Uploaded by Shrutika Agarwal

QT LPP

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QT
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Any interior point of the feasible region cannot be an optimal solution for a non-trivial solution.
Any boundary point of the feasible region can be an optimal solution only when the whole
boundary is optimal.
A unique optimal solution can lie only at one of the corner points of the feasible region.
From the computer output, we can easily find:
• objective function value
• values of the decision variables
• reduced costs
• slack/surplus
Computer output also reports how an optimal solution is affected by a change in:
• a coefficient of the objective function
• the right-hand side value of a constraint
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The reduced cost for a decision variable whose value is 0 in the optimal solution is the amount
the variable's objective function coefficient would have to improve (increase for maximization
problems, decrease for minimization problems) before this variable could assume a positive
value.
The reduced cost for a decision variable with a positive value is 0.
A feasible region may be unbounded and yet there may be optimal solutions. This is common in
minimization problems and is possible in maximization problems.
Special Cases
Alternative Optimal Solutions In the graphical method, if the objective function line is parallel to a
boundary constraint in the direction of optimization, there are alternate optimal solutions, with all
points on this line segment being optimal.
Infeasibility
A linear program which is overconstrained so that no point satisfies all the constraints is said
to be infeasible.
Unbounded
Primal and Dual
Corresponding to every primal constraint is a decision variable in the dual and vice versa.
Corresponding to every primal decision variable is a constraint in the dual and vice versa.
In general, if the Primal is read “Horizontally”, the Dual can be read “Vertically” from the same LP – in
other words one is a transpose of the other. More specifically, we can use the Tucker’s Tableau if the
Primal is written in the Standard Form.
Min z2 = 240y1 + 180y2 + 300y3
s.t.
3y1 + 3y2 + y3 ≥ 25
2y1 + y2 + 3y3 ≥ 18
y1, y2, y3 ≥ 0
The Standard Form
In the Standard Form, If the objective function is a maximising function, write all constraints as “less
than or equal to” constraints If the objective function is a minimising function, write all constraints as
“greater than or equal to” constraints
Write the Dual of the
following LP:
Duality Results
Result 1: For any feasible solution of the Primal and any feasible solution of the Dual, the value of the
objective function of the primal (maximising) problem is always less than or equal to the value of the
objective function of the dual (minimising) problem.
Result 2: If we have the optimal solutions of the primal and the dual, the two corresponding objective
function values are equal.
Result 3: If we have a feasible solution of the primal and a feasible solution of the dual and further, if the
value of the two objective functions corresponding to these solutions are equal then the two feasible
solutions are also optimal for the primal and the dual respectively.
Result 4: In the primal optimal, if any constraint is non-binding, the corresponding dual variable is
equal to zero.
Result 5: In the primal optimal, if any decision variable is strictly positive the corresponding dual
constraint must be binding.
Result 6: If any of the primal constraint is an equality, the corresponding dual variable is unrestricted in
sign.
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