Monday, April 11, 2011
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
Friday, April 1, 2011
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.
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.
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