CSC 445/545 Notes 2.2: Solving the dual problem by the 2-phase method

Phase 1: Before pivoting to make feasible.

Y4 =  -4.00 + 1.00 Y1 + 5.00 Y2 - 1.00 Y3 + 1.00 Y0 
Y5 =  -1.00 - 1.00 Y1 + 1.00 Y2 + 2.00 Y3 + 1.00 Y0 
Y6 =  -5.00 - 1.00 Y1 + 3.00 Y2 + 3.00 Y3 + 1.00 Y0 
Y7 =  -3.00 + 3.00 Y1 + 8.00 Y2 - 5.00 Y3 + 1.00 Y0 
-------------------------------------------------
z  =  -0.00 + 0.00 Y1 + 0.00 Y2 + 0.00 Y3 - 1.00 Y0 

Y0  enters.   Y6  leaves.  z = -0.000000 

The initial dictionary:

Y4 =   1.00 + 2.00 Y1 + 2.00 Y2 - 4.00 Y3 + 1.00 Y6 
Y5 =   4.00 + 0.00 Y1 - 2.00 Y2 - 1.00 Y3 + 1.00 Y6 
Y0 =   5.00 + 1.00 Y1 - 3.00 Y2 - 3.00 Y3 + 1.00 Y6 
Y7 =   2.00 + 4.00 Y1 + 5.00 Y2 - 8.00 Y3 + 1.00 Y6 
-------------------------------------------------
z  =  -5.00 - 1.00 Y1 + 3.00 Y2 + 3.00 Y3 - 1.00 Y6 

Y2  enters.   Y0  leaves.  z = -5.000000 

After   1 pivot:
Y4 =   4.33 + 2.67 Y1 - 6.00 Y3 + 1.67 Y6 - 0.67 Y0 
Y5 =   0.67 - 0.67 Y1 + 1.00 Y3 + 0.33 Y6 + 0.67 Y0 
Y2 =   1.67 + 0.33 Y1 - 1.00 Y3 + 0.33 Y6 - 0.33 Y0 
Y7 =  10.33 + 5.67 Y1 -13.00 Y3 + 2.67 Y6 - 1.67 Y0 
-------------------------------------------------
z  =  -0.00 + 0.00 Y1 + 0.00 Y3 + 0.00 Y6 - 1.00 Y0 

The optimal solution: -0.00
Y1  =  0.00  Y2  =  1.67  Y3  =  0.00  Y4  =  4.33  
Y5  =  0.67  Y6  =  0.00  Y7  = 10.33  


Recall that the objective function is:

Maximize -1 Y1  -55 Y2  -3 Y3

The initial dictionary:
 
Y4 =   4.33 + 2.67 Y1 - 6.00 Y3 + 1.67 Y6 
Y5 =   0.67 - 0.67 Y1 + 1.00 Y3 + 0.33 Y6 
Y2 =   1.67 + 0.33 Y1 - 1.00 Y3 + 0.33 Y6 
Y7 =  10.33 + 5.67 Y1 -13.00 Y3 + 2.67 Y6 
-------------------------------------------------
z  =  -91.67 -19.33 Y1 +52.00 Y3 -18.33 Y6 

Y3  enters.   Y4  leaves.  z = -91.666664 

After   1 pivot:
Y3 =   0.72 + 0.44 Y1 - 0.17 Y4 + 0.28 Y6 
Y5 =   1.39 - 0.22 Y1 - 0.17 Y4 + 0.61 Y6 
Y2 =   0.94 - 0.11 Y1 + 0.17 Y4 + 0.06 Y6 
Y7 =   0.94 - 0.11 Y1 + 2.17 Y4 - 0.94 Y6 
-------------------------------------------------
z  =  -54.11 + 3.78 Y1 - 8.67 Y4 - 3.89 Y6 

Y1  enters.   Y5  leaves.  z = -54.111111 

After   2 pivots:
Y3 =   3.50 - 0.50 Y4 - 2.00 Y5 + 1.50 Y6 
Y1 =   6.25 - 0.75 Y4 - 4.50 Y5 + 2.75 Y6 
Y2 =   0.25 + 0.25 Y4 + 0.50 Y5 - 0.25 Y6 
Y7 =   0.25 + 2.25 Y4 + 0.50 Y5 - 1.25 Y6 
-------------------------------------------------
z  =  -30.50 -11.50 Y4 -17.00 Y5 + 6.50 Y6 

Y6  enters.   Y7  leaves.  z = -30.500008 

After   3 pivots:
Y3 =   3.80 + 2.20 Y4 - 1.40 Y5 - 1.20 Y7 
Y1 =   6.80 + 4.20 Y4 - 3.40 Y5 - 2.20 Y7 
Y2 =   0.20 - 0.20 Y4 + 0.40 Y5 + 0.20 Y7 
Y6 =   0.20 + 1.80 Y4 + 0.40 Y5 - 0.80 Y7 
-------------------------------------------------
z  =  -29.20 + 0.20 Y4 -14.40 Y5 - 5.20 Y7 

y4  Enters.   Y2  leaves.  z = -29.199991 

After   4 pivots:
Y3 =   6.00 -11.00 Y2 + 3.00 Y5 + 1.00 Y7 
Y1 =  11.00 -21.00 Y2 + 5.00 Y5 + 2.00 Y7 
Y4 =   1.00 - 5.00 Y2 + 2.00 Y5 + 1.00 Y7 
Y6 =   2.00 - 9.00 Y2 + 4.00 Y5 + 1.00 Y7 
-------------------------------------------------
z  =  -29.00 - 1.00 Y2 -14.00 Y5 - 5.00 Y7 

The optimal solution: -29.00
Y1  = 11.00  Y2  =  0.00  Y3  =  6.00  Y4  =  1.00  
Y5  =  0.00  Y6  =  2.00  Y7  =  0.00