In this class, we addressed problem 14 from Chapter 2. So first step, we expressed the problem in STANDARD FORM:
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1 In this class, we addressed problem 14 from Chapter 2. So first step, we expressed the problem in STANDARD FORM: Now that we have done that, we want to plot our constraint lines, so we can find our feasible region. Since we have three constraints (for material 1, material 2 and material 3) we will plot three lines. The easiest way to plot our graphs is to find both the x- and y- intercepts on each of the three lines and connect them. For the Material 1 constraint, we first set f to zero, and solve the equation: So we have one point (0, 40). Next we set s equal to zero and solve for f:
2 Here then we have another point on the graph: (50, 0). For the material two constraint, we determined that, since f is not part of the equation, it will just be a horizontal line, where s = 25: The line happens to be horizontal on our plot, just because we are choosing to put the s amount on the y (vertical) axis
3 For the material 3 constraint, we found our two points: (0, 70) and (35, 0): So to summarize, we have enough points to draw our three constraint lines. Here are the lines plotted:
4 You can see that there are three lines, labeled M1, M2 and M3. These lines represent the constraints imposed by the limitations of the amounts of material 1, 2, and 3, respectively. The arrows next to the label indicate the side where solutions will be feasible. (I ve blocked out some things for simplicity). After all of the lines are plotted, we can shade in and label the feasible region. The feasible region is the region where any solution will be feasible, according to all of the constraints. We have done so. Now we will find a FEASIBLE (not the optimal) solution line by selecting an ARBITRARY amount, and solve the solution. For instance we chose $1200 as our arbitrary amount. In a sense, we are asking if it is feasible to contribute $1200 to the profit, given the constraints. The line is calculated in a similar way to above, by finding both the x- and y- intercept points:
5 Here, we have found two points for our feasible solution line: (0, 40) and (30, 0) This line is plotted on the graph, in red. We can see that part of the line is inside of the feasible region, so it is feasible that we can make $1200 given these constraints. Since part of the line is outside of the feasible region, not all of the solutions along that line are feasible. Now on to the optimal solution. Remember that the optimal solution is ALWAYS where a line intercepts with another line, or possibly where a line meets an axis. First, we can approximate the optimal solution using the graphical method. We do this by taking a ruler, and, while keeping it parallel to the feasible line, slide it away from the origin (0, 0). Keep in mind, this is a maximization model, so we slide it away from the origin. If this was a minimization problem, we would slide it TOWARDS the origin. Here is an example where the feasible solution line has slid to the optimal solution.
6 Optimal solution line (approximated) Feasible line 40f + 30s = $1200 Now we can approximate an optimal solution. By eyeballing the graph, we can see that the optimal solution is to produce about 25 tons of fuel additive and about 20 tons of solvent base. But there is a more accurate way to find a solution. For instance it may be too close to call, compared to another possible optimal solution. You may need to solve for both, and see what the more optimal solution is. Essentially, the optimal solution is always at an intersect of two lines. In this case, it is at the intersect of the constraint lines for materials 1 and 3. The way to solve for the optimal amounts of fuel additive (f) and solvent base (s) is the same method used for determining the intersect between two lines. This method consists of three steps, but can be done in a variety of ways (ie there is no one right way to do it, but there is only one correct answer). So take the equations for the two lines, and perform the following three steps: 1. For the first equation (for instance the material 1 constraint), isolate one of the terms (for instance f ). this will give you what f equals, in terms of s 2. In the second equation (for instance the material 3 constraint), substitute the results of the first step for one of the terms (for instance f ), to solve for the other (for instance s ). this will give you the value of s 3. Back to the first equation (for instance the material 1 constraint), substitute in the value for s and solve for f
7 So here is what we did. We solved for f in the first equation. The result was f = 50 5/4s. It has to go through a couple of calculations, and these calculations will not be the same for other ones, but this happens to be the method we used. Here is the original equation: Then we isolated the f term: So we want to turn that 2/5f into just f. So we divided everything by two: Then we multiplied by five: So now we have completed the first step, because we know what f equals (although we don t yet know s, but we still have to do some more calculation). Start of step two: we put this result into the second equation, wherever we see f. Here is the original equation: So we substitute f with the results from the first phase. It is shown in the brackets:
8 Now we got rid of the brackets by multiplying the 3/5 with each of the two terms in the brackets: This step just solved the one 150/5 terms, because it reduces to 30. This just arranges the s terms on one side, and the integers on the other: is -9, so we do this calculation: We take the ¾ and 3/10 terms, and make them like terms. The lowest common denominator of 4 and 10 is 20, so we change the terms of these fractions to 20. For the ¾, we multiply the top term 3 by 5 to get 15, because we also have to multiply 5 to turn 4 into 20. For the 3/10, we multiply the top 3 by 2 to get 6, because we multiply 10 by 2 to get 20: = 9, so we get 9/20 for the fraction: Cancel out the negatives:
9 And then we know that s = 20. This is the end of the second step. For the third step, we will sub the 20 in the s terms from the FIRST equation: Which gives us Moving the 10 to the right side gives us : Therefore: Third step is done. Now we have s = 20 and f = 25. Check out the graph that we drew. Our estimate was actually pretty accurate. 20 tons of solvent base, and 25 tons of fuel additive, is the optimal amount we can produce, while satisfying all of the constraints. One last step: now we want to find out how much profit contribution we will make when we produce the optimal amount: We go back to the optimization equation that we first created in the beginning, and substitute in 20 for s and 25 for f, and we get a profit contribution of $1600.
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