BUS606: Operations and Supply Chain Management

Our proposed two-stage procedure with sequential generation of feasible solution and selection of final solution can be best described by the following algorithm.

1. Calculate the theoretical minimum number of workstations, $N_{\min }$, following the formula $\sum_{i=1}^{K} \frac{\mu_{i}}{C} \leq N_{\min } \leq \sum_{i=1}^{K} \frac{\mu_{i}}{C}+1$. Calculate the minimum cycle time, $C_{\text {min }}$, using the relationship, $C_{\min }=\left[\sum_{i=1}^{K} \frac{\mu_{i}}{N_{\min }}+1\right]$.

2. Set the trial cycle time $C_{t}$ at $C_{\min }$.

3. Prepare the list of all unvisited tasks - call it List $U$.

4. Prepare List $R$ from the tasks of List $U$ with no immediate predecessor or whose immediate predecessors have been visited. The tasks of $R$ are ready for selection.

5. Prepare List $A$ from the task of List $R$ having assembly time less than that of cycle time and is allowable for inclusion.

6. Randomly select a task from List $A$ and reset the cycle time as {$C_{t}$ assembly time}.

7. If cycle time is less than the assembly time, then open a new workstation. Reinitialize cycle time to its original value and repeat the above steps until all nodes are visited.

8. After getting the complete distribution of tasks to workstations, calculate $R_{A L}$, the reliability of the assembly line.

9. After each run, the new reliability value $R_{A L}$ is compared with the previous $R_{A L}$ value. If the new $R_{A L}$ value is greater than the previous value, the new solution is stored as the basis for next comparison.

10. Increase the cycle time by one unit until it crosses $C$ value. If $C$ value is crossed, go to step 12.

11. Repeat step 2 to 10.

12. Check whether all the work elements have been assigned to specified number of work stations. If not, increase the value of $N_{min}$ by 1 and go to step 2.

13. Print the best solution in terms of overall maximum reliability.