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Linear Programming Supplements (Optional). Standard Form LP (a.k.a. First Primal Form). Strictly ≤. All x j 's are non-negative. Transforming Problems into Standard Form. Min c T x  Max - c T x Max ( c T x + constant)  Max c T x

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Linear Programming Supplements (Optional)

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Linear programming supplements optional

Linear ProgrammingSupplements(Optional)

Standard form lp a k a first primal form

Standard Form LP (a.k.a. First Primal Form)

Strictly ≤

All xj's are non-negative

Transforming problems into standard form

Transforming Problems into Standard Form

  • Min cTxMax -cTx

  • Max (cTx + constant)Max cTx

  • Replace a constraint like ∑aijxj≥bi by -∑aijxj≤ -bi

  • Replace a constraint like ∑aijxj=bi by ∑aijxj≤bi and -∑aijxj≤ -bi

  • If xjis allowed to take on negative value, replace xi by the difference of two nonnegative variables, says xi = ui – vi, where ui ≥ 0 and vi ≥ 0.

Example of transforming a problem into standard form

Example of transforming a problem into Standard Form

Replace x1 by u1– v1

Dual problem

Dual Problem

Every primal LP problem in the form

Maximize cTx

subject to Ax≤b, x ≥ 0

has a corresponding dual problem in the form

Minimize bTy

subject to ATy ≥ c, y ≥ 0

Theorem on Primal and Dual Problems

If x satisfies the constraints of the primal problem and y satisfies the constraints of its dual, then cTx≤bTy.

Consequently, if cTx=bTy, then x and y are solutions of the primal problem and the dual problem respectively.

Dual problem1

Dual Problem

Duality Theorem

If the original problem has a solution x*, then the dual problem has a solution y*; furthermore, cTx*=bTy*.

If the original primal problem contains much more constraints than variables (i.e., m >> n), then solving the dual problem may be more efficient. (Less constraints implies less corner points to check)

The dual problem also offers a different interpretation of the problem (Maximize profit == Minimize cost)

Matlab lp solver linprog

MATLAB LP solver – linprog()

Partial help manual generated by MATLAB:

X=LINPROG(f,A,b) attempts to solve the linear programming problem:

min f'*x subject to: A*x <= b


X=LINPROG(f,A,b,Aeq,beq) solves the problem above while additionally

satisfying the equality constraints Aeq*x = beq.

X=LINPROG(f,A,b,Aeq,beq,LB,UB) defines a set of lower and upper

bounds on the design variables, X, so that the solution is in

the range LB <= X <= UB. Use empty matrices for LB and UB

if no bounds exist. Set LB(i) = -Inf if X(i) is unbounded below;

set UB(i) = Inf if X(i) is unbounded above.

X=LINPROG(f,A,b,Aeq,beq,LB,UB,X0) sets the starting point to X0. This option is only available with the active-set algorithm. The default interior point algorithm will ignore any non-empty starting point.

Matlab example

MATLAB example

% Turn into minimization problem

c = [ -150 -175 ]';

A = [ 7 11; 10 8; 1 0; 0 1 ];

b = [77 80 9 6]';

LB = [0 0]';

% There is no equality constraints

xmin = linprog(c, A, b, [], [], LB)

Optimization terminated.

xmin =



Integer lp problem

Integer LP Problem

  • If the variables can only take integer values, we cannot take the integers closest to the solution of the corresponding LP problem as the solution.

  • Integer Programming (IP) or Integer Linear Programming (ILP) problems are NP-hard problems.

    • Some of the algorithm for solving IP problems include branch-and-bound, branch-and-cut.

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