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Lecture 15 Special cases by Dr. Arshad Zaheer

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2 Four Special cases in Simplex

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3 Simplex Algorithm – Special cases There are four special cases arise in the use of the simplex method. 1.Degeneracy 2.Alternative optimal 3.Unbounded solution 4.infeasible solution

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4 Degeneracy ( no improvement in objective) Degeneracy: It is situation when the solution of the problem degenerates. Degenerate Solution: A Solution of the problem is said to be degenerate solution if value or values of basic variable(s) become zero It occurs due to redundant constraints.

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This is in itself not a problem, but making simplex iterations form a degenerate solution, give rise to cycling, meaning that after a certain number of iterations without improvement in objective value the method may turn back to the point where it started.

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6 Degeneracy – Special cases (cont.) Example: Max f = 3x 1 + 9x 2 Subject to: x 1 + 4x 2 ≤ 8 X 1 + 2x 2 ≤ 4 X 1, x 2 ≥ 0

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7 The solution: Step 1. write inequalities in equation form Let S 1 and S 2 be the slack variables X 1 + 4X 2 + s 1 = 8 X 1 + 2X 2 + s 2 = 4 X 1, X 2,s 1,s 2 ≥ 0 Let X1=0, X2=0 f=0, S1=8, S2=4

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8 Initial TableauBasis X1X1X1X1 X2X2X2X2 S1S1S1S1 S2S2S2S2RHSRatio S1S1S1S /4=2 S2S2S2S /2=2 f Entering VariableLeaving Variable S 1 and S 2 tie for leaving variable( with same minimum ratio 2) so we can take any one of them arbitrary. X 2 will enter the basis because of having minimum objective function coefficient.

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9 Tableau 1Basis X1X1X1X1 X2X2X2X2 S1S1S1S1 S2S2S2S2RHSRatio X2X2X2X21/411/4028 S2S2S2S2½0-1/210 0 min f-3/409/4018 Here basic variable S 2 is 0 resulting in a degenerate basic solution.

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10 Tableau 2 Same objective function no change and improvement ( cycle)Basis X1X1X1X1 X2X2X2X2 S1S1S1S1 S2S2S2S2RHS X2X2X2X2011/2-1/22 X1X1X1X11020 f003/23/218 Same objective

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This is redundant constraint because its presence or absence does not affect, neither on optimal point nor on feasible region. Feasible Region

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12 Alternative optimal If the f-row value for one or more non basic variables is 0 in the optimal tableau, alternate optimal solution exists. When the objective function is parallel to a binding constraint, objective function will assume same optimal value. So this is a situation when the value of optimal objective function remains the same. We have infinite number of such points

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13 Example: Max 2x x 2 ST x 1 + 2x 2 ≤ 5 x 1 + x 2 ≤ 4 x 1, x 2 ≥0

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14 The solution Max 2x 1 + 4x 2 Let S 1 and S 2 be the slack variables x 1 + 2x 2 + s 1 = 5 x 1 + x 2 + s 2 = 4 x 1, x 2, s 1, s 2 ≥0 Initial solution Let x 1 = 0, x 2 = 0, f=0, s 1 = 5, s 2 = 4

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15 Initial TableauBasisX1X2S1S2RHSRatio S1S1S1S /2=2.5 S2S2S2S /1=4 f Entering Variable Leaving Variable

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16 Optimal solution is 10 when x2=5/2, x1=0. How do we know that alternative optimal exist ?BasisX1X2S1S2RHSRatio X2X2X2X21/211/205/25 S2S2S2S21/20-1/213/2 3 min f002010

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17 By looking at f-row coefficient of the nonbasic variable. The coefficient for x1 is 0, which indicates that x1 can enter the basic solution without changing the value of f. Optimal sol. f=10 x1=0, x2=5/2 Basi s X1X2S1S2RHSRatio X2X2X2X21/211/205/25 S2S2S2S21/20-1/213/23 f Entering Variable Leaving Variable

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18 The second alternative optima is: The new optimal solution is 10 when x1=3, x2=1BasisX1X2S1S2RHS X2X2X2X20111 X1X1X1X11023 f002010

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19 Any point on BC represents an alternate optimum with f=10 Objective Function As the objective function line is parallel to line BC so the optimal solution lies on every point on line BC

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20 In practice alternate optimals are useful as they allow us to choose from many solutions experiencing deterioration in the objective value.

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21 When determining the leaving variable of any tableau, if there is no positive ratio (all the entries in the pivot column are negative and zeros), then the solution is unbounded.

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Example Max 2x 1 + x 2 Subject to x 1 – x 2 ≤10 2x 1 ≤ 40 x 1, x 2 ≥0 Unbounded Solution – Special cases (cont.)

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Solution Max 2x 1 + x 2 Let S 1 and S 2 be the slack variables x 1 – x 2 +s 1 =10 2x 1 +0x 2 + s 2 =40 x 1, x 2, s 1,s 2 ≥0 Initial Solution: x 1 = 0, x 2 =0, f=0, s 1 =10, s 2 =40 Unbounded Solution – Special cases (cont.)

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24 BasisX1X2S1S2RHSRatio S1S1S1S min S2S2S2S f-2000 Unbounded Solution – Special cases (cont.)

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25 BasisX1X2S1S2RHSRatio X1X1X1X S2S2S2S f Unbounded Solution – Special cases (cont.)

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The values of non basic variable are either zero or negative. The values of non basic variable are either zero or negative. So, solution space is unbounded So, solution space is unbounded 26 BasisX1X2S1S2RHSRatio X1X1X1X1100½20- X2X2X2X2011/210- f00-33/250 Unbounded Solution – Special cases (cont.)

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27 Unbounded Solution – Special cases (cont.) Graphical Solution 2x 1 ≤ 40 x 1 – x 2 ≤10 Unbounded Solution Space Optimal Point Objective function

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A problem is said to have infeasible solution if there is no feasible optimal solution is available

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Max f=3x 1 +2x 2 S.T. 2x 1 +x 2 ≤ 2 3x 1 +4x 2 ≥ 12 x 1, x 2 =0

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THANK YOU

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