In an $$M/M/1$$ queuing system, the number of arrivals in an interval of length $$T$$ is a Poisson random variable (i.e., the probability of there being $$n$$ arrivals in an interval of length $$T$$ is $${{{e^{ - \lambda T}}{{\left( {\lambda T} \right)}^n}} \over {n!}}$$). The probability density function $$f(t)$$ of the inter-arrival time is given by

Consider the Linear programme $$(LP)$$
Max $$4x$$ + $$6y$$
Subject to
$$\eqalign{ & \,\,\,\,\,\,\,\,\,\,\,3x + 2y \le 6 \cr & \,\,\,\,\,\,\,\,\,\,\,2x + 3y \le 6 \cr & \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,x,y \ge 0 \cr} $$

The dual for the $$LP$$ is

Consider the Linear programme $$(LP)$$
Max $$4x$$ + $$6y$$
Subject to
$$\eqalign{ & \,\,\,\,\,\,\,\,\,\,\,3x + 2y \le 6 \cr & \,\,\,\,\,\,\,\,\,\,\,2x + 3y \le 6 \cr & \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,x,y \ge 0 \cr} $$

After introducing slack variables $$s$$ and $$t$$, the initial basic feasible solution is represented by the table below (basic variables are $$s=6$$ $$t=6,$$ and the objective function value is $$0$$).

After some simplex iterations, the following table is obtained

From this, one can conclude that

For the standard transportation linear programme with $$m$$ sources and $$n$$ destinations and total supply equaling total demand, an optimal solution (lowest cost) with the smallest number of non-zero $${X_{ij}}$$ values (amounts from source $$i$$ to destination $$j$$) is desired. The best upper bound for this number is