Duality
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Duality Duality Presentation Transcript

  • DUALITY
  • Duality Theory
    • Every LP problem (called the ‘Primal’) has associated with another problem called the ‘Dual’.
    • The ‘Dual’ problem is an LP defined directly and systematically from the original (or Primal) LP model.
    • The optimal solution of one problem yields the optimal solution to the other.
    • Duality ease the calculations for the problems, whose number of variables is large.
  • Rules for converting Primal to Dual
    • If the Primal is to maximize, the dual is to minimize.
    • If the Primal is to minimize, the dual is to maximize.
    • For every constraint in the primal, there is a dual variable.
    • For every variable in the primal, there is a constraint in the dual.
    View slide
  • Dual Problem Primal LP : Max z = c 1 x 1 + c 2 x 2 + ... + c n x n subject to: a 11 x 1 + a 12 x 2 + ... + a 1n x n ≤ b 1 a 21 x 1 + a 22 x 2 + ... + a 2n x n ≤ b 2 : a m1 x 1 + a m2 x 2 + ... + a mn x n ≤ b m x 1 ≥ 0, x 2 ≥ 0,…….x j ≥ 0,……., x n ≥ 0. Associated Dual LP : Min. z = b 1 y 1 + b 2 y 2 + ... + b m y m subject to: a 11 y 1 + a 21 y 2 + ... + a m1 y m ≥ c 1 a 12 y 1 + a 22 y 2 + ... + a m2 y m ≥ c 2 : a 1n y 1 + a 2n y 2 + ... + a mn y m ≥ c n y 1 ≥ 0, y 2 ≥ 0,…….y j ≥ 0,……., y m ≥ 0. View slide
  • Example Primal Max. Z = 3x 1 +5x 2 Subject to constraints: x 1 < 4 y 1 2x 2 < 12 y 2 3x 1 +2x 2 < 18 y 3 x 1 , x 2 > 0 The Primal has: 2 variables and 3 constraints. So the Dual has: 3 variables and 2 constraints Dual Min. Z’ = 4y 1 +12y 2 +18y 3 Subject to constraints: y 1 + 3y 3 > 3 2y 2 +2y 3 > 5 y 1 , y 2 , y 3 > 0 We define one dual variable for each primal constraint.
  • Example Primal Min.. Z = 10x 1 +15x 2 Subject to constraints: 5x 1 + 7x 2 > 80 6x 1 + 11x 2 > 100 x 1 , x 2 > 0
  • Solution Dual Max.. Z’ = 80y 1 +100y 2 Subject to constraints: 5y 1 + 6y 2 < 10 7y 1 + 11y 2 < 15 y 1 , y 2 > 0
  • Example Primal Max. Z = 12x 1 + 4x 2 Subject to constraints: 4x 1 + 7x 2 < 56 2x 1 + 5x 2 > 20 5x 1 + 4x 2 = 40 x 1 , x 2 > 0
  • Solution
    • The equality constraint 5x 1 + 4x 2 = 40 can be replaced by the following two inequality constraints:
    • 5x 1 + 4x 2 < 40
    • 5x 1 + 4x 2 > 40 -5x 1 - 4x 2 < -40
    • The second inequality 2x 1 + 5x 2 > 20 can be changed to the less-than-or-equal-to type by multiplying both sides of the inequality by -1 and reversing the direction of the inequality; that is,
    • -2x 1 - 5x 2 < -20
  • Cont… The primal problem can now take the following standard form: Max. Z = 12x 1 + 4x 2 Subject to constraints: 4x 1 + 7x 2 < 56 -2x 1 - 5x 2 < -20 5x 1 + 4x 2 < 40 -5x 1 - 4x 2 < -40 x 1 , x 2 > 0
  • Cont… Min. Z’ = 56y 1 -20y 2 + 40y 3 – 40y 4 Subject to constraints: 4y 1 – 2y 2 + 5y 3 – 5y 4 > 12 7y 1 - 5y 2 + 4y 3 – 4y 4 > 4 y 1 , y 2 , y 3 , y 4 > 0 The dual of this problem can now be obtained as follows:
  • Example Primal Min.. Z = 2x 2 + 5x 3 Subject to constraints: x 1 + x 2 > 2 2x 1 + x 2 +6x 3 < 6 x 1 - x 2 +3x 3 = 4 x 1 , x 2 , x 3 > 0
  • Solution Primal in standard form : Max.. Z = -2x 2 - 5x 3 Subject to constraints: -x 1 - x 2 < -2 2x 1 + x 2 +6x 3 < 6 x 1 - x 2 +3x 3 < 4 - x 1 + x 2 - 3x 3 < -4 x 1 , x 2 , x 3 > 0
  • Cont… Dual Min. Z’ = -2y 1 + 6y 2 + 4y 3 – 4y 4 Subject to constraints: -y 1 + 2y 2 + y 3 – y 4 > 0 -y 1 + y 2 - y 3 + y 4 > -2 6y 2 + 3y 3 - 3y 4 > -5 y 1 , y 2 , y 3 , y 4 > 0
  • DUAL SIMPLEX METHOD
  • Introduction Suppose a “basic solution” satisfies the optimality condition but not feasible, then we apply dual simplex method. In regular Simplex method, we start with a Basic Feasible solution (which is not optimal) and move towards optimality always retaining feasibility. In the dual simplex method, the exact opposite occurs. We start with a “optimal” solution (which is not feasible) and move towards feasibility always retaining optimality condition.The algorithm ends once we obtain feasibility.
  • Dual Simplex Method
    • To start the dual Simplex method, the following two conditions are to be met:
    • The objective function must satisfy the optimality conditions of the regular Simplex method.
    • All the constraints must be of the type  .
  • Example Min. Z = 3x 1 + 2x 2 Subject to constraints: 3x 1 + x 2 > 3 4x 1 + 3x 2 > 6 x 1 + x 2 < 3 x 1 , x 2 > 0
  • Cont… Step I: The first two inequalities are multiplied by –1 to convert them to < constraints and convert the objective function into maximization function. Max. Z’ = -3x 1 - 2x 2 where Z’= -Z Subject to constraints: -3x 1 - x 2 < -3 -4x 1 - 3x 2 < -6 x 1 + x 2 < 3 x 1 , x 2 > 0
  • Cont… Let S 1 , S 2 , S 3 be three slack variables Model can rewritten as: Z’ + 3x 1 + 2x 2 = 0 -3x 1 - x 2 +S 1 = -3 -4x 1 - 3x 2 +S 2 = -6 x 1 + x 2 +S 3 = 3 Initial BS is : x 1 = 0, x 2 = 0, S 1 = -3, S 2 = -6, S 3 = 3 and Z=0.
  • Cont…
    • Initial Basic Solution is Optimal (as the optimality condition is satisfied) but infeasible.
    • Choose the most negative basic variable. Therefore, S 2 is the departing variable.
    • Calculate Ratio = |Z row / S 2 row| (S 2 < 0)
    • Choose minimum ratio. Therefore, x 2 is the entering variable.
    - - - 2/3 3/4 - Ratio 3 1 0 0 1 1 0 S 3 -6 0 1 0 -3 -4 0 S 2 -3 0 0 1 -1 -3 0 S 1 0 0 0 0 2 3 1 Z Sol. S 3 S 2 S 1 x 2 x 1 Z Coefficients of: Basic Variable
  • Cont… Therefore, S 1 is the departing variable and x 1 is the entering variable. - 2 - - 1/5 - Ratio 1 1 1/3 0 0 -1/3 0 S 3 2 0 -1/3 0 1 4/3 0 x 2 -1 0 -1/3 1 0 -5/3 0 S 1 4 0 2/3 0 0 1/3 1 Z Sol. S 3 S 2 S 1 x 2 x 1 Z Coefficients of: Basic Variable
  • Cont… Optimal Solution is : x 1 = 3/5, x 2 = 6/5, Z= 21/5 6/5 1 2/5 -1/5 0 0 0 S 3 6/5 0 -3/5 4/5 1 0 0 x 2 3/5 0 1/5 -3/5 0 1 0 x 1 21/5 0 3/5 1/5 0 0 1 Z Sol. S 3 S 2 S 1 x 2 x 1 Z Coefficients of: Basic Variable
  • Example Max. Z = -x 1 - x 2 Subject to constraints: x 1 + x 2 < 8 x 2 > 3 -x 1 + x 2 < 2 x 1 , x 2 > 0
  • Cont… Let S 1 , S 2 , S 3 be three slack variables Model can rewritten as: Z + x 1 + x 2 = 0 x 1 + x 2 + S 1 = 8 -x 2 + S 2 = -3 -x 1 + x 2 + S 3 = 2 x 1 , x 2 > 0 Initial BS is : x 1 = 0, x 2 = 0, S 1 = 8, S 2 = -3, S 3 = 2 and Z=0.
  • Cont… Therefore, S 2 is the departing variable and x 2 is the entering variable. - - - 1 - - Ratio 2 1 0 0 1 -1 0 S 3 -3 0 1 0 -1 0 0 S 2 8 0 0 1 1 1 0 S 1 0 0 0 0 1 1 1 Z Sol. S 3 S 2 S 1 x 2 x 1 Z Coefficients of: Basic Variable
  • Cont… Therefore, S 3 is the departing variable and x 1 is the entering variable. - - - - 1 - Ratio -1 1 1 0 0 -1 0 S 3 3 0 -1 0 1 0 0 x 2 5 0 1 1 0 1 0 S 1 -3 0 1 0 0 1 1 Z Sol. S 3 S 2 S 1 x 2 x 1 Z Coefficients of: Basic Variable
  • Cont… Optimal Solution is : x 1 = 1, x 2 = 3, Z= -4 1 -1 -1 0 0 1 0 x 1 3 0 -1 0 1 0 0 x 2 4 0 2 1 0 0 0 S 1 -4 1 2 0 0 0 1 Z Sol. S 3 S 2 S 1 x 2 x 1 Z Coefficients of: Basic Variable