We see that Flair Furniture’s two constraint equations can be expressed as follows:

The numbers $\left(2, 1, 1, 0\right)$ in the first row represent the coefficients of the first equation— namely, $2T+1C+1S_1+0S_2.$ The numbers $\left(4, 3, 0, 1\right)$ in the second row are the algebraic equivalent of the constraint $4T+3C+0S_1+1S_2.$

As suggested earlier, we begin the initial solution procedure at the origin, where $T=0$ and $C=0.$ The values of the other two variables must then be nonzero, so $S_1=100$ and $S_2=240.$ These two slack variables constitute the initial solution mix; their values are found in the quantity (or right-hand-side [RHS]) column. Because T and C are not in the solution mix, their initial values are automatically equal to 0.

This initial solution is a basic feasible solution and is described in vector, or column, form as

Variables in the solution mix, which is called the basis in LP terminology, are referred to as basic variables. In this example, the basic variables are $S_1$ and $S_2.$ Variables that are not in the solution mix or basis and that have been set equal to zero (T and C, in this case) are called ­nonbasic ­variables. Of course, if the optimal solution to this LP problem turned out to be $T=30,$ $C=40,$ $S_1=0,$ and $S_2=0,$ or

then T and C would be the final basic variables, and $S_1$ and $S_2$ would be the nonbasic variables. Notice that for any corner point, exactly two of the four variables will equal zero.

Many students are unsure as to the actual meaning of the numbers in the columns under each variable. We know that the entries are the coefficients for that variable. Under T are the coefficients $\left(\begin{array}{c}2\\ 4\end{array}\right),$ under C are $\left(\begin{array}{c}1\\ 3\end{array}\right),$ under $S_1$ are $\left(\begin{array}{c}1\\ 0\end{array}\right),$ and under $S_2$ are $\left(\begin{array}{c}0\\ 1\end{array}\right).$

But what is their interpretation? The numbers in the body of the simplex tableau (see Table M7.1) can be thought of as substitution rates. For example, suppose we now wish to make T larger than 0—that is, to produce some tables. For every unit of the T product introduced into the current solution, 2 units of $S_1$ and 4 units of $S_2$ must be removed from the solution. This is so because each table requires 2 hours of the currently unused painting department slack time, $S_1.$ It also takes 4 hours of carpentry time; hence, 4 units of variable $S_2$ must be removed from the solution for every unit of T that enters. Similarly, the substitution rates for each unit of C that enters the current solution are 1 unit of $S_1$ and 3 units of $S_2.$

Another point that you are reminded of throughout this module is that for any variable ever to appear in the solution mix column, it must have the number 1 someplace in its column and 0s in every other place in that column. We see that column $S_1$ contains $\left(\begin{array}{c}1\\ 0\end{array}\right),$ so variable $S_1$ is in the solution. Similarly, the $S_2$ column is $\left(\begin{array}{c}0\\ 1\end{array}\right),$ so $S_2$ is also in the solution.2

We now continue to the next step in establishing the first simplex tableau. We add a row to reflect the objective function values for each variable. These contribution rates, called $C_j,$ appear just above each respective variable, as shown in the following table:

The unit profit rates are not just found in the top $C_j$ row: in the leftmost column, $C_j$ indicates the unit profit for each variable currently in the solution mix. If $S_1$ were removed from the solution and replaced—for example, by C—$50 would appear in the $C_j$ column just to the left of the term C.

We can complete the initial Flair Furniture simplex tableau by adding two final rows. These last two rows provide us with important economic information, including the total profit and the answer as to whether the current solution is optimal.

We compute the $Z_j$ row values of the initial solution in Table M7.1 by multiplying the 0 contribution value of each number in the $C_j$ column by each number in that row and the jth column and then summing. The $Z_j$ value for the quantity column provides the total contribution (gross profit, in this case) of the given solution:

The $Z_j$ values for the other columns (under the variables $T, C, S_1,$ and $S_2$) represent the gross profit given up by adding one unit of this variable into the current solution. Their calculations are as follows:

Thus,

We see that there is no profit lost by adding one unit of T (tables), C (chairs), $S_1,$ or $S_2.$

The $C_j-Z_j$ number in each column represents the net profit—that is, the profit gained minus the profit given up—that will result from introducing 1 unit of each product or variable into the solution. It is not calculated for the quantity column. To compute these numbers, simply subtract the $Z_j$ total for each column from the $C_j$ value at the very top of that variable’s column. The calculations for the net profit per unit (the $C_j-Z_j$ row) in this example follow:

It is obvious to us when we compute a profit of $0 that the initial solution is not optimal. By examining the numbers in the $C_j-Z_j$ row of Table M7.1, we see that the total profit can be increased by $70 for each unit of T (tables) and by $50 for each unit of C (chairs) added to the solution mix. A negative number in the $C_j-Z_j$ row would tell us that profits would decrease if the corresponding variable were added to the solution mix. An optimal solution is reached in the simplex method when the $C_j-Z_j$ row contains no positive numbers. Such is not the case in our initial tableau.