In linear programming, all model parameters are assumed to be constant; but in real life situations, the decision environment is always dynamic.

Therefore, it is important for the management to know how profit would be affected by an increase or decrease in the resource level, by a change in the technological process, and by a change in the cost of raw materials.

Such an investigation is known as **sensitivity
analysis or post-optimality analysis**. The results of sensitivity analysis
establish upper and lower bounds for input parameter values within which
they can vary without causing violent changes in the current optimal
solution.

The mechanics of *sensitivity testing* are explained with the help of
following example.

Luminous Lamps produces three types of lamps - A, B, and C. These lamps are processed on three machines - X, Y, and Z. The full technology and input restrictions are given in the following table.

Product | Machine | Profit per unit | ||
---|---|---|---|---|

M1 | M2 | M3 | ||

A | 10 | 7 | 2 | 12 |

B | 2 | 3 | 4 | 3 |

C | 1 | 2 | 1 | 1 |

Available Time | 100 | 77 | 80 |

Solution.

The linear programming model for this problem can stated as:

Maximize z = 12x_{1} + 3x_{2} + x_{3}

subject to

10x_{1} + 2x_{2} + x_{3 }≤ 100

7x_{1} + 3x_{2} + 2x_{3} ≤ 77

2x_{1}+ 4x_{2} + x_{3} ≤ 80

x_{1}, x_{2}, x_{3} ≥
0

The optimal solution to this problem is given below.

Final Table

c_{j} |
12 | 3 | 1 | 0 | 0 | 0 | ||
---|---|---|---|---|---|---|---|---|

c_{B} |
Basic variables B |
x_{1} |
x_{2} |
x_{3} |
x_{4} |
x_{5} |
x_{6} |
Solution values b (= X _{B}) |

12 | x_{1} |
1 | 0 | -1/16 | 3/16 | -1/8 | 0 | 73/8 |

3 | x_{2} |
0 | 1 | 13/16 | -7/16 | 5/8 | 0 | 35/8 |

0 | x_{6} |
0 | 0 | -17/8 | 11/8 | -9/4 | 1 | 177/4 |

z_{j} |
12 | 3 | 27/16 | 15/16 | 3/8 | 0 | ||

z_{j}-c_{j} |
0 | 0 | 11/16 | 15/16 | 3/8 | 0 |

An optimal policy is x_{1 }=73/8, x_{2} = 35/8, x_{3} = 0.

The associated optimal value of the objective function is 981/8.

Changes In Contribution Rate

First we investigate whether a previously determined **optimal solution** remains optimal if the contribution rate is changed. An increase in
c_{j} of a variable would mean that resources from other products
should be diverted to this more profitable product. The reverse is true
for a minimization problem.

Changes in c_{j} of a non basic variable

A non basic variable can be brought into the basis only if its contribution
rate becomes attractive. Hence, we need to determine the upper limit
of the profit contribution (c_{j}) of each non basic variable.
The reverse is true for a minimization problem.

From the above final simplex table, we note that profit contribution
for product C is Re 1, which is not greater than its z_{j}.
Thus, to bring x_{3} into the basis, its profit contribution
rate c_{j} must exceed Rs. 27/16 to make z_{j}-c_{j} value negative or zero (i.e., z_{j}-c_{j} ≤
0)

Specifically:

If c_{j* }- c_{j} > (z_{j}-c_{j}),
then a new optimal solution must be derived.

If c_{j* }- c_{j} = (z_{j}-c_{j}),
then alternative optimal solutions exist

If c_{j* }- c_{j} < (z_{j}-c_{j}),
then current optimal solution remains unchanged.

In our case c_{3} = 1 and z_{3}-c_{3} = 11/16, then

c_{3}* - 1 ≥ 11/16

c_{3}* ≥ 11/16 + 1 = 27/16

x_{3} can be introduced into the basis if its contribution
rate c_{3} increases upto atleast Rs. 27/16. If it increases beyond
that then the current solution will no longer be optimal.

Change in c_{j} of a Basic Variable

Let us consider the case of product A (x_{1} column), and
divide each z_{j}-c_{j} entry in the index row (for
non basic variable) by the corresponding coefficients in the x_{1} row as shown below.

- Minimum (z_{j}-c_{j} / y_{1j}; y_{1j} > 0) ≤
Δ1 ≤
Minimum (z_{j}-c_{j} / -y_{1j}; y_{1j} < 0)

Referring to the final simplex table, we observe that corresponding
to the non basic variables x_{3} & x_{5}, y_{13, } y_{15} < 0 Hence,

Minimum | 11/16 ------- -(-1/16) |
, | 3/8 ----- -(-1/8) |

= Minimum (11, 3) = 3

Corresponding to the non basic variable x_{4}, y_{14} > 0. Hence,

Minimum | 15/16 ------- 3/16 |

= 5

Hence,

- 5 ≤ c_{1}* - 12 ≤
3, i.e., 7 ≤ c_{1}* ≤
15

Thus, the optimal solution is insensitive so long as the changed profit
coefficient c_{1}* varies between Rs. 7 and Rs. 15.

Change In Available Resources

Now we investigate whether a previous optimal solution remains feasible
if the available resources change. For long-term planning it is important
to know the bounds within which each available resource (e.g., machine
hours) can vary without causing violent changes in the current optimal
solution. To illustrate, divide each quantity in the X_{B} column
by the corresponding coefficient in the x_{4} column of table.

X_{B} |
x_{4} |
X_{B }/ x_{4} |
---|---|---|

73/8 | 3/16_{} |
146/3 |

35/8 | -7/16 | -10 |

177/4 | 11/8 | 354/11 |

The least positive ratio (354/11) indicates to how the number of hours
of machine M_{1} can be decreased. The least negative ratio
(-10) indicates to how much the number of hours of machine M_{1} can be increased.

Calculating the range

Lower limit = 100 - 354/11 = 746/11

Upper Limit = 100 - (-10) = 110.

Hence, the range of hours for M_{1} is 746/11 to 110. By the
same way, the range of hours for Machine M_{2} & M3 can
be calculated.

Change In Technological Coefficients

Changes in the technological coefficients reflect potential change
either in the efficiency of manpower or in the state of technology.
To illustrate, x_{3} is non basic in the optimal solution of
this example. Consider the possibility that its coefficient in second
constraint is altered by Δ. Then, the
associated dual is

W_{1} + (2 + Δ)W_{2} + W_{3} ≥ 1

Substituting the current optimal values of the dual variables from
the final simplex table.

15/16 + (2 + Δ) X 3/8 + 0 ≥
1

or 27 + 6 Δ ≥ 16

or Δ ≥
- 11/6

Therefore, if Δ is smaller than - 11/6,
you should enter x_{3} into the basis.