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 METHODSecond 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)

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 formMax -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 0 125 1 1 0 -1 0 0 1 350 2 1 0 0 1 0 0 600 What next ???? What are basic variables????

INITIAL SIMPLEX TABLE

Basis CB x1 x2 s1 s2 s3 a1 a2 -2 -3 0 0 0 -M -M a1 -M 1 0 -1 0 0 1 0 125

a2 -M 1 1 0 -1 0 0 1 350

s3 0 2 1 0 0 1 0 0 600

Zj -2M -M M -M 0 -M -M -475 Cj-Zj -2+2M -3+M -M -M 0 0 0 Which variable need to go basis ????

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 x2 s1 s2 s3 a2 -2 -3 0 0 0 -M x1 -2 1 0 -1 0 0 0 125

a2 -M 0 1 1 -1 0 1 225

s3 0 0 1 2 0 1 0 350

Zj -2 -M 2- M M 0 -M -250-225M

Cj-Zj 0 -3+M -2+M -M 0 0 Which variable need to go basis ????

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 s2 s3 -2 -3 0 0 0 x1 -2 1 0 -0 1 1 250

x2 -3 0 1 0 -2 -1 100

s1 0 0 0 1 1 1 125

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

Basic CB x1 x2 s1 s2 s3 s4 a4 50 40 0 0 0 0 -M X2 40 0 1 8/25 0 -3/25 0 0 12 S2 0 0 0 -8/25 1 3/25 0 0 8 X1 50 1 0 -5/25 0 5/25 0 0 30 a4 -M 0 0 -3/25 0 -2/25 -1 1 8 Zj 50 40 (70+3M)/

25 0 (130+2M)/25 M -M 1980-8M

Cj-Zj 0 0 (-70-3M)/25 0 (-130-2M)/

25 -M 0 Feasible Solution ???????

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. A Maximization Problem is unbounded if it is possible to make the value of optimal solution as large as possible without violating the constraints.

2. A surplus variable can be interpreted as amount of the basis variable over the minimum amount required.

3. 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

Range of optimality for an objective function 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

FINAL SIMPLEX OF THE PROBLEM

Basic cb X1 x2 s1 s2 s3

50 40 0 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 -14/5 0 -26/5

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)

Basic cb X1 x2 s1 s2 s3

C1 40 0 0 0 X2 40 0 1 8/25 0 -3/25 12 s2 0 0 0 -8/25 1 3/25 8 X1 C1 1 0 -5/25 0 5/25 30 Zj c1 40 (64-c1)/5 0 (c1-24)/5

480+30c1

Cj-Zj 0 0 c1-64/5 0 24-c1/5

Applying cj-zj <=0 -> c1-64/5 <=0 C1-64<=0 -- C1<=64 24-c1<=0-- 24<=C1 thus 24<=C1<=64

DISCUSSION

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 X2 40 0 1 8/25 0 -3/25 12 s2 0 0 0 -8/25 1 3/25 8 X1 30 1 0 -5/25 0 5/25 30 Zj 30 40 34/5 0 6/5 1380

Cj-Zj 0 0 -34/5 0 -6/5 HOW???

DISCUSSION

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 X2 40 0 1 8/25 0 -3/25 12 s2 0 0 0 -8/25 1 3/25 8 X1 20 1 0 -5/25 0 5/25 30 Zj 20 40 44/5 0 4/5

1080 Cj-Zj 0 0 -44/5 0 4/5

RANGE OF OPTIMALITY FOR BASIC VARIABLE

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$

FINAL SIMPLEX OF THE PROBLEM

Basic cb X1 x2 s1 s2 s3

50 40 0 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 -14/5 0 -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.

SETTING NON BASIC VAIABLES

B1 a1j 0B2 a2j 0B3 + delB* a3j >= 0..BN anj 0

Current solution or l Last column of the final column corresponding to slack variable tableu

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

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.

SLOVING WITH SIMPLEX

u1 u2 u3 s1 s2 a1 a2 Basic Cb -150 -20 -300 0 0 -M -M

why- A1 -M 3 0 8 -1 0 1 0

50 A2 -M 5 1 5 0 -1 0 1

40

Zj -8M -M -13M M M -M -M -90M

Cj-1 -150+8M -20+M -300+13M –M –M 0 0

FINAL SIMPLEX

u1 u2 u3 s1 s2 Basic Cb -150 -20 -300 0 0 u3 -300 0 -3/25 1 -5/25 3/25 26/5 u1 -150 1 8/25 0 5/25 -8/25 14/5

Zj -150 -12 -300 30 12 Cj-1 0 -8 0 -30 -12

-1980 Assembly Time $2.80 Portable Display= 0 Warehouse space = $5.20

FINAL TABLEAU