Page 78 - DCOM303_DMGT504_OPERATION_RESEARCH

Unit 3: Linear Programming Problem – Simplex Method

3.3.2 Change in the Right-hand Side Constraints Values and Effect on Notes
Optimal Solution

Suppose an additional 40 kgs of material 3 is available, the right-hand side constraint increases
from 150 to 190 kgs.
Now, if the problem is solved, we get the optimal values as
x = 23.61, x = 16.11 and Z = 1427.78
1 2 max
From this, we can infer that an additional resources of 40 kgs increases the profit by = 1427.78 –
1250 = ` 177.78
Therefore, for one kg or one unit increase, the profit will increase by

= 177.78/40
= ` 4.44
Dual price is the improvement in the value of the optimal solution per unit increase in the right-
hand side of a constraint. Hence, the dual price of material 3 is ` 4.44 per kg.
Increase in material 2 will simply increase the unused material 2 rather than increase in objective
function. We cannot increase the RHS constraint values or the resources. If the limit increases,
there will be a change in the optimal values.
The limit values are given in Table 2.10, i.e., Min RHS and Max RHS values.
For example, for material 3, the dual price ` 4.44 applies only to the limit range 150 kgs to 262.50
kgs.
Where there are simultaneous changes in more than one constraint RHS values, the
100 Per cent Rule must be applied.

Reduced Cost

Cost of consumed  Profit per unit 
   
Reduced cost/unit of activity = resources per unit   of activity 


   
 of activity   
If the activity's reduced cost per unit is positive, then its unit cost of consumed resources is
higher than its unit profit, and the activity should be discarded. This means that the value of its
associated variable in the optimum solution should be zero.
Alternatively, an activity that is economically attractive will have a zero reduced cost in the
optimum solution signifying equilibrium between the output (unit profit) and the input (unit
cost of consumed resources).
In the problem, both x and x assume positive values in the optimum solution and hence have
1 2
zero reduced cost.
Considering one more variable x with profit ` 50
3
Z = 40x + 30x + 50x
max 1 2 3
Subject to constraints,
4x + 5x + 6x  175 (a)
1 2 3
2x + 1x  50 (b)
2 3

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