Greater-than-or-equal-to $(\ge )$ constraints, such as constraint 1 as just described, require a different approach than do the less-than-or-equal-to $(\le )$ constraints we saw in the Flair Furniture problem. They involve the subtraction of a surplus variable rather than the addition of a slack variable. The surplus variable tells us how much the solution exceeds the constraint amount. Because of its analogy to a slack variable, surplus is sometimes simply called negative slack. To convert the first constraint, we begin by subtracting a surplus variable, $S_1,$ to create an equality:

If, for example, a solution to an LP problem involving this constraint is $X_1=20,$ X₂ = 8, and $X_3=5,$ the amount of surplus could be computed as follows:

There is one more step, however, in preparing a $\ge$ constraint for the simplex method.

There is one small problem in trying to use the first constraint (as it has just been rewritten) in setting up an initial simplex solution. Since all “real” variables such as $X_1,$ $X_2,$ and $X_3$ are set to 0 in the initial tableau, $S_1$ takes on a negative value:

All variables in LP problems, be they real, slack, or surplus, must be nonnegative at all times. If $S_1=\mathrm{-}210,$ this important condition is violated.

To resolve the situation, we introduce one last kind of variable, called an artificial variable. We simply add the artificial variable, $A_1,$ to the constraint, as follows:

Now, not only the $X_1, X_2,$ and $X_3$ variables but the $S_1$ surplus variable as well may be set to 0 in the initial simplex solution. This leaves us with $A_1=210.$

Let’s turn our attention to constraint 2 for a moment. This constraint is already an equality, so why worry about it? To be included in the initial simplex solution, it turns out that even an equality must have an artificial variable added to it:

The reason for inserting an artificial variable into an equality constraint deals with the usual problem of finding an initial LP solution. In a simple constraint such as number 2, it’s easy to guess that $X_1=0, X_2=30$ would yield an initial feasible solution. But what if our problem had 10 equality constraints, each containing seven variables? It would be extremely difficult to sit down and “eyeball” a set of initial solutions. By adding artificial variables, such as $A_2,$ we can provide an automatic initial solution. In this case, when $X_1$ and $X_2$ are set equal to $0, A_2=900.$

Artificial variables have no meaning in a physical sense and are nothing more than computational tools for generating initial LP solutions. If an artificial variable has a positive (nonzero) value, then the original constraint where this artificial variable was added has not been satisfied. A feasible solution has been found when all artificial variables are equal to zero, indicating all constraints have been met. Before the final simplex solution has been reached, all artificial variables must be gone from the solution mix. This matter is handled through the problem’s objective function.

Whenever an artificial or surplus variable is added to one of the constraints, it must also be included in the other equations and in the problem’s objective function, just as was done for slack variables. Since artificial variables must be forced out of the solution, we can assign a very high $C_j$ cost to each. In minimization problems, variables with low costs are the most desirable ones and the first to enter the solution. Variables with high costs leave the solution quickly, or never enter it at all. Rather than set an actual dollar figure of $10,000 or $1 million for each artificial variable, however, we simply use $M to represent a very large number.3 Surplus variables, like slack variables, carry a zero cost. In maximization problems, we use negative M.

If a problem had an objective function that read

and constraints such as the two mentioned previously, the completed objective function and constraints would appear as follows: