Hi k9queen!
Here are the answers to your questions (at last!)
1. Let's first write the objective function and the constraints. Let's
call X to the number of suits against surgeons and Y to the number of
suits against professor. Since the company earns 5 per suit against
surgeons and 1 per suit against professors, we have that the company
must maximize the following function:
Max 5X + Y
Regarding the constraints, the company has a total of 30
person-months. The professor suits take 6 person-months and the
surgeon suits take 10 person-months to prepare. Therefore, one
constraint is:
10X + 6Y <= 30
Similarly, regarding witnesses, the other constraint is:
3X + 5Y <= 15.
Finally, it's clear that both quantities should be positive. In order
to solve this problem graphically we must draw the constraints and
iso-profit curves in a two-dimension graph. Let's first get the
equations for the lines that define the constraints. One of them is:
10X + 6Y <= 30
6Y <= 30 - 10X
Y <= 5 - (10/6)X (1)
So if we draw the line Y=5-(10/6)X, all points below this line
(because of the <=) satisfy this constraint. The other constraint
defines the line
3X + 5Y <= 15
5Y <= 15 - 3X
Y <= 3 - (3/5)X (2)
Again, all points below this line satisfy this constraint. Since there
are no more constraints, we have that all points that satisfy BOTH
constraint are the *feasible* points. We must choose among this
feasible points when maximizing the functions; otherwise we would be
violating one or both the constraints.
Finally, we have to find the equation for the iso-profit lines.
Iso-profit lines represent different combinations X and Y that
generate the same profit. For example, let's see what combinations of
X and Y gives a profit of 30. To do this, we set:
5X + Y = 30
Y = 30 - 5X
All the points that belong to this line (for example, X=1 and Y=25; or
X=5 and Y=5) generate the same profit (30). Now let the profit level
be unspecified and call it U. We have that isoprofit lines are of the
form:
5X + Y = U
Y = U - 5X
As you can see, all isoprofit lines (in this problem, of course) have
the same slope (-5) and have different intercepts (U).
Now, the problem is to maximize profits. Notice that the isoprofit
lines that correspond to high profits are up and to the right (they
have a higher intercept). The problem is then to choose the "highest"
isoprofit line possible such that at least one of its points is
feasible. Let's see it graphically:
http://www.angelfire.com/alt/elmarto/googleanswers/283068q1.jpg
The blue line represents the first restriction. The green line
represents the second one. The "shaded" are contains the feasible
points: these points are below both restriction lines. Red lines are
the iso-profit lines. As you can see, a profit of 10 is feasible (one
would only need to choose X and Y along this line and inside the
shaded area) but it's not optimum: we can choose a line more rightward
(which represents a higher profit) and still pass through the
feasibility set. Of course, we can't go on indefinitely. The isoprofit
line of 20 profit doesn't pass through the feasibility set. Therefore,
it's not possible to obtain a profit of 20. As you can see. The
isoprofit line of 15 is the best we can do. It touches the feasibility
set only at X=3 and Y=0. This means that the answer to this problem is
that the firm must file 3 surgeon suits and 0 professor suits, for
total profits of 15.
2. The procedure is exactly the same here. There is however a
difference in where the feasibility set is and which way the optimum
is. Let's see the problem. Call X the number of packages from Duffin
House and Y the number of packages from Gorman Press. The function to
minimize is then:
50X + 150Y
Since the firm needs 2,500 mystery novel, and both suppliers offer
packages which contain 5 mystery novels each, we have that one of the
restriction is:
5X + 5Y >= 2500 (1)
Also, the firm needs 3,500 romance novels. Duffin's packages come with
5 romance novels, while Gorman's come with 10. Therefore,
5X + 10Y >= 3500 (2)
Finally, the firm has agreed to buy at least 25% of the total number
of packages to Gorman. So we have that:
Y/(X+Y) >= 1/4
4Y >= X+Y
3Y >= X (3)
Now, as before, let's rewrite the restrictions so we get the lines in the graph.
Restriction (1) becomes:
Y >= 500 - X
Restriction (2):
Y >= 350 - (1/2)X
Restriction (3):
Y >= X/3
Here we have the first difference with the previous problem. The
feasible points are *above* the constraint lines and not below,
because of the >= sign. But, just as before, the feasible set is the
set of points that satisfy ALL of the restrictions. Let's also compute
the equuation for the iso-cost lines. We have:
50X + 150Y = U
Y = U/150 - (1/3)X
Notice that, clearly, they are now iso-cost and not iso-profit lines.
This brings the second difference with the previous problem. Here we
have that it's "better" to choose isocost lines that are as low as
possible. Check the graph for this question
http://www.angelfire.com/alt/elmarto/googleanswers/283068q2.jpg
There is no shaded are because it would be to messy to understand.
Keep in mind then that the feasible points are the ones that are above
all 3 restrictions. As you can see in the graph, it's not possible to
go below $42,000 in cost. Trying to get a lower cost violates
restrictions (2) and (3). Therefore $42,000 is the minimum possible
cost. As you can see, the optimum combination of X and Y is where
restriction (2) intersects restriction (3). So let's find X and Y by
finding this intersection (of course, in order to come up with the
42,000 number, I first figured out X and Y from the intersection and
then calculated their associated cost).
Restriction (2): Y = 350 - (1/2)X
Retsriction (3): Y = (1/3)X
From this system of 2 equations and 2 unkowns, we get X=420, Y=140.
That's the optimum in this problem.
3). The problem (objective function and constraints) can be found at
the link politicalguru-ga provides. Let's see why the answer is the
one provided there, which is to send 4 from Little Rock to Detroit and
5 from Urbana to St. Louis.
a) We have to minimize cost and we must send at least 4 to Detroit and
5 to St. Louis. Since sending people representatives is costly,
clearly this constraint will be binding; that is, we'll want to send
EXACTLY 4 people to Detroit and EXACTLY 5 to St. Louis. Now, the
accomodation costs are irrelevant to the choice, because they will
have to be paid the same either if the representatives come from
Urbana or Little Rock. Therefore, we must send representatives from
Urbana to the city to which it's cheaper to travel (St. Louis) and the
same with the people from Little Rock. Since in both places there are
6 people, and we must send 4 to one city and 5 to the other one, it's
feasible to do this. So the solution is the one mentioned before.
b) Changing the hotel price would indeed change the total cost, but it
wouldn't affect the decision of sending 4 from Little Rock to Detroit
and 5 from Urbana to St. Louis. The idea is that it doesn't matter
where the representatives come from, the accomodation cost will have
to be paid anyway. Changing the air fares can change the answer,
though. For example, if traveling from Urbana to both two cities were
cheaper than flying from Little Rock. It would be best to send all 6
representatives from Urbana, some to one city and some to the other
one. This would take advantage of the low air fares from this city.
Something similar could be said if travelling from Little Rock were
very cheap.
You may want to visit the following links to learn more about solving
linear programming problems graphically:
http://www.ms.ic.ac.uk/jeb/or/morelp.html
http://jwilson.coe.uga.edu/EMT668/EMAT6680.F99/McCallum/emat6690/Linear%20Programming/Linear%20Programming.html
Google search strategy
linear programming graphically
://www.google.com/search?sourceid=navclient&ie=UTF-8&oe=UTF-8&q=linear+programming+graphically
I hope this helps! If you have any doubt regarding my answer, please
don't hesitate to request a clarification before rating it. Otherwise
I await your rating and final comments.
Cheers!
elmarto |
