Lost-customers approximation of semi-open queueing networks with backordering -- An application to minimise the number of robots in robotic mobile fulfilment systems

Abstract

We consider a semi-open queueing network (SOQN), where a customer requires exactly one resource from the resource pool for service. If there is a resource available, the customer is immediately served and the resource enters an inner network. If there is no resource available, the new customer has to wait in an external queue until one becomes available ("backordering"). When a resource exits the inner network, it is returned to the resource pool and waits for another customer. In this paper, we present a new solution approach. To approximate the inner network with the resource pool of the SOQN, we consider a modification, where newly arriving customers will decide not to join the external queue and are lost if the resource pool is empty "lost customers". We prove that we can adjust the arrival rate of the modified system so that the throughputs in each node are pairwise identical to those in the original network. We also prove that the probabilities that the nodes with constant service rates are idling are pairwise identical too. Moreover, we provide a closed-form expression for these throughputs and probabilities of idle nodes. To approximate the external queue of the SOQN with backordering, we construct a reduced SOQN with backordering, where the inner network consists only of one node, by using Norton's theorem and results from the lost-customers modification. In a final step, we use the closed-form solution of this reduced SOQN, to estimate the performance of the original SOQN. We apply our results to robotic mobile fulfilment systems (RMFSs). Instead of sending pickers to the storage area to search for the ordered items and pick them, robots carry shelves with ordered items from the storage area to picking stations. We model the RMFS as an SOQN, analyse its stability and determine the minimal number of robots for such systems using the results from the first part.