1
CHEM-E7155 Production planning and
control
Exercise 1
Simplex method
2
Problem 1.
Minimize the function f(x) = - x1 + x2 under following
restrictions:
2x1 - x2 ³ - 2
-x1 + 3x2 ³ - 2
-x1 - x2 ³ - 4
x1, x 2 ³ 0
Use the Simplex-algorithm.
3
Problem 1 - Solution
• Step 1. Problem is transformed:
– the goal function is multiplied by (–1). f*(x)=x1-x2 is to
be maximized then.
– The constraints are also multiplied by (-1). The direction
of the inequality sign changes:
• -2x1 + x2 £ 2
x1 - 3x2 £ 2
x1 + x2 £ 4
x1, x2 ³ 0
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Problem 1 - Solution
• Step 2.
• The constraints are changed to equality constraints
by adding the slack variables x3, x4 and x5.
-2x1 + x2 + x3 = 2
x1 - 3x2 + x4 = 2
x1 + x2 + x5 = 4
x1, x 2, x 3, x 4, x 5 ³ 0
The goal function is changed to f* - x1 + x2 = 0
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Problem 1 - Solution
• Step 3. A simplex tableau is derived:
basic
line
variables
Right
f* x1 x2 x3 x4 x5
side
(0) f*
1 -1 1 0 0 0 0
(1) x3
0 -2 1 1 0 0 2
(2) x4
0 1 -3 0 1 0 2
(3) x5
0 1 1 0 0 1 4
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Problem 1 - Solution
• Step 4.
• Optimality test: Are all the coefficients in the line
goal function (line 0) >=0 ? If not, the optimum is
not yet reached.
• Pivot column is the column where the goal function
has the most negative coefficient. x1 seems to be it
à It will be the next basic variable. Question is,
which basic variable will be swapped with it?
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Problem 1 - Solution
• The search for the pivot row: Every element
in the right side column will be divided by the
corresponding pivot column element (if the
element is >0). The lowest value will be 2/1
= 2.
• The pivot row is the second row. The pivot
element value is 1 and x4 will be swapped
with x1. First the operations leading to this
must be accomplished.
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Problem 1 - Solution
Pivot row is (2), column is (x1) and the pivot value is 1.
Row Basic
f*
variables
x1
x2
x3
x4
x5
Right
side
(0)
f*
1
-1
1
0
0
0
0
*1
(1)
x3
0
-2
1
1
0
0
2
*2
(2)
x4
0
1
-3
0
1
0
2
*1
(3)
x5
0
1
1
0
0
1
4
*(-1)
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Problem 1 - Solution
• Step 5.
• The elements in the pivot column are desired to be
zeros, except the pivot element which is desired to
be 1. This is done by Gaussian elimination
operations. The pivot row is first multiplied by 2
and it is added to line (1). After that pivot row is
mulplied by (-1) and it is added to row (3). Finally
the pivot row is added to the goal row.
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Problem 1 - Solution
Row
Basic
f*
variables
x1
x2
x3
x4
x5
Right
side
(0)
f*
1
0
-2
0
1
0
2
*1/2
(1)
x3
0
0
-5
1
2
0
6
*5/4
(2)
x1
0
1
-3
0
1
0
2
*3/4
(3)
x5
0
0
4
0
-1
1
2
*1/4
• Optimality check: In the row (0) every
variable isn’t positive. Iteration continues.
New pivot column is x2. Pivot row is the third
row since 2/4=2
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Problem 1 - Solution
Row
Basic
f*
variables
x1
x2
x3
x4
x5
Right
side
(0)
f*
1
0
0
0
1/2
1/2
3
(1)
x3
0
0
0
1
3/4
5/4
17/2
(2)
x1
0
1
0
0
1/4
3/4
7/2
(3)
x2
0
0
1
0
-1/4
1/4
1/2
• Optimality check: In the row (0) every
variable is positive or zero; the optimum is
reached. Optimum is the right side of the
goal row (0): f*=3. The variables which are
not basic variables are zeros.
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Problem 1 - Solution
• Basic variables are x1 = 3.5, x2 = 0.5 and x3 =
8.5
• Because the minimization was turned to
maximization by multiplying the goal
function by (-1) the optimum is also
multiplied by (-1). F* = -3
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Problem 2
• Maximize f = x1 - 7x2 + 3x3 under following
restrictions:
• 2x1 + x2 - x3 £ 4 (resource 1)
4x1 - 3x2 £ 2 (resource 2)
-3x1 + 2x2 + x3 £ 3 (resource 3)
x1 , x 2 , x 3 ³ 0
• Use Simplex method. Define the shadow prices for
every resource and describe their importance
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Problem 2 – The Solution
•
Slack variables x4, x5 and x6 are introduced to the constraints:
2x1 + x2 – x3 +x4 = 4
4x1 – 3x2 +x5 = 2
-3x1 + 2x2 + x3 +x6 = 3
Row
Basic
f
variables
x1
x2
x3
x4
x5
x6
Right
side
(0)
f
1
-1
7
-3
0
0
0
0
*3
(1)
x4
0
2
1
-1
1
0
0
4
*1
(2)
x5
0
4
-3
0
0
1
0
2
*0
(3)
x6
0
-3
2
1
0
0
1
3
*1
• Pivot column is x3, à Gaussian elim. à x6 is swapped
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Problem 2 – The Solution
• The next iteration is as follows (Pivot column
is x1. The pivot row is the second row. x5 is
swapped with x1):
Row
Basic
f
variables
x1
x2
x3
x4
x5
x6
Right
sides
(0)
f
1
-10
13
0
0
0
3
9
*5/2
(1)
x4
0
-1
3
0
1
0
1
7
*1/4
(2)
x5
0
-3
0
0
1
0
2
*1/4
(3)
x3
0
4
-3
2
1
0
0
1
3
*3/4
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Problem 2 – The Solution
• Table after iteration:
Row
Basic
variables
f
x1
x2
x3
x4
x5
x6
Right
sides
(0)
f
1
0
11/2
0
0
5/2
3
14
(1)
x4
0
0
9/4
0
1
1/4
1
15/2
(2)
x1
0
1
-3/4
0
0
1/4
0
1/2
(3)
x3
0
0
-1/4
1
0
3/4
1
9/2
• Now optimum is reached since all the goal
function’s coefficients are positive!
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Problem 2 – The Solution
• The basic variables are x1 = 0.5, x3 = 4.5, x4 = 15/2
• Non-basic variables are all zero
• Coefficients in the goal function’s row are the
shadow prices
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Problem 3
MAX
+ +2
Subject to constraints:
•
+2
20
2 +4 +2
60
• 2 +3 +
50
• , ,
0
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Problem 4
MAX = 2 + 4 + 3
Subject to constraints:
• 3 +4 +2
60
• 2 + +2
40
•
+3 +2
80
• , ,
0