solution Cj$6.$15.$0.$0.$0. BasisQty X1X1X1X1 X2X2X2X2 S1S1S1S1 S2S2 S3S3S3S3 $0. S1S1S1S1141010 $0. S2S2S2S2605001-10 $15. X2X2X2X21001001 Zj$150.$0.$15.$0.$0.$15. Cj-Zj---$6.$0.$0. COMPUTING THE Cj - Zj ROW

59. Toward 2 nd Feasible Solution Cj$6.$15.$0.$0.$0. BasisQty X1X1X1X1 X2X2X2X2 S1S1S1S1 S2S2 S3S3S3S3 $0. S1S1S1S1141010 $0. S2S2S2S2605001-10 $15. X2X2X2X21001001 Zj$150.$0.$15.$0.$0.$15. Cj-Zj---$6.$0.$0.$0. COMPUTING THE Cj - Zj ROW

60. Toward 2 nd Feasible Solution Cj$6.$15.$0.$0.$0. BasisQty X1X1X1X1 X2X2X2X2 S1S1S1S1 S2S2 S3S3S3S3 $0. S1S1S1S1141010 $0. S2S2S2S2605001-10 $15. X2X2X2X21001001 Zj$150.$0.$15.$0.$0.$15. Cj-Zj---$6.$0.$0.$0.-$15. COMPUTING THE Cj - Zj ROW

61. The 2 nd Feasible Solution Cj$6.$15.$0.$0.$0. BasisQty X1X1X1X1 X2X2X2X2 S1S1S1S1 S2S2 S3S3S3S3 $0. S1S1S1S1141010 $0. S2S2S2S2605001-10 $15. X2X2X2X21001001 Zj$150.$0.$15.$0.$0.$15. Cj-Zj---$6.$0.$0.$0.-$15. THE COMPLETED Cj – Zj ROW

62. The 2 nd Feasible Solution Cj$6.$15.$0.$0.$0. BasisQty X1X1X1X1 X2X2X2X2 S1S1S1S1 S2S2 S3S3S3S3 $0. S1S1S1S1141010 $0. S2S2S2S2605001-10 $15. X2X2X2X21001001 Zj$150.$0.$15.$0.$0.$15. Cj-Zj---$6.$0.$0.$0.-$15.

63. The 2 nd Feasible Solution Cj$6.$15.$0.$0.$0. BasisQty X1X1X1X1 X2X2X2X2 S1S1S1S1 S2S2 S3S3S3S3 $0. S1S1S1S1141010 $0. S2S2S2S2605001-10 $15. X2X2X2X21001001 Zj$150.$0.$15.$0.$0.$15. Cj-Zj---$6.$0.$0.$0.-$15.

64. Simplex Linear Programming 2 nd FEASIBLE SOLUTION X1X1X1X1 BLACK AND WHITE TVs 0 X2X2X2X2 COLOR TVs 10 S1S1S1S1 CHASSIS 14 S2S2S2S2 LABOR HOURS 60 S3S3S3S3 COLOR TUBES 0 TOTAL PROFIT = $150.00

65. Simplex Linear Programming 2 nd FEASIBLE SOLUTION Z ( total profit ) = $150.00 X 1 ( B + W TVs ) = 0 NOT IN THE BASIS S 3 ( Color Tubes ) = 0 NOT IN THE BASIS

66. Toward 3rd Feasible Solution Cj$6.$15.$0.$0.$0. BasisQty X1X1X1X1 X2X2X2X2 S1S1S1S1 S2S2 S3S3S3S3 $0. S1S1S1S1141010 $0. S2S2S2S2605001-10 $15. X2X2X2X21001001 Zj$150.$0.$15.$0.$0.$15. Cj-Zj---$6.$0.$0.$0.-$15. WE BRING BLACK + WHITE TELEVISIONS (X1) INTO THE SOLUTION X1 BECOMES THE PIVOT COLUMN

67. Toward 3rd Feasible Solution Cj$6.$15.$0.$0.$0. BasisQty X1X1X1X1 X2X2X2X2 S1S1S1S1 S2S2 S3S3S3S3 $0. S1S1S1S1141010 $0. S2S2S2S2605001-10 $15. X2X2X2X21001001 Zj$150.$0.$15.$0.$0.$15. Cj-Zj---$6.$0.$0.$0.-$15. WE COMPUTE EACH ROW’S RATIO OF QUANTITY DIVIDED BY ITS PIVOT COLUMN COEFFICIENT THE PIVOT COLUMN 14 12 0

68. Toward 3rd Feasible Solution Cj$6.$15.$0.$0.$0. BasisQty X1X1X1X1 X2X2X2X2 S1S1S1S1 S2S2 S3S3S3S3 $0. S1S1S1S1141010 $0. S2S2S2S2605001-10 $15. X2X2X2X21001001 Zj$150.$0.$15.$0.$0.$15. Cj-Zj---$6.$0.$0.$0.-$15. THE LOWEST POSITIVE RATIO DENOTES THE PIVOT ROW THE PIVOT COLUMN 14 12 0 PIVOTROW MEANS WE CAN ONLY PRODUCE 12 B+W TVs THIS IS THE LIMITED RESOURCE

69. Toward 3rd Feasible Solution Cj$6.$15.$0.$0.$0. BasisQty X1X1X1X1 X2X2X2X2 S1S1S1S1 S2S2 S3S3S3S3 $0. S1S1S1S1141010 $6. X1X1X1X1605001-10 $15. X2X2X2X21001001 Zj$150.$0.$15.$0.$0.$15. Cj-Zj---$6.$0.$0.$0.-$15. REAL VARIABLE X1 REPLACES SLACK VARIABLE S2 IN THE BASIS THE PIVOT COLUMN PIVOTROW TO PRODUCE 12 B+W TVs MEANS THAT ALL REMAINING LABOR HOURS MUST BE CONSUMED, MAKING S2 = 0, AND FORCING IT TO LEAVE THE BASIS

70. Toward 3rd Feasible Solution Cj$6.$15.$0.$0.$0. BasisQty X1X1X1X1 X2X2X2X2 S1S1S1S1 S2S2 S3S3S3S3 $0. S1S1S1S1141010 $6. X1X1X1X1605001-10 $15. X2X2X2X21001001 Zj$150.$0.$15.$0.$0.$15. Cj-Zj---$6.$0.$0.$0.-$15. THE PIVOT NUMBER IS LOCATED AT THE INTERSECTION OF THE PIVOT ROW AND PIVOT COLUMN THE PIVOT COLUMN PIVOTROW

71. Toward 3rd Feasible Solution Cj$6.$15.$0.$0.$0. BasisQty X1X1X1X1 X2X2X2X2 S1S1S1S1 S2S2 S3S3S3S3 $0. S1S1S1S1141010 $6. X1X1X1X1605001-10 $15. X2X2X2X21001001 Zj$150.$0.$15.$0.$0.$15. Cj-Zj---$6.$0.$0.$0.-$15. THE PIVOT COLUMN PIVOTROW THE PIVOT NUMBER MUST ALWAYS BE “1”

72. Toward 3rd Feasible Solution Cj$6.$15.$0.$0.$0. BasisQty X1X1X1X1 X2X2X2X2 S1S1S1S1 S2S2 S3S3S3S3 $0. S1S1S1S1141010 $6. X1X1X1X112100.2-2.0 $15. X2X2X2X21001001 Zj$150.$0.$15.$0.$0.$15. Cj-Zj---$6.$0.$0.$0.-$15. THE PIVOT COLUMN PIVOTROW DIVIDE THE ENTIRE PIVOT ROW BY “5”

73. Toward 3 rd Feasible Solution ROW TRANSFORMATION Old 1 st Row S1S1 141010 Pivot Row X1X1 12100.2- 2.0 New 1 st Row S1S1 2001-.21 SUBTRACT THE PIVOT ROW ( X 1 ) FROM THE FIRST ROW ( S 1 ) HERE, THERE WAS NO NEED TO MULTIPLY THE PIVOT ROW BY ANY NUMBER IN ORDER TO FORCE A ZERO COEFFICIENT BELOW THE PIVOT NUMBER Pivot Number

74. Toward 3 rd Feasible Solution ROW TRANSFORMATION Old 3 rd Row X2X2 1001001 Pivot Row X1X1 12100.2- 2.0 New 3 rd Row X2X2 1001001 SUBTRACT THE PIVOT ROW ( X 1 ) FROM THE THIRD ROW ( X 2 ) HERE, THERE WAS NO NEED TO TRANSFORM ROW “3” BECAUSE IT ALREADY HAD A ZERO COEFFICIENT BELOW THE PIVOT NUMBER Pivot Number

75. Toward 3rd Feasible Solution Cj$6.$15.$0.$0.$0. BasisQty X1X1X1X1 X2X2X2X2 S1S1S1S1 S2S2 S3S3S3S3 $0. S1S1S1S12001-.21 $6. X1X1X1X112100.2-2.0 $15. X2X2X2X21001001 Zj Cj-Zj INSERTING THE TRANSFORMED ROWS INTO THE MATRIX

76. Toward 3rd Feasible Solution Cj$6.$15.$0.$0.$0. BasisQty X1X1X1X1 X2X2X2X2 S1S1S1S1 S2S2 S3S3S3S3 $0. S1S1S1S12001-.21 $6. X1X1X1X112100.2-2.0 $15. X2X2X2X21001001 Zj$222. Cj-Zj COMPUTING THE Zj ROW $0. $72. $150. MULTIPLY EACH BASIS VARIABLE’S “Cj” BY ITS QUANTITY AND ADD THE RESULTS

77. Toward 3rd Feasible Solution Cj$6.$15.$0.$0.$0. BasisQty X1X1X1X1 X2X2X2X2 S1S1S1S1 S2S2 S3S3S3S3 $0. S1S1S1S12001-.21 $6. X1X1X1X112100.2-2.0 $15. X2X2X2X21001001 Zj$222.$6. Cj-Zj COMPUTING THE Zj ROW $0. $6. $0. MULTIPLY EACH BASIS VARIABLE’S “Cj” BY ITS VARIABLE COEFFICIENTS AND ADD THE RESULTS

78. Toward 3rd Feasible Solution Cj$6.$15.$0.$0.$0. BasisQty X1X1X1X1 X2X2X2X2 S1S1S1S1 S2S2 S3S3S3S3 $0.S12001-.21 $6.X112100.2-2.0 $15. X2X2X2X21001001 Zj$222.$6.$15. Cj-Zj COMPUTING THE Zj ROW $0. $0. $15. MULTIPLY EACH BASIS VARIABLE’S “Cj” BY ITS VARIABLE COEFFICIENTS AND ADD THE RESULTS

79. Toward 3rd Feasible Solution Cj$6.$15.$0.$0.$0. BasisQty X1X1X1X1 X2X2X2X2 S1S1S1S1 S2S2 S3S3S3S3 $0. S1S1S1S12001-.21 $6. X1X1X1X112100.2-2.0 $15. X2X2X2X21001001 Zj$222.$6.$15.$0. Cj-Zj COMPUTING THE Zj ROW $0. $0. $0. MULTIPLY EACH BASIS VARIABLE’S “Cj” BY ITS VARIABLE COEFFICIENTS AND ADD THE RESULTS

80. Toward 3rd Feasible Solution Cj$6.$15.$0.$0.$0. BasisQty X1X1X1X1 X2X2X2X2 S1S1S1S1 S2S2S2S2 S3S3S3S3 $0. S1S1S1S12001-.21 $6. X1X1X1X112100.2-2.0 $15. X2X2X2X21001001 Zj$222.$6.$15.$0.$1.2 Cj-Zj COMPUTING THE Zj ROW $0. $1.2 $0. MULTIPLY EACH BASIS VARIABLE’S “Cj” BY ITS VARIABLE COEFFICIENTS AND ADD THE RESULTS

81. Toward 3rd Feasible Solution Cj$6.$15.$0.$0.$0. BasisQty X1X1X1X1 X2X2X2X2 S1S1S1S1 S2S2 S3S3S3S3 $0. S1S1S1S12001-.21 $6. X1X1X1X112100.2-2.0 $15. X2X2X2X21001001 Zj$222.$6.$15.$0.$1.2$3.0 Cj-Zj COMPUTING THE Zj ROW $0. $15. MULTIPLY EACH BASIS VARIABLE’S “Cj” BY ITS VARIABLE COEFFICIENTS AND ADD THE RESULTS -$12.

82. Toward 3rd Feasible Solution Cj$6.$15.$0.$0.$0. BasisQty X1X1X1X1 X2X2X2X2 S1S1S1S1 S2S2 S3S3S3S3 $0. S1S1S1S12001-.21 $6. X1X1X1X112100.2-2.0 $15. X2X2X2X21001001 Zj$222.$6.$15.$0.$1.2$3.0 Cj-Zj THE COMPLETED Zj ROW

83. Toward 3rd Feasible Solution Cj$6.$15.$0.$0.$0. BasisQty X1X1X1X1 X2X2X2X2 S1S1S1S1 S2S2 S3S3S3S3 $0. S1S1S1S12001-.21 $6. X1X1X1X112100.2-2.0 $15. X2X2X2X21001001 Zj$222.$6.$15.$0.$1.2$3.0 Cj-Zj

84. Toward 3rd Feasible Solution Cj$6.$15.$0.$0.$0. BasisQty X1X1X1X1 X2X2X2X2 S1S1S1S1 S2S2 S3S3S3S3 $0. S1S1S1S12001-.21 $6. X1X1X1X112100.2-2.0 $15. X2X2X2X21001001 Zj$222.$6.$15.$0.$1.2$3.0 Cj-Zj$0. COMPUTING THE Cj – Zj ROW ON A COLUMN BY COLUMN BASIS, SUBTRACT Zj FROM Cj

85. Toward 3rd Feasible Solution Cj$6.$15.$0.$0.$0. BasisQty X1X1X1X1 X2X2X2X2 S1S1S1S1 S2S2 S3S3S3S3 $0. S1S1S1S12001-.21 $6. X1X1X1X112100.2-2.0 $15. X2X2X2X21001001 Zj$222.$6.$15.$0.$1.2$3.0 Cj-Zj$0.$0. COMPUTING THE Cj – Zj ROW ON A COLUMN BY COLUMN BASIS, SUBTRACT Zj FROM Cj

86. Toward 3rd Feasible Solution Cj$6.$15.$0.$0.$0. BasisQty X1X1X1X1 X2X2X2X2 S1S1S1S1 S2S2 S3S3S3S3 $0. S1S1S1S12001-.21 $6. X1X1X1X112100.2-2.0 $15. X2X2X2X21001001 Zj$222.$6.$15.$0.$1.2$3.0 Cj-Zj$0.$0.$0. COMPUTING THE Cj – Zj ROW ON A COLUMN BY COLUMN BASIS, SUBTRACT Zj FROM Cj

87. Toward 3rd Feasible Solution Cj$6.$15.$0.$0.$0. BasisQty X1X1X1X1 X2X2X2X2 S1S1S1S1 S2S2 S3S3S3S3 $0. S1S1S1S12001-.21 $6. X1X1X1X112100.2-2.0 $15. X2X2X2X21001001 Zj$222.$6.$15.$0. $1.2 $1.2$3.0 Cj-Zj$0.$0.$0.-$1.2 COMPUTING THE Cj – Zj ROW ON A COLUMN BY COLUMN BASIS, SUBTRACT Zj FROM Cj

88. Toward 3rd Feasible Solution Cj$6.$15.$0.$0.$0. BasisQty X1X1X1X1 X2X2X2X2 S1S1S1S1 S2S2 S3S3S3S3 $0. S1S1S1S12001-.21 $6. X1X1X1X112100.2-2.0 $15. X2X2X2X21001001 Zj$222.$6.$15.$0. $1.2 $1.2 $3.0 $3.0 Cj-Zj$0.$0.$0.-$1.2-$3.0 COMPUTING THE Cj – Zj ROW ON A COLUMN BY COLUMN BASIS, SUBTRACT Zj FROM Cj

89. The 3rd Feasible Solution Cj$6.$15.$0.$0.$0. BasisQty X1X1X1X1 X2X2X2X2 S1S1S1S1 S2S2 S3S3S3S3 $0. S1S1S1S12001-.21 $6. X1X1X1X112100.2-2.0 $15. X2X2X2X21001001 Zj$222.$6.$15.$0. $1.2 $1.2 $3.0 $3.0 Cj-Zj$0.$0.$0.-$1.2-$3.0 THE COMPLETED Cj – Zj ROW

90. The 3rd Feasible Solution Cj$6.$15.$0.$0.$0. BasisQty X1X1X1X1 X2X2X2X2 S1S1S1S1 S2S2 S3S3S3S3 $0. S1S1S1S12001-.21 $6. X1X1X1X112100.2-2.0 $15. X2X2X2X21001001 Zj$222.$6.$15.$0. $1.2 $1.2 $3.0 $3.0 Cj-Zj$0.$0.$0.-$1.2-$3.0

91. The 3rd Feasible Solution Cj$6.$15.$0.$0.$0. BasisQty X1X1X1X1 X2X2X2X2 S1S1S1S1 S2S2 S3S3S3S3 $0. S1S1S1S12001-.21 $6. X1X1X1X112100.2-2.0 $15. X2X2X2X21001001 Zj$222.$6.$15.$0. $1.2 $1.2 $3.0 $3.0 Cj-Zj$0.$0.$0.-$1.2-$3.0

92. The 3rd Feasible Solution Cj$6.$15.$0.$0.$0. BasisQty X1X1X1X1 X2X2X2X2 S1S1S1S1 S2S2 S3S3S3S3 $0. S1S1S1S12001-.21 $6. X1X1X1X112100.2-2.0 $15. X2X2X2X21001001 Zj$222.$6.$15.$0. $1.2 $1.2 $3.0 $3.0 Cj-Zj$0.$0.$0.-$1.2-$3.0 THERE ARE NO POSITIVE NUMBERS IN THE Cj – Zj ROW THEOPTIMALSOLUTION

93. Simplex Linear Programming 3 rd AND OPTIMAL FEASIBLE SOLUTION X 1 = 12 X 2 = 10 S 1 = 2 S 2 = 0 S 3 = 0 Z = $222.00 twelve Produce twelve black and white televisions ten Produce ten color televisions two There are two chassis left over no There are no labor hours left over no There are no color tubes left over $222.00 Total profit realized is $222.00 A RECORD SHOULD BE KEPT OF WHAT EACH VARIABLE REPRESENTS

94. The 3rd Feasible Solution Cj$6.$15.$0.$0.$0. BasisQty X1X1X1X1 X2X2X2X2 S1S1S1S1 S2S2S2S2 S3S3S3S3 $0. S1S1S1S12001-.21 $6. X1X1X1X112100.2-2.0 $15. X2X2X2X21001001 Zj$222.$6.$15.$0. $1.2 $1.2 $3.0 $3.0 Cj-Zj$0.$0.$0.-$1.2 -$3.0 -$3.0 THE SLACK VARIABLE SHADOW PRICES ALWAYS FOUND IN THE Cj – Zj ROW

95. Simplex Linear Programming 3 rd AND OPTIMAL FEASIBLE SOLUTION SHADOW PRICES S 1 = $0.00 S 2 = - $1.20 S 3 = - $3.00 WILLING TO PAY WE WOULD BE WILLING TO PAY ZERO DOLLARS UP TO ZERO DOLLARS FOR ADDITIONAL CHASSIS WOULD BE WILLING TO PAY WE WOULD BE WILLING TO PAY $1.20 UP TO $1.20 FOR ADDITIONAL LABOR HOURS WE WOULD BE WILLING TO PAY $3.00 UP TO $3.00 FOR ADDITIONAL COLOR TUBES THIS PARTICULAR SIMPLEX METHOD DISPLAYS NEGATIVE VALUES FOR POSITIVE SHADOW PRICES. THIS IS CALLED THE “MIRROR EFFECT”.

96. Simplex Linear Programming under QM for WINDOWS

97. Applied Management Science for Decision Making, 2e © 2014 Pearson Learning Solutions

112. Template

125. Simplex Linear Programming Maximization