Introduction to Operations Research

1. Duality Theory Introduction to Operations Research

2. Since dual problems are also a linear programming problem, it also has CP solutions. • Since dual problems have function constraints in ≥ form, must augment by subtracting surplus variables from functional constraints. • Use surplus variable: zj – cj for each constraint j • After augmenting dual problem: (y1, y2, ..., ym, z1-c1, z2-c2, ..., zn-cn) Augmenting the Dual Problem

3. Complementary Basic Solution Property – Each basic solution in the primal problem has a complementary basic solution in the dual problem s.t. Z = W • Complementary Slackness Property – Given the association between variables, the variables in the primal basic solution and the complementary dual basic solution satisfy the complementary slackness relationship: Primal/Dual Problems

4. Wyndor Glass Co. Example Maximize Z = 3x1 + 5x2 Subject to: x1 + x3= 4 2x2 + x4 = 12 3x1 + 2x2 + x5 = 18 Minimize W = 4y1 + 12y2 + 18y3 Subject to: y1 + 3y3– (z1 – c1)= 3 2y2 + 2y3 – (z2 – c2) = 5

5. Wyndor Glass Co. Example Dual Solution Optimal Solution Primal Solution

6. Wyndor Glass Co. Example

7. Complementary Optimal Basic Solutions Property – Each optimal basic solution in the primal problem has a complementary optimal basic solution in the dual problem such that Z = W. Complementary Basic Solutions Satisfies Condition for Optimality? Yes No Feasible? Yes Optimal Sub-Optimal No Super-Optimal Neither Feasible nor Optimal

8. The approach here is similar to when we dealt with non-standard formulations in the context of the simplex method. • There is one exception: we do not add artificial variables. We handle “=“ constraints by writing them as “<=“ constraints. • This is possible here because we do not require here that the RHS is non-negative. NONSTANDARD LP

9. Greater than or equal to constraints: • Multiply through the inequality constraint by -1 to make it a less than constraint: x1 + 2x2 ≥ 5 (x1 + 2x2 ≥ 5)(-1) -x1– 2x2 ≤ –5 Nonstandard lp

10. Equality constraints: • Convert the equality constraint to a pair of inequality constraints: x1 + 2x2 = 5 x1 + 2x2 ≤ 5 x1 + 2x2 ≤ 5 and x1 + 2x2 ≥ 5 -x1– 2x2 ≤ –5 Nonstandard lp

11. Unrestricted variables: • Replace the variable unrestricted in sign, by the difference of two nonnegative variables: x1 + 2x2 ≤ 5 x1 ≥ 0, x2 unrestricted x1 + 2(x2 – x2) ≤ 5 Nonstandard lp

12. Example Maximize Z = x1+ x2 + x3 Subject to: 2x2– x3≥4 x1– 3x2 + 4x3 = 5 x1– 2x2≤ 18 x1, x2≥ 0, x3 unrestricted Maximize Z = x1+ x2 + (x3 – x3) Subject to: 2x2– (x3 – x3)≥4 x1– 3x2 + 4(x3 – x3) = 5 x1– 2x2 ≤ 18 – 2x2 + x3 – x3≤ –4 x1– 3x2 + 4x3 – 4x3 ≤ 5 –x1 +3x2–4x3 + 4x3 ≤ –5 x1– 2x2 ≤ 18

13. Example Primal Problem Maximize Z = x1 + x2 + x3 – x3 Subject to: – 2x2 + x3 – x3≤ –4 x1– 3x2 + 4x3 – 4x3 ≤ 5 –x1 +3x2–4x3 + 4x3 ≤ –5 x1– 2x2 ≤ 18 Dual Problem Minimize W = –4y1+5y2–5y3+3y4 Subject to: y2– y3 + y4≥ 1 –2y1– 3y2 + 3y3 – 2y4≥ 1 –y1– 4y2+ 4y3≥–1 y1+4y2 –4y3≥ 1

14. Example Dual Problem Minimize W = –4y1+5y2+3y3 Subject to: y2+ y3≥ 1 –2y1– 3y2– 2y3≥ 1 y1+4y2 = 1 y1, y3≥ 0, y2unrestricted Dual Problem Minimize W = –4y1+5y2–5y3+3y4 Subject to: y2– y3 + y4≥ 1 –2y1– 3y2 + 3y3 – 2y4≥ 1 –y1– 4y2+ 4y3≥–1 y1+4y2 –4y3≥ 1 Equality constraint (y2– y3) –3(y2– y3) –4(y2 – y3) 4(y2 – y3) Unrestricted variable (y2 – y3)

15. An equality constraint in the primal generates a dual variable that is unrestricted in sign. • An unrestricted in sign variable in the primal generates an equality constraint in the dual. Streamlining the conversion Dual Problem Primal Problem opt=min opt=max Constraint i : <= form = form Variable j: xj >= 0 xj urs Variable i : yi >= 0 yi urs Constraint j: >= form = form