Linear Programming Revised Simple Method, Dualit of LP problems and Sensitivit analsis
Introduction Revised simple method is an improvement over simple method. It is computationall more efficient and accurate. Dualit of LP problem is a useful propert that makes the problem easier in some cases Dual simple method is computationall similar to simple method. However, their approaches are different from each other. Primal-Dual relationship is also helpful in sensitivit or post optimalit analsis of decision variables.
Objectives Objectives To eplain revised simple method To discuss about dualit of LP and Primal-Dual relationship To illustrate dual simple method To end with sensitivit or post optimalit analsis 3
Revised Simple method: Introduction Benefit of revised simple method is clearl comprehended in case of large LP problems. In simple method the entire simple tableau is updated while a small part of it is used. The revised simple method uses eactl the same steps as those in simple method. The onl difference occurs in the details of computing the entering variables and departing variable. 4
Revised Simple method 5 Consider the following LP problem (with general notations, after transforming it to its standard form and incorporating all required slack, surplus and artificial variables) ( ) ( ) Z c + c + c33 + LLL + cnn + Z = 0 c + c + c + LLL + c = b i 3 3 n n ( ) c + c + c + LLL + c = b M M M M M M c + c + c + LLL+ c = b ( ) j 3 3 n n l m m m3 3 mn n m As the revised simple method is mostl beneficial for large LP problems, it will be discussed in the contet of matri notation.
6 Revised Simple method: Matri form Matri notation with : subject to : z Minimize T 0 X B AX X C = = = n M X = c n c c M C m b b b = B M = 0 0 0 M 0 = mn m m n n c c c c c c c c c L M O M M L L A where
Revised Simple method: Notations Notations for subsequent discussions: ck Column vector corresponding to a decision variable k is ck. M cmk X S is the column vector of basic variables C S is the row vector of cost coefficients corresponding to X S, and 7 S is the basis matri corresponding to X S
Revised Simple method: Iterative steps. Selection of entering variable For each of the nonbasic variables, calculate the coefficient (WP - c), where, P is the corresponding column vector associated with the nonbasic variable at hand, c is the cost coefficient associated with that nonbasic variable and W = C S S -. For maimization (minimization) problem, nonbasic variable, having the lowest negative (highest positive) coefficient, as calculated above, is the entering variable. 8
Revised Simple method: Iterative steps 9. Selection of departing variable a) A new column vector U is calculated as U = S - B b) Corresponding to the entering variable, another vector V is calculated as V = S - P, where P is the column vector corresponding to entering variable. c) It ma be noted that length of both U and V is same (= m). For i =,, m, the ratios, U(i)/V(i), are calculated provided V(i) > 0. i = r, for which the ratio is least, is noted. The r th basic variable of the current basis is the departing variable. If it is found that V(i) < 0 for all i, then further calculation is stopped concluding that bounded solution does not eist for the LP problem at hand.
Revised Simple method: Iterative steps 3. Update to new Basis Old basis S, is updated to new basis S new, as S new = [ E S - ] - where E 0 L η L 0 0 0 L η L 0 0 η = M M L r L M M M M O M L M M M M L M O M M 0 0 L η L 0 m 0 0 L η L 0 m and ( i) () r V V ηi = V () r for i r for i = r 0 r th column
Revised Simple method: Iterative steps S is replaced b S new and steps through 3 are repeated. If all the coefficients calculated in step, i.e., is positive (negative) in case of maimization (minimization) problem, then optimum solution is reached The optimal solution is X S =S - B and z = CX S
Dualit of LP problems Each LP problem (called as Primal in this contet) is associated with its counterpart known as Dual LP problem. Instead of primal, solving the dual LP problem is sometimes easier in following cases a) The dual has fewer constraints than primal Time required for solving LP problems is directl affected b the number of constraints, i.e., number of iterations necessar to converge to an optimum solution, which in Simple method usuall ranges from.5 to 3 times the number of structural constraints in the problem b) The dual involves maimization of an objective function It ma be possible to avoid artificial variables that otherwise would be used in a primal minimization problem.
Finding Dual of a LP problem 3 Primal Maimization Minimization i th variable j th constraint i > 0 Minimization Maimization i th constraint j th variable Dual Inequalit sign of i th Constraint: if dual is maimization if dual is minimization contd. to net slide
Finding Dual of a LP problem contd. Primal i th variable unrestricted j th constraint with = sign RHS of j th constraint Cost coefficient associated with i th variable in the objective function Dual i th constraint with = sign j th variable unrestricted Cost coefficient associated with j th variable in the objective function RHS of i th constraint constraints Refer class notes for pictorial representation of all the operations 4
Dual from a Primal 5
Finding Dual of a LP problem contd. Note: Before finding its dual, all the constraints should be transformed to less-than-equal-to or equal-to tpe for maimization problem and to greater-than-equal-to or equal-to tpe for minimization problem. It can be done b multipling with - both sides of the constraints, so that inequalit sign gets reversed. 6
Finding Dual of a LP problem: An eample 7 Maimize Primal Subject to + 3 Z + 4000 0 6000 000 unrestricted Minimize Subject to 3 + Dual Z = 6000 000 + 4000 = 4 3 3 + 3 = 0 0 3 0 Note: Second constraint in the primal is transformed to before constructing the dual. 3 4 + 000
Primal-Dual relationships If one problem (either primal or dual) has an optimal feasible solution, other problem also has an optimal feasible solution. The optimal objective function value is same for both primal and dual. If one problem has no solution (infeasible), the other problem is either infeasible or unbounded. If one problem is unbounded the other problem is infeasible. 8
Dual Simple Method Simple Method verses Dual Simple Method. Simple method starts with a nonoptimal but feasible solution where as dual simple method starts with an optimal but infeasible solution.. Simple method maintains the feasibilit during successive iterations where as dual simple method maintains the optimalit. 9
Dual Simple Method: Iterative steps Steps involved in the dual simple method are:. All the constraints (ecept those with equalit (=) sign) are modified to less-than-equal-to sign. Constraints with greater-than-equal-to sign are multiplied b - through out so that inequalit sign gets reversed. Finall, all these constraints are transformed to equalit sign b introducing required slack variables.. Modified problem, as in step one, is epressed in the form of a simple tableau. If all the cost coefficients are positive (i.e., optimalit condition is satisfied) and one or more basic variables have negative values (i.e., non-feasible solution), then dual simple method is applicable. 0
Dual Simple Method: Iterative steps contd. 3. Selection of eiting variable: The basic variable with the highest negative value is the eiting variable. If there are two candidates for eiting variable, an one is selected. The row of the selected eiting variable is marked as pivotal row. 4. Selection of entering variable: Cost coefficients, corresponding to all the negative elements of the pivotal row, are identified. Their ratios are calculated after changing the sign of the elements of pivotal row, i.e., Cost Coefficients ratio = Elements of pivotal The column corresponding to minimum ratio is identified as the pivotal column and associated decision variable is the entering variable. row
Dual Simple Method: Iterative steps contd. 5. Pivotal operation: Pivotal operation is eactl same as in the case of simple method, considering the pivotal element as the element at the intersection of pivotal row and pivotal column. 6. Check for optimalit: If all the basic variables have nonnegative values then the optimum solution is reached. Otherwise, Steps 3 to 5 are repeated until the optimum is reached.
Dual Simple Method: An Eample Consider the following problem: Minimize subject to Z = 3 4 + 4 + 3 + + 4 3
Dual Simple Method: An Eample contd. After introducing the surplus variables the problem is reformulated with equalit constraints as follows: Minimize subject to Z = 3 4 + + 4 3 + + + + 3 4 5 6 = = 4 = = 4
Dual Simple Method: An Eample contd. Epressing the problem in the tableau form: 5
Dual Simple Method: An Eample contd. Successive iterations: 6
Dual Simple Method: An Eample contd. Successive iterations: 7
Dual Simple Method: An Eample contd. Successive iterations: 8 As all the b r are positive, optimum solution is reached. Thus, the optimal solution is Z = 5.5 with = and =.5
9 Solution of Dual from Primal Simple 0,, 4 4 5 to subject 4 0 6 ' Minimize 3 3 3 3 3 + + + + + = Z Primal Dual 0,, 4 5 0 4 6 to subject 4 Maimize 3 3 3 3 3 + + + + = Z 3 Z
Sensitivit or post optimalit analsis Changes that can affect onl Optimalit Change in coefficients of the objective function, C, C,.. Re-solve the problem to obtain the solution Changes that can affect onl Feasibilit Change in right hand side values, b, b,.. Appl dual simple method or stud the dual variable values Changes that can affect both Optimalit and Feasibilit Simultaneous change in C, C,.. and b, b,.. Use both primal simple and dual simple or re-solve 30
Sensitivit or post optimalit analsis A dual variable, associated with a constraint, indicates a change in Z value (optimum) for a small change in RHS of that constraint. where Δ Z = Δb j i j is the dual variable associated with the i th constraint, Δb i is the small change in the RHS of i th constraint, ΔZ is the change in objective function owing to Δb i. 3
Sensitivit or post optimalit analsis: An Eample Let, for a LP problem, ith constraint be + 50 and the optimum value of the objective function be 50. RHS of the i th constraint changes to 55, i.e., i th constraint changes to + 55 Let, dual variable associated with the i th constraint is j, optimum value of which is.5 (sa). Thus, Δb i = 55 50 = 5 and j =.5 3 So, ΔZ = j Δb i =.55 =.5 and revised optimum value of the objective function is 50 +.5 = 6.5.
Thank You 33