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Gomory’s Cutting Plane MethodPRESENTER

RAJESH PIRYANI

SOUTH ASIAN UNIVERSITY

Outline1. Why Integer Programming

2. Introduction to All Integer Linear Programming Problem (AILP) and Mixed Integer Linear Programming Problem (MILP)

3. Common Approach for solving AILP

4. Introduction to Gomory’s Cutting Plane Method

5. Derivation of Gomory’s Cutting Plane Method

6. Gomory’s Cutting Plane Method Algorithms

7. Explaination of Gomory’s Cutting Plane Method Algorithm with Example

5/8/2015 HTTPS://SITES.GOOGLE.COM/SITE/PIRYANIRAJESH/ 2

Why Integer ProgrammingProduction Problem◦ Items being produced may be in complete units

◦ E.g. TV Sets of 21” and 29”

◦ Therefore fractional number of item have no meaning


IPP ExpressionPROBLEM DEFINITION

𝑀𝑎𝑥 𝑧 = 𝑗=1𝑛 𝑐𝑗𝑥𝑗

subject to

𝑗=1𝑛 𝑎𝑖𝑗𝑥𝑗 = 𝑏𝑖 (𝑖 = 1,… ,𝑚)

𝑥𝑗 ≥ 0 (j=1,…,n)

and

𝑥𝑗 𝑖𝑛𝑡𝑒𝑔𝑒𝑟 𝑓𝑜𝑟 𝑗1∁ 𝑗

where j={1,2, … ,n}

DEFINITION

All Integer LPP (AILP):- If all variable takeinteger values only. (if 𝒋𝟏 = 𝒋)

(slack & surplus variable take integer value)

Mixed Integer LPP (MILP):- If some but not all variable of the problem are constrained Integer values.


IPP ExampleEXAMPLE OF AILP

𝑀𝑎𝑥 𝑧 = 4𝑥1 + 3𝑥2

subject to

𝑥1 + 𝑥2 ≤ 8

2𝑥1 + 𝑥2 ≤ 10

𝑥1, 𝑥2 ≥ 0

and

𝑥1 𝑎𝑛𝑑 𝑥2 𝑖𝑛𝑡𝑒𝑔𝑒𝑟

𝑥1 and 𝑥2 are non-negative integer

slack variable

𝑥3 = 8 − 𝑥1 − 𝑥2 & 𝑥4 = 10 − 2𝑥1 − 𝑥2

are also non-negative integer

if we consider 2nd constraints is given as:

2𝑥1 + 𝑥2 ≤ 10;

𝑥4 = 10 − 2𝑥1 − 𝑥2

Then this problem no more AILP. But MILP


Common Approach (Rounding off)PROBLEM

𝑀𝑎𝑥 𝑧 = 21𝑥1 + 11𝑥2

subject to

7𝑥1 + 4𝑥2 ≤ 13

𝑥1, 𝑥2 ≥ 0

and


The Feasible Set of discrete points

0,0 , 0,1 , 1,0 , 1,1 , 0,2 , 0,3 .

Lies inside feasible region, can be visualize in figure

Optimal Soln. of ILP 𝑥1∗ = 0, 𝑥2

∗ = 3, 𝑧∗ = 33Optimal Soln. of LLP (𝑥1

∗ = 13/7, 𝑥2∗ = 0, 𝑧∗ = 39)

Rounding LLP Soln. (𝑥1∗ = 2, 𝑥2

∗ = 0, 𝑧∗ = 42), two obj. fn are not close in any meaningful sense

Rounding off is not correct approach to solve ILP’s5/8/2015 HTTPS://SITES.GOOGLE.COM/SITE/PIRYANIRAJESH/ 6

Common Approach (Convex Hull)PROBLEM

𝑀𝑎𝑥 𝑧 = 21𝑥1 + 11𝑥2

subject to

7𝑥1 + 4𝑥2 ≤ 13

𝑥1, 𝑥2 ≥ 0

and


The Feasible Set of the given ILP is non convex, its convex hull is apolytope whose corner points meet the integer requirements.

0,0 , 0,1 , 1,0 , 1,1 , 0,2 , 0,3 .

Lies inside feasible region, can be visualize in figure


∗ = 3, 𝑧∗ = 33


Common Approach (Convex Hull)PROBLEM (ILP EQUIVALENT TO SOLVING LPP)

𝑀𝑎𝑥 𝑧 = 𝑗=1𝑛 𝑐𝑗𝑥𝑗

subject to

(𝑥1, … , 𝑥𝑛) ∈ 𝑆,

Where S is the polytope

𝑗=1

𝑛

𝑎𝑖𝑗𝑥𝑗 = 𝑏𝑖(𝑖 = 1,… ,𝑚)

𝑥𝑗 ≥ 0 (j=1,…,n)

and

𝑥𝑗 𝑖𝑛𝑡𝑒𝑔𝑒𝑟 𝑓𝑜𝑟 𝑗 ∈ 𝐽1∁ 𝐽 = {1,… , 𝑛}.


∗ = 3, 𝑧∗ = 33This method is perfectly valid except that there are certain practical difficulties in getting the convex hull. When Euclidean space is more than two or three


Gomory’s Cutting Plane method for AILPPROBLEM (ILP EQUIVALENT TO SOLVING LPP)

𝑀𝑎𝑥 𝑧 = 𝑐𝑇𝑥

subject to

𝐴𝑥 = 𝑏,

𝑥 ≥ 0

𝑥 𝑖𝑛𝑡𝑒𝑔𝑒𝑟

𝐴, 𝑏 𝑎𝑛𝑑 𝑐 are integer,

The objective function is automatically constrained to be integer.

Let

(𝑳𝑷)𝟏→ 𝑨𝒔𝒔𝒐𝒄𝒊𝒂𝒕𝒆𝒅 𝑳𝑷𝑷 𝒇𝒐𝒓 𝑨𝑰𝑳𝑷

𝒙(𝟏) → 𝑶𝒑𝒕𝒊𝒎𝒂𝒍 𝑺𝒐𝒍𝒖𝒕𝒊𝒐𝒏

(if all constrained are integer then it is optimal solution.

Else according to Gomory,

A new constrained 𝒑𝑻𝒙 ≤ 𝒅 append to new (𝑳𝑷)𝟏 to get a new (𝑳𝑷)𝟐

The basic purpose of the cut constrained◦ Delete a part of the feasible region 𝑺𝟏◦ Don’t delete the points which have integer

coordinates

Finitely many cut constrained will be needed to solve the given AILP.


Gomory’s Cutting Plane method for AILPDERIVATION OF THE GOMORY’S CUT CONSTRAINT

𝑀𝑎𝑥 𝑧 = 𝑐𝑇𝑥

subject to

𝐴𝑥 = 𝑏,

𝑥 ≥ 0 (Eq. 1)

𝑨 = 𝑩:𝑹 , 𝒙 = 𝒄𝒐𝒍 𝒙𝑩, 𝒙𝑹 , 𝒄 = 𝒄𝒐𝒍(𝒄𝑩 , 𝒄𝑹)𝐴𝑥 = 𝑏

𝐵: 𝑅𝑥𝐵𝑥𝑅= 𝑏

𝐵𝑥𝐵 + 𝑅𝑥𝑅 = 𝑏𝑥𝐵 = 𝐵

−1𝑏 − 𝐵−1𝑅𝑥𝑅

𝒙𝑩𝒊 = 𝒚𝒊𝟎 − 𝒋∈𝑹𝒚𝒊𝒋𝒙𝒋 𝒇𝒐𝒓(𝒊 = 𝟏,… ,𝒎) (Eq.2)

𝑧 = 𝑐𝑇𝑥= 𝑐𝐵𝑇𝑥𝐵 + 𝑐𝑅

𝑇𝑥𝑅= 𝑐𝐵𝑇(𝐵−1𝑏 − 𝐵−1𝑥𝑅) + 𝑐𝑅

𝑇𝑥𝑅= 𝑐𝐵𝑇 𝐵−1𝑏 − (𝑐𝐵

𝑇𝐵−1𝑅 − 𝑐𝑅𝑇)𝑥𝑅

Which can be written as

𝒙𝑩𝟎 = 𝒚𝟎𝟎 − 𝒋∈𝑹𝒚𝟎𝒋𝒙𝒋 (Eq. 3)

where

𝒙𝑩𝟎 = 𝒛, 𝒚𝟎𝟎 = 𝒄𝑩𝑻 𝑩−𝟏𝒃 & 𝒚𝟎𝒋 = 𝒛𝒋 − 𝒄𝒋

𝒙𝑩𝒊 = 𝒚𝒊𝟎 −

𝒋∈𝑹

𝒚𝒊𝒋𝒙𝒋 𝒇𝒐𝒓 𝒊 = 𝟎, 𝟏,… ,𝒎 (𝑬𝒒. 𝟒)

Where 𝒊 = 𝟎 refers to objective function and 𝒊 =𝟏,… ,𝒎 refers to the m constraints.

𝒚𝟎𝟎 → 𝒄𝒖𝒓𝒓𝒆𝒏𝒕 𝒐𝒃𝒋. 𝒇𝒖𝒏𝒄𝒕𝒊𝒐𝒏𝒚𝒊𝟎 𝒊 = 𝟏,… ,𝒎 → 𝒄𝒖𝒓𝒓𝒆𝒏𝒕 𝒃. 𝒇. 𝒔.


Gomory’s Cutting Plane method for AILPDERIVATION OF THE GOMORY’S CUT CONSTRAINT

𝑨𝑰𝑳𝑷 𝑹𝒆𝒑𝒓𝒔𝒆𝒏𝒕𝒂𝒕𝒊𝒐𝒏 𝑀𝑎𝑥 𝑧 = 𝑐𝑇𝑥

subject to

𝐴𝑥 = 𝑏,

𝑥 ≥ 0, 𝑥 𝑖𝑛𝑡𝑒𝑔𝑒𝑟 (Eq. 0)𝑨𝒔𝒔𝒐𝒄𝒊𝒂𝒕𝒆𝒅 𝑳𝑷𝑷𝑀𝑎𝑥 𝑧 = 𝑐𝑇𝑥

subject to

𝐴𝑥 = 𝑏, 𝑥 ≥ 0 (Eq. 1)

𝒙𝑩𝒊 = 𝒚𝒊𝟎 −

𝒋∈𝑹

𝒚𝒊𝒋𝒙𝒋 𝒇𝒐𝒓 𝒊 = 𝟎, 𝟏, … ,𝒎 (𝑬𝒒. 𝟒)

This holds for any feasible solution of LPP (Eq. 1) and (Eq. 0)

If for any real number a

Fractional part 𝒇𝒂 = 𝒂 − 𝒂[𝒂] → 𝒈𝒓𝒆𝒂𝒕𝒆𝒔𝒕 𝒊𝒏𝒕𝒆𝒈𝒆𝒓 𝒇𝒖𝒏𝒄𝒕𝒊𝒐𝒏

𝟎 ≤ 𝒇𝒂 < 𝟏

For 𝒂 = −𝟏 𝒇𝒂 = 𝟎

But 𝒂 = −𝟏. 𝟔,𝒇𝒂 = −𝟏. 𝟔 − −𝟏. 𝟔 = −𝟏. 𝟔 − −𝟐 = 𝟎. 𝟒


Gomory’s Cutting Plane method for AILP𝒙𝑩𝒊 = 𝒚𝒊𝟎 −

𝒋∈𝑹

𝒚𝒊𝒋𝒙𝒋 𝒇𝒐𝒓 𝒊 = 𝟎, 𝟏, … ,𝒎 (𝑬𝒒. 𝟒)

𝒋∈𝑹

𝒚𝒊𝒋 𝒙𝒋 +

𝒋∈𝑹

𝒚𝒊𝒋 − 𝒚𝒊𝒋 𝒙𝒋 + 𝒙𝑩𝒊 = 𝒚𝒊𝟎 + (𝒚𝒊𝟎 − [𝒚𝒊𝟎])

i.e.

𝒋∈𝑹

[𝒚𝒊𝒋] 𝒙𝒋 + 𝒙𝑩𝒊 − 𝒚𝒊𝟎 = 𝒚𝒊𝟎 − 𝒚𝒊𝟎 −

𝒋∈𝑹

𝒚𝒊𝒋 − 𝒚𝒊𝒋 𝒙𝒋

i.e.

𝒋∈𝑹

[𝒚𝒊𝒋] 𝒙𝒋 + 𝒙𝑩𝒊 − 𝒚𝒊𝟎 = 𝒇𝒊𝟎 −

𝒋∈𝑹

𝒇𝒊𝒋𝒙𝒋 (𝑬𝒒. 𝟓)

(Eq. 5) holds for all feasible points of points LPP (Eq. 0) and for the given AILP.

Therefore the R.H.S must also be integer.


Gomory’s Cutting Plane method for AILP𝑓𝑖0 −

𝑗∈𝑅

𝑓𝑖𝑗𝑥𝑗

𝑖 = 0 included because for the AILP, the objective function is also constrained to be integer.

𝑁𝑜𝑤 𝑓𝑖𝑗 ≥ 0 𝑎𝑛𝑑 𝑥𝑗 ≥ 0 𝑓𝑜𝑟 𝑗 ∈ 𝑅. 𝑇ℎ𝑒𝑟𝑒𝑓𝑜𝑟𝑒

𝑗∈𝑅 𝑓𝑖𝑗𝑥𝑗 ≥ 0 (Eq. 6)

𝑓𝑖0 < 1 𝑎𝑛𝑑 𝐸𝑞. 6 𝑔𝑖𝑣𝑒𝑠

𝑓𝑖0 − 𝑗∈𝑅 𝑓𝑖𝑗𝑥𝑗 < 1 is an integer

𝒇𝒊𝟎 − 𝒋∈𝑹𝒇𝒊𝒋𝒙𝒋 ≤ 𝟎 (Eq. 7)

The inequality (Eq. 7) is satisfied by every integer feasible point of the given AILP.


Gomory’s Cutting Plane method for AILPIf current b.f.s. 𝒙𝑩 is not an integer. It doesn’t meet the requirement of AILP.

In that case, inequality is not satisfied.

“ It certainly deletes a part of the feasible region of the associated LLP ( at least the currentb.f.s. 𝒙𝑩 and may be more points) but does not delete any feasible point with integer co-ordinates. Hence it is valid cut constraint and it is called Gomory’s cut constraint”

−𝑓𝑖0= 𝑠𝑖 −

𝑗∈𝑅

𝑓𝑖𝑗𝑥𝑗

Append this to associated LPP, (𝐿𝑃)1 to get the new LPP (𝐿𝑃)2 Therefore we solve(𝐿𝑃)2 𝑎𝑛𝑑 𝑟𝑒𝑝𝑒𝑎𝑡 𝑡ℎ𝑒 𝑝𝑟𝑜𝑐𝑒𝑑𝑢𝑟𝑒.


Stepwise DescriptionStep 1: Solve the associated LLP, say (𝑳𝑷)𝟏,by the simplex method. Set 𝒌 = 𝟏

Step 2:

◦ If the optimal solution obtained at Step 1 is integer

◦ Stop

◦ Otherwise go to Step3

Step 3: For any updated constraint 𝒊 whose 𝒚𝒊𝟎 value is fractional (including 𝒊 = 𝟎, i.e. obj. fun.)◦ Generate Gomory’s cut constraint as given at (6.13).

◦ Select the value of 𝒊, 𝟎 ≤ 𝒊 ≤ 𝒎 for which 𝒇𝒊𝟎 value is maximum.

◦ Theoretically we can choose any i for which 𝒇𝒊𝟎 > 𝟎 but the maximum of 𝒇𝒊𝟎 is chosen with the hope that it may give adeeper cut

Step 4: Append the Gomory’s cut constraint derived at Step 3 above the (𝑳𝑷)𝒌 to get the new LPP(𝑳𝑷)𝒌+𝟏 .

◦ Solve by the dual simplex method and return to Step2


Theorem“The number of Gomory’s cut constraints needed to solve anyinstance of all integer linear programming (AILP) problem isalways finite”

As the no. of cut constraints needed is always finite, we are solvingonly finitely many LPP to get an optimal solution of the given AILP.

But unfortunately, even for a problem of “average” size, the no. ofcut constraints needed may be ‘too many’ as AILP belongs to theclass of Hard Problem.


ExampleCONSIDER THE INTEGER LPP

𝑀𝑎𝑥 𝑧 = 5𝑥1 + 2𝑥2

subject to

2𝑥1 + 2𝑥2 ≤ 9

3𝑥1 + 𝑥2 ≤ 11

𝑥1, 𝑥2 ≥ 0

𝑥1, 𝑥2integer

THE GIVEN ILP IS EQUIVALENT TO

𝑀𝑎𝑥 𝑧 = 5𝑥1 + 2𝑥2 + 0𝑥3 + 0𝑥4subject to

2𝑥1 + 2𝑥2 + 𝑥3 = 9

3𝑥1 + 𝑥2 + 𝑥4 = 11

𝑥1, 𝑥2, 𝑥3, 𝑥4 ≥ 0

all integer

𝑥3 = 9 − 2𝑥1 − 2𝑥2 and

𝑥4 = 11 − 3𝑥1 − 𝑥2

𝑥1, 𝑥2 𝑎𝑟𝑒 𝑖𝑛𝑡𝑒𝑔𝑒𝑟, 𝑠𝑜 𝑥3, 𝑥4 𝑎𝑟𝑒 𝑎𝑙𝑠𝑜 𝑖𝑛𝑡𝑒𝑔𝑒𝑟


𝒙𝟏 𝒙𝟐 𝒙𝟑 𝒙𝟒

𝒛 0 -5 -2 0 0

𝒙𝟑 9 2 2 1 0

𝒙𝟒 11 3 1 0 1

𝒙𝟏 𝒙𝟐 𝒙𝟑 𝒙𝟒

𝒛 18.33 0 -0.33 0 1.67

𝒙𝟑 1.67 0 1.33 1 -0.67

𝒙𝟏 3.67 1 0.33 0 0.33

ExampleFIRST ITERATION (LP1)

𝑀𝑎𝑥 𝑧 = 5𝑥1 + 2𝑥2 + 0𝑥3 + 0𝑥4subject to

2𝑥1 + 2𝑥2 + 𝑥3 = 9

3𝑥1 + 𝑥2 + 𝑥4 = 11

𝑥1, 𝑥2, 𝑥3, 𝑥4 ≥ 0

all integer

𝑥3 = 9 − 2𝑥1 − 2𝑥2 and

𝑥4 = 11 − 3𝑥1 − 𝑥2

𝑥1, 𝑥2 𝑎𝑟𝑒 𝑖𝑛𝑡𝑒𝑔𝑒𝑟, 𝑠𝑜 𝑥3, 𝑥4 𝑎𝑟𝑒 𝑎𝑙𝑠𝑜 𝑖𝑛𝑡𝑒𝑔𝑒𝑟

𝒙𝟏 𝒙𝟐 𝒙𝟑 𝒙𝟒

𝒛 18.75 0 0 0.25 1.5

𝒙𝟐 1.25 0 1 0.75 -0.5

𝒙𝟏 3.25 1 0 -0.25 0.5

(𝑧𝑗−𝑐𝑗)


Geometrical Representation


Geometrical Representation


Finding Gomory’s Cut ConstraintFINDING GOMORY’S CUT

Gomory cut constraint by choosing value of i for which

𝑓𝑖0 𝑖𝑠 𝑝𝑜𝑠𝑖𝑡𝑖𝑣𝑒 𝑚𝑜𝑠𝑡. In table we can see

𝒇𝟎𝟎 𝒊𝒔 𝒎𝒐𝒔𝒕 𝒑𝒐𝒔𝒊𝒕𝒊𝒗𝒆

𝒇𝒊𝟎 − 𝒋∈𝑹𝒇𝒊𝒋𝒙𝒋 ≤ 𝟎

𝑓00 − 𝑓03𝑥3 − 𝑓04𝑥4 ≤ 0

0.75 − 0.25𝑥3 − 0.5𝑥4 ≤ 0

0.75 − 0.25𝑥3 − 0.5𝑥4 + 𝑠1 = 0

−𝟎. 𝟕𝟓 = 𝒔𝟏 − 𝟎. 𝟐𝟓𝒙𝟑 − 𝟎. 𝟓𝒙𝟒

𝒙𝟏 𝒙𝟐 𝒙𝟑 𝒙𝟒

𝒛 18.75 0 0 0.25 1.5

𝒙𝟐 1.25 0 1 0.75 -0.5

𝒙𝟏 3.25 1 0 -0.25 0.5

New Constrained in 𝒙𝟏𝒂𝒏𝒅 𝒙𝟐𝑓𝑜𝑟𝑚𝑥3 = 9 − 2𝑥1 − 2𝑥2𝑥4 = 11 − 3𝑥1 − 𝑥20.25𝑥3 + 0.5𝑥4 ≥ 0.75

0.25 9 − 2𝑥1 − 2𝑥2 + 0.5(11 − 3𝑥1 − 𝑥2) ≥ 0.75𝟐𝒙𝟏 + 𝒙𝟐 ≤ 𝟕

(𝑧𝑗−𝑐𝑗) ≥ 0

𝑓00

𝑓10

𝑓20

𝟎 ≤ 𝒇𝒂 < 𝟏


Change in Geometry due to cut






Second Iteration (LP2)DUAL SIMPLEX METHOD

𝑀𝑎𝑥 𝑧 = 5𝑥1 + 2𝑥2 + 0𝑥3 + 0𝑥4 + 0𝑠1subject to

2𝑥1 + 2𝑥2 + 𝑥3 = 9

3𝑥1 + 𝑥2 + 𝑥4 = 11

−0.25𝑥3 − 0.5𝑥4 + 𝑠1 = −0.75

𝑥1, 𝑥2, 𝑥3, 𝑥4, 𝑠1 ≥ 0

𝒙𝟏 𝒙𝟐 𝒙𝟑 𝒙𝟒 𝒔𝟏

𝒛 18.75 0 0 0.25 1.5 0

𝒙𝟐 1.25 0 1 0.75 -0.5 0

𝒙𝟏 3.25 1 0 -0.25 0.5 0

𝒔𝟏 -0.75 0 0 -0.25 -0.5 1

𝒙𝟏 𝒙𝟐 𝒙𝟑 𝒙𝟒 𝒔𝟏

𝒛 18 0 0 0 1 1

𝒙𝟐 -1 0 1 0 -2 3

𝒙𝟏 4 1 0 0 1 -1

𝒙𝟑 3 0 0 1 2 4

𝒙𝟏 𝒙𝟐 𝒙𝟑 𝒙𝟒 𝒔𝟏

𝒛 17.5 0 0.5 0 0 2.5

𝒙𝟒 0.5 0 -0.5 0 1 -1.5

𝒙𝟏 3.5 1 0.5 0 0 0.5

𝒙𝟑 2 0 1 1 0 -1

By taking constraint directly in the last Tableau


Finding Second CutFINDING GOMORY’S CUT

Gomory cut constraint by choosing value of i for which

𝑓𝑖0 𝑖𝑠 𝑝𝑜𝑠𝑖𝑡𝑖𝑣𝑒 𝑚𝑜𝑠𝑡. In table we can see

𝒇𝟎𝟎 = 𝒇𝟏𝟎 = 𝒇𝟐𝟎 = 𝟎. 𝟓 𝒊𝒔 𝒎𝒐𝒔𝒕 𝒑𝒐𝒔𝒊𝒕𝒊𝒗𝒆

So w can generate cut through 𝒛 𝒐𝒓 𝒙𝟒 𝒐𝒓 𝒙𝟏 we choose𝒊 = 𝟎 𝒂𝒏𝒅 𝒅𝒆𝒓𝒊𝒗𝒆 𝒄𝒐𝒏𝒔𝒕𝒓𝒂𝒊𝒏𝒕

𝒇𝒊𝟎 − 𝒋∈𝑹𝒇𝒊𝒋𝒙𝒋 ≤ 𝟎

0.5 − 0.5𝑥2 − 0.5𝑠1 ≤ 0

0.5 − 0.5𝑥2 − 0.5𝑠1 + 𝑠2 = 0

−0.5 = 𝑠2 − 0.5𝑥2 − 0.5𝑠1

𝒙𝟏 𝒙𝟐 𝒙𝟑 𝒙𝟒 𝒔𝟏

𝒛 17.5 0 0.5 0 0 2.5

𝒙𝟒 0.5 0 -0.5 0 1 -1.5

𝒙𝟏 3.5 1 0.5 0 0 0.5

𝒙𝟑 2 0 1 1 0 -1

New Constrained in 𝒙𝟏𝒂𝒏𝒅 𝒙𝟐𝑓𝑜𝑟𝑚𝑥3 = 9 − 2𝑥1 − 2𝑥2𝑥4 = 11 − 3𝑥1 − 𝑥2𝟐𝒙𝟏 + 𝒙𝟐 ≤ 𝟕𝒔𝟏 = 𝟕 − 𝟐𝒙𝟏 − 𝒙𝟐0.5𝑥2 + 0.5𝑠1 ≥ 0.5

0.5𝑥2 + 0.5(𝟕 − 𝟐𝒙𝟏 − 𝒙𝟐) ≥ 0.5𝒙𝟏 ≤ 𝟑

(𝑧𝑗−𝑐𝑗) ≥ 0

𝑓00

𝑓10

𝑓20

𝑓30

𝟎 ≤ 𝒇𝒂 < 𝟏








Third Iteration (LP)3DUAL SIMPLEX METHOD

𝑀𝑎𝑥 𝑧 = 5𝑥1 + 2𝑥2 + 0𝑥3 + 0𝑥4 + 0𝑠1subject to

2𝑥1 + 2𝑥2 + 𝑥3 = 9

3𝑥1 + 𝑥2 + 𝑥4 = 11

−0.25𝑥3 − 0.5𝑥4 + 𝑠1 = −0.75

−0.5 = 𝑠2 − 0.5𝑥2 − 0.5𝑠1𝑥1, 𝑥2, 𝑥3, 𝑥4, 𝑠1 ≥ 0

Optimal Solution: (𝒙𝟏∗ = 𝟑, 𝒙𝟐

∗ = 𝟏, 𝒛∗ = 𝟏𝟕)

𝒙𝟏 𝒙𝟐 𝒙𝟑 𝒙𝟒 𝒔𝟏 𝒔𝟐

𝒛 17.5 0 0.5 0 0 2.5 0

𝒙𝟒 0.5 0 -0.5 0 1 -1.5 0

𝒙𝟏 3.5 1 0.5 0 0 0.5 0

𝒙𝟑 2 0 1 1 0 -1 0

𝒔𝟐 -0.5 0 -0.5 0 0 -0.5 1

𝒙𝟏 𝒙𝟐 𝒙𝟑 𝒙𝟒 𝒔𝟏 𝒔𝟐

𝒛 17 0 0 0 0 2 1

𝒙𝟒 1 0 0 0 1 -1 -1

𝒙𝟏 3 1 0 0 0 0 1

𝒙𝟑 1 0 0 1 0 -2 2

𝒔𝟐 1 0 1 0 0 1 -2

By taking constraint directly in the last Tableau


Optimal point Geometrical


References

