Transcript Chapter 4: Linear Programming Sensitivity Analysis

CHAPTER 5:
LINEAR PROGRAMMING:
THE SIMPLEX METHOD
SIMPLEX METHOD CONTINUED……
Minimization Problem
 First Approach

 Introduce
the basis variable
 To solve minimization problem we simple reverse
the rule that is we select the variable with most
negative cj-zj to select new basic variable in the
next iteration
 The stopping rule is also changed ; the iteration is
stopped when every value is the cj-zj row is zero or
positive.
SIMPLEX METHOD
Second Approach
 Change
the minimization problem to
maximization problem by multiplying the
objective function with -1
 Solve the problem using the simplex method
 Stop the iteration when cj-zj row is either
negative or zero.
 Multiply the objective function value to -1 in the
last iteration to convert the maximization to
original minimization problem
MINIMIZATION PROBLEM (SIMPLEX METHOD)
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Management want to minimize the cost of producing
two products to a demand constraint of product A,a
minimum total production quantity requirement and a
constraint on available process time
Min 2x1+3x2
S.t
1x1 >=125 (Demand for Product A)
1x1+1x2>=350(Total Production)
2x1+1x2<=600 (Processing Time)
X1,x2 >=0
SIMPLEX METHOD
 To solve this minimization problem we multiply
the objective solution by -1 to convert in to
maximization problem
 Max -2x1-3x2
 S.t
 1x1 >=125 (Demand for Product A)
 1x1+1x2>=350(Total Production)
 2x1+1x2<=600 (Processing Time)
 X1,x2 >=0
 Standardized form ?????
SIMPLEX METHOD FOR <= CONSTRAINT
Standard form
Max -2x1-3x2+0s1+0s2+0s3-Ma1-Ma2
1x1 -1s1+1a1=125
1x1+1x2-1s2+1a2=350
2x1+1x2+1s3=600
x1.,x2,s1,s2,s3,a1,a2 >=0
Initial Simplex Table ????
INITIAL SIMPLEX TABLE
x1 x2 s1 s2 s3 a1 a2
 -2
-3 0 0
0 -M -M
1
0 -1 0
0 1
0
1
1
0 -1 0 0
1
2
1 0 0
1 0
0
 What next ????
 What are basic variables????

125
350
600
INITIAL SIMPLEX TABLE
Basis CB
a1
a2
s3
-M
-M
0
x1 x2 s1 s2 s3 a1 a2
-2 -3 0 0
0 -M -M
1
0 -1 0
0 1
0
1
1
0 -1 0 0
1
2
1 0 0 1 0 0
Zj
-2M -M M -M 0 -M -M
 Cj-Zj
-2+2M -3+M -M -M 0 0 0
 Which variable need to go basis ????

125
350
600
-475
X1 need to go basis because -2+2M is largest
 A1 need to go nonbasis because 125/1=125;
 350/1=350; 600/2=300 hence a1 is smallest
 Also a11=1;a21=0;a31=0 as need for basis
variable
 Row operation : Subtract row 1 from row 2
 Multiply row 1 by 2 and subtract from row 1
 a1 is removed as it is now nonbasic and need tobe
zero.

INITIAL SIMPLEX TABLE (IST ITERATION)
Basis CB
x1
a2
s3
-2
-M
0
x1 x2 s1 s2 s3 a2
-2 -3 0 0
0 -M
1
0 -1 0
0
0
0
1
1 -1
0 1
0
1 2 0 1 0
Zj
-2 -M 2- M M 0 -M
 Cj-Zj
0 -3+M -2+M -M 0 0
 Which variable need to go basis ????

125
225
350
-250-225M
STANDARD FORM WITH SLACK VARIABLES

Now S1 will be basis and a2 will be non basis
hence removed from tableau continue with two
other iterations.
INITIAL SIMPLEX TABLE (IST ITERATION)
Basis CB
x1
x2
s1
-2
-3
0
x1 x2 s1 s2 s3
-2 -3 0 0
0
1
0 -0 1
1
0
1
0 -2
-1
0
0
1 1
1
250
100
125
Zj
-2 -3 0 4 1
-800
 Cj-Zj
0
0 0 -4 -1
 Optimal solution is +800 becaz it is minimization
problem

SPECIAL CASES: INFEASIBILTY
Occurs when no solution can be found that
satisfies all constraints. Identified by positive
value of artificial variable in the solution
 Max 50x1+ 40x2
 S.t
 3x1+5x2<=150 (Assembly Time of two product)
 1x2
<=20 (portable display)
 8x1+5x2 <=300 (warehouse space)
 1x1+1x2 >=50 (Minimum total production)

SOLUTION AFTER TWO ITERATION
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Basic CB x1
x2 s1 s2
50
40 0
0
X2 40
0
1 8/25 0
S2 0
0
0 -8/25 1
X1 50
1
0
-5/25 0
a4 -M
0
0
-3/25 0
Zj
50 40 (70+3M)/25 0
Cj-Zj
0
0 (-70-3M)/25 0
Feasible Solution ???????
s3
s4 a4
0
0
-M
-3/25 0
0 12
3/
0
0
8
25
5/
0
0 30
25
-2/25 -1 1 8
(130+2M)/ M -M 1980-8M
25
(-130-2M)/
25 -M 0
INFEASIBILITY
Presence of a4 =8 in the solution means it is
not feasible although cj-zj is non negative.
 X1=30;x2=12 x1+x2=42 <50 hence violates
the fourth constraint of at least 50 units, a4 =8
indicates that constraint 4th is violates by 8
unit
 S1 and s3=0 means warehouse space and
assembly time constraint are binding as not
enough spare house space and time is
available hence minimum combined total
production of 50 units is lowered by 8 units.

INFEASIBILITY

If more time or space is not allotted than
management will have to relax total production
by 8 units
UNBOUNDEDNESS
1.
2.
3.
A Maximization Problem is unbounded if it is
possible to make the value of optimal solution as
large as possible without violating the
constraints.
A surplus variable can be interpreted as amount
of the basis variable over the minimum amount
required.
If a solution is unbounded then we can increases
the minimum amount of a basic variable as
much as we want and the objective function will
have no upper bound provided the basis variable
has positive coefficient in the objective function
UNBOUNDED SOLUTION
In simplex tableau we recognized the
unbounded solution is that all aij are less than
or equal to zero in column associated with
incoming variables.
 Max 20x1 +10x2
 S.t 1x1 >=2
 1x2<=5
 X1,x2>=0

CHAPTER 6:SIMPLEX –BASED SENSITVITY
ANALYSIS
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Range of optimality for an objective function

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Is the range of that coefficient for which the current optimal
solution will remain optimal (keeping all other coefficients
constant). However the objective function value may change.
The range of optimality for the basic variable
defines the objective function coefficient values for which
current variable will remain the part of the basic feasible
solution.
Range of optimality for nonbasic variable defines
objective function coefficient values for which that
variable remain nonbaisc
OBJECTIVE FUNCTION COEFFICIENTS
AND RANGE OF OPTIMALITY
 Change
the objective function coefficient to
ck in the cj row.
 If xk is basic, then also change the objective
function coefficient to ck in the cB column
and recalculate the zj row in terms of ck.
 Recalculate the cj - zj row in terms of ck.
 Determine the range of values for ck that
keep all entries in the cj - zj row less than or
equal to 0.
REVISED PROBLEM WITH >0 CONSTRAINT
Max 50x1+ 40x2
 S.t
 3x1+5x2+ <=150 (Assembly time)
 1x2+
<=20 (Portable display)
 8x1+5x2 <=300 (warehouse capacity)
 X1 is number of units of Desktop PC
 X2 is number of the portable display
 X1,x2>=0
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FINAL SIMPLEX OF THE PROBLEM
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Basic cb
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X2
s2
X1
Zj
Cj-Zj
40
0
50
X1 x2 s1
50 40 0
0 1 8/25
0 0 -8/25
1 0 -5/25
50 40 14/5
-14/
0 0
5
s2
0
0
1
0
0
0
s3
0
-3/
25
3/
25
5/
25
26/
5
-26/
5
12
8
30
1980
The range of optimality for an objective function coefficient is
determined by those coefficient values that maintain
 Cj-zj<=0
RANGEL OF OPTIMALITY OF THE PROFICT
CONTRIBUTION PER UNIT OF DESKTOP (X1)
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Basic cb
X1 x2
s1
s2
s3
C1 40
0
0
0
8/
-3/
X2
40
0 1
0
25
25
-8/
3/
s2
0
0
0
1
25
25
-5/
5/
X1 C1
1
0
0
25
25
Zj
c1
40 (64-c1)/5
0 (c1-24)/5
c1-64/
24-c1/
Cj-Zj
0
0
0
5
5
Applying cj-zj <=0 -> c1-64/5 <=0
C1-64<=0 -- C1<=64
24-c1<=0-- 24<=C1 thus 24<=C1<=64
12
8
30
480+30c1
DISCUSSION
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Suppose increase in material cost reduces profit
contribution per unit of desktop to 30
Range of optimality of X1 indicates still the solution
x2=12;s1=0;s3=0 would be an optimal solution. The final
tableau for x1=30 verifies this.
Basic cb
X1 x2
s1
s2
s3
30 40
0
0
0
8/
-3/
X2
40
0
1
0
12
25
25
-8/
3/
s2
0
0
0
1
8
25
25
-5/
5/
X1
30
1
0
0
30
25
25
Zj
30 40 34/5
0 6 /5
1380
-34/
Cj-Zj
0
0
0
-6/5 HOW???
5
DISCUSSION
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Current solution is not optimal as can be seen from cj-zj
row
S3 is positive means it is not optimal solution .
Basic cb
X1 x2
s1
s2
s3
20 40
0
0
0
8/
-3/
X2
40
0
1
0
12
25
25
-8/
s2
0
0
0
1 3/25
8
25
-5/
5/
X1 20
1
0
0
30
25
25
Zj
20 40 44/5
0 4 /5
1080
-44/
Cj-Zj
0
0
0 4/5
5
RANGE OF OPTIMALITY FOR BASIC VARIABLE
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Basic cb X1 x2 s1
s2 s3
50 40 cs1
0 0
X2
40 0 1 8/25 0 -3/25
12
s2
0 0 0 -8/25 1 3/25
8
X1
50 1 0 -5/25 0 5/25
30
Zj
50 40 14/5 0 26/5
1980
Cj-Zj
0 0
cs1-14/5 0 -26/5
Cs1-14/5<=0 cs1<=14/5
SIMPLEX METHOD
Dual Price
 Increase in objective per unit increase in RHS
of constraint. In Simplex method they are
identified in zj row row of final simplex method
 Value of slack variables in final
 S1=14/5=2.80;s2=0;s3=26/5=5.20
 The dual price for assembly constraint mean
that one 1 unit increase in assembly hours
increases the objective function by 2.80$
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FINAL SIMPLEX OF THE PROBLEM
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Basic cb
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X2
s2
X1
Zj
Cj-Zj
40
0
50
X1 x2 s1
s2
50 40 0
0
8/
0 1
0
25
0 0 -8/25
1
-5/
1 0
0
25
50 40 14/5
0
-14/
0 0
0
5
s3
0
-3/
12
25
3/
8
25
5/
30
25
26/
1980
5
-26/
5
Zj corresponding to S2 is 0 means dual price for
portable display is zero. Also S2 is basic
(slack)variable = 8 which shows there are 8
unused display left so increasing them will not
effect objective function
 If a slack variable is a basic variable in optimal
solution then its dual price will be zero.
 Dual price for >= constraint is given by negative of
zj entry for surplus variable
 Dual Price for = constraint is determined by zj
values for corresponding artificial variables

RANGE OF FEASIBILTY
Is the Range of RHS of constraint that does
not make current basis solution infeasible.
 This means range of elements of b column
matrix that does not make the solution
infeasible.
 Using the dual price concept the increase in
one unit of element in b vector appeared in
objective function is called dual price. Now we
wish to know the range of this b column matrix.

STANDARD FORM WITH SLACK VARIABLES
If we want to know the range of say b1 than we
need to use the corresponding slack of final
tableu. s1. WHY
 IT is because the coefficient in this matrix show
corresponding decrease in the basic variable or
in other word how many units of basic variable
will be driven out from solution or alternatively
decreasing b column.
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SETTING NON BASIC VAIABLES

B1
B2
B3
.
.
BN
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Current solution or l
Last column of the final
tableu
+ delB*
a1j
a2j
a3j
>=
anj
column corresponding to slack variable
0
0
0
0

Example
DUALITY
Is a two different perspective of the same
problem. The original linear problem is called
as primal and to each primal problem a
corresponding optimization problem is
associated which is termed as dual problem
 Primal Problem
 the objective function is a linear combination
of n variables and m constraints to maximize
the value of the objective function subject to
the constraints

DUALITY
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Primal problem
A per unit value of each product is given and it is
determined how much of each product is produced
to maximized the value of total production. the
constraint require amount of each resourced used
to be less than or equal to amount available.
Dual Problem
Is resource valuation problem. In dual problem
availability of each resource is given and we
determine the per unit value of each constraint
such that total resources used is minimized. In other
word we determine what is the value of unit
consumption of the available resource.
GENERAL RULES FOR CONSTRUCTING DUAL
The number of variables in the dual problem is
equal to the number of constraints in the
original (primal) problem. The number of
constraints in the dual problem is equal to the
number of variables in the original problem.
 If the original problem is a max model, the dual
is a min model; if the original problem is a min
model, the dual problem is the max problem.
 Convert the problem in to conical form

GENERAL RULES FOR CONSTRUCTING DUAL
Conical form for Maximization problem
 All constraint should be less than or equal to
constraint.
 Conical form for Minimization problem
 All constraint should be greater than or equal to
constraint.

Decision variable in primal becomes constraint
in dual problem
 the first constraint in dual will be associated
with the first decision variable second with
second and so.
 The RHS of the constraint in the primal
becomes the coefficient of objective function in
the dual

The objective function coefficient of the primal
becomes RHS of the constraint in the dual.
 the constraint coefficient ith primal variable
becomes coefficient of ith constraint in the
dual. In other words the row becomes column
or the ‘A’ matrix is transposed.

PROPERTIES OF DUAL PROBLEM
If the dual problem has an optimal solution ,
the primal has an optimal solution and vice
versa. Both have same value of optimal
solution in term of objective function
 The optimal values of the primal decision
variables in the final simplex tableau are given
by zj entries for surplus variables. The optimal
value of the slack variable are given by
negative of cj-zj entries for uj variables.

EXAMPLE
Primal Problem
 Max 50x1+ 40x2
 S.t
 3x1+5x2<=150 (Assembly Time of two product)
 1x2
<=20 (portable display)
 8x1+5x2 <=300 (warehouse space)
 Is it in conical Form???
Dual Problem
 Min 150u1+ 20u2+300u3
 S.t
 3u1+8u3>=50
 5u1+u2+5u3>=40
 U1,u2,u3>=0
 u1 is associated with assembly time constraint
 u2 is associated with portable display constraint
 u3 with warehouse space constraint.
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SLOVING WITH SIMPLEX
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Basic Cb
A1
-M
A2
-M
Zj
Cj-1
u1
u2
-150 -20
3
0
5
1
u3
s1 s2
-300 0 0
8
-1 0
5
0 -1
a1
-M
1
0
a2
-M
0
1
why50
40
-8M -M
-13M M M -M -M -90M
-150+8M -20+M -300+13M –M –M 0 0
FINAL SIMPLEX
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Basic Cb
u3
-300
u1
-150
u1
u2
u3
s1
s2
-150 -20 -300 0
0
3/
0
-3/25 1
-5/25
25
8/
5/
-8/
1
25 0
25
25
Zj
-150 -12
Cj-1
0
-8
Assembly Time $2.80
Portable Display= 0
Warehouse space = $5.20
-300 30
0
-30
12
-12
26/5
14/5
-1980
FINAL TABLEAU