Post on 25-Jun-2020
THE SIMPLEX METHOD
INDR 262-INTRODUCTION TO OPTIMIZATION METHODSMetin Türkay
Department of Industrial EngineeringKoç University, Istanbul
WYNDOR GLASS CO
x1
x2x1 ≤4
2x2≤12
3x1 +2x
2 ≤18
A
B C
D
E
INDR 262 – The Simplex Method Metin Türkay 2
max $ = 3'( + 5'+s.t.
'( ≤ 42'+ ≤ 12
3'( + 2'+ ≤ 18'(, '+ ≥ 0
ADJACENT CPF SOLUTIONS
ØDefinition: Adjacent CPF solutions
For any linear programming problem with ndecision variables, two CPF solutions are
adjacent to each other if they share n-1
constraint boundaries. The two adjacent CPF
solutions are connected by a line segment that
lies on these same shared constraint
boundaries. Such a line is referred to as an edge
on the feasible region.
INDR 262 – The Simplex Method Metin Türkay 3
EXAMPLE: WYNDOR GLASS CO.
Point (x1,x2) Adjacent CPF SolutionsA (0,0) B(0,6)
E(4,0)B (0,6) A(0,0)
C(2,6)C (2,6) B(0,6)
D(4,3)D (4,3) C(2,6)
E(4,0)E (4,0) A(0,0)
D(4,3)
x1
x2
x1 ≤4
2x2≤12
3x1 +2x
2 ≤18
A
B C
D
E
INDR 262 – The Simplex Method Metin Türkay 4
OPTIMALITY
Ø If a CPF solution has no adjacent CPF solutions that are better, then it must be an optimal solution.
INDR 262 – The Simplex Method Metin Türkay 5
OPTIMIZATION ALGORITHM
Ø0. Initialize: choose an initial CPF solution.
Ø1. Optimality Test: evaluate the performance measure at the current solution. If its value is larger than all of its adjacent CPF solutions; the current solution is optimal. Otherwise go to Step 2.
Ø2. Iteration: Move to a better adjacent CPF solution. Evaluate the objective function value. Go to Step 1.
INDR 262 – The Simplex Method Metin Türkay 6
WYNDOR GLASS CO
x1
x2
x1 ≤4
2x2≤12
3x1 +2x
2 ≤18
A
B C
D
E
0. Initialization:Select A as the initial CPF solution
1. Optimality Test:max z=3x1+5x2 can be rearranged asmax z-3x1−5x2=0Since an increase in x1 or x2 results in anincrease in z value, the current solution is not
optimal. 2. Iteration 1:
Both x1 or x2 directions improve the z value.We choose x2 direction since z increases fasterin x2 direction. That will take us to CPF B.z value at CPF solution B is 30.
1. Optimality Test:There are 2 extreme directions that we can follow at CPF solution B.B → A (does not improve the z value, prev. sol.)B → C (improves the z value, while x2 remains
the same, x1 increases)if we further increase x2, feasibility is violated
INDR 262 – The Simplex Method Metin Türkay 7
WYNDOR GLASS CO
x1
x2
x1 ≤4
2x2≤12
3x1 +2x
2 ≤18
A
B C
D
E
2. Iteration 2:The x1 direction improves the z value.That will take us to CPF C.z value at CPF solution C is 36.
1. Optimality Test:There are 2 extreme directions that we can follow at CPF solution C.C → B (does not improve the z value, prev. sol.)C → D (does not improve the z value,
the slope of the constraint is 3/2 while the slope of the objective function is 3/5.So, objective function value decreaseswhen we move in this in direction)
the other two directions violate feasibility.
THE OPTIMAL SOLUTION IS FOUND!z*=36
x1* = 2, x2*=6
INDR 262 – The Simplex Method Metin Türkay 8
CONSIDERATIONSØFocus solely on CPF solutions.ØConduct the search for the optimal solution
iteratively. ØWhenever possible, choose the origin as the initial
CPF solution.ØGiven a CPF solution, it is much quicker to gather
information about its adjacent CPF solutions than its non-adjacent CPF solutions.
INDR 262 – The Simplex Method Metin Türkay 9
CONSIDERATIONSØAfter the current CPF solution is identified, the simplex
method examines each of the vertices of the feasible region
that emerge from this CPF solution. The most promising
vertex is selected for the next iteration.
ØA positive rate of improvement in z implies that the adjacent
CPF solution is better; a negative rate of improvement in zimplies that the adjacent CPF solution is worse. The
optimality test consists of simply of checking whether any of
the vertices give a positive rate improvement in z. If none
do, then the current CPF solution is optimal.
INDR 262 – The Simplex Method Metin Türkay 10
SLACK VARIABLES
Augmented Form of the LP Model
INDR 262 – The Simplex Method Metin Türkay 11
x2
x10 2 4
x1 x3
!" ≤ 4introduce slack variable !%
!" + !% = 4and !% ≥ 0
2!+ ≤ 122!+ + !- = 12
!- ≥ 0
3!" + 2!+ ≤ 183!" + 2!+ + !0 = 18
!0 ≥ 0
max 4 = 3!" + 5!+s.t.
!" + !% = 42!+ + !- = 12
3!" + 2!+ + !0 = 18!", !+, !%, !-, !0 ≥ 0
DEFINITIONS
Ø Augmented Solution: a solution for the original variables that has been augmented by the corresponding values at the slack variables.§ Ex:
Ø Basic Solution: an augmented corner-point solution. § Ex: Points A, B, C, D, E are basic feasible solutions.
• Point A:
INDR 262 – The Simplex Method Metin Türkay 12
!"∗ = 2, !'∗ = 6!) = 2, !* = 2, !+ = 0
!" = 0, !' = 0, !) = 4, !* = 12, !+ = 18
BASIC SOLUTIONS
Ø Each variable is either a basic variable or non-basic variable.
Ø # of basic variables = # of functional constraints.Ø Non-basic variables are set equal to zero.Ø Values of basic variables are obtained by solving
system of equations.Ø If basic variables satisfy non-negativity
constraints, the basic solution is a BFS.
INDR 262 – The Simplex Method Metin Türkay 13
THE SIMPLEX METHOD
Ø The Simplex Method
Initialization
Optimality Test
Iteration
STOP
INDR 262 – The Simplex Method Metin Türkay 14
MODEL REPRESENTATION
Ø Model is rearranged as:
INDR 262 – The Simplex Method Metin Türkay 15
max $ = 3'( + 5'+s.t.
'( + '/ = 42'+ + '2 = 12
3'( + 2'+ + '4 = 18'(, '+, '/, '2, '4 ≥ 0
max $0 $ − 3'( − 5'+ =01 '( + '/ = 42 2'+ + '2 = 12(3) 3'( + 2'+ + '4 = 18
'(, '+, '/, '2, '4 ≥ 0
INITIALIZATION
Ø Since there are 3 equalities, we need to select 3 variables as basic and 2 as non-basic variables.
Ø select x3, x4 and x5 as basic variables
INDR 262 – The Simplex Method Metin Türkay 16
0 " − 3%& − 5%( =01 %& + %- = 42 2%( + %0 = 12(3) 3%& + 2%( + %3 = 18
%&, %(, %-, %0, %3 ≥ 0
%& = 0, %( = 0%- = 4, %0 = 12, %3 = 18
OPTIMALITY TEST
Ø Not Optimal!!!
Ø since ! − 3$% − 5$'=0
if $% or $' increase, ! will also increase.
INDR 262 – The Simplex Method Metin Türkay 17
ITERATION
Ø determine the direction of movementüConsider the objective function (row 0)ü if we choose to move in the direction of x1, the rate of
improvement in z will be 3 (δz/δx1=3)ü if we choose to move in the direction of x2, the rate of
improvement in z will be 5 (δz/δx2=5)üSo, we choose x2ü x2 is called the entering basic variable
INDR 262 – The Simplex Method Metin Türkay 18
ITERATION
Ø determine the limit on the movementüConsider the set of constraintsü x1=0, we will determine the maximum value that x2 will
get (as the entering basic variable) without violating any of the constraints.
1 "# + "% = 42 2") + "* = 12(3) 3"# + 2") + "/ = 18
INDR 262 – The Simplex Method Metin Türkay 19
191.38
THE MINIMUM RATIO
Ø Row1 "# + "% = 4 "% = 42 2") + "* = 12 if ") ≤ #)
) ≤ , the constraint is not violated(3) 3"# + 2") + "= = 18 if ") ≤ #?
) ≤ 9 the constraint is not violated
Ø Select the minimum value as the limit on the movement ⇒ ") = 6This would guarantee feasibility.
Ø These calculations are called the minimum ratio test. Also identify the basic variable associated with the row that is selected in the minimum ratio test as the leaving basic variable ⇒ "* = 0
INDR 262 – The Simplex Method Metin Türkay 20
ITERATION
Ø Solve the new basic feasible solution
ØWe want the column that corresponds to the entering basic variable as:New Old
Row x2 x2(0) 0 -5(1) 0 0(2) 1 2(3) 0 2
1 2
1 3
2 4
1 2 5
1
(0) 3 5 0 (1) 4 (2) 2 12 (3) 3 2 18 ,
z x xx x
x xx x xx x
- - =+ =
+ =+ + =
2 3 4 5, , , 0x x x ³
Pivot element
INDR 262 – The Simplex Method Metin Türkay 21
PIVOTING
Ø Multiply row 2 by (2)-1:
Ø Multiply row 2 by 5 and add to row 0:
1 2
1 3
2 4
1 2
Row Equation (0) 3 5 0 (1) 4
1 (2) 62
(3) 3 2
z x xx x
x x
x x
- - =+ =
+ =
+ 5 18x+ =
1 4
1 3
2 4
1
Row Equation 5 (0) 3 302
(1) 41 (2) 62
(3) 3 2
z x x
x x
x x
x
- + =
+ =
+ =
+ 2 5 18x x+ =INDR 262 – The Simplex Method Metin Türkay 22
PIVOTING
Ø Multiply row 2 by -2 and add to row 3:
Ø BF Solution:1
2
3
4
5
06406
xxxxx
é ù é ùê ú ê úê ú ê úê ú ê ú=ê ú ê úê ú ê úê ú ê úë ûë û
1 4
1 3
2 4
1
Row Equation 5 (0) 3 + 302
(1) 41 (2) 62
(3) 3
z x x
x x
x x
x
- =
+ =
+ =
4 5 6x x- + =
INDR 262 – The Simplex Method Metin Türkay 23
OPTIMALITY TEST
Ø z = 30+3x1− 5/2x4Ø a move in the direction of x1 would increase
the objective function value because it has a positive cost coefficient.
INDR 262 – The Simplex Method Metin Türkay 24
ITERATION #2
Ø x1 is the entering basic variable.Ø Minimum ratio test:
Ø x1=2, x5 is the leaving variable.
Row Equation Ratio
(1) x1 + x3 = 4 x3 = 4− x1 ≥ 0⇒ x1 ≤41= 4
(2) x2 + 12x4 = 6 x2 = 6 ≥ 0 no upper limit on x1
(3) 3x1 − x4 + x5 = 6 x5 = 6−3x1 ≥ 0⇒ x1 ≤63= 2
INDR 262 – The Simplex Method Metin Türkay 25
NEW BF SOLUTION
Ø ero’s:ü Multiply row 3 by (3)-1
ü Multiply row 3 by 3 and add to row 0ü Multiply row 3 by -1 and add to row 1
1 4
1 3
2 4
1
Row Equation 5 (0) 3 302
(1) 41 (2) 62
(3) 3
z x x
x x
x x
x
- + =
+ =
+ =
4 5 6x x- + =
INDR 262 – The Simplex Method Metin Türkay 26
BF SOLUTION
Ø New BF solution:
Ø The current BF solution is OPTIMAL!Ø z*=36; x1=2, x2=6, x3=2, x4=x5=0
4 5
3 4 5
2 4
1 4 5
3 (0) 3621 1 (1) 23 31 (2) 621 1 (3) 23 3
z x x
x x x
x x
x x x
+ + =
+ =
+ =
- + =
-
INDR 262 – The Simplex Method Metin Türkay 27
TABLEAU FORMAT
Ø All of the information on the iterations of the simplex method can be represented in a systematic form
Ø The most natural representation uses the matrix form (in numerical values) of the objective function and the constraints
INDR 262 – The Simplex Method Metin Türkay 28
TABLEAU FORMAT
xB z x1 ... xn+1 ... RHS Ratio
z 1 −c1 ... 0 ... 0 ---
xn+1 0 a11 ... 1 ... b1
xn+2 0 a21 ... 0 ... b2
... 0 ... ... 0 ... ...
INITIAL TABLEAUX
Basic Variables
Objective FunctionVariable
Original Variables
Slack Variables
Right Hand Side
Table Header
Row 0
Row 1
Row 2
Row ...
Ratio Calculation
Update the Tableaux for each iteration of the Simplex method.INDR 262 – The Simplex Method Metin Türkay 29
WYNDOR GLASS CO.
xB z x1 x2 x3 x4 x5 RHS Ratio
z 1 −3 −5 0 0 0 0 ---
x3 0 1 0 1 0 0 4 4/0=∞
x4 0 0 2 0 1 0 12 12/2=6
x5 0 3 2 0 0 1 18 18/2=9
Entering Basic Variable, also the Pivot Column(the variabe with the most negative Row 0 coefficient)
Leaving Basic Variable, also the Pivot Row(the row with the minimum ratio) Pivot Element
(all ero’s will be defined with this element)
INITIAL TABLEAUX
INDR 262 – The Simplex Method Metin Türkay 30
WYNDOR GLASS CO.
xB z x1 x2 x3 x4 x5 RHS Ratio
z 1 −3 0 0 5/2 0 30 ---
x3 0 1 0 1 0 0 4 4/1=4
x2 0 0 1 0 1/2 0 6 6/0=∞
x5 0 3 0 0 −1 1 6 6/3=2
ITERATION (TABLEAUX) 1:Update the initial tableaux with the following ero’s:New Row 2 = (1/2) Old Row 2New Row 0 = (+5) New Row 2 + Old Row 0New Row 1 = (0) New Row 2 + Old Row 1New Row 3 = (-2) New Row 2 + Old Row 3
INDR 262 – The Simplex Method Metin Türkay 31
WYNDOR GLASS CO.
xB z x1 x2 x3 x4 x5 RHS Ratio
z 1 0 0 0 3/2 1 36 ---
x3 0 0 0 1 1/3 −1/3 2
x2 0 0 1 0 1/2 0 6
x1 0 1 0 0 −1/3 1/3 2
ITERATION (TABLEAUX) 2:Update the initial tableaux with the following ero’s:New Row 3 = (1/3) Old Row 3New Row 0 = (+3) New Row 3 + Old Row 0New Row 1 = (-1) New Row 3 + Old Row 1New Row 2 = (0) New Row 3 + Old Row 2
OPTIMAL SOLUTION: z*=36, x1*=2, x2*=6INDR 262 – The Simplex Method Metin Türkay 32
ANOTHER EXAMPLE
1 2 3
1 2 3
1 3
1 2 3
1 2 3
1 2 3
max =5 4 3s.t. +3 3 3 2 2 2 4 2 +3 2 , , 0
z x x x
x x xx xx x xx x xx x x
+ +
+ £- + £
+ + £+ £³
1 2 3
1 2 3 4
1 3 5
1 2 3 6
max s.t. 5 4 3 =0 +3 3 3 2 2 2 4
z
z x x xx x x xx x xx x x x
- - -
+ + =- + + =
+ + + =
1 2 3 7
1 2 3 4 5 6 7
2 +3 2 , , , , , , 0
x x x xx x x x x x x
+ + =
³
INDR 262 – The Simplex Method Metin Türkay 33
ANOTHER EXAMPLExB z x1 x2 x3 x4 x5 x6 x7 RHS Ratio
z 1 −5 −4 −3 0 0 0 0 0 ---
x4 0 1 3 1 1 0 0 0 3 3/1=3
x5 0 −1 0 3 0 1 0 0 2 ---
x6 0 2 1 2 0 0 1 0 4 4/2=2
x7 0 2 3 1 0 0 0 1 2 2/2=1
z 1 0 7/2 −1/2 0 0 0 5/2 5 ---
x4 0 0 3/2 1/2 1 0 0 −1/2 2 2/(1/2)=4
x5 0 0 3/2 7/2 0 1 0 1/2 3 3/(7/2)=6/7
x6 0 0 −2 1 0 0 1 -1 2 2/1=2
x1 0 1 3/2 1/2 0 0 0 1/2 1 1/(1/2) =2
INDR 262 – The Simplex Method Metin Türkay 34
ANOTHER EXAMPLE
xB z x1 x2 x3 x4 x5 x6 x7 RHS
z 1 0 26/7 0 0 1/7 0 18/7 38/7
x4 0 0 9/7 0 1 −1/7 0 −4/7 11/7
x3 0 0 3/7 1 0 2/7 0 1/7 6/7
x6 0 0 −17/7 0 0 −2/7 1 −8/7 8/7
x1 0 1 9/7 0 0 −1/7 0 3/7 4/7
OPTIMAL SOLUTION: z*=38/7, x1*=4/7, x2*=0, x3*=6/7
INDR 262 – The Simplex Method Metin Türkay 35
COMPLICATIONS
Ø Bounds• Positive Upper Bounds• Negative Values
Ø Ties:• Entering Basic Variable• Leaving Basic Variable – Degeneracy
Ø No Leaving Basic Variable – UnboundednessØ Multiple Optimal SolutionsØ Origin IS NOT Basic Feasible
INDR 262 – The Simplex Method Metin Türkay 36
BOUNDSØ Bounds on Decision Variables:
üLower Bounds (! ≥ !#):use variable substutition,
Solve the problem and perform backsubstutitionExample:
üUpper Bounds (! ≤ !%):treat them as functional constraints (for now!!!)
1 2
1 2
1 2
max =3 2s.t. 2 6 0, 1
z x x
x xx x
+
+ £³ ³
'1 2
'1 2
'1 2
max =3 2( 1)s.t. 2( 1) 6 , 0
z x x
x xx x
+ +
+ + £³
'1 2
'1 2
'1 2
max =3 2 2s.t. 2 4 , 0
z x x
x xx x
+ +
+ £³
INDR 262 – The Simplex Method Metin Türkay 37
max ) = +,-s.t.
1- ≤ 2- ≥ -3
max ) = +,(-5 + -3)s.t.
1(-5 + -3) ≤ 2-5 ≥ 0
- = -5 + -3
BOUNDSüNegative Values are Allowed (−∞ ≤ $ ≤ ∞):
use variable substutition,
Solve the problem and perform backsubstutitionExample:
1 2
1 2
1 2
1 2
max =2 3s.t. 2 2 2 8 0, -
z x x
x xx x
x x
+
- £+ £
³ ¥ £ £ ¥
INDR 262 – The Simplex Method Metin Türkay 38
1 2 2
1 2 2
1 2 2
1 2 2
max =2 3 3s.t. 2 2 2 2 2 8 , , 0
z x x x
x x xx x xx x x
+ -
+ -
+ -
+ -
+ -
- + £+ - £
³
% = %' − %(
max , = -.%s.t.
2% ≤ 3−∞ ≤ % ≤ ∞
max , = -.(%' − %()s.t.
2(%' − %() ≤ 3%', %( ≥ 0
TIESØ Two Types of Ties:
üTie for Entering Basic VariableüTie for Leaving Basic Variable
Ø Tie for Entering Basic Variable: This situation is observed when two or more variables have the same most negative (for max problems) row 0 coefficient.
Ø Options:1. Select one of the variables in tie randomly as
the entering basic variable2. Select one of the variables in tie with the
smallest variable index as the entering basic variable
3. More advanced rules that requires probing or application of the lexicographic rule
INDR 262 – The Simplex Method Metin Türkay 39
DEGENERACY
Ø Tie for Leaving Basic Variable: This situation is observed when two or more rows have the same minimum ratio. In this case, we have to conduct at least one DEGENERATE iteration where the objective function value does not change after conducting iteration. There are 2 possible outcomes when DEGENERACY is encountered.
1. The objective function improves after 1 or more iterations,
2. The objective function does not improve during a number of iterations where the basic feasible solutions circle back to the iteration where the degeneracy is encountered for the first time resulting in an infinite loop of iterations.
INDR 262 – The Simplex Method Metin Türkay 40
DEGENERACY EXAMPLE1 2
1 2
1 2
1 2
1 2
max =3 2s.t. 4 8 4 3 12 4 8 , 0
z x x
x xx xx xx x
+
- £+ £+ £
³
1 2
1 2 3
1 2 4
1 2 5
1 2 3 4 5
max s.t. 3 2 0 4 8 4 3 12 4 8 , , , , 0
z
z x xx x xx x xx x xx x x x x
- - =- + =
+ + =+ + =
³
xB z x1 x2 x3 x4 x5 RHS Ratio
z 1 −3 −2 0 0 0 0 ---
x3 0 4 −1 1 0 0 8 8/4=2
x4 0 4 3 0 1 0 12 12/4=3
x5 0 4 1 0 0 1 8 8/4=2
TIE
INDR 262 – The Simplex Method Metin Türkay 41
DEGENERACY EXAMPLE
xB z x1 x2 x3 x4 x5 RHS Ratio
z 1 0 −11/4 3/4 0 0 6 ---
x1 0 1 −1/4 1/4 0 0 2 ---
x4 0 0 4 −1 1 0 4 4/4=1
x5 0 0 2 −1 0 1 0 0/2=0
TIE fromprev. iter.
z 1 0 0 −5/8 0 11/8 6 ---
x1 0 1 0 1/8 0 1/8 2 2/(1/8)=16
x4 0 0 0 1 1 −2 4 4/1=4
x2 0 0 1 −1/2 0 1/2 0 ---
INDR 262 – The Simplex Method Metin Türkay 42
DEGENERACY EXAMPLE
xB z x1 x2 x3 x4 x5 RHS Ratio
z 1 0 0 0 5/8 1/8 68/8 ---
x1 0 1 0 0 −1/8 3/8 3/2
x3 0 0 0 1 1 −2 4
x2 0 0 1 0 1/2 −1/2 2
OPTIMAL SOLUTION: z*=68/8, x1*=3/2, x2*=2
We were lucky to solve this problem with 1 degenerate iteration only. Most problems require more degenerate iterations.
INDR 262 – The Simplex Method Metin Türkay 43
UNBOUNDEDNESS
1 2
1 2
1
1 2
max =2s.t. 10 2 40 , 0
z x x
x xxx x
+
- ££³
1 2
1 2 3
1 4
1 2 3 4
max s.t. 2 0 10 2 40 , , , 0
z
z x xx x xx xx x x x
- - =- + =
+ =³
INDR 262 – The Simplex Method Metin Türkay 44
UNBOUNDEDNESS
xB z x1 x2 x3 x4 RHS Ratio
z 1 −2 −1 0 0 0 ---
x3 0 1 −1 1 0 10 10/1=10
x4 0 2 0 0 1 40 40/2=20
z 1 0 −3 2 0 20 ---
x1 0 1 −1 1 0 10 ---
x4 0 0 2 −2 1 20 20/2=10
z 1 0 0 −1 3/2 50 ---
x1 0 1 0 0 1/2 20
x2 0 0 1 −1 1/2 10
NO LEAVING VARIABLE, UNBOUNDED OBJ. FN. VALUE
Question: If the feasible region is unbounded, is the objective function value ALWAYS unbounded?
INDR 262 – The Simplex Method Metin Türkay 45
MULTIPLE OPTIMA
1 2
1 2
1 2
1 2
max = 2s.t. 2 4 2 4 6 , 0
z x x
x xx xx x
+
+ £+ £
³
1 2
1 2 3
1 2 4
1 2 3 4
max s.t. 2 0 2 4 2 4 6 , , , 0
z
z x xx x xx x xx x x x
- - =+ + =+ + =
³
INDR 262 – The Simplex Method Metin Türkay 46
MULTIPLE OPTIMAxB z x1 x2 x3 x
4RHS Ratio
z 1 −1 −2 0 0 0 ---
x3 0 2 1 1 0 4 4/1=4
x4 0 2 4 0 1 6 6/4=3/2
z 1 0 0 0 1/2 3 ---
x3 0 3/2 0 1 −1/4 5/2 (5/2)/(3/2)=5/3
x2 0 1/2 1 0 1/4 3/2 (3/2)/(1/2)=3
z 1 0 0 0 1/2 3
x1 0 1 0 2/3 −1/6 5/3
x2 0 0 1 −1/3 1/3 2/3
Ø A row 0 coefficient of a non-basic variable with a value of 0 at the optimal solution indicates multiple optimal solutions.
Ø The objective function is parallel to a binding constraints at the optimal solution and two adjacent CPF solutions on the binding constraint are optimal.
INDR 262 – The Simplex Method Metin Türkay 47
OPTIMAL SOLUTIONS: z*=3, 1:x1*=0, x2*=3/22:x1*=5/3, x2*=2/3
INTERESTING PROBLEM
1 2
1 2
1 2 3
1 2 3
1 2 3
max =3s.t. 2 5 2 7 3 5 20 , , 0
z x x
x xx x xx x xx x x
+
+ £+ - £
+ - £
³
1 2
1 2 4
1 2 3 5
1 2 3 6
max s.t. 3 0 2 5 2 7 3 5 20
z
z x xx x xx x x xx x x x
- - =+ + =+ - + =+ - + =
1 2 3 4 5 6 , , , , , 0x x x x x x ³
xB z x1 x2 x3 x4 x5 x6 RHS Ratio
z 1 −3 −1 0 0 0 0 0 ---
x4 0 1 2 0 1 0 0 5 5/1=5
x5 0 1 1 −1 0 1 0 2 2/1=2
x6 0 7 3 −5 0 0 1 20 20/7
INDR 262 – The Simplex Method Metin Türkay 48
INTERESTING PROBLEMxB z x1 x2 x3 x4 x5 x6 RHS Ratio
z 1 0 2 −3 0 3 0 6 ---
x4 0 0 1 1 1 −1 0 3 3/1=3
x1 0 1 1 −1 0 1 0 2 ---
x6 0 0 −4 2 0 −7 1 6 6/2=3
z 1 0 5 0 3 0 0 15 ---
x3 0 0 1 1 1 −1 0 3
x1 0 1 2 0 1 0 0 5
x6 0 0 −6 0 −2 −5 1 0
WHAT IS THE VERDICT?INDR 262 – The Simplex Method Metin Türkay 49