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Problem 4. Consider the following standard LP_max 2X1 3x2s.t_2x1 + Xz + X3 = 16 X1 + 3x2 ~ X4 = 20X1 + xz = 10 X1'- ~,X4201) Use big M method to solve this LP ...

Question

Problem 4. Consider the following standard LP_max 2X1 3x2s.t_2x1 + Xz + X3 = 16 X1 + 3x2 ~ X4 = 20X1 + xz = 10 X1'- ~,X4201) Use big M method to solve this LP by adding three artificial variables, i.e-, to solve max 2X1 + 3xz M(a1 + 42 + 43)S.t_2x1 + Xz + X3 + 41 = 16 X1 + 3xz ~ X4 + a2 20 X1 + Xz + 43 = 10 X1' ,X4,C1,42,43 Z 0(20 points) 2) Use big M method to solve this LP by adding TWO artificial variables as follows max 2x1 + 3xz M(az + 43) S.t_ 2x1 +xz + X3 = 16X1 + 3x2 ~ X4 + 42

Problem 4. Consider the following standard LP_ max 2X1 3x2 s.t_ 2x1 + Xz + X3 = 16 X1 + 3x2 ~ X4 = 20 X1 + xz = 10 X1'- ~,X420 1) Use big M method to solve this LP by adding three artificial variables, i.e-, to solve max 2X1 + 3xz M(a1 + 42 + 43) S.t_ 2x1 + Xz + X3 + 41 = 16 X1 + 3xz ~ X4 + a2 20 X1 + Xz + 43 = 10 X1' ,X4,C1,42,43 Z 0 (20 points) 2) Use big M method to solve this LP by adding TWO artificial variables as follows max 2x1 + 3xz M(az + 43) S.t_ 2x1 +xz + X3 = 16 X1 + 3x2 ~ X4 + 42 = 20 X1 + Xz + @3 = 10 X1' ,X4,42,43 > 0



Answers

Use a graphing calculator, Excel, or other technology to solve the following linear programming problems.
Maximize $\quad z=2.0 x_{1}+1.7 x_{2}+2.1 x_{3}+2.4 x_{4}+2.2 x_{5}$
subject to: $\quad 12 x_{1}+10 x_{2}+11 x_{3}+12 x_{4}+13 x_{5} \leq 4250$
$\quad 8 x_{1}+8 x_{2}+7 x_{3}+18 x_{4}+5 x_{5} \leq 4130$
$\quad 9 x_{1}+10 x_{2}+12 x_{3}+11 x_{4}+8 x_{5} \leq 3500$
$\quad 5 x_{1}+3 x_{2}+4 x_{3}+5 x_{4}+4 x_{5} \leq 1600$
with $\quad x_{1} \geq 0, \quad x_{2} \geq 0, \quad x_{3} \geq 0, \quad x_{4} \geq 0, \quad x_{5} \geq 0$

For this problem. We have been asked to find the dual problem for the given maximization Question, uh, that I've copied here for you to look at. We're supposed to maximize Z equals four X up one plus three, except two plus two, except three. And we want to know what the dual problem is for this. Now, duality says that every standard maximization problem has a corresponding associate ID standard minimization problem that have the same answer to it now we're not going to solve it. We just want to see what the dual pro problem is. So in order to set that up first, we're going to make a matrix with all of these coefficients. Okay, so let's start. We do this in green. Let's start with our conditions on. We're gonna put just the coefficients into an augmented matrix. Okay, so our first line one except 11 except to one except three less than or equal to five. So those were my coefficients and my constant terms. Next, I have 1104 There is no, except three. So we do have to put a zero there, and our last one is 213 and 15. Hey, now we're going to complete this matrix. The bottom row is going to be our maximization equation itself. I'm not going to be putting in negatives like we sometimes do with simplex method. Just the coefficients for three and to equals zero. Okay, now, next step, we're going to transpose this matrix. That means Rose become columns. Columns become rose. So take a look at this. First row here. 1115 We're now going to write that as a column instead of a row. 1115 Next second row 1104 We're going to write that as the second column. Okay, third row, 213 15. Third row becomes our third column, and our fourth row becomes our fourth column right now. From this, we can get our dual problem of a minimization function. So let's write this down here. We want to minimize. Okay? Are minimization function. We're going to get by looking at the coefficients on our bottom row here, so it's gonna be five. And because we have switched from maximization, the minimization I don't want to use X is I'm gonna use wise is my variables so five y sub one plus four y sub two plus 15 y sub three. And the convention your textbook is using is using a w here instead of a Z again showing that we are in a different, uh, problem set right now. Okay, Now, for our constraints were going to be subject thio the following constraints and we're gonna find thes by reading across our new columns. Kent, we're gonna use wise as our variables. So I have why sub one plus wise up to Plus to wise up three. Now, if if you look when we did our maximization we use less than or equal to for minimization we're going to switch that and use greater than or equal to four. Okay, second row y sub one plus wise up to plus wise up three greater than or equal to three and our last one. Why someone no y sub choose. Plus why Sub three is greater than or equal to two. And as always, we're going to say that thes variables do have to be greater than Sirrah. So this is our do a pro problem for our given maximization question,

For this problem, we are asked to maximize the function P equals two X plus three Y. Subject to the constraints that 0.1, X plus 0.2 Y is greater than or equal to one two, X plus Y is greater than or equal to 10 and both X and Y are greater than or equal to zero. So we start off with our plot of the region of feasibility where I'll note that the region is going to be everything outside of that area, who funky pen, Everything outside of that area near the origin. So what that indicates to us then is that we have an unbounded region. Now this wouldn't be a problem or an unbounded region on one side at least. This wouldn't be a problem if we were looking to minimize our function. But no, we are looking to were asked to maximize it, But there's nothing that stops us from taking x equals 52 billion and y equals 5007 90 732 trillion. So there's no bound on how large we can make X and Y. So we have to say no optimal solution. And it's because we have an unbounded domain or an unbounded feasibility region, unbounded feasible region

For this problem, we are given the function P equals X plus two Y. And we are asked to maximize that While subject to the constraints that 30 X plus 20 Y. It's less than or equal to 600. 0.1 X plus 0.4 Y is less than or equal to 40.2 X plus 0.3 Y is less than or equal to 4.5 And both X&Y are greater than or equal to zero. So to begin we have our plot of our region of feasibility here as well as our corners which I will label as A. B, C and D. So having our corners we want to evaluate our function P at each one of those points. So at point A will have P is going to equal zero. At point B will have P is going to equal zero plus two times 10. So 20 a point C P is going to equal 12 Plus two times 7. So that's 12 plus 14. That is going to equal 26 A point D. We'll have that P is going to equal just 20. So we can see that the optimal value is 26. Found at the .127

For this problem, we are asked to maximize P equals X plus Y. Subject to the constraints that X plus two, Y is less than or equal to +92, X plus Y is less than or equal to nine and both X and Y are greater than zero. So here we have a plot of the region of feasibility here. So we can see our corner points. Now, all we need to do is calculate the values of our corner points. So we can clearly see that at 00. We have a value of zero. Actually let me just label those corner points first of all. And I will go let's l go clockwise, starting at 00. That's a that's be that's see that's D. So we can see that appoint A. We have that P is going to equal uh zero At Point B. We have that P is going to equal 4.5. We have that at point C. He is going to equal uh not three P is going to equal six and at point D P is going to equal 4.5. So we can see that the maximum is going to be P equals six. At that point C.


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