Download Notes on Sensitivity - Linear Programming and Network Flows | MATH 444 and more Exams Mathematics in PDF only on Docsity!
10 Sensitivity.
10.1 Issues.
The solution of a linear programming problem yields an optimal feasible point. It is often necessary to be concerned about how changes in the data A b c , , affect the solution. It is
particularly troubling if tiny changes in the data radically alter the optimal solution. The search for these so-called sensitive values is therefore a worthwhile pursuit. Simple examples of sources of change in the data are market fluctuations and measurement errors. In contrast to these uncontrolled changes are controlled changes in the data. Examples include changes in the available resources, price alterations, and efficiency modifications. Sensitivity analysis is carried out to get quantitative information about the effect of changes in the data.
10.2 Changes in the resources.
Let the original b be replaced by b. Use the notation ∆ b = b − b so that the new constraint has right-hand side b + ∆ b. If the basic variables are kept the same, then the final tableau changes to ( ) ( )
1 1 1 T TB 1 TB 1 TB 1
B A B B^ b^ b c c B A c B c B b b
− −^ − − − −
The only way this tableau can fail to be a final tableau is if B −^1 ( b + ∆ b )≥ 0 is violated. Let ej be the m × 1 matrix with all entries equal to zero except in the j th ' row where the entry is a one. Write ∆ b = ∆ b e 1 1 (^) + " + ∆ b em m where ∆ b (^) j = bj − bj. It follows that
B −^1 ( ∆ b ) = ∆ b B 1 −^1 ( e 1 ) + " + ∆ b Bm −^1 ( em ) where B −^1 ( e j ) is the j th ' column in B −^1. The
requirement B −^1 ( b + ∆ b )≥ 0 is equivalent to a system of linear inequalities in the ∆ b (^) j.
Example Consider the problem
1 2 3 1 2 3 1 2 3 1 2 3
max 2 when 2 3 12 , , 0
x x x x x x x x x x x x
The final tableau is given by 3 0 5 1 1 22 2 1 3 0 1 12 − 3 0 − 5 0 − 2 − 24
To determine how much b can vary without a change of basic variables look at the condition B −^1 ( b + ∆ b )≥ 0. In this example
B 0 1
− = ^
^.
It follows that
1 2
12^ b^ 0 b 1 0
+^ ∆^ +^ ∆^ ≥
From this it follows that ∆ b 2 ≥ − 12. The admissible region looks like:
Nothing is sensitive. Suppose ∆ b 1 = − 7 and ∆ b 2 = − 8. It follows that 3 b (^) 4
and 1 1 1 3 7 B b 0 1 4 4
− = ^ ^ ^ =^
The new optimal solution is given by x 1 (^) = 0, x 2 (^) = 4, x 3 = 0. A quick check shows that this
solution is in the new feasible set. The new value of the primal objective value is equal to the new value of the dual objective.
10.3 Buying resources.
Interpret b as a limit on the available resources. Suppose the cost, in dollars per unit, of resource j is given by αj. Assume that there is an additional amount of β dollars to buy more resources.
Let β (^) j be the amount spent on resource j. It is natural to ask how the additional money should be spent. The change in resource j is given by ∆ b (^) j = βj / αj. The change in objective value is
given by
Deduce that μ ˆ T = ^3 0 5 and ∆ c T ≤ ^3 0 5 , with ( ∆ c ) 2 = 0 , will preserve the
optimality conditions. To test the limits, change the primal objective to 4 x 1 (^) + 2 x 2 (^) + 6 x 3. The dual constraints 1 2 1 2 1 2
λ λ λ λ λ λ
are satisfied with no slack by the dual solution λ ˆ T^ = ^0 2 . Conversely, change the primal
objective to 5 x 1 (^) + 2 x 2 (^) + x 3. The optimal primal solution changes from x ˆ T = ^0 12 0 to
x ˆ T = ^6 0 0 with dual solution λ ˆ T^ = ^0 5 / 2. A quick application of the Verification
Theorem proves this. It is also clear that the previous optimal solution is no longer optimal.
10.5 Variation in basic variable cost.
In this case the entire bottom row of the tableau is affected. Both
( c + ∆ c ) T − ( cB + ∆ c B )^ T B −^1 A ≤ 0 ,
and
− ( cB + ∆ c B )^ T B −^1 ≤ 0
must hold. Example Consider the problem 1 2 1 2 (^1 2 1 ) 1 2
max 80 70 when (^2 6 ) , 0
x x x x x x (^) x x x x
^ +^ ≤
The final tableau is given by 1 0 1/ 3 1 0 14 0 1 1/ 3 2 0 4 0 0 4 / 3 10 1 20 0 0 10 / 3 60 0 1400
Rewrite the optimality inequalities as −∆ c B TB^ −^1 ≤ c BTB −^1 and ∆ c T − ∆ c B TB^ −^1 A ≤ c BTB^ −^1 A − cT. In the current example they correspond to
1 2
c c
− ∆^ ∆ ^ ^ − ≤^
and
1 2 1 2
c c c c
∆ ∆ − ∆ ∆ ^ ≤
The last matrix inequality is trivially satisfied. The first matrix inequality is equivalent to 1 2 1 2
c c c c
The admissible region looks like:
There is an alternative way to derive the same relationship. Since the problem only involves two variables, graph the feasible set. The solution is at the vertex ( 14, 4 ), which is at the intersection of the lines 6 x 1 (^) + 3 x 2 = 96 and x 1 (^) + x 2 = 18. The slope is − 2 and − 1 , respectively. To avoid changing the optimal point, the slope of the level lines corresponding to the objective
( 80 + ∆ c 1 ) x 1 + ( 70 + ∆ c 2 ) x 2 must satisfy
1 2
c c
− ≤ − +^ ∆ ≤ −
This is equivalent to 70 + ∆ c 2 (^) ≤ 80 + ∆ c 1 (^) ≤ 140 + 2 ∆ c 2 , which is equivalent to the previous pair of inequalities.
10.6 Variations in efficiency.
The tableau 1 1 1 T TB 1 TB 1 TB 1
B A B B b c c B A c B c B b
− − − − −^ − −^ − −
reveals that variations in the left-hand side of the constraints impact the entire tableau. The reason for this is that the matrix B depends on the matrix A. It is cumbersome to track the propagation of variations in A into the matrix B −^1. From a practical point of view it is comforting to know that the entries in A often are related to hardware and technology as opposed to market supply or demand. It follows that the entries in A are less susceptible to variations. In large-scale problems a different concern is the possibility of data entry errors. In a
constraints
1 1 ,^0
T m n
A c α α λ^ c + λ
≥^ ^ ≥
are satisfied. It follows that AT λ ≥ c , λ ≥ 0 so λ is feasible in the dual of the old setting. By the
Fundamental Theorem the objectives are equal in the new setting and hence
1
T (^) ˆ T (^) n 0 T x c x c c (^) + λ b
^ ^
By the Verification Theorem, the solution λ is optimal in the dual of the old setting. This is a contradiction since none of the optimal solutions in the dual of the old setting are feasible in the new setting. Hence, the introduction of a new product should only be considered if cn (^) + 1 > α λ 1 ˆ 1^ + "+ α λm ˆ m for all optimal solutions in the dual of the old setting.
Example A manufacturer produces a standard cage in a two-step process. In the first step the cage is machine-assembled. It takes four hours to assemble one cage. In the second step the cage is hand- painted. It takes two hours to paint one cage. The manufacturer has rented the assembly machine for 120 hours each week. A paint crew is hired and works 60 hours each week. The manufacturer considers introducing a collapsible cage that takes ten hours to assemble and nine hours to paint. Each standard cage built is sold for $200 and each collapsible cage is sold for $800. The initial tableau in the original setting is 4 1 0 120 2 0 1 60 200 0 0 0
The final tableau in the original setting is 1 1/ 4 0 30 0 1/ 2 1 0 0 50 0 6000
An optimal solution to the dual in the original setting is given by λ ˆ T^ = ^50 0 . This solution is
not feasible in the new setting since 10 λ ˆ 1^ + 9 λ ˆ 2 = 500 < 800. To conclude that the new product should be introduced is a mistake. Since the ratios in the initial tableau are equal, it is possible to pivot on the second row and get 0 1 2 0 1 0 1/ 2 30 0 0 100 6000
This time λ ˆ T^ = ^0 100 . This solution is feasible in the new setting since
10 λ ˆ 1^ + 9 λ ˆ 2 = 900 ≥ 800. The Verification Theorem guarantees that x ˆ 1 = 30, x ˆ 2 = 0 is optimal
since both objectives have value 6000. The new setting has initial tableau 4 10 1 0 120 2 9 0 1 60 200 800 0 0 0
The next tableau is given by 16 / 9 0 1 10 / 9 160 / 3 2 / 9 1 0 1/ 9 20 / 3 200 / 9 0 0 800 / 9 16000 / 3
The final tableau is given by 0 8 1 2 0 1 9 / 2 0 1/ 2 30 0 100 0 100 6000
or 1 0 9 /16 5 / 8 30 0 1 1/ 8 1/ 4 0 0 0 25 / 2 225 / 3 6000
Observe how the same optimal primal solution corresponds to two different dual solutions. This example illustrates some of the difficulties encountered in the presence of degeneracy , i.e., when one of more of the basic variables have value zero.
