CSC 445/545 Notes 3.1: Primal/Dual- Feasible, Infeasible, Unbounded Combinations

Primal: Dual:
Max cT x Min bT y
subject to subject to 
    A x <= b           
    AT  y >= c 
    x >= 0
 
    y >= 0

(a) For all primal feasible solutions x and for all dual feasible solutions y: cT x <= bT y.

This is sufficent to imply that:

(b) The Duality Theorem states that if either the primal or the dual has an optimal solution, then they both have optimal solutions having the same objective function values.
By the Fundamental Theorem of Linear Programming, all linear programs either have an optimal solution, are infeasible, or are unbounded.
Which combinations can occur?

Primal\Dual Optimal Infeasible Unbounded
Optimal 
   YES (b)   
   NO  (b)   
   NO  (b)
Infeasible 
   NO  (b) 
       (c) 
   YES (a)
Unbounded 
   NO  (b) 
   YES (a) 
   NO  (a)

(a) For all primal feasible solutions x and for all dual feasible solutions y: cT x <= bT y.

(b) The Duality Theorem.

(c) Can both the primal and the dual be infeasible?

Consider 

Maximize   2 X1  - X2
subject to
           X1  - X2  <=  1
          -X1  + X2  <= -2
X >= 0

The dual is:
   
Minimize   1 Y1  -  2 Y2
subject to
           Y1  - Y2  >=  2
          -Y1  + Y2  >= -1
Y >= 0

The dual in standard form:

Maximize   - Y1  +  2 Y2
subject to
           - Y1  + Y2  <=  -2
             Y1  - Y2  <=   1
Y >= 0
Solving the primal problem: 

Phase 1: Before pivoting to make feasible.
X3 =   1 - 1 X1 + 1 X2 + 1 X0 
X4 =  -2 + 1 X1 - 1 X2 + 1 X0 
-------------------------------------------
z  =   0 + 0 X1 + 0 X2 - 1 X0 

X0  enters.   X4  leaves.  z =  0 

The initial dictionary:
X3 =   3 - 2 X1 + 2 X2 + 1 X4 
X0 =   2 - 1 X1 + 1 X2 + 1 X4 
-------------------------------------------
z  =  -2 + 1 X1 - 1 X2 - 1 X4 

X1  enters.   X3  leaves.  z = -2 

After   1 pivot:
X1 =   1.50 + 1 X2 - 0.50 X3 + 0.50 X4 
X0 =   0.50 + 0 X2 + 0.50 X3 + 0.50 X4 
-------------------------------------------
z  =  -0.50 + 0 X2 - 0.50 X3 - 0.50 X4 

Original problem is INFEASIBLE.
Solving the dual problem: 

Phase 1: Before pivoting to make feasible.
Y3 =  -2 + 1 Y1 - 1 Y2 + 1 Y0 
Y4 =   1 - 1 Y1 + 1 Y2 + 1 Y0 
-------------------------------------------
z  =   0 + 0 Y1 + 0 Y2 - 1 Y0 

Y0  enters.   Y3  leaves.  z =  0 

The initial dictionary:
Y0 =   2 - 1 Y1 + 1 Y2 + 1 Y3 
Y4 =   3 - 2 Y1 + 2 Y2 + 1 Y3 
-------------------------------------------
z  =  -2 + 1 Y1 - 1 Y2 - 1 Y3 

Y1  enters.   Y4  leaves.  z = -2 

After   1 pivot:
Y0 =   0.50 + 0 Y2 + 0.50 Y3 + 0.50 Y4 
Y1 =   1.50 + 1 Y2 + 0.50 Y3 - 0.50 Y4 
-------------------------------------------
z  =  -0.50 + 0 Y2 - 0.50 Y3 - 0.50 Y4 

Original problem is INFEASIBLE.