Linear Programming: The Simplex Method. Chapter 5. Learning Objectives. The Objectives of this chapter:. Convert LP constraints to equalities with slack, surplus, and artificial variables Set up and solve LP problems with simplex tableaus
The Objectives of this chapter:
Why should we study the simplex method?
Maximize profit = $70T + $50C (objective function)
subject to 2T + 1C ≤ 100 (painting hours constraint)
4T+ 3C ≤ 240 (carpentry hours constraint)
T, C ≥ 0 (nonnegativity constraint)How To Set Up The Initial Simplex Solution
T= number of tables produced
C = number of chairs produced
S1 = slack variable representing unused hours in the painting department
S2 = slack variable representing unused hours in the carpentry department
2T + 1C + S1 = 100
4T + 3C + S2 = 240
2T + 1C + S1 = 100
2(40) + 1(10) + S1 = 100
S1 = 10
2T + 1C + 1S1 + 0S2 = 100
4T + 3C + 0S1 + 1S2 = 240
T, C, S1, S2≥ 0
Maximize profit = $70T + $50C + $0S1 + $0S2
Number of Chairs
| | | | |
0 20 40 60 80
Number of TablesFinding an Initial Solution Algebraically
Corner points for the Flair Furniture Company problem
B = (0, 80)
2T+ 1C≤ 100
C = (30, 40)
4T+ 3C≤ 240
D = (50, 0)
(0, 0) A
Production mix column
Real variables columns
Profit per unit column
Profit per unit row
Constraint equation rows
Gross profit row
Net profit rowThe First Simplex Tableau
The first tableau is called a simplex tableau
=The First Simplex Tableau
=The First Simplex Tableau
The ZjandCj– Zjrows
Zj (gross profit) = (Profit per unit of S1) (Number of units of S1)
+ (profit per unit of S2) (Number of units of S2)
= $0 100 units + $0 240 units
= $0 profit
Zj = (Profit per unit of S1) (Substitution rate in row 1)
+ (profit per unit of S2) (Substitution rate in row 2)
Zj (for column T) = ($0)(2) + ($0)(4) = $0
Zj (for column C) = ($0)(1) + ($0)(3) = $0
Zj (for column S1) = ($0)(1) + ($0)(0) = $0
Zj (for column S2) = ($0)(0) + ($0)(1) = $0
Number above or below pivot number
Corresponding number in the new row, that is, the row replaced in step 3
– xFive Steps of the Simplex Method for Maximization Problems
(New row numbers) = (Numbers in old row)
Step 1. Select the variable with the largest positive Cj - Zj value to enter the solution next. In this case, variable T with a contribution value of $70.
For the S1 row
For the S2 rowThe Second Simplex Tableau
Step 2. Select the variable to be replaced. Either S1 or S2 will have to leave to make room for T in the basis. The following ratios need to be calculated.
We choose the smaller ratio (50) and this determines the S1 variable is to be replaced. This corresponds to point D on the graph in Figure 9.2.
Step 3. We can now begin to develop the second, improved simplex tableau. We have to compute a replacement for the pivot row. This is done by dividing every number in the pivot row by the pivot number. The new version of the pivot row is below.
Step 4. Completing the rest of the tableau, the S2 row, is slightly more complicated. The right of the following expression is used to find the left side.
The T column contains and the S2 column
contains , necessary conditions for variables to
be in the solution. The manipulations of steps 3 and 4 were designed to produce 0s and 1s in the appropriate positions.
0The Second Simplex Tableau
Step 5. The final step of the second iteration is to introduce the effect of the objective function. This involves computing the Cj - Zj rows. The Zj for the quantity row gives us the gross profit and the other Zj represent the gross profit given up by adding one unit of each variable into the solution.
Zj (for T column) = ($70)(1) + ($0)(0) = $70
Zj (for C column) = ($70)(0.5) + ($0)(1) = $35
Zj (for S1 column) = ($70)(0.5) + ($0)(–2) = $35
Zj (for S2 column) = ($70)(0) + ($0)(1) = $0
Zj (for total profit) = ($70)(50) + ($0)(40) = $3,500
Step 1. Variable C will enter the solution as its Cj - Zj value of 15 is the largest positive value. The C column is the new pivot column.
Step 2. Identify the pivot row by dividing the number in the quantity column by its corresponding substitution rate in the C column.
These ratios correspond to the values of C at points F and C in Figure 9.2. The S2 row has the smallest ratio so S2 will leave the basis and will be replaced by C.
Step 3. The pivot row is replaced by dividing every number in it by the pivot point number
The new C row is
Step 4. The new values for the T row may now be computed
Step 5. The Zj and Cj - Zj rows can now be calculated
Zj (for T column) = ($70)(1) + ($50)(0) = $70
Zj (for C column) = ($70)(0) + ($50)(1) = $50
Zj (for S1 column) = ($70)(1.5) + ($50)(–2) = $5
Zj (for S2 column) = ($70)(–0.5) + ($50)(1) = $15
Zj (for total profit) = ($70)(30) + ($50)(40) = $4,100
And the net profit per unit row is now
T = 30 tables
C = 40 chairs
S1 = 0 slack hours in the painting department
S2 = 0 slack hours in the carpentry department
profit = $4,100 for the optimal solution