1. ISS500E-Quantitiative Decision Tehniques
Solving Minimization Problems with Simplex Algorithm //
Sensitivity Analysis

2. Big M Method : A version of the simplex algorithm that first finds a bfs (basic feasible solution) by adding artificial variables to the problem The objective function must be modified to ensure that the artificial variables are all equal to zero at the conclusion of the simplex algorithm.
- If the LP is min problem, add (for each artificial variable) Mai to the objective function. If the LP is max problem, add (for each artificial variable) -Mai to the objective function. Here, M represents a very large positive number.
-Also add the sign restriction Ai (ai) ≥ 0
-Remember that (4M-2) is more positive than (3M +900), and (-6M-5) is more negative than (-5M- 40).
(Winston, 2004, pp.172,174)
Artificial Variable: A variable that has no meaning in a physical sense but acts as a tool to help generate an initial LP solution. They are needed in ≥ and = constraint (Render, 2003, pp.366).

3. SOLVING MINIMIZATION PROBLEMS
Minimization problems are quite similar to the maximization problems tackled earlier. The significant difference involves the Cj - Zj row. Our objective is to minimize cost, and a negative Cj - Zj value indicates that the total cost will decrease if that variable is selected to enter the solution. Thus, the new variable to enter the solution in each Table will be the one with a negative Cj - Zj that gives the largest improvement. We choose the variable that decreases costs the most. In minimization problems, an optimal solution is reached when all the numbers in the Cj - Zj row are 0 or positive – just the opposite from maximization case. All other simplex steps remain the same (Render, 2003, pp.366) .

4. Steps for solving LP minimization problems *
I. Formulate the LP problem’s objective function and constraints.
II. Include slack variables in each less-than-or-equal-to constraint, artificial variables in each equality constraint, and both surplus and artificial variables in each greater-than-or-equal-to constraint. Then add all these variables to the problem’s objective function.
III. Develop an initial simplex Table with artificial and slack variables in the basis. Compute the Zj and Cj - Zj values for this table.
*( Render et al., 2003, pp. 353)

5. IV. Follow these five steps until an optimal solution has been reached:
1. Choose the variable with the negative Cj - Zj indicating the
greatest improvement to enter the solution. This is the pivot
column.
2. Determine the row to be replaced by selecting the one with
the smallest (nonnegative) quantity -to-pivot column
substitution rate ratio. This is the pivot row.
3. Calculate the new values for the pivot row.
4. Calculate the new values for the other row(s).
5. Calculate the Zj and Cj - Zj values for the table. If there are
any Cj - Zj numbers less than 0, return to step1. If there are no
Cj - Zj numbers that are less than 0, an optimal solution has
been reached.

6. The Muddy River Chemical Company Example*:
The Muddy River Chemical Corporation must produce exactly 1000 pounds of a special mixture of phosphate and potassium for a customer. Phosphate costs $5 per pound and potassium costs $6 per pound. No more than 300 pounds of phosphate can be used, and at least 150 pounds of potassium must be used. The problem is to determine the least –cost blend of the two ingredients.
*Render et al., 2003, pp

7. The Muddy River Chemical Company Example*:
X1 : number of pounds of phosphate
X2 : number of pounds of potassium
The problem is:
Z min = 5X1 + 6X2
subject to :
X2 ≥ 150
X1 + X2 = 1000
X1 ≤ 300
X1 , X2 ≥ 0
*Render et al., 2003, pp

8. Converting the constraints and objective function
Z min = 5X1 + 6X E1 + MA1 + MA2 + 0S1
X2 -E1 + A1 = 150
X1 + X2 +A2 = 1000
X1 + S1 = 300
X1, X2, S1, E1, A1, A2 ≥ 0
Table1. Initial Simplex Table for the Muddy River Chemical Corporation Problem 5 6 M X1 X2 E1 A1 A2 S1 1 -1
150
1000
300 Zj $M 2M -M
1150M
Cj - Zj
5-M
6-2M

9. 5 6 M X1 X2 E1 A1 A2 S1 1 -1 150 850 300 Zj M-6 6-M 850M+900 Cj - Zj
Table2. Second Simplex Table for the Muddy River Chemical Corporation Problem 5 6 M X1 X2 E1 A1 A2 S1 1 -1
150
850
300 Zj
M-6
6-M
850M+900
Cj - Zj
5-M
2M-6

10. Table3. Third Simplex Table for the Muddy River Chemical Corporation Problem5 6 M X1 X2 E1 A1 A2 S1 1 -1
150
550
300 Zj
M-6
6-M
5-M
550M+2400
Cj - Zj
2M-6
M-5

11. Table4. Fourth and Optimal Solution to the Muddy River Chemical Corporation Problem5 6 M X1 X2 E1 A1 A2 S1 1 -1
700
550
300 Zj
5700
Cj - Zj
M-6

12. The optimal solution:
X2=700, X1= 300, E1=550,
S1= A1=A2 = 0
Z= $5700

13. Interpreting the Simplex Tableau
Example 2:The Flair Furniture Company produces inexpensive tables and chairs. The production process for each is similar in that both require a certain number of hours of carpentry work and a certain number of labor hours in the painting and vanishing department. Each table takes 4 hours of carpentry and 2 hours in the painting and varnishing shop. Each chair requires 3 hours of carpentry and 1 hour in the painting and varnishing. During the current production period, 240 hours of carpentry time are available and 100 hours in the painting and varnishing time are available. Each table sold yields a profit of $7; each chair produced is sold for a $ 5 profit. Flair Furniture’s problem is to determine the best possible combination of tables and chairs to manufacture in order to reach the maximum profit. The firm would like this production mix situation formulated as an LP problem.

14. X1: number of tables produced
X2: number of chair produced
Mathematical statement of the LP problem
Zmax= 7X1 + 5X2 (objective function)
s.t.
2X1 + X2 ≤ 100 (Painting hours constraint)
4X1 + 3X2 ≤ 240 (Carpentry hours constraint)
X1 , X2 ≥0 (nonnegativity constraints)
standard form:
Zmax= 7X1 + 5X2 + 0S1+0 S2 (objective function)
2X1 + X2 + S1 = 100 (Painting hours constraint)
4X1 + 3X2 + S2 = 240 (Carpentry hours constraint)
X1 , X2 ,S1, S2 = 0 (nonnegativity constraints)

15. Table 1. Initial Simplex Tableau7 5 X1 X2 S1 S2
0 S1 2 1
100
0 S2 4 3
240 Zj
Cj-Zj

16. Table 2. Second Simplex Tableau7 5
Solution Mix X1 X2 S1 S2
Quantity
7 X1
1 (12)
1/2 (16)
1/2 (4)
0 (8)
50 (1)
0 S2
0 (13)
1 (17)
-2 (5)
1 (9)
40 (2) Zj
7 (14)
7/2 (18)
7/2 (6)
0 (10)
350 (3)
Cj-Zj
0 (15)
3/2 (19)
-7/2 (7)
0 (11)

17. Quantity Column: (1),(2)
A column in the simplex tableau that gives the numeric value of each variable in the solution mix column.
Basis: The set of variables that are in the solution, have positive, nonzero values, and are listed in the solution mix column. They are also called basic variables.
Solution mix :A column in the simplex tableau that contains all the basic variables in the solution.
Nonbasic variables: Variables not in the solution mix or basis. Nonbasic variables are equal to zero.
(3): The Zj value for the quantity column provides the total contribution of the given solution

18. Substitution Rates : (4),(5),(8),(9),(12),(13),(16),(17)
The coefficients of the central body of each simplex table. They indicate the number of units of each basic variable that must be removed from the solution if a new variable is entered.
A negative substitution rates means that if 1 unit of a column variable is added to the solution, the value of the corresponding solution (row) variable will be increased.
A positive substitution rates means that if 1 unit of a column variable is added to the solution, the row variable will decrease by the rate.

19. Zj Row (6),(10),(14),(18)
The row containing the figures for gross profit or loss given up by adding one unit of variable into the solution.
Cj - Zj Row : (19),(15),(11),(7)
The row containing the net profit or loss that will result from introducing one unit of the variable indicated in that column into the solution.
A negative number in the Cj - Zj row : profits ( in this case) will decrease if the corresponding variable is added to the solution mix.
A positive number in the Cj - Zj row : profits ( in this case) will increase if the corresponding variable is added to the solution mix.

20. 7 5 X1 X2 S1 S2 7 X1 1 3/2 -1/2 30 5X2 -2 40 Zj 1/2 410 Cj-Zj -3/2
Table 3. Third and Optimal Simplex Tableau 7 5 X1 X2 S1 S2
7 X1 1
3/2
-1/2 30
5X2 -2 40 Zj
1/2
410
Cj-Zj
-3/2

21. Sensitivity Analysis
One of the assumptions of LP is the certainty assumption. It is assumed that the model parameters (cj, bi, aij) are known with certainty.
It is rare that a manager can determine the model parameters of a linear programming problem with absolute certainty.
In reality the model parameters are usually simply estimates and, such subject to uncertainty.
The values of the parameters may change. For example, the selling price of the products, the variable cost per a unit of product or the available amount of resources may change.
As a results, it may often be desirable for a manager / decision maker to observe the effects of parameter changes (which would reflect uncertainty) on the optimal solution of the problem.

22. The analysis of parameter changes and their effects on linear programming solutions is referred to as sensitivity or postoptimality, analysis.
In other words, sensitivity analysis is the study of parameter changes and the sensitivity of the optimal solution to these changes.
We discuss how three types of changes in LP’s parameters change the optimal solution:
Changing the objective function coefficient of a nonbasic variable.
Changing the objective function coefficient of a basic variable.
Changing the right- hand side of a constraint

23. Changing the right- hand side of a constraint
A change in the right- hand side of a constraint will affect the right-hand side of the constraints in the optimal tableau. As long as the right hand side of each constraint in the optimal tableau remains nonnegative, the current basis remains feasible and optimal. If at least one right-hand side in the optimal tableau becomes negative, then the current basis is no longer feasible and therefore no longer optimal.
If the right-hand side of ith constraint is increased by Δ, then the right-hand side of the optimal tableau is given by

24. Original right hand side of the optimal tableau + (column i of B-1)Δ
For the current basis to remain optimal, we require that the right-hand side of each constraint in the optimal tableau remain nonnegative. This means that the current basis will remain optimal if and only if the following hold:
Original right hand side of the optimal tableau + (column i of B-1)
Δ ≥ 0
B-1 : For any simplex tableau , B-1 is the m x m consisting of the columns in the current tableau that correspond to the initial tableau’s set of basic variables (taken in the same order)

25. Shadow price of the ith constraint is the amount by which the optimal z- value is improved (increased in a max problem and decreased in a min problem) if we increase bi by 1 ( from bi to bi +1). This assumes that after the right hand side of constraint i has been changed to bi +1, the current basis remains optimal.

26. Changing the right- hand side of a constraint
Example 1.Zmax= 4X1 + 5X2
X1 + 2X2  10 (labor,hr)
6X1 + 6X2  36 (material,lb)
8X1 + 4X2  40 (space,ft2)
X1 , X2  0
In many LP Problems we can interpret the RHS values as the resource available
For instance, in the problem the first RHS represents the available amount of labor time; the second the available amount of material; the third the available amount of storage space.
Table1 represents the initial basic solution of the problem, Table 2 optimal basic solution of the problem.

27. Table 1. Initial Simplex Tableau4 5 X1 X2 S1 S2 S3
0 S1 1 2 10
0 S2 6 36
0 S3 8 40 Zj
Cj-Zj

28. 4 5 X1 X2 S1 S2 S3 5 X2 1 -1/6 4 X1 -1 1/3 2 0 S3 -2 8 Zj 1/2 28 Cj-Zj
Table 2.Optimal Simplex Tableau 4 5 X1 X2 S1 S2 S3
5 X2 1
-1/6
4 X1 -1
1/3 2
0 S3 -2 8 Zj
1/2 28
Cj-Zj
-1/2

29. Changing the RHS of constraint 1(labor constraint)
A change in the right- hand side of a constraint will affect the right-hand side of the constraints in the optimal tableau. As long as the right hand side of each constraint in the optimal tableau remains nonnegative, the current basis remains feasible and optimal. If at least one right-hand side in the optimal tableau becomes negative, then the current basis is no longer feasible and therefore no longer optimal.

30. Consider a general increase in the labor constraint by Δ hours
Consider a general increase in the labor constraint by Δ hours. The problem constraint become:
Zmax= 4X1 + 5X2
X1 + 2X2  10 +Δ (labor constraint)
6X1 + 6X2  36 (material constraint)
8X1 + 4X2  40 (space constraint)
X1 , X2  0
-Substitute these constraints into the table
-obtain a modified table (Table3) showing the initial basic solution with the changes.
-Observe that in the table , the coefficients of Δ added to quantity column are the same as the values in the S1 column

31. Table 3. Initial Simplex Tableau (b1=10+Δ)4 5 X1 X2 S1 S2 S3
0 S1 1 2 10
+1Δ
0 S2 6 36
+ 0Δ
0 S3 8 40 Zj
Cj-Zj
If we proceed to solve the problem in a similar way with original problem, we obtain following solution.

32. Table 4. Optimal Simplex Tableau (b1=10+Δ)5 X1 X2 S1 S2 S3
5 X2 1
-1/6
+1Δ
4 X1 -1
1/3 2
-1Δ
0 S3 -2 8
+ 4Δ Zj
1/2
28+Δ
Cj-Zj
-1/2

33. Note that the coefficients of Δ in the RHS column(quantity) have remained the same as the values in the S1 column.
This means that it is unnecessary to examine all simplex tables to determine the general effects of Δ on the optimal solution.
We can compute the range of feasibility by working with the final simplex tableu.
As a general rule , it is only necessary to increase (decrease) the final RHS values by multiple of Δ, where the multiple is given by the column coefficients of the slack variable (Si) for bi

34. multiply each value in that column by ∆
In other words, we can determine the upper and lower limit of ∆ by working final simplex table. Shortly;
determine the column of final simplex tableau corresponding to the slack variable associated with constraint i
multiply each value in that column by ∆
add these products to the corresponding RHS values of the final simplex tableau
calculate the range of feasibility for bi by solving the following inequalities (satisfy the feasibility condition)column of the final simplex tableau corresponding to the slack variable associated with constraint i
 0
Current solution +∆
column of the final simplex tableau corresponding to the slack variable associated with constraint i
 0

35. The new values should be  0
The new values should be  0. Because these solution values will remain feasible as long as they are nonnegative (for both max. and min problem).

36. In our example
determine the column of final simplex tableau corresponding to the slack variable associated with constraint 1. This is S1column
multiply each value in that column by ∆
add these products to the corresponding RHS values of the final simplex tableau
∆ S1column rhs = new values with ∆
= ∆
= ∆
= ∆

37. calculate the range of feasibility for bi by solving the following inequalities (satisfy the feasibility condition) 4
+1Δ ≥
X2 will remain as a basic variable as long as its value is grater than or equal to 0 2
-1Δ
X1 will remain as a basic variable as long as its value is grater than or equal to 0 8
+4Δ
S3 will remain as a basic variable as long as its value is grater than or equal to 0

38. Thus we can say that the current basis will remain feasible therefore optimal as long as the RHS in optimal tableau remains nonnegative. 4
+1Δ ≥ ∆ -4 2
-1Δ ≤ 8
+ 4Δ -2
4≤-2 ≤ Δ ≤ 2,
(-2 is more constraining than –4, is discarded)
The range for ∆ is :-2 ≤ Δ ≤ 2,
For -2 ≤ Δ ≤ 2, the current basis remain feasible and therefore optimal.
The range of feasibility, ≤ b1 ≤10+2
For 8 ≤ b1 ≤12 , the current basis remain feasible and therefore optimal.

39. For the ≥ and = type constraints ,
Shadow price of the ith constraint is the amount by which the optimal z- value is improved (increased in a max problem and decreased in a min problem) if we increase bi by 1 (from bi to bi +). This assumes that after the right hand side of constraint i has been changed to bi +1, the current basis remains optimal
For the ≥ and = type constraints ,
use the column of the optimal simplex tableau corresponding to the artificial variable associated with the constraint i
The procedure for calculating the range of feasibility for these constraints is same.
column of the final simplex tableau corresponding to the artificial variable associated with constraint i
Current solution +∆
 0

40. All procedure for calculating the range of feasibility for the constraints for a minimization problem is same!!!
- The firm is considering hiring an extra labor on a part time basis. Say its cost is $ 0.7 per hour. Should the firm do this???
Yes.
Because
The shadow price of labor resource is $1 per hour.
The firm will net $0.3 (1-0.7) for every additional hour of labor
Note : For 8 ≤ b1 ≤12, shadow price will remain valid.

41. Optimal simplex tableaus with a Δ change in b1
Table 5. Optimal Simplex Tableau Δ = 2 (b1=12) 4 5 X1 X2 S1 S2 S3
5 X2 1
-1/6 6
4 X1 -1
1/3
0 S3 -2 16 Zj
1/2 30
Cj-Zj
-1/2

42. Table 6. Optimal Simplex Tableau Δ = -1 (b1=9)4 5 X1 X2 S1 S2 S3
5 X2 1
-1/6 3
4 X1 -1
1/3
0 S3 -2 Zj
1/2 27
Cj-Zj
-1/2

43. Table 7. Optimal Simplex Tableau Δ = 2 (b1=11)4 5 X1 X2 S1 S2 S3
5 X2 1
-1/6
4 X1 -1
1/3
0 S3 -2 12 Zj
1/2 29
Cj-Zj
-1/2

44. Changing the objective function coefficient[1]
A change in the objective function coefficient can affect the optimality of the original solution. Hence in performing sensitivity analysis of a cj parameter we are interested in determining the range for cj over which the present solution will remain optimal. Changes in the profit or cost contributions in the objective function can occur for a basic or non basic variables.
Example 2
Zmax= 6X1 + 4X2 ( profit)
2X1 + 6X2  12 (labor)
4X1 + 4X2  16 (material)
X1 , X2  0
Table 8. Final Optimal Tableau 6 4 X1 X2 S1 S2
0 S1 1
-1/2
6 X1
1/4 Zj
3/2 24
Cj-Zj -2
-3/2

45. i. Changing the objective function coefficient of a nonbasic variable
Zmax= 6X1 + (4 + Δ)X2
2X1 + 6X2  12
4X1 + 4X2  16
X1 , X2  0
When xj is a nonbasic variable it is necessary to observe only the final simplex tableau to determine the range values for cj. Table 8 is the optimal solution tableau for the example problem. Since x2 is nonbasic, the only rows affected by a change in c2 are the cj and cj-zj rows. Thus a Δ change in c2 results in the modified tableau in Table 9.

46. Tablo 9. A Final Tableau with a Δ Change in C2[1]6
4+ Δ2 X1 X2 S1 S2
0 S1 4 1
-1/2
6 X1
1/4 Zj
3/2 24
Cj-Zj
-2 + Δ2
-3/2
The cj-zj row value in the x2 column is computed by subtracting 6 from 4+ Δ2 , which yields
-2 + Δ2.
This solution will remain optimal if all cj-zj ≤ 0 ( for maximization problem).
Thus, to determine Δ2, we make following computation: -2
+1Δ ≤
∆≤ 2
The current solution is optimal as long as X2’s profit coefficient does not increase by more than 2.

47. It is unbounded in the left hand side , so lower limit is -∞
It is unbounded in the left hand side , so lower limit is -∞. It means that you can decrease the value of c2 to negative infinity without affecting the solution.
[-∞ ≤ c2 ≤6] (the range of variable of c2 for which the current basis remains optimal) It is called as the range of insignificance.
If the value of objective function coefficient for the variable is outside this range, the current solution will not be optimal.
Expl. Say C2=8 : Its value in the row will be positive and current solution is not optimal. Then it will selected as a entering variable , and then optimal solution will change.
For any non basic variable the reduced cost for the variable is the amount by which its objective function coefficient must be improved before that variable will be a basic variable in some optimal solution (Winston, 2004, pp. 343).
The reduced cost of a nonbasic variable is the amount by which the value of Z will decrease if the increase the value of nonbasic variable by 1 (while all the other nonbasic variables remain equal to 0( Winston, 2004, pp. 343).

48. ii. Changing the objective function coefficient of a basic variable.
Zmax= (6 + Δ1)X1 + 4X2
2X1 + 6X2  12
4X1 + 4X2  16
X1 , X2  0
Table 13. Initial Simplex Tableau with a Δ change in C1 (C1=6 + Δ1)
6+ Δ1 4 X1 X2 S1 S2
0 S1 2 6 1 12
0 S2 16 Zj
Cj-Zj

49. Table 14. Final Simplex Tableau (C1=6 + Δ1)
6+ Δ1 4 X1 X2 S1 S2
0 S1 1
-1/2
(6+ Δ1)X1
1/4 Zj
3/2+ Δ1/4
24+ 4Δ1
Cj-Zj
-2 - Δ1
-3/2- Δ1/4
A change in objective function coefficient for a basic variable can affect the Cj-Zj values of all nonbasic variables.
Since X1 is a basic variable, the C1 value is included in the Cj column. and Cj row.
As a result, C1=6 + Δ becomes a multiple of the Zj row values.
Therefore, the effect of the Δ change in the final tableau can be determined simply by inserting Cj +Δ in place of Cj column in the final tableau and calculating the Cj-Zj values normally.

50. The results in Cj-Zj row values for nonbasic variables, which include Δ values.
In order for the current solution to remain optimal, these Cj-Zj values must remain negative or zero (max. problem).
Note : For min problem, in order for the current solution to remain optimal, the Cj-Zj values must remain positive or zero
Thus, to determine Δ2, we make following computations: -2
-1Δ ≤
∆≥-2
The optimal solution will not change unless X1’s profit coefficient decreases by at least 2, which is a change of ∆= -2. Hence, variable X2 will not enter the basis unless the profit per a unit product 1 drops from 6 to 4.
-3/2
- Δ1/4
∆≥-6
The optimal solution will not change unless X1’s profit coefficient decreases by at least 6, which is a change of ∆= -6. Hence, variable S2 will not enter the basis unless the profit per a unit product 1 drops from 6 to 0.

51. So,
-6≤-2 ≤ Δ (-2 is more constraining than –6, -6 is discarded).
It is unbounded in the right hand side, upper limit is ∞. It means that you can increase the value of c1 to positive infinity without affecting the solution.
[4 ≤ c1 ≤∞] (the range of variables of c1 for which the current basis remains optimal). It is called as range of optimality.
If the value of objective function coefficient for the variable is outside this range, the current solution will not be optimal.

52. Optimal simplex tableau with a Δ change in cj
Table 15. Optimal Simplex Tableau
Δ2= -1, (C2=3) 6 3 X1 X2 S1 S2
0 S1 4 1
-1/2
6 X1
1/4 Zj
3/2 24
Cj-Zj -3
-3/2

53. Table 16. Optimal Simplex Tableau
Δ2=1, (C2=5) 6 5 X1 X2 S1 S2
0 S1 4 1
-1/2
6 X1
1/4 Zj
3/2 24
Cj-Zj -1
-3/2

54. Table 17 Optimal Simplex Tableau
Δ1= -1, (C1=5) 5 4 X1 X2 S1 S2
0 S1 1
-1/2
5 X1
1/4 Zj
5/4 20
Cj-Zj -1
-5/4

55. Table 18. Optimal Simplex Tableau
Δ1=3, (C1=9) 9 4 X1 X2 S1 S2
0 S1 1
-1/2
9 X1
1/4 Zj
9/4 36
Cj-Zj -5
-9/4

56. Example 4. A linear programming problem has been formulated and solved
Example 4. A linear programming problem has been formulated and solved.The optimal simplex tableau for this is shown in Table 19.
a). Determine the range of variables of Cj for which current basis remains optimal.
b). Determine the range of values for bi within which current basis remains optimal.
Zmin= 14X1 + 10X2+4X3 + 8X4
X  40
X1 + X2 + X3 + X4 = 85
2X1 + 3X2 + 4X3 + 5X4  320
3X1 + 4X2 + 5X3 + 6X4  410
X1 , X2, X3 , X4  0

57. Table 19. Optimal Simplex Tableau14 10 4 8 M X1 X2 X3 X4 E1 A1 A2 S3 S4
4 X3 1
3/2 -1
1/2 75
14 X1
-1/2 2
0 E1 35
0 S4 5 Zj 9 24 -5
440
Cj-Zj
M-24

58. Optimal simplex tableau with a Δ change in bi
Table 20. Optimal Simplex Tableau for Example 4(b1=39) 14 10 4 8 M X1 X2 X3 X4 E1 A1 A2 S3 S4
4 X3
1/2 1
3/2 -1 75
14 X1
-1/2 2
0 E1 36
0 S4 5 Zj 9 24 -5
440
Cj-Zj
M-24

59. Table 21. Optimal Simplex Tableau for Example 4(b2=84)14 10 4 8 M X1 X2 X3 X4 E1 A1 A2 S3 S4
4 X3
1/2 1
3/2 -1 76
14 X1
-1/2 2
0 E1 36
0 S4 6 Zj 9 24 -5
416
Cj-Zj
M-24 5

60. Table 22. Optimal Simplex Tableau for Example 4(b3=319)14 10 4 8 M X1 X2 X3 X4 E1 A1 A2 S3 S4
4 X3
1/2 1
3/2 -1
74½
14 X1
-1/2 2
10½
0 E1
34½
0 S4 6 Zj 9 24 -5
445
Cj-Zj
M-24 5

61. Table 23. Optimal Simplex Tableau for Example 4(b4=409)14 10 4 8 M X1 X2 X3 X4 E1 A1 A2 S3 S4
4 X3
1/2 1
3/2 -1 75
14 X1
-1/2 2
0 E1 35
0 S4 Zj 9 24 -5
440
Cj-Zj
M-24 5

62. Example 5. A linear programming problem has been formulated and solved
Example 5. A linear programming problem has been formulated and solved. The optimal simplex tableau for this is shown in Table 24.
a). Determine the range of variables of Cj for which current basis remains optimal.
b). Determine the range of values for bi within which current basis remains optimal.
Zmax= 14X1 + 10X2+4X3 + 8X4
X  40
X1 + X2 + X3 + X4 = 85
2X1 + 3X2 + 4X3 + 5X4  320
3X1 + 4X2 + 5X3 + 6X4  410
X1 , X2, X3 , X4  0

63. Table 24. Optimal Simplex Tableau14 10 4 8 -M X1 X2 X3 X4 E1 A1 A2 S3 S4
4 X3 1 -1 40
14 X1 45
0 S3 3 2 -2 70
0 S4 -3 75 Zj
-10
790
Cj-Zj -4 -6
-M+10
-M-14