Question 6.2: Using the information contained in the table shown, do each ......
Using the information contained in the table shown, do each of the following:
1. Draw a precedence diagram.
2. Assuming an eight-hour workday, compute the cycle time needed to obtain an output of 400 units per day.
3. Determine the minimum number of workstations required.
4. Assign tasks to workstations using this rule: Assign tasks according to greatest number of following tasks. In case of a tie, use the tiebreaker of assigning the task with the longest processing time first.
5. Compute the resulting percent idle time and efficiency of the system.
| Task | Immediate Predecessor |
Task Time (in minutes) |
| a | — | 0.2 |
| b | a | 0.2 |
| c | — | 0.8 |
| d | c | 0.6 |
| e | b | 0.3 |
| f | d, e | 1.0 |
| g | f | 0.4 |
| h | g | \underline{\ 0.3} |
| Σt = 3.8 | ||
1. Drawing a precedence diagram is a relatively straightforward task. Begin with activities with no predecessors. We see from the list that tasks a and c do not have predecessors. We build from here(Figure 1).
2. {\mathrm{Cycle~time}}\ =\ {\frac{{\mathrm{Operating~time}}}{{\mathrm{Desired~output\ rate}}}}\ =~{\frac{480~{\mathrm{minutes~per~day}}}{400\ {\mathrm{units~per~day}}}}=1.2\ \text{minutes per cycle}
3. N_{\mathrm{min}}\;=\;{\frac{\Sigma t}{\mathrm{Cycle}\,\mathrm{time}}}\,=\,{\frac{3.8\,\mathrm{minutes~per\ unit}}{1.2\,\mathrm{minutes\ per\ cycle\ per\ station}}}\;=\;3.17\ \mathrm{stations\ (round\ to~4)}
4. Beginning with station 1, make assignments following this procedure: Determine from the precedence diagram which tasks are eligible for assignment. Then determine which of the eligible tasks will fit the time remaining for the station. Use the tiebreaker if necessary. Once a task has been assigned, remove it from consideration. When a station cannot take any more assignments, go on to the next station. Continue until all tasks have been assigned.
These assignments are shown in the following diagram. Note: One should not expect that heuristic approaches will always produce optimal solutions; they merely provide a practical way to deal with complex problems that may not lend themselves to optimizing techniques. Moreover, different heuristics often yield different answers(Figure 2).
5. \mathrm{Percent~idle~time}\;=\;\frac{1.0\;\mathrm{min.}}{4\times1.2\;\mathrm{min.}}\times100\;=\;20.83\%
Efficiency = 100% – 20.83% = 79.17%
| Station | Time Remaining |
Eligible | Will Fit | Assign (task time) |
Revised Time Remaining |
Idle |
| 1 | 1.2 | a, c * | a, c * | a (0.2) | 1 | |
| 1 | c, b ** | c, b ** | c (0.8) | 0.2 | ||
| 0.2 | b, d | b | b (0.2) | 0.0 | ||
| 0 | e, d | None | — | 0.0 | ||
| 2 | 1.2 | e, d | e, d | d (0.6) | 0.6 | |
| 0.6 | e | e | e (0.3) | 0.3 | ||
| 0.3^{***} | f | None | — | 0.3 | ||
| 3 | 1.2 | f | f | f (1.0) | 0.2 | |
| 0.2 | g | None | — | 0.2 | ||
| 4 | 1.2 | g | g | g (0.4) | 0.8 | |
| 0.8 | h | h | h (0.3) | 0.5 | ||
| 0.5 | — | — | — | \underline{~~0.5~~} | ||
| 1.0 min |
*Neither a nor c has any predecessors, so both are eligible. Task a was assigned since it has more followers.
**Once a is assigned, b and c are now eligible. Both will fit in the time remaining of 1.0 minute. The tie cannot be broken by the “most followers” rule, so the longer task is assigned.
***Although f is eligible, this task will not fit, so station 2 is left with 0.3 minute of idle time per 1.2-minute cycle.
