GRAPHICAL
LP SOLUTION
The
graphical procedure includes two steps:
1. Determination of the feasible solution space.
2. Determination of the optimum solution from among
all the feasible points in the solution space.
The
procedure uses two examples to show how maximization and minimization objective
functions are handled.
2. Solution of a Minimization Model
Example
2.2-2 (Diet Problem)
Ozark
Farms uses at least 800 lb of special feed daily. The special feed is a mixture
of corn and soybean meal with the following compositions:
The
dietary requirements of the special feed are at least 30% protein and at most
5% fiber. Ozark Farms wishes to determine the daily minimum-cost feed mix.
Because
the feed mix consists of corn and soybean meal, the decision variables of the
model are defined as
x_{1}
= lb of corn in the daily mix
x_{2} = lb of soybean meal in the daily mix
The
objective function seeks to minimize the total daily cost (in dollars) of the
feed mix and is thus expressed as
Minimize
z= 3x_{1} +9x_{2}
The
constraints of the model reflect the daily amount needed and the dietary
requirements. Because Ozark Farms needs at least 800 Ib of feed a day, the associated
constraint can be expressed as
x_{1}+x_{2}
> =800
As for
the protein dietary requirement constraint, the amount of protein included in x_{1}
lb of corn and x_{2} lb of soybean meal is (.09x_{1} + .6x_{2}) lb. This quantity should equal at least 30% of the
total feed mix (x_{1} + x_{2}) lb-that is,
.09x_{1}
+ .6x_{2} >= .3(x_{1}+x_{2})
In a
similar manner, the fiber requirement of at most 5% is constructed as
.02x_{1}
+ .06x_{2} <= .05(x_{1}+x_{2})
The
constraints are simplified by moving the terms in x_{1} and x_{2}
to the left-hand side of each inequality, leaving only a constant on the
right-hand side. The complete model thus becomes
Minimize
z= .3x_{1} + .9x_{2}
Subject
to
x_{1}+x_{2}
>= 800
.21x_{1}
- .30x_{2} <= 0
.03x1 - .0lx_{2}
>= 0
x1,x2
>=0
Figure
2.3 provides the graphical solution of the model. Unlike those of the Reddy
Mikks model (Example 2.2-1), the second and third constraints pass through the
origin. To plot the associated straight lines, we need one additional point,
which can be obtained by assigning a value to one of the variables and then
solving for the other variable. For example, in the second constraint, x_{1}
= 200 will yield .21 x 200 - .3x_{2} = 0, or x_{2} = 140. This means that the
straight line .21x_{1} - .3x2
= 0 passes through (0,0) and (200,140). Note also that (0,0) cannot be used as
a reference point for constraints 2 and 3, because both lines pass through the
origin. In-stead, any other point [e.g., (100, 0) or (0,100)] can be used for
that purpose.
Solution:
Because
the present model seeks the minimization of the objective function, we need to
reduce the value of z as much as
possible in the direction shown in Figure 2.3. The optimum solution is the
intersection of the two lines x_{1} + x_{2} = 800 and .21x_{1} - .3X_{2} = 0, which yields X_{1} =
470.591b and X_{2} = 329.41 lb. The associated minimum cost
of the feed mix is z = .3 X 470.59 + .9 x 329.42 = $437.65
per day.
Remarks. We need to take note of the way
the constraints of the problem are constructed. Because the model is minimizing
the total cost, one may argue that the solution will seek exactly 800 tons of
feed. Indeed, this is what the optimum solution given above does. Does this
mean then that the first constraint can be deleted altogether simply by
including the amount 800 tons
in the
remaining constraints? To find the answer, we state the new protein and fiber
constraints as
.09x_{1} + .6x_{2} ≥ 3 x 800
.02x_{1} + .06x_{2}
≤ 05 X 800
or
.09x_{1} + .6x_{2} ≥ 240
.02x_{1} + .06x_{2} ≤ 40
The new
formulation yields the solution x_{1} = 0, and x_{2} = 400 lb (verify with TORA!), which does not satisfy
the implied
requirement fOT SOO Ib of feed. This means that the constraint x_{1}
+x_{2 }≥ 800 must
be used explicitly and that the protein and fiber constraints must remain exactly
as given originally.
Along the
same line of reasoning, one may be tempted to replace x_{1} + x_{2 }≥ 800 with x_{1} + x_{2
}= 800. In the
present example, the two constraints yield the same answer. But in gen-eral
this may not be the case. For example, suppose that the daily mix must include
at least 500 lb of corn. In this case, the optimum solution will call for using
500 lb of corn and 350 Ib of soybean (verify with TORA!), which is equivalent
to a daily feed mix of 500 + 350 =
850 lb. Imposing the equality constraint a priori will lead to the conclusion
that the problem has no feasible solution (verify with TORA!). On the other
hand, the use of the inequality is inclusive of the equality case, and hence
its use does not prevent the model from producing exactly 800 Ib of feed mix,
should the remaining constraints allow it. The conclusion is that we should not
"pre-guess" the solution by imposing the additional equality
restriction, and we should always use in-equalities unless the situation
explicitly stipulates the use of equalities.
PROBLEM
SET 2.2B
1. Identify
the direction of decrease in z in
each of the following cases:
Minimize
z = 4x_{I} - 2x_{2}'
Minimize
z = -3x_{1}
+ x_{2}'
Minimize
z = - x_{1}
- 2x_{2}'
2. For
the diet model, suppose that the daily availability of corn is limited to 450
lb. Identify the new solution space, and determine the new optimum solution.
3. For
the diet model, what type of optimum solution would the model yield if the feed
mix should not exceed 800 Ib a day? Does the solution make sense?
4. John
must work at least 20 hours a week to supplement his income while attending
school. He has the opportunity to work in two retail stores. In store 1, he can
work between 5 and 12 hours a week, and in store 2 he is allowed between 6 and
10 hours. Both stores pay the same hourly wage. In deciding how many hours to
work in each store, John wants to base his decision on work stress. Based on
interviews with present employees, John estimates that, on an ascending scale
of 1 to 10, the stress factors are 8 and 6 at stores 1 and 2, respectively.
Because stress mounts by the hour, he assumes that the total stress for each
store at the end of the week is proportional to the number of hours he works in
the store. How many hours should 10hn work in each store?
5. OilCo
is building a refinery to produce four products: diesel, gasoline, lubricants,
and jet fuel. The minimum demand (in bbl/day) for each of these products is
14,000,30,000, 10,000, and 8,000, respectively. Iran and Dubai are under
contract to ship crude to OilCo. Because of the production quotas specified by
OPEC (Organization of Petroleum Ex-porting Countries) the new refinery can receive
at least 40% of its crude from Iran and the remaining amount from Dubai. OilCo
predicts that the demand and crude oil quotas will remain steady over the next
ten years.
The
specifications of the two crude oils lead to different product mixes: One barrel
of Iran crude yields .2 bbl of diesel, .25 bbl of gasoline,. l bbl of
lubricant, and .15 bbl of jet fuel. The corresponding yields from Dubai crude
are .1, .6, .15, and .1, respectively.
Oileo
needs to determine the minimum capacity of the refinery (in bbll day).
6. Day
Trader wants to invest a sum of money that would generate an annual yield of at
least $10,000. Two stock groups are available: blue chips and high tech, with
average an-nual yields of 10% and 25%, respectively. Though high-tech stocks provide
higher yield, they are more risky, and Trader wants to limit the amount
invested in these stocks to no more than 60% of the total investment. What is
the minimum amount Trader should in-vest in each stock group to accomplish the
investment goal?
7. An
industrial recycling center uses two scrap aluminum metals, A and B, to
produce a special alloy. Scrap A contains 6% aluminum, 3%
silicon, and 4 % carbon. Scrap B has 3% aluminum, 6% silicon, and
3% carbon. The costs per ton for scraps A and Bare $100 and $80, respectively. TIle
specifications of the special alloy require that (1) the aluminum content must
be at least 3% and at most 6%, (2) the silicon content must lie between 3% and
5%, and (3) the carbon content must be between 3% and 7%. Determine the
opti-mum mix of the scraps that should be used in producing 1000 tons of the
alloy.
8. TO RA Experiment. Consider the Diet Model and let the objective
function be given as
Minimize z = .8x_{1} + .8x_{2}
Use TORA to show that the optimum
solution is associated with two
distinct corner points and that both points yield the same objective value. In
this case, the problem is said to have alternative
optima. Explain the conditions leading to this situation and show that, in
effect, the problem has an infinite number of alternative optima, then provide
a formula for determining all such solutions.
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