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SIMPLEX METHOD
SIMPLEX METHOD
SIMPLEX METHOD
1. INTRODUCTION The graphical method is
usually inefficient or impossible The Simplex method used. All constrains must be
expressed in the linear form:
SIMPLEX METHOD
2. Standard Maximum FormThe objective function is to be maximizedAll variables are nonnegative (xi ≥ 0,i = 1, 2, 3 …)All constraints involve ≤The constants on the right side in the constraints are all nonnegative (b ≥ 0)
SIMPLEX METHOD
3. Setting Up Problem3. Setting Up Problem
Convert x1+x2 10 in tox1+x2+x3=10x3 : slack variable
1 2 3
1 2 3
1 2 3
1 2 3
1 2 3
: 2 34 100
2 1503 2 3200, 0, 0
Maximize z x x xsubject to x x x
x x xx x x
with x x x
Example:
1 2 3
1 2 3 4
1 2 3 5
1 2 3 6
1 2 3 4 5 6
: 2 34 100
2 1503 2 3200, 0, 0, 0, 0, 0
Maximize z x x xsubject to x x x x
x x x xx x x x
with x x x x x x
Slack variables
3. Setting Up Problem3. Setting Up ProblemRestate the following linear programming by introducing slack vars
1 2 3
1 2 3 4
1 2 3 5
1 2 3 6
1 2 3 4 5 6
: 2 34 100
2 1503 2 3200, 0, 0, 0, 0, 0
Maximize z x x xsubject to x x x x
x x x xx x x x
with x x x x x x
1 1 4 1 0 0 0 1001 2 1 0 1 0 0 1503 2 1 0 0 1 0 3202 3 1 0 0 0 1 0
x1 x2 x3 x4 x5 x6 z
Constraint 1
Constraint 2
Constraint 3
Objective Function
Indicators
1 2 32 3 0x x x z
3. Setting Up Problem3. Setting Up Problem
3. Setting Up Problem3. Setting Up Problem
Make the initial simplex tableauIndicators: the number in the bottom row
of the initial simplex tableau, except for the last element (1) and 0 on the right
Goal: To find a solution in which all the variables are nonnegative and z is as larger as possible.
Quotients
1 1 4 1 0 0 0 1001 2 1 0 1 0 0 1503 2 1 0 0 1 0 3202 3 1 0 0 0 1 0
x1 x2 x3 x4 x5 x6 z
Most negative indicator
100 /1 100
150 / 2 75
320 / 2 160
Smallest
1 1 4 1 0 0 0 1001 1 11 0 0 0 752 2 23 2 1 0 0 1 0 3202 3 1 0 0 0 1 0
x1 x2 x3 x4 x5 x6 z
4. Selecting the Pivot4. Selecting the Pivot
4. Selecting the Pivot4. Selecting the Pivot Change a particular nonzore to 1,
then all other elements in that column are changed to 0
In example 2: select the most negative one. The column contains that number is pivot column
01000132320010012315000101211000001411
5. Pivoting5. Pivoting1 1 4 1 0 0 0 1001 1 11 0 0 0 752 2 23 2 1 0 0 1 0 3202 3 1 0 0 0 1 0
x1 x2 x3 x4 x5 x6 z
1 7 10 1 0 0 252 2 21 1 11 0 0 0 752 2 22 0 0 0 1 1 0 1701 1 30 0 0 1 2252 2 2
x1 x2 x3 x4 x5 x6 z
R1 = R1 – R2R3 = R3 – 2*R2R4 = R4 + 3*R2
Selecting the new Pivot1 7 10 1 0 0 252 2 21 1 11 0 0 0 752 2 22 0 0 0 1 1 0 1701 1 30 0 0 1 2252 2 2
x1 x2 x3 x4 x5 x6 z
Most negative indicator: Pivot column
Quotients25 / 0.5 50
75 / 0.5 150
170 / 2 85
SmallestPivotrow
1 0 7 2 1 0 0 501 1 11 0 0 0 752 2 22 0 0 0 1 1 0 1701 1 30 0 0 1 2252 2 2
x1 x2 x3 x4 x5 x6 z
Pivoting again1 0 7 2 1 0 0 501 1 11 0 0 0 752 2 22 0 0 0 1 1 0 1701 1 30 0 0 1 2252 2 2
x1 x2 x3 x4 x5 x6 z
1 0 7 2 1 0 0 500 1 3 1 1 0 0 500 0 14 4 1 1 0 700 0 4 1 1 0 1 250
x1 x2 x3 x4 x5 x6 z
R2 = R2 – 0.5*R1R3 = R3 – 2*R1R4 = R4 + 0.5*R2
No negative indicator: Stop!!!!!
The final simplex tableau
6. Reading solution6. Reading solution
Solution:The maximum value of z is 250,
where x1=50, x2=50 and x3=0
1 0 7 2 1 0 0 500 1 3 1 1 0 0 500 0 14 4 1 1 0 700 0 4 1 1 0 1 250
x1 x2 x3 x4 x5 x6 z
3 4 54 250x x x z 3 4 5250 4z x x x
3 4 50, 0 0x x and x Maximize value of z
1 50x
2 50x
6 70x
Basic variables Non-basic variables
6. Reading solution (cont)6. Reading solution (cont)
In any simplex tableau: Basic variables: The variables
corresponding to the column one element is 1
Non-basic variables: the variables corresponding other columns.
25010114007001141400500011310500012701
7. Simplex Method7. Simplex Method
1. Determine the objective function.2. Write all necessary constraints.3. Convert each constraint into an
equation by adding slack variables.4. Set up the initial simplex tableau.5. Locate the most negative indicator.
If there are two such indicators, choose one. This indicator determines the pivot column.
7. Simplex Method (cont)7. Simplex Method (cont)
6. Use The Positive Entries In The Pivot Column To Form The Quotients Necessary For Determining The Pivot. If There Are No Positive Entries In The Pivot Column, No Maximum Solution Exists. If 2 quotients are equally the smallest, let either determines the pivot.
.Simplex Method (cont).Simplex Method (cont)
7. Multiply every entry in the pivot row by the reciprocal of the pivot to change the pivot to 1. The use row operations to change all other entries in the pivot column to 0 by adding suitable multiplies of the pivot to the other rows.
7. Simplex Method (cont)7. Simplex Method (cont)
8. If the indicators are all positive or 0, this is the final tableau. If not, go back to step 5 above and repeat the process until a tableau with no negative indicators is obtained.
9. Determine the basic and non-basic variables and read the solution from the final tableau. The maximum value of the objective function is the number in the lower right corner of the final tableau.
