Graphical model - Linear Programming Task 2 - Speech or Presentation Example

Graphical Model – Linear Programming Task 2 College Graphical Model – Linear Programming Task 2 Task:  A. There are three constraints nutrient C, flavor additive and color additive hat are plotted on the attached “Graph 1”. The supply of three ingredients for each production period is limited to 30 units of nutrient C, 72 units of flavor additive and 90 units of color additive. The formula for a case of Brand X calls for 4 units of nutrient C, 12 units of flavor additive, and 6 units of color additive.

The Brand Y formula per case requires 4 units of nutrient C, 6 units of flavor additive, and 15 units of color additive.Therefore, equations for the linear constraints areNutrient constraint: 4X + 4Y ≤ 30 (Maximum constraint)Flavor constraint: 12X + 6Y ≤ 72 (Maximum constraint)Color additive constraint: 6X + 15Y ≤ 90 (Maximum constraint)Non-negativity Constraints: X ≥ 0 and Y ≥ 0B. The contribution to profit from the Brand X and Brand Y are $40 and $30, respectively. Therefore, objective function is toMaximize: 40X + 30YThe equation for the purple objective function (profit line) passing from two points (0, 8) and (6, 0) is given by:Y – Y1 = {(Y2 – Y1)/(X2 – X1)}(X – X1)Y – 8 = {(0 – 8)/(6 – 0)}(X – 0)Y = -(4/3)X + 8Table 1 gives the combination of cases of Brand X and Brand Y that lies on the purple objective function (profit line) and the total contribution to profit.

Table 1: The combination of cases of Brand X and Brand Y and the total contribution to profitXY = -(4/3)X + 8Total Contribution to Profit = 40X + 30Y08$24016.67$24025.33$24034$24042.67$24051.33$24060$240As shown in table 1, for any combination of cases of Brand X and Brand Y that lies on the purple objective function, the total contribution to profit is always $240.Therefore, the total contribution to profit if the company produces a combination of cases of Brand X and Brand Y that lies on the purple objective function (profit line) will be $240.C. Shaded area in figure 1 shows the feasible solution.

Figure 1: Corner (extreme) PointsThe profit will be maximum at the corner (extreme) points of feasible solution. The two corner points are (0, 6) and (6, 0). The other corner points are given by the solution of system of equations for ‘Nutrient and Flavor’ and ‘Nutrient and Color’.The system of equations for Nutrient and Flavor are 4X + 4Y = 30 and 12X + 6Y = 72 that gives solution as (4.5, 3).The system of equations for Nutrient and Color are 4X + 4Y = 30 and 6X + 15Y = 90 that gives solution as (2.5, 5).Thus, four corner points are (0, 6), (2.5, 5), (4.5, 3) and (6, 0).

Table 2 shows the optimal solution and the total contribution to profit at the corner (extreme) points of feasible solution.Table 2: The optimal solution and the total contribution to profitXYTotal Contribution to Profit = 40X + 30Y 06$1802.55$2504.53$270 (Maximum)60$240The maximum total contribution to profit is for 4.5 cases of Brand X and 3 cases of Brand Y and equals to $270. Thus, 4.5 cases of Brand X and 3 cases of Brand Y I recommend should be produced during each production period for optimum production if Company A wants to generate the greatest amount of profit.D. Total contribution to profit for 4.

5 cases of Brand X and 3 cases of Brand Y = 40X + 30Y = 40(4.5) + 30(3) = $270The total contribution to profit that would be generated by the production level of 4.5 cases of Brand X and 3 cases of Brand Y I recommended is $270.