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In a satisfiable system of linear inequalities

a11x1+···+a1nxn≤b1:am1x1+···+amnxn≤bm

we describe the inequality as forced-equal if it is satisfied with equality by every solution x = (x1,...,xn)of the system. Equivalently,Piajixi≤bj is not forced-equal if there exists an x that satisfies the whole system and such that Piajixi≤bj.

For example, in

x1+x2≤2-x1-x2≤-2x1≤1-x2≤0

Short Answer

Expert verified

Describe the jthinequality is forced equal solution of x=(x1...xn)and get correct answer base on different formula to countthat satisfies the whole system and such that ƛj+∑iaj,i(∑(J∈x)xi1≤∑ia(j,i)xi∑bj

Step by step solution

01

Maximum Flow Diagram of Following Network

a) The claim hold trivially of all constants are forced-equal, so assume at least one constraint is not forced-equal.

Let I be the set of constraints that are not forced equal, so that Ie I it there is some feasible solution.

we show that xI=(X1I...XnI)defined by

xI=(X1I...XnI)such Ithat is satisfied without equality.

X1=∑1∈I1χiIcharacteristic solution indeed for any fired I≤j≤m, the constraint is satisfied by ‘x’ since summing the left – hand side and the right – hand side of the inequality over all χigives us

02

Formula base Calculation

∑iaj,i(∑(J∈x)xi1≤|I|bj

That implies,
∑iaj,i(∑(J∈x)xi1≤i∑a(j,i)xi≤bj

Furthermore, if 1th constraint is not forced – equal. So that

∑iaj,ixiI<bj

∑iaj,i(∑J∈xxi1≤|I|bj

So,
∑iaj,ixiI<bjand the constraint is satisfied without equality.

Thus, ‘x’ is a feasible solution where every
J∈Iis satisfied without equality.

By the algorithm maintains a set of I constraints, initialized to be the empty set. And iterates.Through each constraint j=1,2,....m.

Consider thej iteration of the algorithm and let I denote the j constraint. We define a linear program, obtained from the original set of constraints as follows.

03

Conclusion

We introduct a new variable ƛi and replace the ‘j’ constraint with the following constraint :

ƛj+∑iaj,i(∑J∈xxi1≤∑iaj,ixi≤bj

We also add an objective function : max λi. we then find a optimal solution to the resulting LP.

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Most popular questions from this chapter

A quadratic programming problem seeks to maximize a quadratic objective function (with terms like 3x12or5x1x2) subject to a set of linear constraints. Give an example of a quadratic program in two variables x1, x2 such that the feasible region is nonempty and bounded, and yet none of the vertices of this region optimize the (quadratic) objective.

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maximize 5x+3y

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Plot the feasible region and identify the optimal solution.

The Canine Products company offers two dog foods, Frisky Pup and Husky Hound, that are made from a blend of cereal and meat. A package of Frisky Pup requires 1 pound of cereal and 1.5pounds of meat, and sells for \(7. A package of Husky Hound uses 2 pounds of cereal and 1 pound of meat, and sells for \)6. Raw cereal costs\(1per pound and raw meat costs\)2per pound. It also costslocalid="1658981348093" \(1.40to package the Frisky Pup and localid="1658981352345" \)0.60to package the Husky Hound. A total of localid="1658981356694" 240,000pounds of cereal and pounds of meat are available each month. The only production bottleneck is that the factory can only package 110,000bags of Frisky Pup per month. Needless to say, management would like to maximize profit.

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You are given the following points in the plane:

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.You want to find a lineax+by=c that approximately passes through these points (no line is a perfect fit). Write a linear program (you don’t need to solve it) to find the line that minimizes the maximum absolute error,max1≤i≤7|axi+byi−c|

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