5. advance topics in lp

ADVANCED OPERATIONS

RESEARCH

By: -Hakeem–Ur–Rehman IQTM–PU 1

RA OADVANCED TOPICS IN LINEAR

PROGRAMMING

ADVANCE TOPICS IN LINEAR

PROGRAMMING

2

Duality in Linear Programming

The Dual Simplex Method

Important & Efficient Computational

Technique: The Revised Simplex Method / Simplex

Method with Multipliers

Bounded Variables LP Problem

Parametric Analysis

DUALITY IN LINEAR PROGRAMMING

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It is interesting that every linear programming model has two forms: The Primal and The Dual.

The original form of a linear programming model is called the Primal.

The dual is an alternative model form derived completely from the primal. The dual is useful because it provides the decision maker with an alternative way of looking at a problem.

DUALITY (Cont…)

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DUAL OF AN LP PROBLEM:The dual is derived completely from the primal; for easily getting the dual from the primal; divide the linear programming problems into two forms:

1. Normal Linear Programming Problem2. Non–Normal Linear Programming Problem

1. NORMAL LINEAR PROGRAMMING PROBLEM:

We call normal maximum LP problem is an LP problem in which all the variables are required to be non–negative and all the constraints are less than or equal to (≤) form.

For Example: a normal maximum LP problem has the following form:

Maximum: Z = C1X1+C2X2+ - - - +CnXn

Subject to:

a11X1+a12X2+ - - - +a1nXn ≤ b1

a21X1+a22X2+ - - - +a2nXn ≤ b2

-

-

-

am1X1+am2X2+ - - - +amnXn ≤ bm

Where, Xj ≥ 0, j = 0, 1, 2, - - -,n.

While, a normal minimum LP problem is an LP problem in which all the variables are required to be non–negative and all the constraints are greater than or equal to (≥) form.

For Example: a normal minimum LP problem has the following form:

Minimum: Z = C1X1+C2X2+ - - - +CnXn

Subject to: a11X1+a12X2+ - - - +a1nXn ≥ b1

a21X1+a22X2+ - - - +a2nXn ≥ b2

---

am1X1+am2X2+ - - - +amnXn ≥ bm

Where, Xj ≥ 0, j = 0, 1, 2, - - -,n.

STEPS FOR PRIMAL PROBLEM TO DUAL PROBLEM IN CASE OF NORMAL LP PROBLEM:

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Step–1: If the primal problem is in a maximization form then the dual problem will be in minimization form. But if the primal problem is in a minimization form then the dual problem will be in maximization form.Step–2: The number of decision variables in the dual problem is equal to the number of constraints in the primal problem.Step–3: The quantities (numerical values) which appear on the right hand side (RHS) of the constraints of the primal problem become the coefficients of the decision variables in the objective function of the dual problem. Step–4: The number of constraints in the dual problem is equal to the number of variables in the primal problem. Step–5: The coefficients of the variables in the constraints of the primal problem which appear from left to right be placed from top to bottom in the constraints of the dual problem. Step–6: If the primal problem has less than or equal to (≤) type constraints then the dual problem will have greater than or equal to (≥) type constraints; But if the primal problem has greater than or equal to (≥) type constraints then the dual problem will have less than or equal to (≤) type constraints. Step–7: The coefficients of the objective function of the primal problem which appears on the right hand side (RHS) of the constraints of the dual problem. Step–8: Non–negativity restriction will also apply to decision variable of dual problem.

DUALITY (Cont…)

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PRIMAL MAXIMIZATION / DUAL MINIMIZATION Maximum: Z = CTXSubject to: AX ≤ b X ≥ 0

Minimum: Z = bTYSubject to: ATY ≥ C Y ≥ 0

PRIMAL MINIMIZATION / DUAL MAXIMIZATIONMinimization: Z = CTXSubject to: AX ≥ b X ≥ 0

Maximization: Z = bTYSubject to: ATY ≤ C Y ≥ 0

DUALITY (Cont…)

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1) Find the dual of the following problem:Maximize: Z = 3X1 + 4X2

Subject to:2X1 + 3X2 ≤ 164X1 + 2X2 ≤ 16 X1, X2 ≥ 0

Dual problem in case of normal LP problem;

Minimize: Z = 16Y1 + 16Y2

Subject to: 2Y1 + 4Y2 ≥ 3 3Y1 + 2Y2 ≥ 4 Y1, Y2 ≥ 0

2) Find the dual of the following problem:Maximize: Z = 30X1 + 35X2 +50X3

Subject to: 6X1 + 3X2 + 7X3 ≤ 53 5X1 + 3X2 + 4X3 ≤ 19

2X1 + 4X2 + 5X3 ≤ 11 X1, X2, X3 ≥ 0

Dual problem in case of normal LP problem; Minimize: Z = 53Y1 + 19Y2 + 11Y3

Subject to: 6Y1 + 5Y2 + 2Y3 ≥ 30 3Y1 + 3Y2 + 4Y3 ≥ 35 7Y1 + 4Y2 + 5Y3 ≥ 50

Y1, Y2, Y3 ≥ 0Find the dual of the following problem:

Minimize: Z = 40X1 + 200X2 Subject to: 4X1 + 40X2 ≥ 160 3X1 + 10X2 ≥ 60

8X1 + 10X2 ≥ 80 X1, X2 ≥ 0

Dual problem in case of normal LP problem; Maximize: Z = 160Y1 + 60Y2 + 80Y3

Subject to: 4Y1 + 3Y2 + 8Y3 ≤ 40 40Y1 + 10Y2 + 10Y3 ≤ 200 Y1, Y2, Y3 ≥ 0

STEPS FOR PRIMAL PROBLEM TO DUAL PROBLEM IN CASE OF NON–NORMAL LP PROBLEM: Step–1: If the given LP problem is in non–normal form then we first convert it into normal form, for converting the non–normal LP problem into normal LP problem adopt the following procedure:

IN CASE OF MAXIMIZATION PROBLEMConstraints / Variable Type Procedure

1. If Less than or equal to (≤) type2. If greater than or equal to (≥) type3. If equal to (=) type

4. Unrestricted–in–sign Decision Variable (Xi)

No Change is requiredConvert the ‘≥’ type inequality into ‘≤’ type by multiplying it by ‘–1’. a. Convert the equality into two inequalities in which one having

‘≥’ sign while other having ‘≤’ sign. b. Convert the ‘≥’ type inequality into ‘≤’ type by multiplying it by

‘–1’. Any unrestricted in sign decision variable can be rewritten ‘X i’ as the difference Xi = Xi

/ – Xi// of two non–negative decision variables

Xi/, Xi

//.

IN CASE OF MINIMIZATION PROBLEMConstraints / Variable Type Procedure

1. If Less than or equal to (≤) type2. If greater than or equal to (≥) type3. If equal to (=) type

4. Unrestricted–in–sign Decision Variable (Xi)

Convert the ‘≤’ type inequality into ‘≥’ type by multiplying it by ‘–1’. No Change is requireda. Convert the equality into two inequalities in which one having ‘≥’

sign while other having ‘≤’ sign. b. Convert the ‘≤’ type inequality into ‘≥’ type by multiplying it by ‘–

1’. Any unrestricted in sign decision variable can be rewritten ‘X i’ as the difference Xi = Xi

/ – Xi// of two non–negative decision variables X i

/, Xi//.

DUALITY (Cont…)

STEPS FOR PRIMAL PROBLEM TO DUAL PROBLEM IN CASE OF NON–NORMAL LP PROBLEM (Cont…):

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Step–2: Restate the primal problem after taking step–1. Step–3: If the primal problem is in a maximization form then the dual problem will be in minimization form. But if the primal problem is in a minimization form then the dual problem will be in maximization form.Step–4: The number of decision variables in the dual problem is equal to the number of constraints in the primal problem.Step–5: The quantities (numerical values) which appear on the right hand side (RHS) of the constraints of the primal problem become the coefficients of the decision variables in the objective function of the dual problem. Step–6: The number of constraints in the dual problem is equal to the number of variables in the primal problem. Step–7: The coefficients of the variables in the constraints of the primal problem which appear from left to right be placed from top to bottom in the constraints of the dual problem. Step–8: If the primal problem has less than or equal to (≤) type constraints then the dual problem will have greater than or equal to (≥) type constraints; But if the primal problem has greater than or equal to (≥) type constraints then the dual problem will have less than or equal to (≤) type constraints. Step–9: The coefficients of the objective function of the primal problem which appears on the right hand side (RHS) of the constraints of the dual problem. Step–10: Non–negativity restriction will also apply to decision variable of dual problem.

DUALITY (Cont…)

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Find the dual of the following problem:Maximize: Z = 6X1 + 4X2 + 6X3 + X4 Subject to:

4X1 + 5X2 + 4X3 + 8X4 = 21 3X1 + 7X2 + 8X3 + 2X4 ≤ 48

X1, X2, X3, X4 ≥ 0

DUALITY (Cont…)As the given LP problem is in the form of maximum

non–normal form so we need to convert it into maximum normal LP problem in which we need all the constraints in the form of less than or equal to ‘≤’. So,

Constraint–1: 4X1 + 5X2 + 4X3 + 8X4 = 21; Now we convert it into two inequalities that are: 4X1 + 5X2 + 4X3 + 8X4 ≤ 21 4X1 + 5X2 + 4X3 + 8X4 ≥ 21Inequality having ‘≥’ sign; convert it into ‘≤’ form by multiplying with ‘–1’; we get: –4X1 – 5X2 – 4X3 – 8X4 ≤ –21

Now restating the primal as below:Maximize: Z = 6X1 + 4X2 + 6X3 + X4 Subject to: 4X1 + 5X2 + 4X3 + 8X4 ≤ 21 –4X1 – 5X2 – 4X3 – 8X4 ≤ –21 3X1 + 7X2 + 8X3 + 2X4 ≤ 48

X1, X2, X3, X4 ≥ 0

Note: Regarding the step–4 of the algorithm, the number of decision variables in the dual problem is equal to the number of constraints in the primal problem. But we can see that in the given original primal there are two ‘2’ constraints and four ‘4’ variables while in dual there are four ‘4’ constraints and three ‘3’ variables which shows the inconsistency. So in order to remove such an inconsistency: let (Y1 – Y2) = Y/.

Now follow the remaining steps for converting the primal into dual; we will get the dual of the given problem which is: Minimize: Z = 21Y/ + 48Y3

Subject to: 4Y/ + 3Y3 ≥ 6 5Y/ + 7Y3 ≥ 4 4Y/ + 8Y3 ≥ 6 8Y/ + 2Y3 ≥ 1Where Y3 ≥ 0; Y/ (unrestricted in sign)

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Find the dual of the following problem:Maximize: Z = 6X1 + 8X2Subject to:

2X1 + 3X2 ≤ 16 4X1 + 2X2 ≥ 16 2X1 + X2 = 16 X1, X2 ≥ 0

DUALITY (Cont…)

As the given LP problem is in the form of maximum non–normal form so we need to convert it into maximum normal LP problem in which we need all the constraints in the form of less than or equal to ‘≤’. So,

Now restating the primal as below:Maximize: Z = 6X1 + 8X2Subject to:

2X1 + 3X2 ≤ 16 –4X1 – 2X2 ≤ –16

2X1 + X2 ≤ 16 –2X1 – X2 ≤ –16

X1, X2 ≥ 0

DUAL:Minimize: Z = 16Y1 –16Y2 + 16Y3 –16Y4Subject to:

2Y1 – 4Y2 + 2 Y3 – 2Y4 ≥ 6 3Y1 – 2Y2 + Y3 – Y4 ≥ 8 Y1, Y2, Y3, Y4 ≥ 0

Minimize: Z = 16Y1 –16Y2 + 16Y/

Subject to: 2Y1 – 4Y2 + 2Y/ ≥ 6 3Y1 – 2Y2 + Y/ ≥ 8

Where, Y1, Y2 ≥ 0; Y/ unrestricted in sign

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DUALITY (Cont…)

Find the dual of the following problem: Maximize: Z = 12X1 + 15X2 + 9X3

Subject to: 8X1 + 16X2 + 12X3 ≤ 25 4X1 + 8X2 + 10X3 ≥ 80 7X1 + 9X2 + 8X3 = 105 X1, X2, X3 ≥ 0

Minimize: Z = 25Y1 – 80Y2 + 105Y/

Subject to: 8Y1 – 4Y2 + 7Y/ ≥ 1216Y1 – 8Y2 + 9Y/ ≥ 15

12Y1 – 10Y2 + 8Y/ ≥ 9Where, Y1, Y2 ≥ 0; Y/ unrestricted in sign

Find the dual of the following problem:Maximize: Z = 3X1 + X2 + X3 – X4Subject to: X1 + X2 + 2X3 + 3X4 ≤ 5 X3 – X4 ≥ –1

X1 – X2 = –1 X1, X2, X3, X4 ≥

0

Minimize: Z = 5Y1 + Y2 – Y/ Subject to:

Y1 + Y/ ≥ 3 Y1 – Y/ ≥ 1 2Y1 – Y2 ≥ 1

3Y1 + Y2 ≥ –1Where, Y1, Y2 ≥ 0; Y/ (unrestricted in sign)

INTERPRETATION OF THE PRIMAL–DUAL SOLUTION RELATIONSHIP:

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PRIMAL PROBLEM: Solve the given LP problem.

Minimize: Z = 40X1 + 200X2Subject to:

4X1 + 40X2 ≥ 160 3X1 + 10X2 ≥ 60

8X1 + 10X2 ≥ 80 X1 , X2 ≥ 0

DUALITY (Cont…)

Contribution Per Unit Cj –40 –200 0 0 0 –M –M –MRatioCBi

Basic Variables (B)

Quantity(Qty) X1 X2 S1 S2 S3 A1 A2 A3

–M A1 160 4 40* –1 0 0 1 0 0 4 ←–M A2 60 3 10 0 –1 0 0 1 0 6–M A3 80 8 10 0 0 –1 0 0 1 8

Total Profit (Zj) –300M –15M –60M M M M –M –M –MNet Contribution (Cj – Zj) –40+15M –200+60M

↑–M –M –M 0 0 0

Maximize: Z = –40X1 – 200X2 + 0S1 + 0S2 + 0S3 – MA1 – MA2 – MA3

Subject to:4X1 + 40X2 – S1 + A1 = 1603X1 + 10X2 – S2 + A2 = 608X1 + 10X2 – S3 + A3 = 80

Where: X1, X2, X3, S1, S2, S3, A1, A2, A3 ≥ 0


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DUALITY (Cont…)


Basic Variables (B)


–200 X2 4 1/10 1 –1/40 0 0 1/40 0 0 40–M A2 20 2 0 1/4 –1 0 –1/4 1 0 10–M A3 40 7* 0 1/4 0 –1 –1/4 0 1 40/7 ←

Total Profit (Zj) –800–60M –20–9M –200 5 – M/2 M M –5+M/2 –M –MNet Contribution (Cj – Zj) –20+9M ↑ 0 –5+M/2 –M –M 5–3M/2 0 0


B.V. (B)


–200 X2 24/7 0 1 –1/35 0 1/70 1/35 0 –1/70 240–M A2 60/7 0 0 5/28 –1 (2/7)* –5/28 1 –2/7 30 ←–40 X1 40/7 1 0 1/28 0 –1/7 –1/28 0 1/7 ---

Total Profit: (Zj)

(–6400–60M)/7 –40 –200 (120–

5M)/28 M (20–2M)/7 (–120+ 5M)/28 –M (–20+

2M)/7Net Contribution (Cj – Zj) 0 0 (–120+

5M)/28 –M [(–20+ 2M)/7] ↑

(120–5M)/28 0 (20–

2M)/7


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DUALITY (Cont…)

Contribution Per Unit Cj

–40 –200 0 0 0 –M –M –M

CBiB.V. (B)


–200 X2 3 0 1 –21/560 1/20 0 21/560 –1/20 00 S3 30 0 0 5/8 –7/2 1 –5/8 7/2 –1

–40 X1 10 1 0 1/8 –1/2 0 –1/8 1/2 0Total Profit: (Zj) –1000 –40 –200 5/2 10 0 –M+1/8 –M+1/2 –MNet Contribution (Cj – Zj) 0 0 –5/2 –10 0 –1/8 –1/2 0

Value of ‘Z’ = (–40)(10) + (–200)(3) = –1000Optimal solution is –1000 for Max. ‘Z’; at X1 = 10, X2 = 3. Because Min (Z) = –Max (–Z) = 1000; at X1 = 10, X2 = 3.


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DUALITY (Cont…)

PRIMAL PROBLEM: Solve the given LP problem.Minimize: Z = 40X1 + 200X2

Subject to: 4X1 + 40X2 ≥ 160

3X1 + 10X2 ≥ 60 8X1 + 10X2 ≥ 80 X1 , X2 ≥ 0

DUAL PROBLEM:The Dual of the above primal is written as follows:Maximize: Z = 160Y1 + 60Y2 + 80Y3

Subject to: 4Y1 + 3Y2 + 8Y3 ≤ 40 40Y1 + 10Y2 + 10Y3 ≤ 200

Y1 , Y2, Y3 ≥ 0


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DUALITY (Cont…)

Maximize: Z = 160Y1 + 60Y2 + 80Y3 + 0S1 + 0S2 Subject to: 4Y1 + 3Y2 + 8Y3 + S1 = 40 40Y1 + 10Y2 + 10Y3 + S2 = 200 Y1, Y2, Y3, S1, S2 ≥ 0

Contribution Per Unit Cj 160 60 80 0 0Ratio CBi

Basic Variables (B)

Quantity(Qty) Y1 Y2 Y3 S1 S2

0 S1 40 4 3 8 1 0 100 S2 200 40* 10 10 0 1 5 ←

Total Profit (Zj) 0 0 0 0 0 0Net Contribution (Cj – Zj) 160 ↑ 60 80 0 0 Contribution Per Unit Cj

160 60 80 00

Ratio CBi

Basic Variables (B)

Quantity(Qty) Y1 Y2 Y3 S1 S2

0 S1 20 0 2 7* 1 –1/10 20/7 ←160 Y1 5 1 1/4 1/4 0 1/40 20

Total Profit (Zj) 800 160 40 40 0 4Net Contribution (Cj – Zj) 0 20 40 ↑ 0 –4


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DUALITY (Cont…)


160 60 80 0 0Ratio

CBiBasic Variables

(B)Quantity

(Qty) Y1 Y2 Y3 S1 S2

80 Y3 20/7 0 (2/7)* 1 1/7 –1/70 10 ←160 Y1 30/7 1 5/28 0 –1/28 1/35 24

Total Profit (Zj) 6400/7 160 360/7 80 40/7 24/7Net Contribution (Cj – Zj) 0 60/7 ↑ 0 –40/7 –24/7

Contribution Per Unit Cj 160 60 80 0 0

CBi Basic Variables (B) Quantity(Qty) Y1 Y2 Y3 S1 S2

60 Y2 10 0 1 7/2 1/2 –1/20160 Y1 5/2 1 0 –5/8 –1/8 3/80

Total Profit (Zj) 1000 160 60 110 10 3Net Contribution (Cj – Zj) 0 0 –30 –10 –3

Value of ‘Z’ = (60)(10) + (160)(5/2) = 1000; at Y1 = 5/2, Y2 = 10.


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DUALITY (Cont…)

COMPARISON OF THE PRIMAL AND DUAL OPTIMAL TABLE

Contribution Per Unit Cj –40 –200 0 0 0

CBiB.V. (B)

Quantity(Qty) X1 X2 S1 S2 S3

–200 X2 3 0 1 –21/560 1/20 00 S3 30 0 0 5/8 –7/2 1

–40 X1 10 1 0 1/8 –1/2 0Total Profit: (Zj) –1000 –40 –200 5/2 10 0Net Contribution (Cj – Zj) 0 0 –5/2 –10 0

PRIMAL OPTIMAL SOLUTION (After deleting artificial variable columns)

Contribution Per Unit Cj 160 60 80 0 0

CBi Basic Variables (B) Quantity(Qty) Y1 Y2 Y3 S1 S2

60 Y2 10 0 1 7/2 1/2 –1/20160 Y1 5/2 1 0 –5/8 –1/8 3/80

Total Profit (Zj) 1000 160 60 110 10 3Net Contribution (Cj – Zj) 0 0 –30 –10 –3

DUAL OPTIMAL SOLUTION


DUALITY (Cont…)

COMPARISON OF THE PRIMAL AND DUAL OPTIMAL TABLENote: The respective coefficients of S1, S2 & Y3 in the (Cj – Zj) row (ignoring the sign) from the Dual optimal solution table are 10, 3 & 30.

Now, it is clear that the primal and dual lead to the same solution, even though they are formulated differently.

It is also clear that in the final simplex table of primal problem, the absolute value of the numbers in (C j – Zj) row under the slack variable represents the solutions to the dual problem. In another words it also happens that the absolute value of the (Cj – Zj) values of the slack variable in the optimal dual solution represent the optimal values of the primal ‘X1’ and ‘X2’ variables.

The minimum opportunity cost derived in the dual must always be equal the maximum profit derived in the primal.

PRIMAL–DUAL RELATIONSHIPS:

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DUALITY (Cont…)

WEAK DUALITY THEOREM: If ‘X/’ is a feasible solution for primal problem and ‘Y/’ is a feasible

solution for dual problem, then: (Primal Objective function value) CTX/ ≤ bTY/ (Dual Objective function

value) if the primal is in a maximization case & (Primal Objective function value) CTX/ ≥ bTY/ (Dual Objective function

value) if the primal is in a minimization case This is called the weak duality. The non–negative quantity β = │bTY/ – CTX/│is called the duality gap. So,

when β = 0, we can say that there is no duality gap.

STRONG DUALITY: If ‘X*’ is an optimal solution for primal problem and ‘Y*’ is an optimal

solution for dual problem, then the optimal objective value of the primal is the same as the optimal objective value of the dual: CTX* = bTY*; in this case, the duality gap is zero so this is called the strong duality.

DUAL OF THE DUAL IS PRIMAL

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DUAL SIMPLEX METHOD Dual simplex method, developed by C.E. Lemte, is very similar to

the regular simplex method.

The only differences lies in the criterion used for selecting a variable to enter the basis (Basic Variables) and the leave the basis (Basic Variables). In dual simplex method, we first select the variable to leave the basis (Basic Variables) and then the variable to enter the basis (Basic Variables).

In this method the solution starts from optimum but infeasible and remains infeasible until the true optimum is reached at which the solution becomes feasible.

The advantage of this method is avoiding the artificial and surplus variables introducing in the constraints, as the constraint is in the form of greater than or equal to ‘≥’ converted into less than or equal to ‘≤’.

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DUAL SIMPLEX ALGORITHMConvert the minimization problem, if any, into the

maximization

Convert “≥” type constraints, if any, into “≤” type

Convert “≤” type constraints into equations and setup the initial dual simplex table

Compute “Cj – Zj” values

Are All “Cj – Zj” ≤

0And bi ≥ 0

Optimal Solution Obtained

YES

NOa) Select Key row with the most negative “bi”b) Find ratios of “Cj – Zj” elements to the negative

elements of key row. Select key column with minimum ratio

Mark the key element, perform row operations as in regular simplex method

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DUAL SIMPLEX METHODEXAMPLE : Use Dual Simplex method to solve the LP problem.

Maximize: Z = –3X1 – X2Subject to:

X1+ X2 ≥ 1 2X1 + 3X2 ≥ 2

X1, X2 ≥ 0Step–1: Given problem is already in the form of maximization. So go to the next step. Step–2: Convert the Given constraints into (≤) form by multiplying with ‘–1’ both sides. Maximize: Z = –3X1 – X2

Subject to: –X1 – X2 ≤ –1 –2X1 –3X2 ≤ –2

X1, X2 ≥ 0Step–3: After introducing slack variables (S1, S2) standard form of the given problem is:

Maximize: Z = –3X1 – X2 + 0S1 + 0S2Subject to:

–X1 – X2 + S1 = –1 –2X1 –3X2 + S2 = –2

X1, X2, S1, S2 ≥ 0

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DUAL SIMPLEX METHOD (Cont…) COntribuTion Per Unit Cj –3 –1 0 0CBi

Bas)c VarIables (B)

Quantity(Qty) X1 X2 S1 S2

0 S1 –1 –1 –1 1 00 S2 –2 –2 –3 0 1

Total Profit (Zj) 0 0 0 0 0Net Contribution (Cj – Zj) –3 –1 0 0Step–4: Since all (Cj – Zj) ≤ 0 and all the values in the quantity (Qty)

column less than (<) zero. So, the current solution is not optimum basic feasible solution.

Now, we display the initial simplex table:


–3 –1 0 0

CBi

Basic Variables

(B)Quantity

(Qty) X1 X2 S1 S2

0 S1 –1 –1 –1 1 00 S2 ← –2 –2 –3* 0 1

Total Profit (Zj) 0 0 0 0 0Net Contribution (Cj – Zj) –3 –1 0 0Replacement Ratio 3/2 (1/3) ↑ --- ---

Step–5:

Step–6: Pivot row indicates that outgoing (leaving) variable is ‘S2’ While Pivot column indicates that incoming (entering) variable is ‘X2’. So, we replace the outgoing (leaving) variable ‘S2’ by the incoming (entering) variable ‘X2’ together with its contribution per unit.

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DUAL SIMPLEX METHOD (Cont…)Now, preparing the next duaL simplex table Contribution Per Unit Cj –3 –1 0 0

CBiBasic

Variables (B)Quantity

(Qty) X1 X2 S1 S2

0 S1 ← –1/3 –1/3 0 1 (–1/3)*–1 X2 2/3 2/3 1 0 –1/3

Total Profit (Zj) –2/3 –2/3 –1 0 1/3Net Contribution (Cj – Zj) –7/3 0 0 –1/3Replacement Ratio 7 --- --- 1 ↑

Contribution Per Unit Cj –3 –1 0 0CBi

Basic Variables (B)

Quantity(Qty) X1 X2 S1 S2

0 S2 1 1 0 –3 1–1 X2 1 1 1 –1 0

Total Profit (Zj) –1 –1 –1 1 0Net Contribution (Cj – Zj) –2 0 –1 0Since all (Cj – Zj) ≤ 0 and all the values in the quantity (Qty) column greater than or

equal to zero (≥ 0); so, the current solution is optimum feasible solution. Thus, the optimal feasible solution to the given LP problem is:

Maximum Z = –1; at X1 = 0, X2 = 1.

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DUAL SIMPLEX METHOD (Cont…)

PRACTICE QUESTION:Use Dual Simplex method to solve the LP problem.

Minimize: Z = X1 + 2X2 + 3X3Subject to:

X1 – X2 + X3 ≥ 4 X1 + X2 + 2X3 ≤ 8 X2 – X3 ≥ 2 X1, X2, X3 ≥ 0

REVISED SIMPLEX METHOD / SIMPLEX METHOD WITH

MULTIPLIERS

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This method is a modified version of the Primal Simplex Method that we studied Earlier.

It is designed to exploit the fact that in many practical applications the coefficient matrix {aij} is very sparse, namely most of its elements are equal to zero.

BOTTOM LINE: Don’t update all the columns of the

simplex tableau: update only those columns that you need!

REVISED SIMPLEX METHOD (Cont…)

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Max Z = c x subject to A x ≤ b x ≥ 0

BASIC VARIABLE

VALUES:

Initially constraints become (standard form):


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STEP IN THE REVISED SIMPLEX METHOD1. Write the LP Problem into Standard form.2. Write the Set of Basic Variable “XB” and

Corresponding Basis “B” (Coefficients against Basic Variables in the constraints) Matrix.

3. Determine entering variable, Xj

compute Y = cBB-1 BY= CB Where Bn =B(n-

1)En Where En is a ETA Matrix “Identity Matrix having one column non-identity which equal to (P–Bar)j

compute Cj – Zj = Cj – Y Pj

for all non-basic variables.

Choose largest Positive value (maximization). If none, stop.

4. Determine leaving variable, XB compute P–Bar = B-1Pj Find Ratios. Choose the Smallest positive value.


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EXAMPLE:Max. Z = 6X1 + 8X2Subject to:

X1 + X2 ≤ 102X1 + 3X2 ≤ 25X1 + 5X2 ≤ 35

Xj ≥ 0

Standard LP Form:Max. Z = 6X1 + 8X2 + 0X3 + 0X4 + 0X5Subject to:

X1 + X2 + X3 = 102X1 + 3X2 +X4 = 25X1 + 5X2 + X5 = 35

Xj ≥ 0ITERATION # 0:


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Selection of the Entering Variable from Non–Basic Variables using: (Cj – Zj) = (Cj – YPj)

Selection of the Leaving Variable from Basic Variables using:


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ITERATION # 1:



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REVISED SIMPLEX METHOD (Cont…)ITERATION # 2:


So, X5 will enter the basis


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So, X3 will Leave the Basis.


Because all the values of (Cj – Zj) ≤ 0; So, the current solution is optimum at X1 = 5 and X2 = 5; Z = 6(5)+8(5)=70

QUESTIONS

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5. advance topics in lp

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