Linear programming simplex method

3/19/2020 Dr. Abdulfatah Salem 2
For linear programming problems involving two variables, the graphical solution
method introduced before is convenient.
It is better to use solution methods that are adaptable to computers. One such method
is called the simplex method, The simplex method is an iterative procedure
developed by George Dantzig in 1946. It provides us with a systematic way of
examining the vertices of the feasible region to determine the optimal value of the
objective function.
more than two
variables
problems involving
a large number of
constraints
However, for
problems involving

Standard Minimization Form
• The objective function is to be
minimized.
• All the RHS involved in the
problem are nonnegative.
• All other linear constraints may
be written so that the
expression involving the
variables is greater than or equal
to a nonnegative constant.
Standard Maximization Form
• The objective function is to be
maximized.
• All the RHS involved in the
problem are nonnegative.
• All other linear constraints may
be written so that the expression
involving the variables is less
than or equal to a nonnegative
constant.

1) Be sure that the model is standard model
2) Be sure that the right hand side is +ve value
3) Convert each inequality in the set of constraints to an equation as
follows:
• Slack Variable added to a ≤ constraint to convert it to an equation
(=).
A slack variable represents unused resources A slack variable
contributes nothing to the objective function value.
• Surplus Variable subtracted from a ≥ constraint to convert it to an
equation (=).
A surplus variable represents an excess above a constraint
requirement level. Surplus variables contribute nothing to the
calculated value of the objective function.
Simplex Algorithm

4) Prepare the equations to make the unknowns in the left hand side and
the fixed values in the right hand side
5) Create the initial simplex tableau
6) Select the pivot column. ( The column with the “most negative value”
element in the last row. )
7) Select the pivot row. (The row with the smallest non-negative result
when the last element in the row is divided by the corresponding in
the pivot column)
8) Use elementary row operations to make all numbers in the pivot column
equal to 0 except for the pivot number equal to 1.
9) Check the optimality (entries in the bottom row are zero or positive, If
so, this the final tableau (optimal solution) , If not, go back to step 6
10) The linear programming problem has been solved and maximum solution
obtained, which is given by the entry in the lower-right corner of the
tableau.
Simplex Algorithm (cont.)

= + Artificial (A)
Constraint type Variable to be added
≥ + slack (s)
≤ - Surplus (s) + artificial (A)
Simplex Algorithm (cont.)

The national co. for assembling computer systems. assembles laptops and
printers, Each laptop takes four hours of labor from the hardware department and two hours of labor from
the software department. Each printer requires three hours of hardware and one hour of software. During
the current week, 240 hours of hardware time are available and 100 hours of software time. Each laptop
assembled gives a profit of $70 and each printer a profit of $50. How many printers and laptops should be
assembled in order to maximize the profit?
Example
Solution
ConstraintsprinterlaptopResource
24034hardware
10012software
5070Unit profit $
Max Z = 70x1 + 50x2
4x1 + 3x2 ≥ 240
2x1 + x2 ≥ 100
Z - 70x1 - 50x2 = 0
4x1 + 3x2 + s1 = 240
2x1 + x2 + s2 = 100

4x1 + 3x2 + s1 = 240
2x1 + x2 + s2 = 100
Z - 70x1 - 50x2 = 0
Basic
Variable
X1 X2 S1 S2 Z Rhs
S1 4 3 1 0 0 240
S2 2 1 0 1 0 100
Z -70 -50 0 0 1 0

Basic Variable X1 X2 S1 S2 Z Rhs
S1 4 3 1 0 0 240
S2 2 1 0 1 0 100
Z -70 -50 0 0 1 0
S1 4 3 1 0 0 240
S2 2 1 0 1 0 100
Z -70 -50 0 0 1 0
Ratio
60
50
0
S1 4 3 1 0 0 240
S2 2 1 0 1 0 100
Z -70 -50 0 0 1 0
X1 1 1/2 0 1/2 0 50 /2

S1 4 3 1 0 0 240
X1 1 1/2 0 1/2 0 50
Z -70 -50 0 0 1 0
4
1
-70
❶
❷
1 1/2 0 1/2 0 50
4 2 0 2 0 200
1 ½ 0 1/2 0 50
-70 -35 0 -35 0 -3500
S1
0 1 1 -2 0 40
X1
1 1/2 0 1/2 0 50
Z 0 -15 0 35 1 3500
❸
First
iteration

1
1/2
-15
❶
❷
X2
0 1 1 -2 0 40
X1
1 1/2 0 1/2 0 50
Z 0 -15 0 35 1 3500
0 1 1 -2 0 40
0 1 1 -2 0 40
0 1/2 1/2 -1 0 20
0 -15 -15 30 0 -600
X2 0 1 1 -2 0 40
X1 1 0 -1/2 3/2 0 30
Z 0 0 15 5 1 4100
❸
Second
iteration

BV X1 X2 S1 S2 Z Rhs
X2 0 1 1 -2 0 40
X1 1 0 -1/2 3/2 0 30
Z 0 0 15 5 1 4100
X1 = 30 No of laptops = 30
X2 = 40 No of printers = 40
Z = 4100 The optimal profit = $4100

Solve the following problem using the simplex method
Max Z = 3X1+ 5X2
S.t.
X1  4
2 X2  12
3X1 +2X2  18
X1 , X2  0
Example
Solution
Basic
variables
X1 X2 S1 S2 S3 Z RHS
S1 1 0 1 0 0 0 4
S2 0 2 0 1 0 0 12
S3 3 2 0 0 1 0 18
Z -3 -5 0 0 0 1 0

BV X1 X2 S1 S2 S3 Z RHS
S1 1 0 1 0 0 0 4
X2 0 1 0 1/2 0 0 6
S3 3 2 0 0 1 0 18
Z -3 -5 0 0 0 1 0
0
1
2
-5
0 1 0 1/2 0 0 6
0 0 0 0 0 0 0
0 1 0 1/2 0 0 6
0 2 0 1 0 0 12
0 -5 0 -5/2 0 0 -30
❶
❷
S1 1 0 1 0 0 0 4
X2 0 1 0 1/2 0 0 6
S3 3 0 0 -1 1 0 6
Z -3 0 0 5/2 0 1 30
❸
First
iteration

S1 1 0 1 0 0 0 4
X2 0 1 0 ½ 0 0 6
X1 1 0 0 -1/3 1/3 0 2
Z -3 0 0 5/2 0 1 30
1
0
1
-3
1 0 0 -1/3 1/3 0 2
1 0 0 -1/3 1/3 0 2
0 0 0 0 0 0 0
1 0 0 -1/3 1/3 0 2
-3 0 0 1 -1 0 -6
❶
❷
S1 0 0 1 1/3 -1/3 0 2
X2 0 1 0 1/2 0 0 6
X1 1 0 0 -1/3 1/3 0 2
Z 0 0 0 3/2 1 1 36
❸
Second
iteration

S1 0 0 1 1/3 -1/3 0 2
X2 0 1 0 1/2 0 0 6
X1 1 0 0 -1/3 1/3 0 2
Z 0 0 0 3/2 1 1 36
X1 = 2
X2 = 6
Z = 36
❸

19-Mar-20 Dr. Abdulfatah Salem 17
Max. Z = 13X1+11X2
ST.
4X1 + 5X2 < 1500
5X1 + 3X2 < 1575
X1 + 2X2 < 420
X1, X2 > 0
Example
Solution
X1 + 2X2 + S3 = 420
Z = 13X1+11X2
4X1 + 5X2 < 1500
5X1 + 3X2 < 1575
X1 + 2X2 < 420
Z - 13X1 - 11X2 = 0
4X1 + 5X2 + S1 = 1500
5X1 + 3X2 + S2 = 1575

RHSS3S2S1X2X1BV
150000154S1
157501035X1
42010021S3
0000-11-13Z
RHSS3S2S1X2X1BV
150000154S1
31501/503/51X1
42010021S3
0000-11-13Z
RHSS3S2S1X2X1BV
150000154S1
157501035S2
42010021S3
0000-11-13Z
/5

RHSS3S2S1X2X1BV
150000154S1
31501/503/51X1
42010021S3
0000-11-13Z
31501/503/51
4
1
1
-13
126004/5012/54
31501/503/51
31501/503/51
-40950-13/50-39/5-13
First
iteration
RHSS3S2S1X2X1BV
2400-4/5113/50S1
31501/503/51X1
1051-1/507/50S3
4095013/50-16/50Z

RHSS3S2S1X2X1BV
2400-4/5113/50S1
31501/503/51X1
755/7-1/7010X2
4095013/50-16/50Z
RHSS3S2S1X2X1BV
2400-4/5113/50S1
31501/503/51X1
1051-1/507/50S3
4095013/50-16/50Z
RHSS3S2S1X2X1BV
2400-4/5113/50S1
31501/503/51X1
1051-1/507/50X2
4095013/50-16/50Z
*5/7

RHSS3S2S1X2X1BV
2400-4/5113/50S1
31501/503/51X1
755/7-1/7010X2
4095013/50-16/50Z
19513/7-13/35013/50
453/7-3/3503/50
755/7-1/7010
-240-16/3516/350-16/50
RHSS3S2S1X2X1BV
45-13/7-15/35100S1
270-3/710/35001X1
755/7-1/7010X2
433516/3575/35000Z
755/7-1/7010
13/5
3/5
1
-16/5
Second
iteration

RHSS3S2S1X2X1BV
45-13/7-15/35100S1
270-3/710/35001X1
755/7-1/7010X2
433516/3575/35000Z
RHSBV
45S1
270X1
75X2
4335Z
270X1
75X2
4335Z

Max Z = 2X1 + 4X2
X1 + 4X2 ≥ 12
X1 + 3X2 ≥ 10
Z - 2X1 - 4X2 = 0
X1 + 4X2 + S1 = 12
X1 + 3X2 + S2 = 10
RHSS2S1X2X1BV
120141S1
101031S2
000-4--2Z
Example
Solution
Z -2X1 - 4X2 = 0
X1 + 4X2 ≥ 12
X1 + 3X2 ≥ 10

RHSS2S1X2X1BV
301/411/4X2
101031S2
000-4--2Z
301/411/4
1
3
-4
301/411/4
903/433/4
-120-1-4-1
❶
❷
❸
First
iteration
RHSS2S1X2X1BV
301/411/4X2
11-3/401/4S2
12010-1Z

RHSS2S1X2X1BV
301/411/4X2
44-301X1
12010-1Z
44-301
1/4
1
-1
11-3/401/4
44-301
-4-430-1
❶
❷
❸
Second
iteration
RHSS2S1X2X1BV
2-1110X2
44-301X1
164-200Z

RHSS2S1X2X1BV
2-1110S1
44-301X1
164-200Z
1
-3
-2
2-1110
2-1110
-63-3-30
-42-2-20
RHSS2S1X2X1BV
2-1110S1
101031X1
202020Z
❶
❷
❸
third
iteration

RHSS2S1X2X1BV
2-1110S1
101031X1
202020Z
RHSBV
2S1
10X1
20Z
10X1
0X2
20Z

Example :
Max 20X1 + 8X2
S.T.
9X1 + 4X2 ≤ 36
2X1 - X2 ≤ 2
X1 + X2 ≤ 8
16X1 ≤ 32
Z -20X1 - 8X2 =0
9X1 + 4X2 + S2 = 36
2X1 - X2 + S1 = 2
X1 + X2 + S3 = 8
16X1 + S4 = 32
Max 20X1 + 8X2
S.T.
9X1 + 4X2 ≤ 36
2X1 - X2 ≤ 2
X1 + X2 ≤ 8
16X1 ≤ 32
Solution :

X1 X2 S1 S2 S3 S4 RHS
X1 2 -1 1 0 0 0 2
S2 9 4 0 1 0 0 36
S3 1 1 0 0 1 0 8
S4 16 0 0 0 0 1 32
Z -20 -8 0 0 0 0 0
X1 1 -1/2 1/2 0 0 0 1
S2 9 4 0 1 0 0 36
S3 1 1 0 0 1 0 8
S4 16 0 0 0 0 1 32
Z -20 -8 0 0 0 0 0
First iteration
❶

1 -1/2 1/2 0 0 0 1
1
9
1
16
-20
1 -1/2 1/2 0 0 0 1
9 -9/2 9/2 0 0 0 9
1 -1/2 ½ 0 0 0 1
16 -8 8 0 0 0 16
-20 10 -10 0 0 0 -20
0 17/2 -9/2 1 0 0 27
0 3/2 -1/2 0 1 0 7
0 8 -8 0 0 1 16
0 -18 10 0 0 0 20
❷
❸

X1 1 -1/2 1/2 0 0 0 1
S2 0 17/2 -9/2 1 0 0 27
S3 0 3/2 -1/2 0 1 0 7
X2 0 1 -1 0 0 1/8 2
Z 0 -18 10 0 0 0 20
Second iteration
0 1 -1 0 0 1/8 2
-1/2
17/2
3/2
1
-18
0 -1/2 1/2 0 0 -1/16 -1
0 17/2 -17/2 0 0 17/16 17
0 3/2 -3/2 0 0 3/16 3
0 1 -1 0 0 1/8 2
0 -18 18 0 0 -18/8 -36
❶
❷

1 0 0 0 0 1/16 2
0 0 4 1 1 -17/16 10
0 0 1 0 0 -3/16 4
0 1 -1 0 0 1/8 2
0 0 -8 0 0 18/8 56
X1 1 0 0 0 0 1/16 2
S1 0 0 1 1/4 1/4 -17/64 10/4
S3 0 0 1 0 0 -3/16 4
X2 0 1 -1 0 0 1/8 2
Z 0 0 -8 0 0 18/8 56
Third iteration
❶
❸

0
1
1
-1
-8
0 0 1 1/4 1/4 -17/64 10/4
0 0 O 0 0 0 0
0 0 1 1/4 1/4 -17/64 10/4
0 0 1 1/4 1/4 -17/64 10/4
0 0 -1 -1/4 -1/4 17/64 -10/4
0 0 -8 -2 -2 17/8 -20
1 0 0 0 0 1/16 2
0 0 1 1/4 1/4 -17/64 10/4
0 0 0 -1/4 -2/9 5/64 6/4
0 1 0 ¼ 2/9 -9/64 18/4
0 0 0 2 0 1/8 76
❷

X1 1 0 0 0 0 1/16 2
S1 0 0 1 1/4 1/4 -17/64 10/4
S3 0 0 0 -1/4 -2/9 5/64 6/4
X2 0 1 0 ¼ 2/9 -9/64 18/4
Z 0 0 0 2 0 1/8 76
X1 2
X2 18/4
Z 76
❸

Example :
Solution :
Min Z = X1 - 3X2 + 3X3
S.T.
3X1 - X2 + 2X3 ≤ 7
2X1 + 4X2 ≥ -12
4X1 - 3X2 - 8X3 ≥ -10
X1, X2, X3 ≥ 0
Min Z = X1 - 3X2 + 3X3
S.T.
3X1 - X2 + 2X3 ≤ 7
-2X1 + -4X2 ≤ 12
-4X1 + 3X2 + 8X3 ≤ 10
X1, X2, X3 ≥ 0

RHSS3S2S1X3X2X1BV
70012-13S1
120100-4-2S2
1010083-4S3
0000-33-1Z
3X1 - X2 + 2X3 + S1 = 7
-2X1 - 4X2 + S2 = 12
-4X1 + 3X2 + 8X3 + S3 = 10
Z - X1 + 3X2 - 3X3 = 0
/3
RHSS3S2S1X3X2X1BV
70012-13S1
120100-4-2S2
1010083-4S3
0000-33-1Z
10/31/3008/31-4/3X2

RHSS3S2S1X3X2X1BV
70012-13S1
120100-4-2S2
10/31/3008/31-4/3X2
0000-33-1Z
10/31/3008/31-4/3
-1
-4
1
3
-10/3-1/300-8/3-14/3
-40/3-4/30032/3-416/3
10/31/3008/31-4/3
1010083-4
❶
❷
RHSS3S2S1X3X2X1BV
31/31/30114/305/3S1
61/34/310-32/30-22/3S2
10/31/3008/31-4/3X2
-10-100-1103Z
❸
First
iteration

RHSS3S2S1X3X2X1BV
31/51/503/514/501X1
61/34/310-32/30-22/3S2
10/31/3008/31-4/3X2
-10-100-1103Z
31/51/503/514/501
1
-22/3
-4/3
3
31/51/503/514/501
-682/15-22/150-66/15-305/150-22/3
-124/15-4/150-4/5-56/150-4/3
93/53/509/542/503
RHSS3S2S1X3X2X1BV
31/51/503/514/501X1
987/1542/15166/15000S2
58/519/1504/596/1510X2
-143/5-8/50-9/5-97/500Z
❶
❷
❸
Second
iteration

RHSS3S2S1X3X2X1BV
31/51/503/514/501X1
987/1542/15166/15000S2
58/519/1504/596/1510X2
-143/5-8/50-9/5-97/500Z
The optimal solution will be :
X1 = 31/5
X2 = 58/5
X3 = 0
Z = -143/5

Example :
Solution :
Let P = -Z = -X1 + 3X2 - 3X3
Min Z = Min –P
Min Z = - Max P
Max P = -X1 + 3X2 - 3X3
S.T 3X1 - X2 + 2X3 ≤ 7,
2X1 + 4X2 ≥ -12,
-4X1 + 3X2 + 8X3 ≤ 10
X1, X2, X3 ≥ 0
Min Z = X1 - 3X2 + 3X3
S.T.
3X1 - X2 + 2X3 ≤ 7
2X1 + 4X2 ≥ -12
4X1 - 3X2 - 8X3 ≥ -10
X1, X2, X3 ≥ 0

RHSS3S2S1X3X2X1BV
70012-13S1
120100-4-2S2
1010083-4S3
00003-31P
3X1 - X2 + 2X3 + S1 = 7
-2X1 - 4X2 + S2 = 12
-4X1 + 3X2 + 8X3 + S3 = 10
P X1 - 3X2 + 3X3 = 0
RHSS3S2S1X3X2X1BV
70012-13S1
0100-4-2S2
1010083-4S3
00003-31P
RHSS3S2S1X3X2X1BV
70012-13S1
120100-4-2S2
10/31/3008/31-4/3S3
00003-31P
/3

RHSS3S2S1X3X2X1BV
70012-13S1
120100-4-2S2
10/31/3008/31-4/3X2
00003-31P
-1
-4
1
-3
10/31/3008/31-4/3
-10/3-1/300-8/3-14/3
-40/3-4/300-32/3-416/3
10/31/3008/31-4/3
-10-100-8-34
RHSS3S2S1X3X2X1BV
31/31/30114/305/3S1
76/34/31032/30-22/3S2
10/31/3008/31-4/3X2
10100110-3P
❶
❷
❸
First
iteration

RHSS3S2S1X3X2X1BV
31/51/503/514/501X1
76/34/31032/30-22/3S2
10/31/3008/31-4/3X2
10100110-3P
1
-22/3
-4/3
-3
31/51/503/514/501
31/51/503/514/501
-682/15-22/150-66/15-308/150-22/3
-124/15-4/150-4/5-56/150-4/3
-93/5-3/50-9/5-42/50-3
RHSS3S2S1X3X2X1BV
31/51/503/514/501X1
324/542/15166/15468/1500S2
58/59/1504/596/1510X2
143/58/509/597/500P
❶
❷
❸
Second
iteration

The optimal solution will be :
X1 = 31/5
X2 = 58/5
X3 = 0
P = 143/5
Z = -P = -143/5
RHSS3S2S1X3X2X1BV
31/51/503/514/501S1
324/542/15166/15468/1500S2
58/59/1504/596/1510X2
143/58/509/597/500P

The Dual method
If standard form
The Big-M method
If not standard form

The Dual method
 In solving the linear programming problem, it helps to note that
each maximization problem is associated with a minimization
problem, and vice versa.
 The given problem is called the primal problem, and the related
problem is called the dual problem.
 The solution of maximizing (primal) problem produces the
solution of associated minimizing (dual) problem, and vice-
versa.

The Duality Theorem
 A primal problem has a solution if and only if the corresponding dual
problem has a solution.
 Furthermore, if a solution exists, then:
 The objective functions of both the primal and the dual problem attain
the same optimal value.
 When the dual optimal case reached, the optimal solution to the
primal problem appears under the slack variables in the last row of
the final simplex tableau associated with the dual problem.

The Dual Problem
Primalproblem Dualproblem
M variables N variables
N constraints M constraints
Coefficients of objective function Constants ( RHS)
Constants ( RHS) Coefficients of objective function

transpose
Min P = 8y1 + 16y2
St: y1 + 5y2  9
2y1 + 2y2  10
y1  0, y2  0.
Max Z = 9x1 + 10 x2
St: x1 + 2x2  8
5x1 + 2x2  16
x1  0 , x2  0

Min Z = 6X1 + 8X2
St. 40X1 + 10X2 ≥ 2400
10X1 + 15X2 ≥ 2100
5X1 + 15X2 ≥ 1500
X1 X2 Constant
40 10 ≥ 2400
10 15 ≥ 2100
5 15 ≥ 1500
6 8
S1 S2 S3 Constant
40 10 5 ≤ 6
10 15 15 ≤ 8
2400 2100 1500
Example
Solution

The dual problem (Max) associated with this problem:
Max Z = 2400S1 + 2100S2 + 1500S3
St. 40S1 + 10S2 + 5S3 ≤ 6
10S1 + 15S2 + 15S3 ≤ 8
S1 S2 S3
Constant
40 10 5 ≤ 6
10 15 15 ≤ 8
2400 2100 1500
Dual

The max. standard equations will be
P - 2400S1 - 2100S2 - 1500S3 = 0
40S1 + 10S2 + 5S3 + X1 = 6
10S1 + 15S2 + 15S3 + X2 = 8
RHSPX2X1S3S2S1
BV
600151040X1
8010151510X2
0100-1500-2100-2400P
RHSPX2X1S3S2S1
BV
600151040X1
8010151510X2
0100-1500-2100-2400P
3/20001/401/81/41S1 /40

RHSPX2X1S3S2S1BV
3/20001/401/81/41S1
8010151510X2
0100-1500-2100-2400P
3/20001/401/81/41
1
10
-2400
RHSPX2X1S3S2S1BV
3/20001/401/81/41S1
3/2001/45/45/210X2
-36000-60-300-600-2400P
RHSPX2X1S3S2S1BV
3/20001/401/81/41S1
13/201-1/425/425/20X2
3601060-1200-15000P
First
iteration
❶
❷
❸

RHSPX2X1S3S2S1BV
3/20001/401/81/41S1
13/201-1/425/425/20X2
3601060-1200-15000P
RHSPX2X1S3S2S1BV
3/20001/401/81/41S1
13/2502/25-1/501/210S2
3601060-1200-15000P

RHSPX2X1S3S2S1BV
3/20001/401/81/41S1
13/2502/25-1/501/210S2
3601060-1200-15000P
13/2502/25-1/501/210
1/4
1
-1500
RHSPX2X1S3S2S1BV
13/10001/50-1/1001/81/40S1
13/2502/25-1/501/210S2
-7800-12030-750-15000P
RHSPX2X1S3S2S1BV
-8/1000-1/5014/400001S1
13/2502/25-1/501/210S2
1140112030-45000P
Second
iteration
❶
❷
❸

Write the dual problem (Max) associated with this problem:
Min Z = 6X1 + 8X2
St. 40X1 + 10X2 ≥ 2400
10X1 + 15X2 ≥ 2100
5X1 + 15X2 ≥ 1500
Primal
X1 = 30
X2 = 120
Z = 1140
RHSPX2X1S3S2S1BV
-8/1000-1/5014/400001S1
13/2502/25-1/501/210S2
1140112030-45000P

Min Z = 150X1 + 80X2 + 50X3 + 8X4
St. 12X1 + 8X2 + 2X3 + X4 ≥ 25
10X1 + 4X2 + 4X3 + 2X4 ≥ 20
X1,X2 ,X3,X4 ≥ 0
S1 S2 Constant
12 10 ≤ 150
8 4 ≤ 80
2 4 ≤ 50
1 2 ≤ 8
25 20
X1 X2 X3 X4 Constant
12 8 2 1 ≥ 25
10 4 4 2 ≥ 20
150 80 50 8
Primal
Example
Solution

S1 S2 Constant
12 10 ≤ 150
8 4 ≤ 80
2 4 ≤ 50
1 2 ≤ 8
25 20
Max P = 25S1 + 20S2
St. 12S1 + 10S2 ≥ 150
8S1 + 4S2 ≥ 80
2S1 + 4S2 ≥ 50
S1 + 2S2 ≥ 8
Dual

Max P = 25S1 + 20S2
St. 12S1 + 10S2 ≥ 150
8S1 + 4S2 ≥ 80
2S1 + 4S2 ≥ 50
S1 + 2S2 ≥ 8
P - 25S1 + 20S2 = 0
12S1 + 10S2 + X1 = 150
8S1 + 4S2 + X2 = 80
2S1 + 4S2 + X3 = 50
S1 + 2S2 + X4 = 8
Dual

P - 25S1 - 20S2 = 0
12S1 + 10S2 + X1 = 150
8S1 + 4S2 + X2 = 80
2S1 + 4S2 + X3 = 50
S1 + 2S2 + X4 = 8
RHSX4X3X2X1S1S1
15000011012X1
80001048X2
50010042X3
8100021X4
00000-20-25P

RHSX4X3X2X1S1S1
15000011012X1
80001048X2
50010042X3
8100021X4
00000-20-25P
RHSX4X3X2X1S1S1
96120002412X1
648000168X2
16200042X3
8100021S1
-200-25000-50-25P
8100021
12
8
2
1
-25
First
iteration
❶
❷
❸

RHSX4X3X2X1
S1S1
54-12001-140X1
16-8010120X2
34-210000X3
8100021S1
20025000300P
❸

RHSX4X3X2X1S1S1
54-12001-140X1
16-8010120X2
34-210000X3
8100021S1
20025000300P
Min Z = 150X1 + 80X2 +50X3+ 8X4
St. 12X1 + 8X2 + 2X3 + X4 ≥ 25
10X1 + 4X2 + 4X3 + 2X4 ≥ 20
Z=200
X1 =0
X2 =0
X3 =0
X4 =25
Primal

Max P= 5X1 + 12X2 + 4X3
S.T. X1 +2X2+X3 ≤ 10
2X1 - X2 + 3X3 ≤ 8
X1 , X2 , X3 ≥ 0
Min Z= 10Y1 + 8Y2
S.T. Y1 +2Y2  5
2Y1 - Y2  12
Y1 + 3Y2  4
Y1 , Y2 ≥ 0
Example :
Solution :

RHSY2Y1X3X2X1BV
1001121Y1
8103-12Y2
000-4-12-5P
Max P= 5X1 + 12X2 + 4X3
S.T. X1 +2X2+X3 ≤ 10
2X1 - X2 + 3X3 ≤ 8
X1 , X2 , X3 ≥ 0
P - 5X1 - 12X2 - 4X3 = 0
X1 + 2X2 + X3 + Y1 = 10
2X1 - X2 + 3X3 + Y2 = 8
X2
2
-1
-12
1001121X2 501/21/221/2X2

1001121
1
-1
-12
501/21/211/2
501/21/211/2
-50-1/2-1/2-1-1/2
-600-6-6-12-6
501/21/211/2
1311/27/205/2
6006201
RHSY2Y1X3X2X1BV
501/21/211/2Y1
8103-12Y2
000-4-12-5P
❶
❷
❸

RHSY2Y1X3X2X1BV
501/21/211/2X2
1311/27/205/2Y2
6006201P
Min Z = 10Y1 + 8Y2
S.T. Y1 + 2Y2  5
2Y1 - Y2  12
Y1 + 3Y2  4
Y1 , Y2 ≥ 0
Y1 = 6
Y2 = 0
P = 60

Y1 Y2 Y3 X1 X2 RHS
X1 2 1 3 1 0 8
Y2 1/2 1 1/2 0 1/2 7/2
Z -6 -12 0 0 0 0
1/2 1 1/2 0 1/2 7/2
1
1
-12
1/2 1 1/2 0 1/2 7/2
1/2 1 1/2 0 1/2 7/2
-6 -12 -6 0 -6 -42
3/2 0 5/2 1 -1/2 9/2
1/2 1 1/2 0 1/2 7/2
0 0 6 0 6 42
Y1 Y2 Y3 X1 X2 RHS
X1 3/2 0 5/2 1 -1/2 9/2
Y2 1/2 1 1/2 0 1/2 7/2
Z 0 0 6 0 6 42
X1 X2
0 6
First iteration

Solving the non standard LP model
The Method
Suppose that you add n artificial variables A1 , A2 , …. , An to the LP model
Build the initial tableau representing the LP model with rows representing the
constraints’ equations and last row representing the modified objective function
Ž where :
Model modification
Add M[A1 + A2 + …. + An]to the objective function Eqn. if the model is Min.
Subtract M[A1 + A2 + …. + An] from the objective function Eqn. if the model is Max.
Ž row = Z row + M[A1row + A2row + …. + Anrow] if the model is Min.
Ž row = Z row - M[A1row + A2row + …. + Anrow] if the model is Max.

Solve the following linear programming problem by using the
simplex method:
Min Z = 12 X1 + 20 X2
S.t.
6X1 + 8X2  100
7X1 + 12X2  120
X1, X2  0
The Method
Artificial Variable Technique
Example :

6X1 + 8X2 - S1 + A1 = 100
7X1 + 12X2 - S2 + A2 = 120
Z - 12 X1 - 20 X2 -M(A1 + A2) = 0
Solution:
RHSA2A1X2X1Basic Variable
1000186A1
12010127A2
0-M-M-20-12Z

1000186A1
12010127A2
0-M-M-20-12Z
100M0M8M6MA1
120MM012M7MA2
220MMM20M13MA1 + A2
220M00-20 +20M-12+13MNew Z = old Z + A1 + A2
1000186A1
12010127A2
220M00-20 +20M-12+13MZ

1000186A1
12010127X2
220M00-20 +20M-12 +13MZ
1000186A1
101/12017/12X2
220M00-20 +20M-12 +13MZ
/12

RHSA1X2X1Basic Variable
100186A1
10017/12X2
220M020(M-1)-12 +13MZ
10017/12
8
1
20(M-1)
800856/12
10017/12
200(M-1)020(M-1)35/3(M-1)
20104/3A1
10017/12X2
20M+200001/3(4M-1)Z
❶
❷
❸
First
iteration

20104/3A1
10017/12X2
20M+200001/3(4M-1)Z
RHSX2X1Basic Variable
1501X1
1017/12X2
20M+20001/3(4M-1)Z
*3/4

1501X1
1017/12X2
20M+20001/3(4M-1)Z
1501
1
7/12
1/3(4M-1)
1501
35/407/12
5(4M-1)01/3(4M-1)
1501X1
5/410X2
20500Z
❶
❷
❸
Second
iteration

1501X1
5/410X2
20500Z
This solution is optimal, since there is no positive
value in the last row. The optimal solution is:
X1 = 15, X2 = 5/4 Z = 205

MAX Z = 3X1 – X2
S.T.
2X1 + X2 ≤ 2,
X1 + 3X2 ≥ 3,
X1 ≤ 4
X1, X2 , ≥ 0
Example :
S.T.
2X1 + X2 + S1 = 2
X1 + 3X2 – S2 + A1 = 3
X1 + S3 = 4
MAX Z = 3X1 – X2 - MA1
Solution :

S.T.
2X1 + X2 + S1 = 2
X1 + 3X2 – S2 + A1 = 3
X1 + S3 = 4
MAX Z - 3X1 + X2 + MA1 = 0
RHSA1S3S1X2X1BV
200112S1
310031A1
401001S3
0+M001-3Z

RHSA1S3S1X2X1BV
200112S1
310031A1
401001S3
-3M0001-3M-3-MZ
RHSA1S3S1X2X1BV
200112S1
11/30011/3A1
401001S3
-3M0001-3M-3-MZ
/3

RHSS3S1X2X1BV
20112S1
10011/3X2
41001S3
-3M001-3M-3-MZ
10011/3
1
1
0
1-3M
10011/3
10011/3
00000
1-3M001-3M1/3-M
RHSS3S1X2X1BV
10105/3S1
10011/3X2
41001S3
-1000-10/3Z
❶
❷
❸
First
iteration

RHSS3S1X2X1BV
10105/3S1
10011/3X2
41001S3
-1000-10/3Z
RHSS3X2X1BV
3/5001X1
1011/3X2
4101S3
-100-10/3Z
*3/5

RHSS3X2X1BV
3/5001X1
1011/3X2
4101S3
-100-10/3Z
3/5001
1
1/3
1
-10/3
3/5001
1/5001/3
3/5001
-200-10/3
RHSS3X2X1BV
3/5001X1
4/5010X2
17/5100S3
1000Z
❶
❷
❸
Second
iteration

RHSS3X2X1BV
3/5001X1
4/5010X2
17/5100S3
1000Z
The optimal solution is
X1 = 3/5
X2 = 4/5
Z = 3 X1 – X2 = 1

Min Z = 2 X1 + 3 X2
S.t.
½ X1 + ¼ X2 ≤ 4
X1 + 3X2  20
X1 + X2 = 10
X1, X2  0
Example :
½ X1 + ¼ X2 + S1 = 4
X1 + 3X2 - S2 + A1 = 20
X1 + X2 + A2 = 10
Z – 2 X1 – 3 X2 - MA1 - MA2 = 0
Solution:

Basic variables X1 X2 S1 S2 A1 A2 RHS
S1 1/2 1/4 1 0 0 0 4
A1 1 3 0 -1 1 0 20
A2 1 1 0 0 0 1 10
Z -2 -3 0 0 -M -M 0

Basic variables X1 X2 S1 S2 A1 A2 RHS
S1 1/2 1/4 1 0 0 0 4
A1 1 3 0 -1 1 0 20
A2 1 1 0 0 0 1 10
Z -2 +2M -3 + 4M 0 -M 0 0 30M
Basic variables X1 X2 S1 A2 RHS
S1 1/2 1/4 1 0 4
X2 1/3 1 0 0 20/3
A2 1 1 0 1 10
Z -2 +2M -3 + 4M 0 0 30M
/3

S1 1/2 1/4 1 0 4
X2 1/3 1 0 0 20/3
A2 1 1 0 1 10
Z -2 +2M -3 + 4M 0 0 30M
1/3 1 0 0 20/3
1/4
1
1
-3 + 4M
1/12 1/4 0 0 5/3
1/3 1 0 0 20/3
1/3 1 0 0 20/3
4/3M-1 4M-3 0 0 80/3M-20
S1 5/12 0 1 0 7/3
X2 1/3 1 0 0 20/3
A2 2/3 0 0 1 10/3
Z 2/3M-1 0 0 0 10/3M+20
❶
❷
❸
First
iteration

S1 5/12 0 1 0 7/3
X2 1/3 1 0 0 20/3
A2 2/3 0 0 1 10/3
Z 2/3M-1 0 0 0 10/3M+20
Basic variables X1 X2 S1 RHS
S1 5/12 0 1 7/3
X2 1/3 1 0 20/3
X1 1 0 0 5
Z 2/3M-1 0 0 10/3M+20
*3/2

S1 5/12 0 1 7/3
X2 1/3 1 0 20/3
X1 1 0 0 5
Z 2/3M-1 0 0 10/3M+20
1 0 0 5
5/12
1/3
1
2/3M-1
5/12 0 0 25/12
1/3 0 0 5/3
1 0 0 5
2/3M-1 0 0 10/3M-5
S1 0 0 1 1/4
X2 0 1 0 5
X1 1 0 0 5
Z 0 0 0 25
❶
❷
❸
Second
iteration

X1 = 5
X2 = 5
Z = 25
Basic
variables
X1 X2 S1 RHS
S1 0 0 1 1/4
X2 0 1 0 5
X1 1 0 0 5
Z 0 0 0 25

Min Z = 600X1 + 500X2
ST.
2X1 + X2 ≥ 80
X1 +2X2 ≥ 60
X1, X2 ≥ 0
2X1 + X2 - S1 + A1 = 80
X1 + 2X2 - S2 + A2 = 60
Z - 600X1 - 500X2 - M A1 - M A2 = 0
Example :
Solution:
Basic variables X1 X2 A1 A2 RHS
A1 2 1 1 0 80
A2 1 2 0 1 60
Z -600 -500 -M -M 0

Basic
variables
X1 X2 A1 RHS
A1 2 1 1 80
X2 1 2 0 60
Z -600+3M -500+3M 0 140M
Basic
variables
X1 X2 A1 RHS
A1 2 1 1 80
X2 1/2 1 0 30
Z -600+3M -500+3M 0 140M

Basic variables X1 X2 A1 RHS
A1 2 1 1 80
X2 1/2 1 0 30
Z -600+3M -500+3M 0 140M
1/2 1 0 30
1
1
-500+3M
1/2 1 0 30
1/2 1 0 30
3/2M-250 -500+3M 0 90M-15000
Basic variables X1 X2 A1 RHS
A1 3/2 0 1 50
X2 1/2 1 0 30
Z 3/2M-350 0 0 50M+15000
❶
❷
❸
First
iteration

Basic
variables
X1 X2 A1 RHS
A1 3/2 0 1 50
X2 1/2 1 0 30
New Z 3/2M-350 0 0 50M+15000
Basic
variables
X1 X2 RHS
X1 1 0 100/3
X2 1/2 1 30
New Z 3/2M-350 0 50M+15000
*2/3

Basic
variables
X1 X2 RHS
X1 1 0 100/3
X2 1/2 1 30
New Z 3/2M-350 0 50M+15000
1
1/2
3/2M-350
1 0 100/3
1 0 100/3
1/2 0 50/3
3/2M-350 0 50M-35000/3
Basic variables X1 X2 RHS
X1 1 0 100/3
X2 0 1 40/3
New Z 0 0 80000/3
❶
❷
❸
Second
iteration

X1 = 100/3
X2 = 40/3
Z = 80000/3
Basic
variables
X1 X2 RHS
X1 1 0 100/3
X2 0 1 40/3
New Z 0 0 80000/3

Min Z = 4X1 +X2
S.T:
3X1 +X2 = 3
4X1 +3X2 ≥ 6
X1 +2X2 ≤ 3
X1, X2 ≥ 0.
Min Z = 4X1 +X2 + M(A1 + A2)
S.T:
3X1 +X2 + A1= 3
4X1 +3X2 - S1 + A2 = 6
X1 +2X2 + S2 = 3
Example :

Z - 4X1 - X2 - M(A1 + A2) = 0
3X1 +X2 + A1= 3
4X1 +3X2 - S1 + A2 = 6
X1 +2X2 + S2 = 3
B.V. X1 X2 S1 S2 A1 A2 RHS
A1 3 1 0 0 1 0 3
A2 4 3 -1 0 0 1 6
S2 1 2 0 1 0 0 3
Z -4 -1 0 0 -M -M 0

B.V. X1 X2 S1 S2 A1 A2 RHS
A1 3 1 0 0 1 0 3
A2 4 3 -1 0 0 1 6
S2 1 2 0 1 0 0 3
Z -4 -1 0 0 -M -M 0
B.V. X1 X2 S2 A1 A2 RHS
A1 3 1 0 1 0 3
A2 4 3 0 0 1 6
S2 1 2 1 0 0 3
New Z -4+7M -1+4M 0 0 0 9M

A1 3 1 0 1 0 3
A2 4 3 0 0 1 6
S2 1 2 1 0 0 3
Z -4+7M -1+4M 0 0 0 9M
A1 1 1/3 0 1/3 0 1
A2 4 3 0 0 1 6
S2 1 2 1 0 0 3
Z -4+7M -1+4M 0 0 0 9M

B.V. X1 X2 S2 A2 RHS
X1 1 1/3 0 0 1
A2 4 3 0 1 6
S2 1 2 1 0 3
Z -4+7M -1+4M 0 0 9M
1 1/3 0 0 1
1
4
1
-4+7M
1 1/3 0 0 1
4 4/3 0 0 4
1 1/3 0 0 1
-4+7M -4+7M/3 0 0 -4+7M
❶
❷
❸
X1 1 1/3 0 0 1
A2 0 5/3 0 1 2
S2 0 5/3 1 0 2
Z 0 1+5M/3 0 0 4+2M
First
iteration

X1 1 1/3 0 0 1
A2 0 5/3 0 1 2
S2 0 5/3 1 0 2
Z 0 1+5M/3 0 0 4+2M
B.V. X1 X2 S2 RHS
X1 1 1/3 0 1
X2 0 1 0 6/5
S2 0 5/3 1 2
Z 0 1+5M/3 0 4+2M

B.V. X1 X2 S2 RHS
X1 1 1/3 0 1
X2 0 1 0 6/5
S2 0 5/3 1 2
Z 0 1+5M/3 0 4+2M
1/3
1
5/3
1+5M/3
0 1 0 6/5
0 1/3 0 2/5
0 1 0 6/5
0 5/3 0 2
0 1+5M/3 0 2/5 + 2M
B.V. X1 X2 S2 RHS
X1 1 0 0 3/5
X2 0 1 0 6/5
S2 0 0 1 0
Z 0 0 0 18/5
❶
❷
❸
Second
iteration

B.V. X1 X2 S2 RHS
X1 1 0 0 3/5
X2 0 1 0 6/5
S2 0 0 1 0
Z 0 0 0 18/5
X1 3/5
X2 6/5
Z 18/5

X1 1 1/3 0 0 1
A2 0 5/3 0 1 2
S2 0 5/3 1 0 2
New Z 0 1+5M/3 0 0 4+2M
B.V. X1 X2 A2 RHS
X1 1 1/3 0 1
A2 0 5/3 1 2
X2 0 1 0 6/5
New Z 0 1+5M/3 0 4+2M
Other Solution

B.V. X1 X2 A2 RHS
X1 1 1/3 0 1
A2 0 5/3 1 2
X2 0 1 0 6/5
New Z 0 1+5M/3 0 4+2M
0 1 0 6/5
1/3
5/3
1
1+5M/3
0 1/3 0 2/5
0 5/3 0 2
0 1 0 6/5
0 1+5M/3 0 2/5 + 2M
B.V. X1 X2 A2 RHS
X1 1 0 0 3/5
A2 0 0 1 0
X2 0 1 0 6/5
New Z 0 0 0 18/5
❶
❷
❸
Second
iteration

B.V. X1 X2 A2 RHS
X1 1 0 0 3/5
A2 0 0 1 0
X2 0 1 0 6/5
Z 0 0 0 18/5
X1 3/5
X2 6/5
Z 18/5

Min Z = 3X1 + 4X2
ST.
2X1 + X2 ≤ 16
2X1 + 6X2  40
X1 + X2 = 10
X1, X2  0
Z - 3X1 - 4X2 - MA1 - MA2 = 0
2X1 + X2 + S1 = 16
2X1 + 6X2 - S2 + A1 = 40
X1 + X2 + A2 = 10
RHSA2A1S1X2X1BV
1600112S1
4001062A1
1010011A2
0-M-M0-4-3Z
Example :

RHSA2S1X2X1BV
160112S1
400062X2
101011A2
50M007M - 43M - 3New Z
RHSA2S1X2X1BV
160112S1
40/60011/3X2
101011A2
50M007M - 43M - 3New Z
/6

RHSA2S1X2X1BV
160112S1
40/60011/3X2
101011A2
50M007M - 43M - 3Z
40/60011/3
1
1
1
7M - 4
40/60011/3
40/60011/3
40/60011/3
40/6(7M-4)007M - 41/3(7M – 4)
RHSA2S1X2X1BV
28/30105/3S1
40/60011/3X2
10/31002/3A2
10M/3 + 80/30002M/3-5/3Z
First
iteration
❶
❷
❸

RHSS1X2X1BV
28/3105/3S1
40/6011/3X2
10/3002/3X1
10M/3 + 80/3002M/3-5/3New Z
RHSS1X2X1BV
28/3105/3S1
40/6011/3X2
5001X1
10M/3 + 80/3002M/3-5/3New Z
*3/2

RHSS1X2X1BV
28/3105/3S1
40/6011/3X2
5001X1
10M/3 + 80/3002M/3-5/3Z
5/3
1/3
1
2M/3-5/3
5001
25/3005/3
5/3001/3
5001
10M/3 - 25/3002M/3-5/3
RHSS1X2X1BV
1100S1
5010X2
5001X1
35000Z
Second
iteration
❶
❷
❸

RHSS1X2BV
110S1
501X2
500X1
3500New Z
X1 = 5
X2 = 5
Z = 35

Linear programming simplex method

Linear programming simplex method

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Linear programming simplex method