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Summary of formulas:
= -1
RHS of optimal tableaus constraints = B -1b -1 - j BV j j
Slack variable, s i : i th element of cBV
B -1
-
, i - BV Artificial variable, a i : ( i th element of c BV B -1) + M
RHS of optimal Row 0 = c BV B -1b
1
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We must find B -1 to compute all partsof optimal tableau
We must find c BV B-1
to compute theoptimal tableau Row 0.
2
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Exam le 1: Refer to Exam le 4 Cha ter 3For the following LP, the optimal basis isBV = { x ,s }. Compute the optimal tableau.
min z = 2 x 1 - 5 x 2s.t 3x 1+8x 2 12
2x 1+3x 2 6
x1,x2 0 Solution 1: The standard form
min z = 2x 1 - 5x 2s.t 3x 1+8x 2 +s 1 =12
2x 1+3x 2 +s 2 = 6
3
x1,x2 ,s 1,s 2
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Solution 1: The standard formmin z = - 5x
2+s
2+ 2x
1+s
1 [ ] [ ]2 1
5 0 2 0
x x
z
= +
s.t x 2 + x 1+s 1 =3x 2 + s 2 + 2x 1 = 6x1,x2 ,s 1,s 2 0
2 1
2 1
2 1
8 0 3 1 12
3 1 2 0 6
x x
s s
+ =
2 1
2 10, 0
x x
s s
[ ] [ ]
2
2
5 0 , , 2 0 , BV Bv BNV xc x cs
= = =
1
1
, , ,3 1 2 0 6 NBV
x x B N b
s= = = =
Step 1: Find c BV and B -1cBV=[-5 0]
-108 0 8B = so by using Gauss elimination B
3 1 31
=
4
8
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Step 2: Determine the optimal constraints
Row BV z x 2 s 2 x1 s 1 RHS
0 z 1 0 0
1 x 2 0 1 0
2 s 0 0 1
LHS=formula B -1a j
11 1
1 30 38 8
3 2 7 x B a
= = =
RHS=formula B -1b
1
8 81 1
0 18 8
1 30 128 23 6 3
=
5
1 1 3 0 31
8 8
ss a= = =
8 2
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Thus, the constraints of the optimal tableau are
2 18 8 2 x x
=2 17 3 3
8 8 2s s
6
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Step 3: 10
optimal Row 0 [ ]
1 5 0 03 8
18
BV c B = =
1
LHS (coefficient for NBV) RHS
j BV j jc c B a c=
11 0 2 02 68 8 8 8
15 5
BV c c B b= = = =
1 08 8s
Row BV z x s x s RHS
0 z 1 0 0
1 x 0 1 0 3/8 1/8 3/2
72 s 2 0 0 1 7/8 -3/8 3/2
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any constra nt as a negat ve ,BV is now an infeasible basis.
If a variable in Row 0 may have a negative coefficient,
BV is now a suboptimal basis.
Three types of changes in an LP parameter
1 Chan in the ob ective function coefficient of anonbasic variable.
2) Changing the objective function coefficient of abasic variable.
3) Changing the right hand side (RHS) of a
10
constraint.
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ROW BV z x 1 x2 x3 s 1 s 2 s 3 RHS
0 z 1 0 5 0 0 10 10 280
1 s 1 0 0 -2 0 1 2 -8 24
2 x 3 0 0 -2 1 0 2 -4 8
3 x 1 0 1 1.25 0 0 -0.5 1.5 2
12
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1 21
3 2, , B 0 2 4
0 1/ 2 3 / 2 BV NBV x x x s
x s
= = =
order from optimal tableau
or no nee n or er
The onl nonbasic decision variable is x . Coefficient of x 2 of objective function iscurrently c 2 =30
Question: For what value of c 2 would BV remain optimal?
13
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Question:
+
or w a va ue o c 2 wou rema n op ma
so B remain unchange.
nonnegative. Check Formula: B -1b -1 Does b changed? NO
,values. BV is still feasible (remain unchanged)
14
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Ste 2: Determine the coefficient of NBV inoptimal Row 0Check Formula: =c BVB-1a -c .
BV remain optimal if j 0BV will be subo timal if < 0
The only variable will be change is
12 2 2
6 BV c c B a c
=
[ ] ( ) 0 10 10 2 30 51.5
= + =
15
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BV will remain optimal if5- 0
5 If the coefficient of x 2 is decreased or increased
by 5 or less, (in other word, c 2 30+5=35)
BV will remains optimal. c 2 > , w e su op ma no ongeroptimal). This is because j < 0.
.can increase z by making x 2 as basic variables
16
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For example: c 2=40
1
6
0 10 10 2 40 5c c B a c = = =
1.5 Final (suboptimal) Tablaeu
ROW BV z x 1 x2 x3 s 1 s 2 s 3 RHS
-
1 s 1 0 0 -2 0 1 2 -8 24
2 x 3 0 0 -2 1 0 2 -4 8
3 x 0 1 1.25 0 0 -0.5 1.5 2
17Pivot element
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Optimal Tablaeu
ROW BV z x 1 x2 x3 s 1 s 2 s 3 RHS
1 s 1 0 1.6 0 0 1 1.2 -5.6 27.2
2 x 3 0 1.6 0 1 0 1.2 -1.6 11.2
3 x 2 0 0.8 1 0 0 -0.4 1.2 1.6
18
1 1 , 2 . , 3 .
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4.3.2 Changing the Objective Function
+
c BV will be change to [0 20 60+ ]
nonnegative.Check Formula: B -1bDoes B -1 changed? NODoes b changed? NO
Therefore, RHS of constraints have nonnegativevalues. BV is still feasible (remain unchanged)
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Step 2 : Determine the coefficient of NBV in optimalow
Check Formula: j=c BVB-1a j-c j.BV remain optimal if 0
BV will be suboptimal if j < 0
1 2 8
[ ] [ ]1 0 20 60 0 2 4 0 10 0.5 10 0.5
0 0.5 1.5
BV c B = + = +
1 3 1
12 2 2
, , are , t e r coe c ent n opt ma ta aeu ow must st e
6
0 10 0.5 10 0.5 2 30 5 1.25 BV
s x x
c c B a c = = + = +
2
1.5
sc c
= 1 1
2 2 2th element of 10 0.5 BV s s BV B a c c B = =
20
3 3 3 3th element of 10 0.5s BV s s BV c c B a c c B = = = +
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2
0, 5 1.25 0 4
0, 10 0.5 0 20
c
c
+
3 0, 10 0.5 0 -20/3sc +
Therefore, current basi
1
s remain optimal for -4 20
or 56=60-4 60+20=80c
21
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Let c 1=100(or =40)
1 3 1, , are BV, their coefficient in optimaltablaeu Row 0 must still be 0s x x
1 3 1
2
0
5 1.25 55
sc c c
c
= = =
= + =2
3
10 0.5 10
10 0.5 70s
s
c
c
= = = + =
[ ]1RHS of row 0, 0 10 70 20 360 BV c B b = =
22
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ROW BV z x 1 x2 x3 s 1 s 2 s 3 RHS
Final (suboptimal tableau)
0 z 1 0 55 0 0 -10 70 360
1 s 1 0 0 -2 0 1 2 -8 24
2 x 3 0 0 -2 1 0 2 -4 8
3 x 1 0 1 1.25 0 0 -0.5 1.5 2
23
Pivot element
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ROW BV z x 1 x2 x3 s 1 s 2 s 3 RHS
Optimal tableau
0 z 1 0 45 5 0 0 50 400
1 s 1 0 0 0 -1 1 0 -4 16
2 s 3 0 0 -1 0.5 0 1 -2 4
3 x 1 0 1 0.75 0.25 0 0 0.5 4
z=400, when x 1=4, x 2=0, x 3=0
24
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4.3.3 Changing the RHS of a Constraint
Changing the RHS cannot cause the currentbasic to become suboptimal. As long as the
,remains feasible and optimal.
Su ose we chan e b from 20 to 20+ Step 1: Determine whether RHS of constraint is
nonnegative. -
Does B -1 changed? NO1 0 2 4 20
0 0.5 1.5 8
B b
= +
Therefore,RHS of constraints
24 2 8 2
+ = +
25
.
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Current basis will remain optimal iff
24 2 0 12+
8 2 0 4
2 0.5 0 4
+
-4 4 or 16 =20-4 b2 20+4 =24
26
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If 16 b 24 the current basis remain o timal
(s 1,x3,x1) but the values of decision variables and zchange. uppose 2= w ere + , = so
11
1 2 8 48 28s
= = =
1 0 0.5 1.5 8 1 x
[ ]148
0 10 10 20 2 300 BV z c B b
= = + =
27
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When the current basis is no longer optimal:
Suppose b 2=30 (or =10) RHS of optimal constraints:
11
31 2 8 48 440 2 4 20 10 22
s x B b
= = + =
1
1
. .
Since 3, BV is no longer feasible
of row 0
x
x
RHS
=
[ ]148
0 10 10 20 10 380 BV z c B b
= = + =
We will use dual simplex algorithm to solve LPs whenthe initial tableau has one or more ve RHS and each
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variable in Row 0 has nonnegative coefficient.