Gomorys Cutting Plane Method Step I Ignoring integer
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Gomory’s Cutting Plane Method
Step I: Ignoring integer value requirement, solve the given ILP using Simplex Method Step II: If all the basic variables are have integer values, then we stop, the given optimal solution to LPP is also optimal for the ILP Step III: If any of the basic variables is having fractional value, then generate the cutting plane with the variable having the largest fractional value. Step IV: Add the cutting plane to the bottom of optimal simplex table and find the new optimal solution using the Dual Simplex Method. Steps from II to IV are repeated till all the basic variables are integer valued variables
Example: A Readymade Garment store sells two types of shirts-Zee-shirts and Button down shirts. He makes a profit of $3 and $12 per shirt on Zee-shirts and Button -down shirts respectively. He has two tailors A and B at his disposal for stitching the shirts. Tailors A and B can devote at the most 7 hours and 15 hours per day, respectively. Both of these two shirts are to be stitched by both the tailors. Tailors A and B spend 2 hours and 5 hours respectively in stitching one Zee-shirt and 4 hours and 3 hours respectively on stitching a Buttondown shirt. How many shirts of both types should be stitched in order to maximize the daily profit.
Let x 1 and x 2 denote the numbers Zee-shirts and Button-down shirts to be stitched daily respectively. Maximize Z=3 x 1+12 x 2 Sub to 2 x 1 +4 x 2 ≤ 7 5 x 1+3 x 2 ≤ 15 x 1, x 2 ≥ 0 and integers. Adding slack variables s 1 and s 2, then the problem in the standard form is given below: Maximize Z=3 x 1+12 x 1+0∙s 2 Sub to 2 x 1 +4 x 2 +s 1= 7 5 x 1+ 3 x 2 +s 2=15 x 1, x 2 , s 1, s 2 ≥ 0 and integers.
Let x 1 and x 2 denote the numbers Zee-shirts and Button-down shirts to be stitched daily respectively. Maximize Z=3 x 1+12 x 2 Sub to 2 x 1 +4 x 2 ≤ 7 5 x 1+3 x 2 ≤ 15 x 1, x 2 ≥ 0 and integers. Adding slack variables s 1 and s 2, then the problem in the standard form is given below: Maximize Z=3 x 1+12 x 1+0∙s 2 Sub to 2 x 1 +4 x 2 +s 1= 7 5 x 1+ 3 x 2 +s 2=15 x 1, x 2 , s 1, s 2 ≥ 0 and integers.
The optimal table by using the Simplex Method is given below with optimal solution x 1=0, x 2=7/4 Max Z=21. 3 12 0 0 b Xb y 1 y 2 y 3 y 4 12 x 2 1/2 1 1/4 0 7/4 0 s 2 7/2 0 -3/4 1 39/4 Zj-Cj 3 0 30 0 Z=21 Cb
Since the solution is a non integer one, we construct a Gomory’s fractional cut with the help of the source row. We choose the source row with the largest fractional value of the basic variable. If there is tie we choose arbitrarily. In our case, we have , two basic variable x 2 and s 2 in the optimal table of Simplex, with their integer and fractional values as Xb 1 =IB 1+f. B 1 =9/2=(4+1/2) = (for x 2 ) and Xb 2 =IB 2+f. B 2 =39/4=(9+3/4) (for s 2) Thus, max(f. B 1, f. B 2)=(3/4, 3/4) Since there is a tie, we can select any of x 2 and s 2 and x 2 being the decision variable, we choose x 2 as source row.
Using the source row , we have 7/4=1/2 x 1+x 2+1/4 s 1 (x 2 source row). Writing in terms of integers and fractional values (must be positive) (1+3/4)=(0+1/2)x 1+(1+0)x 2+(0+1/4)s 1 3/4+ (1 -x 2)=1/2 x 1+1/4 s 1 (collecting integers on left hand side) or 3/4 ≤ 1/2 x 1+1/4 s 1 (as we have deleted something from equation) Now, on adding Gomory’s Slack variable to make it an equality), say Sg 1, it becomes 3/4 +sg 1= 1/2 x 1+1/4 s 1 sg 1 -1/2 x 1 -1/4 s 1 =-3/4
Adding this additional constraint to the bottom of the optimal Simplex Table , we get a new table 3 12 0 Cb y 1 y 2 y 3 y 4 y 5 12 x 2 1/2 1 1/4 0 0 7/4 0 s 2 7/2 0 -3/4 1 0 39/4 0 sg 1 -1/2 0 -1/4 0 1 -3/4 Zj-Cj 3 0 0 Z=21 Ratios (Zj-Cj)/(y 3 j) y 3 j<0 -6 - -12 - - Xb 0 0 b
Applying the Dual Simplex Method, we get , Xb 3 0 12 0 b 0 Cb y 1 y 2 y 3 y 4 y 5 12 x 2 0 1 0 0 1 1 0 s 2 0 0 -5/2 1 7 9/2 3 x 1 1 0 -1/2 0 -2 3/2 Zj-Cj 0 0 -3/2 0 Z=33 0
more Gomory’s fractional cut with the help of x 1 as source row with two basic variable x 2 and s 2 in the optimal table of Simplex, with their integer and fractional values as Xb 2 =IB 2+f. B 2 =9/2=(4+1/2) (for s 2) and Xb 3 =IB 3+f. B 3 =3/2=(3+1/2)(for x 1) Thus, max(f. B 2, f. B 3)=(1/2, 1/2) Since there is a tie, we can select any of x 1 and s 2 and x 1 being the decision variable, we choose x 1 as source row and proceed as under 3/2=x 1+1/2 s 1 -2 sg 1 (x 1 source row) (1+1/2)=(1+0)x 1+(0+1/2)s 1+(-2+0)sg 1 1/2+ (1 -x 1+2 sg 1)=1/2 s 1 (collecting integers terms o LHS or 1/2 ≤ 1/2 s 1 on adding Gomory’s Slack variable, sg 2, it becomes 1/2+ sg 2 = 1/2 s 1 sg 2 -1/2 s 1 =-1/2
Applying this cut in to the Simplex Table and applying again Dual Simplex Method, we get , Xb Cb 3 12 0 0 y 1 y 2 y 3 0 0 y 4 y 5 y 6 b 12 x 2 0 0 1 0 s 2 0 1 -5/2 1 7 0 9/2 3 x 1 1 0 1/2 -2 0 3/2 0 sg 2 0 0 -1/2 0 0 1 -1/2 Zj-Cj 0 0 6 Z=33 Ratios(Zj-Cj)/(y 3 j), y 3 j<0 0 6 3/2 3 0
Updating the Table, we get , Xb Cb 3 12 0 0 y 1 y 2 y 3 0 0 y 4 y 5 y 6 b 12 x 2 0 0 0 1 0 s 2 0 1 7 -5 7 3 x 1 1 0 0 0 2 1 1 0 s 1 0 0 2 1 Zj-Cj 0 0 0 6 3 Z=15 0
The table is optimal and has integer solution. Thus the, - the store must stitch 1 Zee- shirt and 1 Button-shirt in order to get the profit of$15. Xb Cb 3 12 0 0 y 1 y 2 y 3 0 0 y 4 y 5 y 6 b 12 x 2 0 0 0 1 0 s 2 0 1 7 -5 7 3 x 1 1 0 0 0 2 1 1 0 s 1 0 0 -2 1 Zj-Cj 0 0 0 6 3 Z=15 0
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