linear programming review

Today in class we reviewed the problems we did before break on Wednesday, Thursday and Friday. Some important things to know that are examples from the lego problem and computer problem (both linear programming) are: objective function-there is always an objective,what you are trying to figure out or accomplish, an equation ex.lego- p=16t + 10c converts in to Z=16 x1 + 10x2 (x little one/ x little two) constraint-limits, almost always the problems we will be given will have a limit ex. how many legos they will ship us or how many computers you can make in a day feasible solution-something possible, there may only be one but they is often many in the problems we are given, does NOT have to be the best solution ex. 3 tables 0 chairs in the lego problem or 0 computers made are both solutions possible with the resources given. optimal solution-the best solution, NOT the amount of $ you made in profit ex. 3 tables and 1 chair made from legos there is not always going to be only two variables x1 and x2. There can be an infinite number of decision variables. non negativity is important to pay attention to as well when we are given limits. ex #37 pg 2-10. x1,x2,x3, and x4 are all greater than or equal to 0

Dealing with constraints in the Computer Flips problem

Today in class, we continued solving Gates Williams computer problem. This problem is in the linear programming unit and we are looking to maximize profit. The objective function for this problem is; Z=200X1+300X2. Z is expressed as the weekly profit, X1 represents the number of Simplex computer built, and X2 represents the number of Omniplex computers built each week.

Individually at our desks we each picked two points in the feasible region of the graph we constructed yesterday, that generated the same weekly profit. For example I picked point (8,8) and point (0,20), they both generated a weekly profit of $4,000. When i connected the two points it created a line called a line of constant profit.

We did it three more times with three different sets of points that generated the same weekly profit. We came to the conclusion that the third line of constant profit is parallel to the boundary lines, meaning they all have the same slope. In this problem the slope of this graph is -2/3, we found that by putting our objective function into slope intercept form (y = mx + b).

The point of this graph that generates the largest weekly profit is (40,0), which generates a weekly profit of $8,000, Each point we choose for our optimal solution had to be a corner point because the intersection of two constraint lines create a corner point and usually generate the most profit.

After we came to this conclusion we were given our first market constraint stating that the company cannot sell more than 20 Simplex computers in any given week. This new constraint gave us a problem for the optimal solution we found earlier, it made the point (40,0) an infeasible solution. After factoring the new constraint we came up with a new optimal solution which is 20 Simplex and 10 Omniplex computers per week. This will generate us a profit of $7,000. That point would maximize our profit while factoring in the new constraint.

Now the students in the sale's department of Computer Flips decide to extend the research on the Omniplex model, creating a new constraint for our problem. The new constraint states that they cannot sell more than 16 Omniplex computers per week. This constraint does not affect our optimal solution because it still lies in a feasible region with the new constraint factored in. So our optimal solution remains the same, it stays at 20 Simplex computers and 10 Omniplex computers per week.

Maximizing Profit at Computer Flips

Today we learned about maximizing profits, using a Junior Achievement Company called Computer Flips as an example.

Computer Flips sells two computer models, the Simplex and the Omniplex. The Simplex has fewer add-ons and can be completed in a shorter time, and the Omniplex has more add-ons and takes twice as long to complete. Students at the company only work a certain amount of time each week, which factors in to how many models can be made.

The production manager of the company must decide the rate of production per week of each computer model in order to maximize the company's weekly profit. The point at which the profit is maximized is called the optimal solution.

In order to make this kind of decision, we use a technique called linear programming.

First we began by making guesses on how many models the company should produce and testing what the profits would be. Although certain combinations resulted in high profits, we found that there wouldn't be enough installation time to actually make that number of each model.

After we found which amounts of the models could be made to maximize profits and also be completed in the given installation time, we needed to write an equation of the amount of weekly profit, z. z is a function of x1 (the weekly production rate for Simplex) and x2 (the weekly production rate for Omniplex). This is called an objective function because the objective is to maximize profit, and the variables are called decision variables because they will help the production manager make his decision.

The equation was written as an inequality that showed the relationship between the installation time required to produce x1 and x2 each week, and the amount of available installation time each week.

To graph the equation, we replaced x1 and then x2 with 0 in order to find the intercepts. Once we had these points, we were able to draw a line on the graph.

The line of the equation creates an area on the graph called the feasible region, which contains all points that represent feasible, or possible, production mixes. They are constrained into this area because there is only a certain amount of installation time each week.

Combinations of the two computer models that will result in the largest profits will be in the feasible region.

Yesterday in class, we started exploring mathematical programming. We started by doing a problem that has a concrete model, and used Lego pieces. We did this to be able to visualize the real world model.

The problem was about a furniture company that makes only two products, tables and chairs. The parts could be used for both tables or chairs. Tables required two large pieces and two small pieces, while the chairs required one large piece and two small pieces. The resources from the company were unlimited, but they would only manufacture as many as they thought it would sell. In the first example, the company could only obtain six large and eight small pieces per day. The profit from each table was $16 and $10 for each chair. In our first question, we had to figure out what the production rates should be. There were four possibilities for production rates and and after making a table and looking at the total profits for each of these options, it was decided the optimal solution was made with 2 tables and 2 chairs. Although all the decision variables are feasible or possible, 2 tables and 2 chairs was the best option.

We then went on to answer a few questions about different amounts of lego pieces available. We were asked if there would be a difference with 7 large pieces instead of 6, and we decided it would. The optimal solution was 3 tables and 1 chair, which would generate $58 instead of $52 in the previous question. But when asked if 9 pieces instead of 8 would make a difference, we concluded it would not and there would not be a new optimal solution. This shows that more pieces is not always necessarily better.

In the last example, we were given 27 small pieces and 18 large pieces per day, and we had to maximize profit. The best way to do this was to have 6 tables and 6 chairs to make $156. We were then given 19 large pieces instead of 18 and we concluded the optimal solution would be 7 tables and 5 chairs. But when given 28 small pieces instead of 27, it would not make a difference and would not change the optimal solution.

In the end, we concluded it is more difficult to solve the second problem than the first because the numbers were bigger. The more pieces available, the higher the numbers get and the more complicated the problem becomes.

Enrique Chooses a College

Today in class, we explored the decision making process that Enrique
Ramirez made along with his friend Anna to decide which college he
should attend. Although Enrique asked for Anna's help, we must notethat
the final decision is made based on what is best for Enrique and not
what Anna thinks, although she is a very good friend for helping her
buddy out. The specific process used is called ther multi-attribute
utility theory (MAUT). Enrique first generated a list containing general criteria that was
important to him in choosing a college. Enrique decided that academics,
location, cost, and social life were the factors most important to him.
  The two then took the criteria and specified a few measures for each
criterion.
For each college, the two then collected data for each measure. They
used data in fields such as SAT scores, what division the school was in
(D1, D2, or D3), ect... They then rescalled each measure to common
units from 0 to 1, with 0 being the worst alternative and 1 being the
best alternative.
After the criterion had been scaled, Anna interviewed Enrique about the
importance of each criteria on a scale of 0 to 100 based on importance.
They then calculated a proportional weight between 0 and 1 for each
measure. Tuition costs were the most important while average daily
temperature was the least important.
Enrique and Anna then calculated a total weighted score for each
college using these rankings. These weights yielded a ranking of
colleges and allowed Enrique to identify the best option based on his
preferences. Although Canisius had the best numbers in the end, Enrique
decided on Clark because his desire for academic excellence outweighed
the large cost. Enrique picked Clark because while Canisius gained a
few more points based on the cost criteria, Clark had more strengths
all together.

Multi-Criteria Decision-Making Example

Today in class we worked through an example of MCDM. In the example Isabelle was trying to decide what cell phone plan to choose. The first step was to identify the criteria and measures. Her criteria were cost and other important factors. Her measures were monthly service charge, number of minutes per month, minimum length of the contract, whether there is unlimited texting, and quality of service. Then she rated them using continuous or categorical measures. The next step is to collect all of the necessary data, and the range of each measure.

After this, all of the measures must be rescaled into a common unit. For this they must be converted using a proportional scale. This scale is between is between zero and one. Zero will always be the worst, and one will always be the best.

After the measures are rescaled, each of the categories can be either totaled or averaged to see which cell phone plan will be the best match. Before a final decision can be made though, the categories must be ranked in order of importance. Isabelle's parents first ranked the categories one through five. Then they put them on a scale of zero to one hundred.

The next step is to total all of the 0-100 rankings. Then to calculate the weight, each ranking is divided by the total. All of the weights should add up to equal one. Next, multiply the weight by the original common unit score, and add the categories up to get a total score.
Finally, the strengths and weaknesses must be determined for the top two options, and then a final decision can be made.

Introduction to Making a Hard Decision

Multi-Criteria Decision-Making (MCDM) is a process for making difficult and complicated decisions.  These complex situations typically have more than one important objective that the decision maker needs to take into consideration. 

For example, when choosing where to locate a new power plant, planners have to weigh the needs of safety by building the plant to withstand various natural disasters.  At the same time, they have to make efforts to keep construction costs down so the plant will be cost-effective.  These two objectives (or criteria), safety and cost, are both important and yet are at odds with each other:  typically one is improved at the expense of the other.

MCDM provides a framework for taking many objectives into account simultaneously.  The decision maker decides on the relative value of each objective (a subjective process, see below).  These values, priorities, and preferences are quantified (transformed into numeric values), which allows mathematics to be used.  When the MCDM process is completed, all the possible alternatives will be given a score and can be ranked in terms of how well they meet the decision maker's objectives.

It is important to note that there is no "right" answer for these types of problems.  Each decision is unique, as are the values, priorities, and preferences of each decision maker.  For example, in choosing which college to attend, "distance from home" is an important factor, but one student may prefer to be far away while another wants to stay close.  These perspectives will be built into the mathematical model of the decision created by MCDM.

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It's not exactly relevant to this section, but here's some more information on Operations Research:
http://hsor.org/

Here's some information specific to Project MINDSET:
http://www.mindsetproject.org/index.php/aboutus