In this lesson, we're going to see what is the impact of changes made to the model parameters. In particular, we're going to see how the optimal solution changes when the modern parameter specifications are different. So let's continue with the example that we had in the last lesson. That is a production optimization problem and see what happens when some of the model parameters are different. In this problem, we have exactly the same set up as the previous one, but now we are allowing one extra hour for machine one. That is instead of machine one running for only eight hours it can run for nine hours. So the objective function remains as before, it remains unchanged. It's only the constraint on machine one that is altered, so you have 2 x1 plus x2 less than equal to 9 in this case. Previously it was eight. We're letting it run for one more hour, so therefore this has to be less than equal 9. The constraint for the machine two doesn't change, the nonnegativity constraint, that is, x 1 is greater than equal to 0, x 2 greater than equal to 0. They also remain unchanged, so let's see where we have an alternative specifications such as this, which is a slightly different problem. The constraint is relaxed by one unit. What's the change in the optimal solution? So here we have the original plot from the previous problem. And this was the region that was the feasible region. But now this particular constraint M1 original, is a bit more relaxed instead of it passing through 0, 8 and 4, 0, it would be passing through some different points. So let's take a look at what those points will be. So the new constraint or the new boundary line for this constraint will be given by, 2 x 1 plus x 2 is equal to 9. So consequently you can calculate the two points through which this will pass. So if you said x 1 equal to 0, then x 2 will be equal to 9 and when you said x 2 equal to 0 then x 1 will be four and a half, that is 9 over 2. So these are two points through which the line 2 x 1 plus x 2 equal to 9 will pass. So 0, 9 is one point, and four and half, 0 is another point. So here you have, four and a half. Somewhere here and 9 over here. So this is 0, 9 and this point here is four and a half, 0. And the new line would pass through these two points and it would be parallel to the previous constraint that we had, since only the intercept is going to change. The slope of this line will be the same because 2 and 1 here remained unchanged, so the slope of the line doesn't change. It's only the intercept that that would change because you have relaxed the constraint by one unit as the machine will run for one hour more. So this is our new constraint, so our feasible region will be this part right here. As shown in the figure. So there is a little bit of extra that you have from the previous version, so in the previous version we had. This is the, this dotted line as the constraint on machine one, but now we have relaxed constraint because machine one runs for one more hour. So this is the solid line, is the constraint for machine one. So now we can solve for this particular point. If you take this two lines. 2 x 1 plus x 2 equal to 9 and the other constraint that remain unchanged, so this is a constraint for machine one. And you have x 1 plus 2 x 2 is equal to 8 for machine two. If you solve these two equations together, you would get x 1 is equal to 10 over 3 and x 2 is equal to 7 over 3. So this point right here now, previously this point was 8 over 3 and 8 over 3 but now the relaxed constraints This point of intersection between the new M1 Line and of M2 boundary line would be 10/3 and 7/3. So this is this other corner point, so here we have the four corner points as before, you have 0, 0, you have 0, 4, you have 10/3 and 7/3 and 4.5, 0. These are the four corner points of the new feasible region. So we can substitute the values of these corner points into the objective function to enumerate and find which corner point would produce the highest value. So the values from the previous set of corner points that we already had, they would remain unchanged. So we had z equal to z at the 0, 0 was equal to 0, then you substitute it in 300x1 + 200x2, which was your objective function. You have z at 0, 0 being 0, z at 0, 4 was going to be 300 x 0 + 200 x 4, which is $800 in total profit. Then z at 4.5, 0, that value now is going to be 300 x 9/2 or 4.5 plus 0, which is going to be 2,700/2. And z at 10/3 and 7/3 is going to be 300 x 10/3 + 200 x 7/3, which comes out to be 4,400/3. So now we see that the value of the objective function at the optimal solution for this new problem, I'll call it z star, new is going to be 4,400/$3. And the Value of the objective function at the optimal solution for the original problem when the machine one was constrained to running for 8 hours at most was 4000/3. So we can see that the optimal value of the objective function has changed by this amount, For a unit change in the capacity of machine one. So machine one capacity change is the denominator, and the numerator is the total amount by which the profit has changed as a result of relaxing machine one's capacity and letting it run for one more hour. So this is going to be 4,400/3- 4000/3, and the change in machine one's capacity or the runtime was 9- 8 is 1. So this is going to be 400/3 which is $133.33. So for one additional hour that machine one runs, my profit would increase by $133.33. In other words, we can write this down as the profit increases by $133.33 for each additional hour that machine one runs. So if you were to let machine one run for one additional hour, that is The machine once original constrained by one hour, you would improve your profit by $133.33. This value is known as the Shadow Price for the constraint on Machine 1. So, This metric that we calculated here, that is the change in the objective functions value at the optimal solution for a unit change in the constraint, is the Shadow Price for that constraint. So here for Machine 1, if you let it run for one more hour, then we would improve our profit by $133.30. And that's the Shadow Price for that constraint. What it tells us, the amount of profit we could have gotten if we could run Machine 1 for one more hour. Now we would do the same kind of calculation for Machine 2.
