Linear Programming

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Linear Programming

• Using the software that comes with the book.

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Lets start with a product mix problem Max
profit 40F 30S, subject to the constraints .4F
.5S 20 .2S 5 .6F .3S 21 and we
have the non-negativity constraints that F and S
0 Here F is the product fuel additive and S is
the product solvent base. Here there are three
constraints to how much can be made of each of F
and S. So, here we have 2 decision variables and
three constraints.
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The software that comes with the text can be used
to solve linear programming problems. At the
start you open the software and click on the
Linear Programming option. You then say how many
variables there are, the number of constraints
(not including the non-negativity constraints),
and say you are looking for a maximum value. Then
you hit ok. You will then be at the input screen
for the problem. Note on the objective function
you have X1 and X2 generically. You can change
these to F and S for our problem. Then you input
the coefficients on the objective function and
for the constraints. In our problem we have less
than or equal to constraints, but you only have
to type in the less than part. Then type in the
constraints right hand side values for example
we had 20 units of material 1. You can save your
work at this time. Then go to solution solve.
The output is on the next slide.
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1600 is the highest profit, given the constraints
F 25 and S 20 are the best values of F and S
and make profit 1600.
Section A
Section B
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Lets see what the section I labeled section A
means. On the computer printout this section is
labeled Objective Coefficient Ranges. What we
are going to do is a form of sensitivity
analysis. On the F variable in our objective
function we had value 40. The lower limit 24 and
the upper limit 60 have the interpretation that
the value of the coefficient could be anywhere
from 24 to 60 and the optimal values of F and S
having values 25 and 20, respectively, would not
change. The profit value could change, but not
these amounts. (This is assuming only the one
coefficient changes.) The profit line has a slope
in the graph. Changing the objective function
coefficient for F changes the slope. But, as
long as the change is in this range the slope
does not change enough to make the optimal values
of F and S change. Similarly the change of the S
coefficient could go between 20 and 50 and the
optimal F and S values would not change.
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Simultaneous changes in the objective function
coefficients If two or more (in our example we
only have two) objective function coefficients
change, then we may use the 100 percent rule to
see if the optimal values of the decision
variables change. In our example F is currently
at 40 and could go up to 60 it could go up 20 F
could go down to as low as 24 - it
could go down 16 and the solution would not
change. Similarly S could go down 10 or up 20
from its current position of 30. Say the
coefficient of F might change to 48. This is
above the current spot of 40. In fact it is 8
above 40 and the allowable limit above 40 is 20
so 8 is (8/20)100 40 above.
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Also say the coefficient of S might become 27.
This is 3 below the current value of 30 and is in
the allowable limit of going down 10 so 3 is
(3/10)100 30 of the allowable decrease. The
two percentage changes added together (one is an
increase and one is a decrease but we just add
the absolute value of the percentages) make
70 The rule if the sum of the allowable
changes is less than 100 the optimal values of
the decision variables will not change. Corollary
to the rule if the sum of the changes is above
100 the optimal solution still may not
change. Note the software with the book permits
editing the problem. For kicks, go to the
problem and change the coefficient of F to 18 and
resolve. Do you get the same solution?
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Change in constraints In our example we had 3
constraints and the limiting values, or what are
called the right hand sides of the constraints,
were 20, 5 and 21, respectively Lets think about
one of these changing at a given time. If a
value changes here, the feasible region of the
solution will change. Essentially the constraint
that we have a change in the right hand side for
will move in a parallel fashion. This will
change the size of the feasible region. Note on
the printout the first constraint has slack value
of zero. This means all 20 units are used. The
dual price of 33.33 means for each unit change in
the value of 20 changes the profit (whatever we
are maximizing, in general) by 33.33. If the
constraint would go up by 1 profit would go up
33.33 and if the constraint would go down by 1
profit would go down by 33.33.
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The point I am making on the previous slide only
holds true when the constraint changes within the
limits listed on the bottom of the output. In
this case the limits are from 14 to 21.5. Note
the second constraint has a slack of 1. This
means the constraint limit of 5 has not all been
used. Consequently, having more will not change
profit because what is on hand is not being
totally used. The 100 percent rule and its
corollary work in this area as well. Lets do a
min problem! Min 2A 3B, subject to the
constraints 1A 125, 1A 1B 350, 2A 1B
600, and the non-negativity constraints. The
output for the problem is on the next slide.
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This is a minimzation problem. Here a positive
dual price means if we had a unit more of the
constraining value cost would go down, by 1 in
this case. A negative dual price means the cost
would go up per unit increase in the constraining
value.
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More than two decision variables Lets look at an
example Max 40F 30S 50C, subject to the
constraints .4F .5S .6C 20 .2S
.1C 5 .6F .3S .3C 21 and the
non-negativity constraints. Lets see the output
on the next slide.
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Note the optimal value of S is zero here. The
reduced cost amount is 12.5. This means that the
coefficient on S in the objective function would
have to go up by at least 12.5 before S has a
value above zero in an optimal solution. Note
If this happens the maximum value will likely
change. By the way, this can happen with only two
decision variables.
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Constraints in Percentage terms Say in our three
decision variable problem that we have another
constraint of the following nature The amount of
S made must be at least 25 of the amount of F
made. Thus S .25F and in the context of the
computer program the constraint should be
written - .25F S 0. I will add this
constraint to the problem and show the output on
the next slide.
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Remember we had as a constraint -.25F S
0. The minus 12.121 means that if S was made to
be 1 unit above .25F, or really constraint -.25F
S 1, Then the maximum value of the objective
would fall by 12.121. Similarly, if S was made
to be 1 unit below .25F the maximum value would
rise by 12.121. Remember all our statements are
in context of being within lower and upper limits.