Answer to Question #269983 in Operations Research for JOE

Question

Grain Handlers Limited has three warehouses, W1, W2 and W3. The table below shows the inventories of rice in the three warehouses.

Warehouse                    W1    W2    W3

Inventory (bags)           260    168    172

The company is required to supply three of its companies C1, C2, and C3 with rice. The requirements of the customers are as follows:

Customer                       C1     C2     C3

Requirement (bags)     280    120    200

The data below shows the cost of transporting one bag of rice from the warehouse to the customers

 

Transportation cost per bag (sh)

 

 

Customer

 

 

C1

C2

C3

 

W1

100

80

120

Warehouse

W2

140

80

140

 

W3

160

120

140

Required

The optimum solution

Answer 1

$\displaystyle \text{Let $x_1$ represent chocolate ice cream and $x_2$ vanilla}\\ \text{The linear program in the given problem is }\\ Maximize: 5x_1 +7x_2\\ \text{Subject to: } 6x_1 +9x_2 \leq 360\\ \qquad \qquad \quad 8x_1 + 5x_2 \leq 400\\\text{The linear program in its standard form is }\\ Maximize: 5x_1 +7x_2+0x_3+0x_4\\ \text{Subject to: } 6x_1 +9x_2 +x_3 = 360\\ \qquad \qquad \quad 8x_1 + 5x_2 + x_4= 400\\ \text{Next we form our first Tableau from our linear program}\\ \begin{matrix} & & x_1 & x_2 & x_3 & x_4 & \\ & & 5 & 7 & 0 & 0 &\\\hline x_3 & 0 & 6 & 9 & 1 & 0 & 360 \\\hline x_4 & 0 & 8 & 5 & 0 & 1 &400\\\hline & & -5 & -7 & 0 & 0 &0\\ \end{matrix}\\ \text{Next, we locate the most negative number in the bottom row(-7), and label the column}\\ \text{where it is found the work column , we then form positive ratios by dividing the elements}\\ \text{in the work column by corresponding elements in the last column. Next we label the }\\ \text{the smallest positive ratio, the pivot element. Using elementary row operations we }\\ \text{reduce the pivot element to 1 and other elements in the work column to 0 to obtain}\\ \text{our tableau 2}\\ \begin{matrix} & & x_1 & x_2 & x_3 & x_4 & \\ & & 5 & 7 & 0 & 0 &\\\hline x_2 & 7 & \frac{2}{3} & 1 & \frac{1}{9} & 0 & 40 \\\hline x_4 & 0 & \frac{14}{3} & 0 & -\frac{5}{9} & 1 &200\\\hline & & -\frac{1}{3} & 0 & \frac{7}{9} & 0 &280\\ \end{matrix}\\ \text{We observe that there is still a negative element in the bottom row, repeating the process }\\ \text{we obtain tableau 2}\\ \begin{matrix} & & x_1 & x_2 & x_3 & x_4 & \\ & & 5 & 7 & 0 & 0 &\\\hline x_2 & 7 & 0 & 1 & \frac{4}{21} & -\frac{1}{7} & 11\frac{3}{7} \\\hline x_1 & 5 & 1 & 0 & -\frac{15}{126} & \frac{3}{14} &42\frac{6}{7}\\\hline & & 0 & 0 & \frac{31}{42} & \frac{1}{14} &294\frac{2}{7}\\ \end{matrix}\\ \text{Since there are no negative elements in the bottom row, the solution is feasible. }\\ \text{The company should produce 5 of chocolate ice cream and 7 of vanilla}\\ \text{The maximum values is 290 to the nearest tenth}$