Some basic linear algebra is essential for this. Have you already studied that a bit? Say you have point $p$ inside your polyhedron of feasible points inside $\mathbb{R}^n$, and let's say you want to maximize the objective function. The objective function is linear, given by some $c \in \mathbb{R}^n$. If $v \in \mathbb{R}^n$ and $c \cdot v = 0$, then $p+v$ has the same objective value as $p$. In other words, one does not improve $p$ by moving it in a direction orthogonal to the objective function. If $v$ has some component parallel to $c$, then $p+v$ will be either strictly better or strictly worse than $p$. Therefore, if you are the point $p$ and you want to improve your objective value, then you should look for a direction $v$ in which you can move which has some positive component parallel to $c$, that is $c \cdot v > 0$. You may not be able to walk exactly in the direction of $c$ because a face of the polyhedron may stand in the way. But you may be able to walk in a direction having positive $c$ component. The only way you could get stuck is that for every possible improving direction, some face blocks you. In other words for every $v$ in the open halfspace $\{v:c\cdot v > 0\}$, the point $p+v$ lies outside the feasible polyhedron. This can only happen if either you already stand at a vertex, or you stand on a face of optimal solutions parallel to the hyperplane $\{v:c\cdot v = 0\}$. If you do not already stand at a vertex, then while staying on the hyperplane, you can walk to a vertex without changing your objective value. In the above explanation, I have used the boundedness of the feasible region: Without boundedness, it might be possible to walk in an improving direction forever, or it might be possible to walk along a face of optimal solutions without ever arriving at a vertex. I have also used convexity: If I do not stand at an optimal solution, then there will always exist an improving direction.
In the previous question, we examined how changing the constants in the constraints changed the optimal solution to a standard maximization problem. Such modification led to a different feasible region and changes to the optimal solution. In this question, we examine how changes in the coefficients of the objective function change the optimal solution. If one of the coefficients of the objective function increases or decreases, the feasible region remains the same. Since the feasible region is defined by the constraints and not the objective function, there are no changes to the corner points. A change in one of the coefficients changes the slope of the isoline, the line that corresponds to setting the objective function equal to a constant. Let’s recall the problem we have been analyzing: Figure 10 – A standard maximization problem and its corresponding feasible region. The objective function takes on the optimal value P = 3450 at the corner point (15, 20). The isoprofit line 70x + 120y = 3450 just touches the feasible region at that point verifying this assertion. Now let’s try increasing and decreasing the coefficient on x in the objective function. Figure 11 – This graph shows the feasible region with the isoprofit line at the optimal solution (gray dashed line) and the isoprofit line for the optimal solution in the modified problem (purple dashed line). The coefficient on x has decreased from 70 to 40. As the coefficient on x is decreased, the isoprofit line for the optimal solution gets less steep. At a value of 40, the isoprofit line passes through (0, 25) and (15, 20). At this point, the isoprofit line forms the border of the feasible region there so all points on the line between (0, 25) and (15, 20) yield the same maximum profit. If the coefficient is decreased to a value less than 40, (15, 20) is no longer part of the optimal solution. Figure 12 -This graph shows the feasible region with the isoprofit line at the optimal solution (gray dashed line) and the isoprofit line for the optimal solution in the modified problem (purple dashed line). The coefficient on x has increased from 70 to 120. As the coefficient on x is increased, the isoprofit line for the optimal solution gets steeper. At a value of 120, the isoprofit line passes through (0, 25) and (25, 10). The isoprofit line forms the border of the feasible region there, so all points on the line between (0, 25) and (25, 10) yield the same maximum profit. If the coefficient is increased to a value greater than 120, (15, 20) is no longer part of the optimal solution. As long as the coefficient on x in the objective function falls between 40 to 120, the optimal solution includes (x, y) = (15, 20). If the coefficient is equal to 40 or 120, any ordered pair on a line connecting (15, 20) to one of the adjacent corner points is optimal. However, if the coefficient on x falls outside of 40 to 120 the optimal solution moves to a new corner point. Recall that the coefficient on x in the objective function is the profit for each frame bag. If the profit per frame bag were to increase by up to $50 (to $120) or decrease by up to $30 (to $40), the optimal solution would not change. This solution, 15 frame bags and 20 panniers, is fairly insensitive to changes in the coefficient on x. Even though the location of the optimal solution does not change over this range, the profit does not change. Notice that we are only varying one coefficient at a time. Varying both coefficients simultaneously is beyond the scope of this presentation. Example 3 Find the Range of Values For Which a Corner Point Remains OptimalThe coefficient on y in the objective function for the linear programming problem can be varied to find a range of values over which (15, 20) is the optimal solution. Over what range of coefficient values will (15, 20) remain optimal? Solution Let’s examine the graphical solution for this linear programming problem. From the earlier discussion on varying the coefficient on x, we observed that changing the coefficient causes the slope of the isoprofit to change. If the slope of the isoprofit line increases so that it coincides with the constraint x + y ≤ 35, the optimal solution is all points on the line connecting (15, 20) and (25, 10). If the slope of the isoprofit line decreases so that it coincides with the constraint x + 3y ≤ 75, the optimal solution is all point on the line connecting (15, 20) and (0, 25). Figure 13 – In a, the coefficient on y in the objective function has been increased from 120 to 210. In b, the coefficient on y has been decreased from 120 to 70. As long as the coefficient on y falls in the range 70 to 210, the solution (15, 20) is optimal. Outside of this range, (15, 20) is no longer optimal. If we can determine the coefficient values that cause the isoprofit line to fall on the binding constraint borders, we know the range of values over which the solution remains optimal. In Figure 13a, the constraint border x + 3y = 35 and isoprofit line 70x + 210y = 5250 coincide. Notice that on each line, the coefficient on y is three times as large as the coefficient on x. Because of this, both lines have the same slope. We can always use this information to find the values of any missing constraints. For instance, to find the coefficient on y in the objective function that makes its slope the same as the slope on x + y = 35, we can observe that the coefficients must be the same. Therefore, the coefficients in the objective function must be the same. Since we have fixed the coefficient on x and are only varying the coefficient on y, we conclude that isoprofit lines for P = 70x + 70y are parallel to the constraint border x + y = 35. In particular, the isoprofit line 70x +70y = 2450 is exactly the same line. Putting this together, we observe that the coefficient on y can range from 70 to 210 and the corner point at (15, 20) is optimal. |