Question 5 of 7: Cycle Time and Production Rate of a Robot Cell

Question 5: Cycle Time and Production Rate of a Robot Cell (20 marks)

Question text not reproduced: the examination questions are © Engineers and Geoscientists BC. Open the official past paper (linked at the top of this page) to read the question, then follow the worked solution below.

Given.

Find. The cell cycle time and hourly production rate with one robot serving the whole route, and the same two quantities when the route is split between two robots at machine C.

Approach. With a single robot and one part in the cell at a time, nothing overlaps: the cycle is the sum of the three machining times, the four loaded transfers with their gripper actions, and the empty return to the input point. With two robots the cell becomes a two-stage line, and its cycle time is set by the slower of the two stages, not by their sum.

1. Confirm the structure of the move table before using it. The table is symmetric and every entry equals 0.3 min times the number of station positions crossed — A to B is 0.3, A to C is 0.6, A to E is 1.2. The five points therefore lie on a straight line at equal spacing, which is what the cell sketch shows and what makes the empty return the longest single move in the cycle.
2. Cost one complete transfer. Every move of a part costs a gripper pull-up at the pick, the travel itself, and a gripper release at the place, $$t_{\text{transfer}} = t_{\text{pull}} + t_{\text{move}} + t_{\text{release}} = 0.1 + t_{\text{move}} + 0.1$$ There are four such transfers per part — A to B, B to C, C to D and D to E — each between adjacent stations and so each with $t_{\text{move}} = 0.3$ min.
3. Add up the four elements of the single-robot cycle. The machining content is $$\sum t_{\text{machine}} = 9.1 + 9.0 + 5.0 = 23.1\ \text{min}$$ the loaded travel is $4(0.3) = 1.2$ min, the gripper actions are $4(0.1+0.1) = 0.8$ min, and the robot must then return empty from E to A, which costs 1.2 min. Hence $$T_c = 23.1 + 1.2 + 0.8 + 1.2$$ $$\boxed{\ T_c = 26.3\ \text{min per part}\ }$$
4. Convert the cycle time to a production rate. With one part completed per cycle, $$R_p = \frac{60}{T_c} = \frac{60}{26.3}$$ $$\boxed{\ R_p = 2.28\ \text{parts per hour}\ }$$ The robot is busy for only $1.2+0.8+1.2 = 3.2$ min of each 26.3 min cycle, a utilisation of 12.2 per cent; the machines are idle for the same reason, each working only its own share of the cycle. That is the diagnosis the second half of the question builds on: the cell is starved because one robot forces the whole route to run in series.
5. Model the two-robot cell as a two-stage line. Robot 1 owns everything from A to C: it picks a part at A, loads B, waits out the 9.1 min on B, unloads B, loads C, and returns empty from C to A ready for the next part. Robot 2 owns everything from C to E: it waits out the 9.0 min on C, moves the part to D, waits out the 5.0 min on D, moves it to E, and returns empty from E to C. The two robots work on different parts at the same time, so the cell delivers one part per cycle where the cycle is the longer of the two loops.
6. Time robot 1's loop. Element by element, $$T_1 = \underbrace{(0.1+0.3+0.1)}_{A \to B} + \underbrace{9.1}_{\text{machine B}} + \underbrace{(0.1+0.3+0.1)}_{B \to C} + \underbrace{0.6}_{C \to A \text{ empty}} = 10.7\ \text{min}$$
7. Time robot 2's loop. Its stage carries two machines, so it is the heavier of the two, $$T_2 = \underbrace{9.0}_{\text{machine C}} + \underbrace{(0.1+0.3+0.1)}_{C \to D} + \underbrace{5.0}_{\text{machine D}} + \underbrace{(0.1+0.3+0.1)}_{D \to E} + \underbrace{0.6}_{E \to C \text{ empty}} = 15.6\ \text{min}$$
8. Take the bottleneck and report the new performance. The cell cycle is set by the slower stage, $$T_c' = \max(T_1, T_2) = \max(10.7,\ 15.6)$$ $$\boxed{\ T_c' = 15.6\ \text{min per part}, \qquad R_p' = \frac{60}{15.6} = 3.85\ \text{parts per hour}\ }$$ The cycle falls by 40.7 per cent and output rises by 68.6 per cent, from 2.28 to 3.85 parts per hour, purely from letting two parts be in process at once. Robot 1 is now idle 31.4 per cent of every cycle, waiting for robot 2's stage to clear.
9. Show that the split the question proposes is the best of the three. The route can only be divided at B, C or D, and the other two divisions are worse because they leave one stage carrying two long machines. Cutting at B gives loops of 0.8 and 25.5 min, a cycle of 25.5 min; cutting at D gives 20.5 and 5.8 min, a cycle of 20.5 min; cutting at C, as proposed, gives 10.7 and 15.6 min and the 15.6 min cycle above. The proposed split is also the most nearly balanced, which is no coincidence — balance is what minimises the maximum. Even so, a floor remains, and it is set by machine B. A single-gripper robot cannot reload B until it has carried B's finished part on to C and fetched the next part from A, a handling round of $(0.1+0.3+0.1) + 0.6 + (0.1+0.3+0.1) = 1.6$ min, so no arrangement of single-gripper robots can cycle faster than $$T_{\min} = 9.1 + 1.6 = 10.7\ \text{min}, \qquad R_p = 5.61\ \text{parts per hour}$$ Robot 1's own loop already equals that floor, so the only ways past 5.61 parts per hour are a second machine B, a double gripper at B, or a shorter operation on B.