a) M = {1, 2, 3, 4} with k = 4 b) M = {4, 6, -3, -10, 3} with k = 5 c) M = {18, -20, 46, 43} with k = 4 d) M = {8, 3, -60, 4, 10} with k = 5
e) M = {14, 19, 243, 34} with k = 4 f) M = {18,-200, 34, 9} with k = 4
2.5 Assembly lines working to full capacity and the assembly line efficiency
One week later the team of Clever Consulting arrives at the production halls of the Car Corp..
There is a hustle and bustle of activity. At the delivery entrance some trucks got stuck, people gesticulate wildly and seem to be very excited. Obviously some suppliers did queue in the wrong place and mixed up the elaborate system of just-in-time deliveries. Is this also an optimization problem for Clever Consulting? Even before the four colleagues find the time to think about it, Herr Wiedner arrives.
Mr. Wiedner: Good afternoon! You must be the team of Clever Consulting. It’s nice to have you here. And who’s the lady I’ve talked to on the phone?
Selina: Good afternoon, Mr. Wiedner. My name is Selina Malik. Thank you very much for the invitation.
Mr. Wiedner: Do you wish to go for a short tour to see the production facilities before we go to my office?
Sebastian: That’s fine! It’s always better to actually see what we’ve been talking about.
Selina: And I can find out which sports car model is the right one for me.
Nadine: How many cars do you produce per day on an average?
Mr. Wiedner: It depends on the order situation, but the assembly line should make around 120 units, that’s the daily output.
On the tour Mr. Wiedner explains how the production process works. Obviously Mr. Wiedner is very proud that the production runs like a clockwork. He explains that a breakdown at any station of the line can of course paralyze the complete assembly line. Such failures are an expensive problem for the Car Corp., since then the machines and workers cannot work productively. Thus at any time there are engineers on site, who can take appropriate actions in case of a failure. As the group arrives at the end of the production street, a brand new blue sports car in the luxury version leaves the assembly line.
Selina: Oh! That’s exactly the one that I want! Mr. Wiedner, I think I know how you could pay for our service.
Mr. Wiedner: Well, let’s talk about that at the end of your work. But for this sports car model you’ll probably have to optimize further processes.
Oliver: It was a good idea to see the production on site. Thank you very much for the tour, Mr.
Wiedner.
Mr. Wiedner: You’re welcome. You really seem to tackle this problem seriously. I’m curious to see your results. If you can help us, we could talk to the business management about a consulting contract between Clever Consulting and the Car Corp.. I think we’ll find some more tasks for your company. In the meantime I’ve talked with our process engineers about the aims. They use the so-called assembly line efficiency as a measure for the utilization. The assembly line efficiency is the ratio between the sum of the processing times and the product of cycle time and number of stations.
In the meantime, the group arrived at Mr. Wiedner’s office. While he talks, Mr. Wiedner writes the following formula on a flipchart:
1 2 1
Assembly line efficiency: p p pn n
ALE C k
− p
+ + + +
= ⋅ .
Herr Wiedner: The assembly line efficiency shows the utilization of the line, it should be as high as possible. The best is if you have a close look at it without ruffle. I don’t want to throw you out, but I’ve an important appointment now.
Nadine: Ok, we don’t want to hold you up any longer, thank you very much for your support.
Herr Wiedner: Oh, and before I forget about it: Our engineer gave me a list of the processing times. It’s for you, you’ll need the figures to get started. The units of the values are minutes.
p1 = 8 p2 = 5 p3 = 1 p4 = 3 p5 = 7 p6 = 4 p7 = 2 p8 = 9 p9 = 2 p10 = 5 p11 = 7
Impressed by the huge production site and ready to start their work, the Clever Consulting team sets off to its office.
Oliver: Did you understand the formula? I didn’t.
Nadine: The best is if we plug in the values that we do already have. Those are the production times in the numerator:
1 2 n1 n 8 5 1 3 7 4 2 9 2 5 7 53. p + p + +p − +p = + + + + + + + + + + =
The sum of the processing times is 53 minutes. Thus for the assembly line efficiency it holds:
ALE 53
=C k
⋅ .
But we neither know the cycle time C nor the number of stations k. Let’s choose an arbitrary division of jobs into stations to simplify matters. Then we can determine the resulting assembly line efficiency.
Nadine is on the right track. An example, even a designed one, often helps to achieve a better understanding of mathematical terms.
Example:
Given is the division of the jobs into three stations: [1, 2, 3, 4] – [5, 6, 7] – [8, 9, 10, 11]; k = 3.
From this we can compute:
- Processing time of the stations
1 1 2 3 4 17; 2 5 6 7 13; 3 8 9 10 11 23.
s =p + p +p +p = s = p + p + p = s = p + p + p +p =
- Cycle time
{
1 2 3} { }
max , , max 17,13,23 23
C= s s s = = .
- Assembly line efficiency
53 53 53
0,768 23 3 69
ALE=C k = = ≈
⋅ ⋅ .
The assembly line efficiency is 0,768 (→E.2.17). Having a look at the processing times of the stations, you can see that this is not a very good value. Station 3 is working at very high capacity.
There the working takes 23 minutes. This station determines the length of the cycle time. In contrast, station 2 only has to work for 13 minutes and then waits ten minutes for the end of the cycle time.
The efficiency of the stations is listed in the diagram below.
0 5 10 15 20 25
Station 1 Station 2 Station 3
Bearbeitungsdauer [Minuten]
Figure 2.9 Diagram of the utilization of the stations
The diagram shows the time sequence of a production process. Here you can see explicitly that station 3 is very heavily used. We obtain a better solution, if we move job 8 from station 3 to station 2, for example.
- Division into stations: [1, 2, 3, 4] – [5, 6, 7, 8] – [ 9, 10, 11]
- Processing time of the stations
1 1 2 3 4 17; 2 5 6 7 8 22; 3 9 10 11 14.
s = p + p +p + p = s =p +p + p + =b s = p + p + p =
- Cycle time
{
1 2 3} { }
max , , max 17,22,14 22
C= s s s = = .
- Assembly line efficiency
53 53 53
0,803 22 3 66
ALE=T k = = ≈
⋅ ⋅ .
0 5 10 15 20 25
Station 1 Station 2 Station 3
Bearbeitungsdauer [Minuten]
Figure 2.10 Diagram after shifting job 8 (→E.2.18)
The cycle time was shortened by one minute. The assembly line efficiency did increase to 0,803.
The efficiency of the machines did improve. Of course this is a success. Nevertheless we do not know if this is the best solution to the problem. The possibilities of finding new solutions is not only limited to the shifting of jobs. The number of stations can be modified as well. The number of the stations was chosen arbitrarily in the example.
Mr. Wiedner told us that the assembly line efficiency should be as high as possible. But how high can it actually get? The cycle time, that is the processing time at one station, multiplied by the number of stations must be at least as long as the sum of the processing times, i.e. it has to hold:
1 2 n1 n
C k⋅ ≥p + p + +p − +p .
Why does this hold? Well, the processing times of the jobs pi are allocated among the stations and are contained in the processing times of the stations sj. Thus it holds:
1 2 n 1 n 1 2 k 1 k
p +p +…+p − +p = + +s s …+s − +s .
Note: The left-hand side of the equation contains n summands, the right-hand side only k summands.
Since C corresponds to the maximum processing time of one station, it holds:
1 2 1
max1, , j k k
j k
C k s k s s s − s
⋅ = = ⋅ ≥ + + + +
… … .
Thus we can conclude:
1 2 n 1 n.
C k⋅ ≥p +p +…+p − +p We will now consider two cases:
Case 1: In the first case both sides of the above inequality have exactly the same value, thus the equation holds:
1 2 n1 n
p + p + + p− + p = ⋅ kC .
In this case, the numerator and denominator for the computation of the assembly line efficiency have the same value, i.e. it follows: ALE = 1.
Case 2: In the second case the inequality holds:
1 2 n1 n
p + p + + p− + p < ⋅ kC .
So the denominator of the assembly line efficiency is always greater than the numerator. This means that the ALE must be smaller than one: ALE < 1.
Taking both cases into account leads to:
1 ALE≤ .
The assembly line efficiency can attain at most a value of one.
If we – as in the first case – obtain an ALE of one, then all jobs are evenly distributed among the stations, then there would be no idle time for the machines. This would be the ideal case, the best solution that can be obtained. We say, the assembly line works at 100% capacity. The assembly line efficiency is a key figure of utilization. As Mr. Wiedner from the Car Corp. said, we have to try to get an assembly line efficiency that is as high as possible. Now we know that it can be at most one.
We can obtain an efficiency of 100% if we build only one single station that deals with all jobs.
The cycle time then is the sum of the processing times: C = p1 + p2 + ... + pn-1 + pn. Let us study this case with example of the Car Corp.:
1 2 1
But this option is not interesting. In principle it means that we do not use an assembly line anymore. All jobs are executed at one single station. In addition, it would not be possible to process several cars at the same time; the output would be very small. Thus the case k = 1 is not relevant in practice. In the following we will forbid k = 1.
The Clever Consulting Team had the same ideas:
Selina: I don’t know; that’s very difficult. There’re so many parameters that we don’t know, the cycle time, the number and the allocation of the stations. So where shall we start?
Oliver: That’s true. But basically all these parameters are related. For example: If I build more stations, then normally the cycle time decreases. Maybe we can use this somehow?
Nadine: Furthermore the number of stations can surely be limited. For example, we do already know that it’s nonsense to build only one station. I could imagine that two are also insufficient if we want to produce a certain minimum amount per day.
Oliver: That’s a good idea! Maybe we can take the minimum output into consideration, we haven’t thought about that before.
Selina: O.K. Mr. Wiedner said 120 units of output per day. Let’s denote the minimum output with Omin and say: Omin = 120. So how many hours per day does the assembly line run?
Sebastian: As far as I know, it can run three shifts, that’s 24 hours. If the order situation is weak, sometimes only two shifts are required, but then the Car Corp. doesn’t have to produce 120 cars.
Nadine: Good. After the lapse of one cycle time C always one car is completed. This means, if the Car Corp. wants to produce at least 120 cars in one day, the cycle time can at most be 24/120 hours. Or 1440/120 minutes. Reducing the fraction gives twelve minutes.
Thus the Clever Consulting Team did find an estimate for the cycle time. The cycle time cannot be longer than this estimate; otherwise 24 hours would not be enough to produce 120 cars.
Therefore this value is called the maximum cycle time Cmax. A cycle time that is longer than Cmax is not allowed.
Before we write down the formula, we denote the production period with P. In our case it is 24 hours or 1440 minutes respectively. But it would also be possible to consider a whole month or just one hour.
The half brackets around the fractions cause that their content is rounded to an integer. The brackets opening upwards in the above equation mean rounding down. Brackets opening downwards mean rounding up. See the following examples:
3,2 4 3,2 3
Fractional numbers are rounded up or down. Integers remain unchanged (→E.2.19-21).
Why is the maximum cycle time rounded down? The cycle time is the sum of the processing times pi, which are integers in our case. The sum of integers is again an integer. Hence a reasonable cycle time is always an integer. Now we have to explain why we round down. Here we study the maximum, i.e. the longest possible cycle time. A longer cycle time is not allowed. If P / Omin yields a fraction, we can only choose smaller integers. Thus Cmax results from rounding down.
The maximum cycle time of course has influence on the number of stations. No station can have a processing time that is longer than the maximum cycle time. If the maximum cycle time was already exceeded by a single job, if e.g. p1 = 15 did hold, then the minimum output could not be achieved. The Car Corp. would have taken on too much with this. The only way to overcome this problem would be to decompose the relevant job into several smaller jobs, until the maximum cycle time is not exceeded anymore. But this possibility would have been examined by the engineering department of the Car Corp..
Sebastian: Very good. Now we have an upper bound for the cycle time. With this it should be possible to limit the number of stations as well. Since the smaller I choose the cycle time, the
more station I have to set up. If I bound the cycle time from above, it means that I need a minimum number of stations. We already know that the cycle time multiplied by the number of stations must be greater than the sum of the processing times:
1 2 n1 n.
C k⋅ ≥p + p + +p − +p
The pi are given. C and k are variables. If we want to know what influence the variable C has on k, we have to rewrite the inequality:
1 2 n1 n
p p p
k C
− p
+ + + +
≥ .
Nadine: Now you can see: If C increases, the right-hand side gets smaller and thus k can be smaller. This behavior can be seen very well if you draw the inequality with the data form our example (see figure 2.11):
k 53
≥ C .
For the moment we will to forget about the constraint that C is an integer. All points (C | k) that lie in the grey hatched area or on the curve are a valid combination of the parameters k and C.
10 20 30 40 50
Taktzeit T
Cycle time C
25 50 75 100 125 150 175 200 Stationsanzahl k
Number of Stations k
Figure 2.11 Area of feasible solutions
Since C is bounded from above by Cmax, the area of the solutions is restricted. The restriction of the area ensures that k cannot be arbitrarily small. Thus the smallest value that k can reach depends on Cmax.
igure 2.12 Restriction by kmin and Cmax
ebastian: Great! With this we can leave out the analysis of all k that are below the bound. This is
formula:
2 4 6 8 10 12 14
10 20 30 40 50
kmin
Tmax
Cmax Stationsanzahl k
Number of Stations k
Taktzeit T
Cycle time C
F
S
much less work than „checking all possibilities“. We should write down a formula for the minimum number of stations kmin.
Sebastian comes up with the following
1 2 1
min
max
n n
k p
C
− +
= p + p + + p .
Selina: I don’t think that this is entirely true. Do plug in our values into the equation.
hen always one half of the worker at the what do we do with the formula? Instead of at least 4,42 stations function with the square brackets that rounds a number up or
orrects the formula by using the brackets that we are already familiar with and finally the Sebastian: Well, kmin is 53 over 12, that is around 4,42.
Selina: And how can the Car Corp. build 4,42 stations? T last station has a lunch break?
Sebastian: Ok, you’re right. So we then need at least five stations.
Oliver: In programming there’s this down.
Oliver c
four colleagues get the right result:
1 2 1
Nadine: While we’re on that subject, we could also bound k from above by a maximum number of stations. After all, k cannot be bigger than the number of jobs n, otherwise we would have to split up the jobs, which isn’t allowed here.
kmax =n
Oliver: Exactly, and for the same reason the cycle time cannot be smaller than the longest processing time of a job.
Sebastian: That’s true. But there exists even another constraint for the minimum cycle time:
Since the number of stations is bounded from above, we cannot choose the cycle time arbitrarily small, otherwise we cannot meet the processing times.
1 2 1
Oliver: Yes, you’re right. So if both constraints are correct, what do we do then?
Selina: Just take the bound that is stronger. The cycle time cannot be smaller than any of the two bounds. This means, the bound with the higher value is the better one.
1 2 1
Selina: For our example it means:
{ }
Sebastian: I suggest that we’re optimistic and go on as if we could get an assembly line efficiency of one.
1 2 1 1 2 1 53
Nadine: Aha, and what’s the advantage of doing so?
Sebastian: Well, for any arbitrary C within our bounds I can calculate a k with the help of this equation that yields an assembly line efficiency of one.
Nadine: What? It’s as simple as that?
Sebastian: Well, it isn’t that simple. In addition, we have to take care that k is an integer, otherwise we cannot use the result (→E.2.24).
Oliver: The best is if we draw the curve, then you’ll see more.
Oliver’s drawing is shown in figure 2.13.
1 2 3 4 5 6 7 8 9 10 11 12 13 represents a combination of the two variables, which yields the highest possible assembly line efficiency of one. The blue point marks a situation where the cycle time was chosen very short and therefore a lot of stations are required. If you, in contrast, set up only a few stations, then it is necessary that the cycle time is chosen long enough. The green point shows an example of this alternative. Both points lie outside of the computed bounds. This means that they could not be realized under the given conditions (→E.2.25-26).
Exercises
E.2.17 The following production times come from a supplier of the Car Corp..
p1 = 3 p2 = 7 p3 = 5 p4 = 13 p5 = 1 p6 = 9 p7 = 3 p8 = 2
Calculate the assembly line efficiency for the following partitions into stations:
a) [1, 2, 3] – [4, 5, 6] – [7, 8] b) [1, 2] – [3, 4] – [5, 6] – [7, 8] c) [1, 2, 3, 4] – [5, 6, 7, 8]
d) [1, 2] – [3, 4, 5] – [6, 7, 8] e) [1, 2, 3] – [4, 5] – [6, 7, 8] f) [1, 2, 3] – [4] – [5, 6] – [7, 8]
Which partition has the best assembly line efficiency?
E.2.18 Draw the diagrams for E.2.16. Describe the different profiles of utilization: Do the stations work at the same capacity? Which stations work at a capacity that is too high, which work at a capacity that is too low? Can the utilization be improved?