Question 10.60: AGV utilization with FCFS dispatching Assuming the transfer ...

AGV utilization with FCFS dispatching

Assuming the transfer station in Figure 10.59 b is labeled station 9, consider the zone defined by stations \{1,6,7,8,9\}. Suppose two job types (labeled A and B) are handled through this zone. Assuming that stations 1 and 9 serve as I/O stations for the zone, suppose the average hourly production volume and the production route for the two job types are given as follows: job A: 6 jobs per hour (1-7-9); job B: 9 jobs per hour (9-6-8-1). The resulting from-to flow chart (including the \Lambda and \lambda values) is shown in Figure 10.60 a.

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Suppose the empty AGV travel speed is fixed at 5 seconds per grid. (A more elaborate travel time scheme, including AGV acceleration/deceleration, and straight-travel speed versus turning speed, can be used instead of our fixed speed of 5 seconds per grid.) The resulting empty travel time matrix (i.e., the \sigma_{i j} values expressed in minutes per trip) is shown in Figure 10.60 b. Assuming the load pickup or deposit time is equal to 9 seconds (0.15 minute), the loaded AGV travel times (i.e., the \tau_{i j} values) are obtained simply by adding the load pickup plus deposit time (0.30 minute) to the corresponding \sigma_{i j} values. (Separate \tau_{i j} values can be used if the loaded AGV travel speed differs significantly from the empty AGV travel speed.)

Using the above flow and travel time values in Equations 10.135 and 10.138, we obtain \alpha_{f}=0.4875 and \bar{s}=0.02057 hour (1.2342 minutes) per move request, respectively. Hence, \rho=0.8023, and \alpha_{e}=0.3148. Note that the busy vehicle would travel empty about 40 \%(0.3148 / 0.8023) of the time, which is a significant portion of time, considering that empty travel is wasted travel or unproductive travel. Unfortunately, this is often the case with the FCFS dispatching rule since the location of the (empty) vehicle, relative to outstanding move requests in the system, is not taken into account when the vehicle is assigned to serve the next move request.

\alpha_{f}=\sum_{i} \sum_{j} f_{i j} \tau_{i j}     (10.135)

\bar{s}=\sum_{k} \sum_{i} \sum_{j \neq i}\left(\frac{\Lambda_{k}}{\Lambda_{T}}\right)\left(\frac{f_{i j}}{\lambda_{T}}\right)\left(\sigma_{k i}+\tau_{i j}\right)     (10.138)

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