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MaMaEuSch
Management Mathematics for European Schools
http://www.mathematik.uni-kl.de/~mamaeusch/
The new assembly line of the Car Corp.
Silvia Schwarze1
Horst W. Hamacher2
MaMaEuSch has been carried out with the partial support of the European Community in the framework of the Sokrates programme. The content does not necessarily reflect the position of the European Community, nor does it involve any responsibility on the part of the European Community.
1 University of Kaiserslautern, Department of Mathematics 2 University of Kaiserslautern, Department of Mathematics
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CHAPTER 2: The new assembly line of the Car Corp.
Keywords from economy - job - work centre - capacity - assembly line efficiency - processing time - assembly line production - output - daily output - cycle time
Keywords from school mathematics - binomial coefficient - exponential growth - factorial - function, function family - equilateral hyperbola - index shift - combinatory - constant - zero - parameter - Pascal’s triangle - permutation - power set - jump discontinuity - random sample, ordered/disordered - subset - inversely proportional - odd function - feasible/optimal solution
Compendium of Chapter 2
The present chapter describes the planning processes in the assembly line production on the basis of an example from the car industry. Production planning always contains combinatorial tasks. Thus the first section describes the development of a combinatorial problem from the economic application. In the following second section the given problem will be analyzed mathematically, new concepts will be derived. The analysis will always be close to the economic application. The third section goes into detail on the relationship between binomial coefficients and Pascal’s triangle. Here in particular technical aspects are of great importance. For the
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interested reader two mathematical proofs were included in this paragraph. The fourth section explains briefly the exponential growth of functions. Here again the connection between theory and practice is shown. Exponential growth of real-world problems is often responsible for difficulties in handling these problems and may cause the failure of solution algorithms. The fifth section then turns again to the production planning. Various technical terms from the field of manufacturing will be introduced. Operating figures and diagrams serve as tools for searching for a solution method. In the final sixth section the problem of integer solutions will be discussed. This makes the given problem more difficult. By using algorithms one can deal with it in the given example and find solutions. The process of mathematical modeling is complex and in practice set backs do occur every now and then. The present chapter will describe this procedure, as far as it is possible in these few words. Starting from the real-world task, the problem will be formalized with well-known or newly developed mathematical methods. The solution methods found have to be tested again and again by going back to the application. In doing so weaknesses of the modeling may become apparent, possibly new approaches will be chosen. In the end of the process of modeling there is a solution that will be presented to the customer. In reality one often settles for a compromise solution, also due to the lack of time. This chapter closes with a satisfactory solution, last but not least to give the pupils a feeling of success, but it also leaves open some further questions.
2.1 A telephone call at the Car Corp.
Although at first glance it doesn’t look like it, but all four colleagues of the Clever Consulting Team are doing their job. Oliver drinks a cup of coffee two and talks about the latest Star Trek film; while Nadine writes long lists where entries like hazels, brown sugar and aubergines can be found. Sebastian browses online at a shop for books, CDs and software and obviously only has eyes for the site with the Import-CDs; meanwhile Selina discusses the latest sports car models with a car dealer on the phone.
Selina: Well, now I have learnt a lot about the production at the Car Corp. and about the planned production line.
Nadine: But it sounded more as if you wanted to buy a new car.
Selina: That’s the case! A blue sports car! I started to chat a bit with the car dealer. Coincidentally the production manager of the Car Corp. was just visiting him. So I have come to know that soon a new sports car model will be produced, for which the Car Corp. plans to start up a new assembly line. Of course I have mentioned that we are experts in the area of production planning.
Nadine: We are what?
Selina: Well, don’t be worried! We’re not that ignorant; remember the cooperation with the company SchokoLeb. Furthermore we’ve always become acquainted with new topics very quickly. Anyway it was worth doing… to show off a bit, because in the end the production manager himself talk to me on the phone!
Sebastian: Did he tell you internal information? That’s certainly important information that you don’t tell everybody. You could have worked for a competitor!
Selina: Well, I had to persuade the production manager a bit. I told him who we are and which projects we’ve already done. It did reassure him to hear that we’ve successfully worked together with other companies in the past. I offered him that we’re going to prepare a concept for the planning of the assembly line.
Nadine: Of course without commitment for him, I suppose?
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Selina: There was no other way to convince him. But this is a big chance. If the Car Corp. is satisfied with our work, they’ll surely let us do further projects, and you cannot image all the things that have to be planned!
Oliver: Now tell us something about the new assembly line!
Selina tells her colleagues in detail about the telephone call with Mister Wiedner, the production manager of the Car Corp.. The company plans to start up a new modern assembly line. A new sports car model will be produced. The production of a car is divided into eleven stages, so-called jobs. The Car Corp. knows how long the single jobs take on the machines. This time span is called the processing time of the job. All jobs are processed along an assembly line, this means that along the assembly line there are stations where the jobs are executed.
Nadine: I see, so for the eleven jobs there are also eleven stations in the assembly line.
Selina: No, it isn’t that easy. Some of the jobs take long, others are very short. The assembly line has to stop at every station and can’t move on before all stations have finished their work. The engineers von Car Corp. call this the die cycle time of the assembly line.
Here we will leave the discussion at Clever Consulting and have a closer look at what was discussed. There are eleven jobs that have to be processed consecutively:
11 10 98 76 5 4 32 1
Figure 2.1 Eleven jobs are necessary for the production of the sports car. The length of the blocks shows the processing time of the jobs
If there is build one station in the assembly line for each job, then eleven sports cars could be (on/)in (auf) the assembly line at the same time, one at each station. The assembly line stops at the stations and moves on when all stations have completed their work. This would mean that the cycle time would be as long as the processing time of the longest job.
Selina has realized right away that this is not a very good idea. Some of the jobs are very short, e.g. job 3. In the associated station the work would be completed much faster than at job 8. Thus the station would not be working at high capacity, the worker in/at this station would have a long idle time between two production/working processes. The management of course does not like it at all. On the one hand the worker and the machine cost a lot of money, both should not be idle. On the other hand it would also be very unfair if the worker in/at station 3 has a rest all the time, while number 8 has to work non-stop. It is easy to understand that this is not a good solution. But what can be done to overcome this? Often the obvious solution is also the best: Stations that are working at low capacity are simply used for further jobs. The aim is that the work is shared equitably by the stations.
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But is it generally possible to conceptualize the notion of equitably in a mathematical way? To resolve this it is important to think about what is meant by an equitable allocation. In everyday language equitably means that everyone is entitled to get the same. It does not matter whether it is an equal sized slice of the cake or the same amount of workload. So in our case an equal distribution of the jobs to the stations would be desirable. Then no station would have to wait for another one to finish. The stations would be working at higher capacity and the cycle time would be short. Now we will join the discussion at Clever Consulting again. Oliver: O.K., I’ve got that, we have to arrange the jobs, preferably in such a way that the processing times of the single stations are of equal length (→E.2.1). Then of course the number of stations will be smaller than eleven. So how many stations do we need?
Selina: Well, that’s exactly what the Car Corp. wants to know from us!
Nadine: Stop! That’s once again too fast for me! Is it in general always possible that the jobs are arranged in such a way that all stations work at the same capacity?
Sebastian: Well, let me think. It’s not that easy to see with eleven jobs. We’d have to check quite some possibilities. Let’s consider an easier example instead: If we just had two production processes, let’s say one with two hours and the other one with three hours of processing time, then it isn’t possible that the two stations work at the same capacity!
Station 1 Station 2
Oliver: Unless we set up just one station that contains both jobs and requires five hours of processing time.
Nadine: All right, that’s the easy way to do it. If we set up only one station with all jobs, then there will be no idle time at the stations. But that’s not a good solution. In this case we would have just one car in the whole assembly line.
Oliver: This would then rather be a stagnant belt than a conveyor belt.....
Sebastian: So except for the simple solution where only one station is set up there doesn’t have to be a solution where all stations work at the same capacity. Thus we should look for a solution where the stations work at a capacity that is as high as possible.
All four agree on this point. Later they have to think about what „as high as possible“ exactly means and how it can be expressed mathematically. In the course of the discussion Selina tells her colleagues another important detail:
Selina: Since for example the car can’t be varnished until the car body is welt together, it is necessary to sort the single jobs. Thus the order of the jobs for the sport car is determined by technical requirements. So at least we don’t have to worry about that, because this sorting was already made by the Engineering Department of the Car Corp..
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Oliver: Don’t these requirements make it much more difficult for us? After all we have to be very careful in our planning that that we don’t change this order.
Sebastian: No, far from it! It will be easier for us, since if the production sequence wasn’t specified, we would have to analyze all possible orders find to the best one. Then there would be many more cases that we have to analyze. So we should be happy about it (→E.2.2).
Exercises E.2.1 Arrange the following processing times in two different ways by keeping the order, such that all stations work at the same capacity: 3, 4, 5, 2, 7, 5, 1, 1. Please note: When you have found a solution, you should have a close look at it.
E.2.2 Arrange the following processing times in at least two different ways, such that all stations work at the same capacity. The number of stations can be chosen arbitrarily. In contrast to exercise E.2.1 the order of the jobs may be changed. Processing times: 1, 1, 2, 2, 3, 3, 4, 4
2.2 How much is much?
Oliver: So how many possibilities of arrangements do exist? Let’s assume that it’s a manageable amount, then we could write a computer program that goes through all possibilities and returns the best solution as output.
Selina: Oh no, we now have to write down all possibilities and count? Count me out!
Nadine: No, of course not. It must somehow be possible to work it out logically. And if we do it this way, we only know the solution to our problem with eleven jobs. It would be nice if we could calculate a solution for any possible number of jobs, this means depending on a number n that represents the number of jobs.
Oliver: I think the best is to sketch the jobs. When we have the problem on paper, it’s easier to come up with an idea how to solve it.
Nadine: All right, here’s some scribbling paper! Here you go!
2 11 8 10 9 7 6 5 4 3 1
Nadine: When we now do a division into stations, the order has to remain unchanged, right?
Sebastian: That’s it.
Nadine: Good, i.e. we can build stations here on the paper by drawing lines between the jobs. That’s quick and we don’t have to write down the numbers over and over again. I can build for example three stations by drawing two lines:
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2 11 8 10 9 7 6 5 4 3 1
Nadine: At random I’ve built the stations [1, 2], [3, 4, 5, 6] and [7, 8, 9, 10, 11]. Whether this is a good solution or not doesn’t matter for now. We just want to see how many possibilities there are to build the stations.
Sebastian: Now we just have to find out how many possibilities there are to put two lines between the jobs. Then we know how many possibilities there are to divide the jobs into three stations. (→E.2.3).
Oliver: Exactly! An excellent idea! We do it for all possible numbers of stations, that’s from one to eleven, and then we’re done! How many possibilities do exist to place two lines?
Sebastian: I have a faint recollection of a formula that we’ve used at university. You can use it to calculate how many possibilities there are to take k elements from a set of n.
Nadine: Ah, I think I know what you mean. It was called „n choose k“ and had to do something with random samples. It was also important that the order of the chosen elements is irrelevant.
Selina: Thus in our case we would have to calculate „2 over 10“, the number of possibilities to select two lines out of ten possibilities.
Sebastian gets his old compendium of formulas to look up the exact definition. In the meantime we’ll concentrate on the definition of „n choose k“ :
We consider the following question: How many possibilities do exist to build a k-element subset from an n-element set if the order within the subset is irrelevant and if it is not allowed to use the elements more than once? Figure 2.2 illustrates this procedure with k = 6 and n = 18.
19 16 15
4 5 6 17183
11 1012 142
7 11 9 2 18 4 1213
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Figure 2.2 Six elements are taken from 18 elements without considering the order of the draw and without repetitions
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Disordered samples without replacement The number of possibilities to draw a disordered sample of k elements fromn elements without replacement is:
( ) ( )( ) ( ) ( ) ( )
1 2 2 1 !1 2 2 1 1 2 1 ! ( )!
nn n n nkk k k n k n k k n k
⋅ − ⋅ − ⋅ ⋅ ⋅ = = ⋅ − ⋅ − ⋅ ⋅ ⋅ ⋅ − ⋅ − − ⋅ ⋅ ⋅ ⋅ −
…… …
.
nk
reads as „n choose k“. This term is a common abbreviation for the
formula shown above. This value is also called Binomial coefficient.
What do the exclamation marks in the binomial coefficient mean? n!, factorial of n, is a common abbreviation for the product of the numbers from 1 to n. Thus it holds:
! 1 2 3 ( 2) ( 1)n n n= ⋅ ⋅ ⋅ ⋅ − ⋅ − ⋅ n .
It reads as „n factorial“. The notion of factorial in only defined for nonnegative natural numbers n. For n = 0 it holds:
0! 1= .
For one does not hurt one’s fingers from typing to calculate the factorial especially for large n, many calculators have this function as a feature. Why is the binomial coefficient defined in such a way? First we want to find out how many possibilities there are to draw k elements from n elements if the order of the draw is important. Thus we analyze the number of ordered samples. This means that e.g. the combinations 1, 2, 3 and 2, 3, 1 are distinguishable and have to be counted separately. For the first draw there exist n possibilities, since each of the n elements can be drawn. For the second draw n – 1 elements are still available. So there are n - 1 possibilities. With each further draw the number of available elements and thus the number of possibilities is reduced by one. For the kth draw there is still a choice of n – k + 1 possibilities. The number of possibilities that can occur with k moves? results from the multiplication of the number of possibilities for the single moves?. This means that the number of possibilities to select k from n elements with considering the order is
( ) ( ) ( )1 2n n n k n k⋅ − ⋅ ⋅ − + ⋅ − +1 .
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When counting the possibilities for an ordered sample, one obtains a much higher value than for a disordered sample, since the order is relevant and thus all possibilities to arrange the chosen elements are counted. A set of k elements can be arranged in k! different ways, they are called permutations. This can be easily seen if you imagine that from a set of k elements all elements, that is k, are taken and the order of the draw is relevant. According to the number formula developed above there exist
( )1 2 1k k k !⋅ − ⋅ ⋅ ⋅ =
possibilities. By drawing with considering the order we have counted each disordered sample k! times. Thus the binomial coefficient is
( ) ( )
( )( ) ( ) ( ) ( )
( ) ( ) ( )1 1 1 1 1 2 1 !
1 2 1 ! 1 2 1 ! !nn n n k n n n k n k n k nkk k k n k n k k n k
⋅ − ⋅ ⋅ − + ⋅ − ⋅ ⋅ − + ⋅ − ⋅ − − ⋅ ⋅ ⋅ = = ⋅ − ⋅ ⋅ ⋅ ⋅ − ⋅ − − ⋅ ⋅ ⋅ ⋅ −
… … …… …
=
(→E.2.4).
In the meantime Sebastian has found the notes about the binomial coefficient in his documents. The four colleagues now try to find out if the binomial coefficient is suitable for their problem.
Sebastian: Well, we have ten positions where we can draw the lines. I want to put two lines. This means, I choose two from then ten positions. That’s the same as if I’d build a disordered sample with 2 elements without replacement. Without replacement because naturally only one line can be drawn in each position, thus each position can appear just once in the sample.
Selina: Yes, but why disordered? We’ve just said that the order of the jobs is important and now suddenly it isn’t any more? I don’t get it!
Sebastian: Disordered just means that it doesn’t matter whether a subset is called [1, 2, 3] or [2, 1, 3]. In this context it’s unimportant whether I choose the second and sixth position for the lines or the sixth and the second. Do you understand what I mean?
Selina: I guess I have to think it over once again. But for now go ahead.
Nadine: I’ll now compute with my calculator how many possibilities there are for two lines.
While Nadine types the numbers in the calculator, we also have a look at the calculation.
10 10! 10! 10 9 2 1 3628800 452 2! (10 2)! 2! 8! 2 1 8 7 2 1 2 40320
⋅ ⋅ ⋅ ⋅ = = = = ⋅ − ⋅ ⋅ ⋅ ⋅ ⋅ ⋅ ⋅ ⋅
……
=
Nadine: There are 45 different possibilities to put two lines in? ten places.
Oliver: Or in other words: There are 45 possibilities to build three stations from eleven jobs (→E.2.5).
Selina: I’m impressed. It’s as simple as that. Ok, and we have to do this ten times and add up the results?
Sebastian: We even have to do it eleven times! There can be drawn zero to ten lines! That’s eleven calculations.
Selina: What? You drive me crazy with your ideas. What does it mean to draw zero lines? That’s nonsense!
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Sebastian: No, think about it! Zero lines means that you don’t divide, thus you build only one station. It then must contain all jobs and there’s just one possibility to set up. Nevertheless you have to include this possibility.
Nadine: Okay, slowly. We should better write this as a formula, otherwise I’ll loose the overview. We have n jobs and want to build k stations, correct?
Selina: Correct.
Nadine: Good, let’s go on. I need k – 1 lines to build k stations. I can put them in n – 1 positions. So I calculate „n – 1 choose k – 1“.
Meanwhile Nadine has put the following formula down on paper:
n – 1 k – 1
n ... number of jobs
k ... number of stations
The number of possibilities to build k stations is
Nadine: And now we combine n of these binomial coefficienk = 1 to k = n and add up. The first summand is for k = 1, for „draw zero lines“ respectively. The second summand is
Number of possibilities to build stations for n jobs:
1 1 1
0 1 2n n n− − −
+ + +
Selina: Your sum looks good! But don’t you have to typesolution. It wouldn’t surprise me if I type it wrong all the time
Sebastian: There’s certainly more in my notes from universsuch formulas, maybe I find something.
Oliver: Anyway, for today I’ve seen enough binomial coeffic
Selina: Good idea. See you tomorrow!
While the experts from Clever Consulting start to make theiat if one really can simplify this sum.
.
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ts. We have to calculate this value for this means for „build one station“, and for „build two stations“ and so on.
nn−
+ − .
a lot in the calculator till you get the (→E.2.6).
ity. Sometimes it’s possible to simplify
ients. I’ll go home now.
r way home, we will have a closer look
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Exercises E.2.3 How many possibilities do exist to build four stations from seven jobs? Represent all stations by lines between numbers.
E.2.4 The set {1, 2, 3, 4, 5} is given.
a) Build all ordered samples 3 from 5. How many do exist?
b) Now build all possibilities for an ordered sample 3 from 3. How many do exist?
c) Build all disordered samples 3 from 5. How many do exist?
Hint: Don’t build new in b) and c), but choose cleverly from a).
d) Calculate the number of disordered samples 3 from 5 as quotient of the results from a) and b) and by the use of the binomial coefficient.
E.2.5.a) Check your results from E.X.3 by using the binomial coefficient.
b) Calculate the following binomial coefficients:
5 47 47 9 10 18 19 4 6 9
, , , , , , , , ,3 0 47 4 6 4 0 2 2 5
E.2.6 How many possibilities do exist to build station from seven jobs (this is from one to seven stations)?
2.3 Binomial coefficients and Pascal’s Triangle
Pascal’s Triangle To become acquainted with the binomial coefficient, we go back a long way in the history of mathematics. We take a look back into the 17th century, when in France the mathematician, theologian and philosopher Blaise Pascal (1623-1662) lived. The so called Pascal’s Triangle is ascribed to him. We’ll now have a closer look at it.
11 1
1 2 11 3 3 1
1 4 6 4 11 5 10 10 5 1
1 6 15 20 15 6 11 7 21 35 35 21 7 1
Figure 2.3 Pascal’s Triangle
The figure above shows Pascal’s Triangle. What is remarkable about this triangle? Well, first of all it is symmetrical about a vertical axis of reflection. Furthermore the left and the right side of the pyramid are lined with ones. When you have a closer look at it, you will perhaps notice that
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directly below the ones, along the sides, the sequence of natural numbers shows up, this is 1, 2, 3, 4,... . How can you build this triangle and what is there to it? Pascal’s Triangle is in fact generated by using a calculation rule and you can extend it by arbitrarily many numbers. The calculation rule is: Each entry in the triangle comes from the sum of the entries that are in the previous row diagonally above the entry. To illustrate this, the calculation rule is shown in an example and is then presented as a mathematical notation.
3 3
11 1
1 2 1
1 11 4 3 36 = +
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Figure 2.4 Example of the calculation rule for Pascal’s Triangle
Furthermore it must be pointed out that the numbering of the rows and row elements in Pascal’s Triangle all begin with zero. As you can see, the second element of the fourth row comes from adding the two elements that are written directly above. For the calculation of the zeroth and the last element of each row you imagine that there is a zero in the empty place above. So the first mystery is already solved: at the sides of the triangle only ones can occur, because for these elements it always results in: 1 + 0 = 1 (→E.2.7).
11 1
1 2 1
1 3 3 01 4 6 4
Figure 2.5 Calculation of the side elements in Pascal’s Triangle
As announced, here is the mathematical notation of the calculation rule:
Calculation rule for the generation of Pascal’s Triangle Let dnk be the kth element in the nth row of the triangle. Then it holds:
1, 1 1,
00
0
11; 1 for all
nk n k n k
n nn
d d ddd d n
− − −= +
== = ∈
.
( E.2.8) The second constraint of this formulation is needed to initialize the triangle, i.e. to enable the triangle to start. Equally the two constraints in the third row are required for the initialization of the single rows. These three conditions are necessary, because otherwise the
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calculation rule would access elements that do not exist. For the formulation above it is important to keep in mind that the counting of the rows and row elements each? start with zero. What does Pascal’s Triangle have to do with the binomial coefficients? The special thing about Pascal’s Triangle is: Each element in the triangle is a binomial coefficient! The element dnk has the value of the binomial coefficient „n choose k“. It is possible to write Pascal’s Triangle in the following form:
00
1 10 1
2 2 20 1 2
3 3 30 1 2
…
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nn
Figure 2.6 Pascal’s Triangle with binomial coefficients
Thus an arbitrary nth row of Pascal’s Triangle can be written in the following way:
. 0 1 1n n n n
k k k − +
(→E.2.9+10) That the elements in Pascal’s Triangle dnk do really correspond to the binomial coefficient „n choose k“, we can check for the first row of Pascal’s Triangle without any problems. But this is not a proof of the general validity of our discovery. To show this, it has to be proven that the binomial coefficients satisfy the calculation rules for Pascal’s Triangle. For the interested reader the proof is shown in the following digression.
Exercises E.2.7 Pascal’s Triangle contains along the sides directly below the ones the sequence of natural numbers. How can this be explained? Does this observation hold for all Pascal’s Triangles (with arbitrarily many rows)?
E.2.8 Generate the zeroth to tenth row of Pascal’s Triangle.
E.2.9 Generate the 13th row of Pascal’s Triangle by using the binomial coefficients.
E.2.10 Calculate the following elements Pascal’s Triangle: d11,2, d11,4, d11,7, d12,3, d12,5 and d12,8.
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Proof*: Binomial coefficients generate Pascal’s Triangle
We want to show that the calculation rule for Pascal’s Triangle generates binomial coefficients „n choose k“. Or in other words, that the binomial coefficients satisfy the calculation rule for Pascal’s Triangle. First of all we repeat the calculation rule for Pascal’s Triangle.
( )( )( )
1, 1 1,
00
0
1
1 2
1; 1 for all 3
nk n k n k
n nn
d d d
d
d d n
− − −= +
=
= = ∈
Let us start with the easiest rule, namely with (2). That this is satisfied by the binomial coefficient, we show by plugging in the values.
00
0 00! 1 10 00! (0 0)! 1
d
= = = ⇒ ⋅ − =
Now for rule (3): We have to show that „n choose 0“ and „n choose n“ are both one for any arbitrary n:
( ) 0
! ! ! 1 for all 0 00! 0 ! 1 ! ! n
n nn n n n dn n n
= = = = ∈ ⇒ ⋅ − ⋅
= ,
( )
! ! ! ! 1 for all ! ! ! 0! ! 1 ! nn
n nn n n n n dn nn n n n n n
= = = = = ∈ ⇒ ⋅ − ⋅ ⋅ = .
Again we were very successful by plugging in the values and calculating the binomial coefficients. A bit more arithmetic is needed to check condition (1). We have to show that the following equation holds:
1 11
n n nk k k
− − = + −
or in detail:
( )
( )( ) ( ) ( )( )
( )( )( )
1 ! 1 !!! ! 1 ! 1 1 ! ! 1 !
n nnk n k k n k k n
− −= +
− − − − − − − k.
We start with the right-hand side of the equation and try to obtain the left-hand side by transformations.
( )( ) ( ) ( )( )
( )( ) ( )( )
( ) ( ) ( ) (( ) ( (
1 11
1 ! 1 !1 ! 1 1 ! ! 1 !
1 2 11 ! !
nn
n nk k
n nk n k k n k
n n n kk k n
− − + = −
− −+ =
− ⋅ − − − ⋅ − −
− ⋅ − ⋅ ⋅ − + ⋅− ⋅ ⋅
) ( )) ( )
( ) ( ) ( ) ( ))
1 2 1 1 2 11 2
1 21 1 2 1
k n k n kk n k
n nk
kn− − ⋅ − ⋅ ⋅⋅ − − ⋅ ⋅ ⋅ − − ⋅ ⋅ ⋅− ⋅ − − ⋅ ⋅ ⋅
− ⋅+
− − ⋅ ⋅ ⋅
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So far only the fractions have been transformed and the factorials have been written in detail. This was done in order to see where the fraction can be reduced. The red terms cancel out and the fraction is reduced to:
( ) ( ) ( )
( )( ) ( ) ( )1 2 1 1 2
1 ! !n n n k n n n
k k− ⋅ − ⋅ ⋅ − + − ⋅ − ⋅ ⋅ −
+−
k.
In the next step we convert the fractions by expansion to the same denominators in order to do the addition, since finally in the end only one factorial should be remaining.
( ) ( ) ( )( )
( ) ( ) ( )
( ) ( ) ( ) ( ) ( ) ( )
1 2 1 1 21 !
1 2 1 1 2!
n n n k n n n kk k
n n n k k n n nk
k
k
− ⋅ − ⋅ ⋅ − + − ⋅ − ⋅ ⋅ −+ =
−
− ⋅ − ⋅ ⋅ − + ⋅ + − ⋅ −
⋅
⋅ ⋅ −
! k⋅
( ) ( ) ( ) ) ( ) (1 2 1 1 2 1n n n k n n k⋅− ⋅ − ⋅ ⋅ − + − ⋅ − ⋅ ⋅ − +
( )( )!
2 12 1n k
n k=
⋅ − ⋅ ⋅ ⋅⋅ − ⋅ ⋅ ⋅
Now we can see that the summands in the numerator contain the same terms. Thus these can be factored out.
( ) ( )
( ) ( ) [ ]
( ) ( )
!1 1
!1
!1
nk nk
n n k k n kk
n n k nk
+ ⋅ −=
− ⋅ ⋅ − + ⋅ + − =
− ⋅ ⋅ − ⋅+
k
Now we are not wide of the mark any more. Expansion of the fraction leads to the desired goal:
( ) ( )
( )
1 1
1 1!1! !
n n n kk
n n nnk k kk n k
⋅ − ⋅ ⋅ − +
− − = ⇒ + = −⋅ −
. nk
With the validation of the last of the three calculation rules we can terminate our proof.
Simplified calculation of the binomial coefficients We have shown the connection between Pascal’s Triangle and the binomial coefficients „n choose k“. But what did we learn from this relationship? On the one hand, we know that binomial coefficients can be generated quite simply with the help of Pascal’s Triangle. When the nth row of Pascal’s Triangle is known, the (n+1)th row can be calculated by simple addition. This computation could be easily done on a sheet of paper or even worked out in one’s head. On the other hand, the original computation of the binomial coefficients leads to multiplications with huge numbers, where normally the calculator is needed (→E.2.11).
The symmetry of Pascal’s Triangle Another characteristic of Pascal’s Triangle can be utilized: the symmetry. As we know, Pascal’s Triangle is symmetrical about the vertical axis. This means, it holds: ,nk n n kd d −= .
16
Or by using the binomial coefficients:
n nk n k
= − .
This property of the binomial coefficients becomes clear when we have a look at an example from everyday life: If you have to choose two from five new CDs, because you can only afford to buy two of them, then this is the same as if you choose the three ones that you are not going to buy:
. 5 52 5 2 3
= = −
5
Relatively easy, by plugging in and transforming, one can show that this rule for binomial coefficients is right:
( ) ( ) ( ) ( )( )
! ! !! ! ! ! ! !
n nn n nk nk n k n k k n k n n k
= = = = k−⋅ − − ⋅ − ⋅ − − .
(→E.2.12)
Sums of binomial coefficients We want to get one final benefit from Pascal’s triangle, since for us this will be the most interesting point. In the last discussion of Clever Consulting Selina was not very happy with the following formula:
1 1 1
0 1 2n n n n
n− − − −
+ + + + 11−
.
If you do not have a modern calculator that provides the sum function, you have to compute each sum separately and you might easily type in the wrong numbers. So it would be nice if we found a possibility to simplify this sum. Let us consider once again the elements in an arbitrary n th row of Pascal’s Triangle:
. 0 1 1 1 1n n n n n n n
k k k n − + − n
The binomial coefficients in these rows are redolent to the summands above. The difference: n – 1 has been replaced by n. Thus the summands in our formula correspond to the elements of the (n – 1)th row of Pascal’s Triangle. Nevertheless, let us consider the nth row. What holds for this row can be transferred to the (n – 1)th row, as long as n > 1. We are interested in the sums of the elements of an arbitrary nth row of Pascal’s Triangle. The best is we first of all consider the first rows as an example:
Sum of row entries
11 1
1 2 11 3 3 1
1 4 6 4 11 5 10 10 5 1
12481632
17
Something is remarkable here: The sum of the entries in the rows is doubled from row to row! Can we already derive a general rule from this observation? No, since the observation could hold for the considered example only, we have to prove the property for the general case. So let us think about what the reason for the doubling of the sums of the rows might be and have a closer look at the generation of Pascal’s triangle. In the following row n = 4 is generated from row n = 3:
3 3 31 13
1 3 3 14 6 14= + = + = +
1
One can see that each entry from the upper row appears exactly twice as summand (and as separate entry at the sides respectively). This behavior can be reconstructed for arbitrary rows. Since each entry from the previous row appears twice in the following row, as summand or as separate entry, the sum of the entries in the row is doubled from one row to the next. Anybody who is interested in a proof of this behavior is invited to study the proof at the end of this chapter. Let us denote sum of the entries in the nth row with RSn, where RSn = dn0 + dn1 + ... + dn,n-1 + dnn. We know the sum of the entries in row zero, it holds: ZS0 = 1. There is no need to know the sum of the entries in the first row, we can calculate it: 1 0 2 1 2 2RS RS= ⋅ = ⋅ = .
The sum of the entries in the third row is calculated in an analogous manner: . 2 1 02 2 2 1 2 2RS RS RS= ⋅ = ⋅ ⋅ = ⋅ ⋅ = 4Then for the nth row it holds:
1 0 02 2 2 2 2 2 for 1n nn n
n
RS RS RS RS n−= ⋅ = ⋅ ⋅ ⋅ ⋅ = ⋅ =… ≥ .
But what about the zeroth row? 20 = 1 = RS0. So this rule also holds for the zeros row and thus in general:
1 2 , 1 2 for 0nn n n n n nnRS d d d d n−= + + + + = ≥… .
So the reduction of the sum formula was successful.
18
Sum of row entries
Row 0 1 1Row 1 1 1 2Row 2 1 2 1 4Row 3 1 3 3 1 8Row 4 1 4 6 4 1 16Row 5 1 5 10 10 5 1 32
Row n
0
0 1
1 2
2 3
3 4
4 5
21 2 2 2 22 2 2 2 24 2 2 2 28 2 2 2 216 2 2 2 2
2n
=
= ⋅ = ⋅ =
= ⋅ = ⋅ =
= ⋅ = ⋅ =
= ⋅ = ⋅ =
= ⋅ = ⋅ =
=
It holds:
20 1 1
nn n n n nk n n
+ + + + + + = −
.
(→E.2.13) Let us go back once again to the original meaning of the binomial coefficient. The number of possibilities to build a k-element subset from an n-element set is „n choose k“. What do we get if we calculate this term for all possible k from zero to n and then sum up? Well, then we have build 0-element, 1-element, 2-element and so on until n-element subsets, i.e. in every possible size. So 2n is the number of all subsets that can be build from an n-element set. The set of all subsets is called power set. The power set has 2n elements. Example: The set {1, 2, 3, 4} has four elements. Thus 24 = 16 subsets can be build:
4 4!0 0! 4!
= = ⋅ 1 0-element ∅ (empty set)
4 4!1 1! 3!
= = ⋅ 4 1-element {1}, {2}, {3}, {4}
4 4!2 2! 2!
= = ⋅ 6 2-element {1, 2}, {1, 3}, {1,4}, {2, 3}, {2, 4}, {3, 4}
4 4!3 3! 1!
= = ⋅ 4 3-element {1, 2, 3}, {1, 2, 4}, {1, 3, 4}, {2, 3, 4}
4 4!4 4! 0!
= = ⋅ 1 4-element {1, 2, 3, 4}
(→E.2.14)
19
Exercises E.2.11 The 17th row of Pascal’s Triangle is:
1 17 136 680 2380 6188 12376 19448 24310 24310 19448 12376 6188 2380 680 136 17 1
Calculate the 18th row of Pascal’s Triangle without using a calculator and by applying addition only. Check your results for the elements d18,4, d18,13, d18,11 and d18,8 with the aid of the calculator, by computing the binomial coefficients.
E.2.12 Which of the following equations and inequalities are right, which are wrong?
8 8 9 9 9 9 17 17 17 17 9 10, , , , ,
3 5 3 5 3 6 10 7 13 5 0 10
9 9 31 43 4 4 5 5 43, , , ,
4 5 0 0 2 4 2 3 1
= = ≠ = ≠ ≠
= = = ≠ ≠
43 23 23, ,
42 19 4
35 35 4 4 43 43 50 50 47 47 13 13, , , , ,
17 18 1 3 23 21 30 10 15 32 4 9
21 21 21 21, ,
15 7 16 6
=
≠ ≠ ≠ = = =
≠ ≠
23 23 22 22 18 18 17 17, , ,
9 19 11 13 9 10 7 10
= ≠ ≠ =
,
,
E.2.13 Check the solution from E.2.5. In addition, give the sum of all elements in the 10th and 13th row of Pascal’ Triangle.
E.2.14 Complete the following table of subsets for the set {1,2,3,4,5}:
Number of subsets
Number of elements per subset
Subset(s)
5 5! 10 0! 5!
= ⋅ = 0-element ∅ (empty set)
20
Proof*: The sums of the entries in the rows of Pascal’s Triangle are doubled
In this short digression it will be shown that the sums of the entries in the rows of Pascal’s Triangle are doubled from one row to the next. To begin with we write down the sum of the entries in the nth row:
0 1 , 10
n
n n n n nn nkk
d d d d d−=
+ + + + =∑ .
From the calculation rule for Pascal’s Triangle we know that dnk can be calculated from the elements of the previous row: dnk=dn-1,k-1+dn-1,k. We plug in this equation and take account of the special cases dn0 and dnn:
. ( )1
0 1, 1 1,0 1
n n
nk n n k n k nnk k
d d d d d−
− − −= =
= + + +∑ ∑The sum can be divided into two sums, for dn0 and dnn we plug in the corresponding values.
1 1
1, 1 1,0 1 1
1 1n n n
nk n k n kk k k
d d d− −
− − −= = =
= + +∑ ∑ ∑ +
=
Let us now have a look at the sums. The second sum almost looks like the sum of the entries in the (n - 1)th row. The difference: It starts with k = 1, while the mentioned sum of the entries in the row starts with k = 0. The first summand dn-1,0 is missing. Since it has the value one, we simply use one of the two ones in the sum. We include this one as dn-1,0 in the second sum and so we get the sum of the entries in the (n – 1) th row.
0
1 1
1, 1 1,0 1
1n n n
nk n k n kk k k
d d d− −
− − −= = =
= + +∑ ∑ ∑In principle, the same happens with the first sum. But in the first sum the index k – 1 in dn-1,k-1 is disturbing. In order to change this, we use a trick that is used in mathematics every now and then in calculations: the index shift. We introduce a new index r. For r it holds: r = k – 1. This means that k can be replaced by r + 1. In the following calculation the effect can be seen. Then the remaining one is included in the sum as dn-1,n-1 in order to complete it.
2
1
1
1, 1 1, 1,0 0
1 1
1, 1, 1,0 0 0
1
, 1,0 0
2
2
n n
nk n n n n kk kn n n
n r n k n kr
n
k kn n
n k n kk k
rd d d d
d d d
d d
−
− − − −= =
− − −
− − −= = =
−
−= =
−
= + +
+ = ⋅
⇒ = ⋅
∑ ∑ ∑
∑ ∑ ∑
∑ ∑
0r=
By transformations we have obtained the required result.
2.4 Exponential growth
The next morning Clever Consulting meets up again to continue to work on the problem of the Car Corp.. But one of the colleagues looks very tired.
Nadine: Gosh Sebastian! What’s wrong with you? Did you party all night?
21
Oliver: Come off it! Not our workaholic! Probably you’ve been sitting over your books all night, right?
Sebastian: Oh, leave me alone. You’d better be glad that at least someone shows dedication! Yesterday I’ve been digging around in my notes from uni.
Oliver: That was clear!
Sebastian: And I’ve found out something! We can simplify our sum formula of yesterday very well.
Selina: Really? Show me!
Sebastian: No, first of all I’ll get a coffee, then you may apologize nicely and maybe then I’ll tell you about my results.
So Sebastian discovered the same simplification that we have developed in detail in the last chapters. After two cups of coffee and being in better spirits Sebastian explains what he found out about the binomial coefficients.
Sebastian: So, and we can use this simplification of the sum formula for our yesterday’s result. Nadine, do you have the notes of yesterday with you?
Having a quick look in her bag, Nadine puts a sheet of paper with the following formula on the table:
1 1 1
0 1 2n n n n
n− − − −
+ + + + 11−
.
Sebastian: You see? Here we build the sum of the elements in the (n – 1) th row, this means our sum is 2n-1.
The number of possibilities to combine n jobs, whose order is given, intostations is:
11 1 1 12
0 1 2 1nn n n n
n n−− − − −
+ + + + = − − .
(→E.2.15)
Oliver: All right, this means we’d have to examine 210 cases fro the Car Corp.. Could you type it in your in your calculator?
Nadine: Already done: 1024 possibilities.
Oliver: Well, that’s still ok. We can write a program for the computer and then it calculates all possibilities. Further on it should show those possibilities, where the stations work at almost the same capacity.
Nadine: Well, that sounds pretty simple. But the number of possibilities growths rapidly, doesn’t it? If we had 20 jobs, then there’d be 524.288 possibilities. And with 100 jobs there’d be... oh, my calculator can’t display the number completely, there’d be 6,338*1029 possibilities! That’s a 30-digits number!
22
The enormous growth that the four colleagues observed is called exponential growth. The notion refers to the fact that in 2n-1 the variable n appears in the exponent. There it accelerates the growth enormously. Let us compare the results for n = 10, n = 20, n = 100 and n = 200.
n 2n-1
10 512 20 524.288
100 6,33825*1029
200 8,03469*1059
A doubling from n = 10 to n = 20 results in a 1024-times higher value of 2n-1. A doubling from n = 100 to n = 200 results in a 1.26765*1030-times higher value of 2n-1. Thus the value of 2n-1 grows very much faster than the variable n.
2 4 6 8 10
100
200
300
400
500
Figure 2.8 Function y = 2n-1
In figure 2.8 one can see very clearly that 2n-1 does not only grow with n, but also that this growths accelerates with increasing n. Such tasks quickly reach a size where they can no longer be handled by using even the latest computer technology. Then the number of possibilities is so large that those cannot be calculated in appropriate time. Today one comes across similar tasks when trying to decipher codes, for example. But in those cases one is glad that these tasks have not been solved yet, since otherwise the encryption systems, like they are used for credit cards or internet connections, would not provide security any more. Oliver: O.K. The size of the task grows rapidly. But nevertheless we can cope with the 1024 possibilities for the Car Corp..
Sebastian: That might be the case, but I don’t like it. There must be a more effective procedure than calculating all possible solutions? I mean, the Car Corp. surely would have hit on it as well!
Nadine: Well, maybe we should summarize everything we know so far, as which values are given and which have to be calculated.
Nadine writes the following list on a scribbling paper:
23
Given: Number of jobs: n Index of the job: i, i=1...n Processing time of job i: p
i
Unknown: Number of stations: k, with index of the station j Processing time of station j: sj Cycle time: C Assignment: jobs to stations
Sebastian: The cycle time C is not known, but we know at least that it is as long as the longest processing time of the stations.
1, ,max jj k
C s=
=…
Selina: What does this notation mean? The max with the j underneath?
Sebastian: It’s not very difficult. It means that among all existing sj, namely from j = 1 to j = k, the one with the maximum value is chosen.
Selina: Ah, o.k. I’ll keep that in mind (→E.2.16).
Nadine: I think we should define a variable that measures the variation from full capacity, so that we can decide which solution is appropriate.
Oliver: That’s a point we have to discuss with the managers of the Car Corp.. How do we know which solution they think is appropriate? O.K., the capacity should be as high as possible, the best is full capacity. But this can’t always be achieved. Is it better if then the idle time of all stations together is as small as possible, so that expenses are saved? Or is it more important that the idle time is divided as equal as possible between the single stations, so that no worker feels treated unfairly? These are two different points of view and certainly there are even more.
Oliver pointed out a very important issue: The definition of a goal. What do Clever Consulting and the Car Corp. want to achieve? Terms like “the capacity should be as high as possible“ are not precise enough. So Selina made an appointment at the Car Corp. to discuss this point. Of course she does not want to miss the chance to see the production of the sports car on site!
24
Exercises
E.2.15 How many possibilities do exist to combine 3, 8, 13, 15 or 20 jobs, whose order is given, into stations?
E.2.16 Please specify for the following sets M with elements mi: 1, ,max ii k
m= …
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.
25
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 1Assembly line efficiency: n np p pALEC 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 1 8 5 1 3 7 4 2 9 2 5 7 53n np p p p−+ + + + = + + + + + + + + + + =
The sum of the processing times is 53 minutes. Thus for the assembly line efficiency it holds:
53ALE
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.
26
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 2 5 6 7 3 8 9 10 1117; 13; 23s p p p p s p p p s p p p p= + + + = = + + = = + + + =
- Cycle time
{ } { }1 2 3max , , max 17,13,23 23C s s s= = = .
- Assembly line efficiency
53 53 53 0,768
23 3 69ALE
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 2 5 6 7 8 3 9 10 1117; 22; 14s p p p p s p p p b s p p p= + + + = = + + + = = + + =
- Cycle time
{ } { }1 2 3max , , max 17,22,14 22C s s s= = = .
27
- Assembly line efficiency
53 53 53 0,803
22 3 66ALE
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 1n nC 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 1 1 2 1n n k kp 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 11, ,
max j k kj kC k s k s s s s−=⋅ = ⋅ ≥ + + + +
…… .
Thus we can conclude: . 1 2 1n nC k p p p p−⋅ ≥ + + + +… We will now consider two cases:
28
Case 1: In the first case both sides of the above inequality have exactly the same value, thus the equation holds: 1 2 1n np p p p C−+ + + + = ⋅ k .
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 1n np p p p C−+ + + + < ⋅ k .
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:
1ALE ≤ .
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
1 2 1
531
53 153 1
n n
n n
C p p p pk
p p p pALEC k
−
−
= + + + =
=+ + +
= =⋅ ⋅
=
.
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.
29
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.
maxmin
1440 12 12120
PCO = = = =
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
5 5 5 5
= = = =
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
30
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 1n nC 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 1n np p pkC
− 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):
53kC
≥ .
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 TCycle time C
25
50
75
100
125
150
175
200
Stationsanzahl kNumber 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.
31
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
T xmaCmax
Stationsanzahl kNumber of Stations k
Taktzeit TCycle time C
F
Smuch 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 1min
max
n npkC
− +=
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? Tlast 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 cfour colleagues get the right result:
32
1 2 1min
max
n np p p pkC
− + + + +=
.
(→E.2.22)
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.
maxk n=
Oliver: Exactly, and for the same reason the cycle time cannot be smaller than the longest processing time of a job.
1, ,
max ii nC p
=≥
…
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
max
n np p pCk
− p + + + +≥
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 1min 1, ,
max
max max , n nii n
p p p pC pk
−
=
+ + + + = …
(→E.2.23) Selina: For our example it means:
{ }
max
min
max
min
1440 /120 1253 512
11
53max 9; max 9;5 911
C
k
k
C
= =
= = =
= = =
.
Oliver: And how do we go on now?
Sebastian: I suggest that we’re optimistic and go on as if we could get an assembly line efficiency of one.
33
1 2 1 1 2 1 531 n n n np p p p p p p pALE kC k C C
− −+ + + + + + + += = ⇒ = =
⋅
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 13eit T
1
2
3
4
5
6
7
8
9
10
11
nsanzahl
12
k StatioNumber of Stations k
14TaktzCycle time C
kmax
kmin
Cmin CmaxFigure 2.13
The curve in figure 2.13 shows the interaction between C and k. Each point (C | k) on this curve 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]
34
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?
E.2.19 Calculate the following expressions:
) 3,9
3, 2
2,9
3
123,99
a = = =
= =
) 19, 2
19,0001
20
20, 0001
43, 6
b = =
= =
=
) 4,9
4,9
326
326
0, 01
c = = = =
− =
E.2.20 Which of the following equations and inequalities are correct, which are wrong?
4 3,99 ; 4 4,99 ; 5,01 4,09 ; 14 14 ; 34,09 35 ;
18,34 19 ; 37,23 37 ; 4 5,999 ; 34,23 34,97 ; 18,01 19,901 ;
4 5,99 ; 43,22 44
≠ = = ≠ ≠ = ≠ = ≠ =
≠ = ; 432,1 432 ; 94,23 95 ; 18,103 19,401≠ = ≠
E.2.21 Calculate the maximum cycle time Cmax for the following pairs of production period P and minimum output Omin:
a) P = 100, Omin = 10 b) P = 98, Omin = 9, c) P = 346, Omin = 45 d) P = 46, Omin = 4,
e) P = 888, Omin = 98, f) P = 435, Omin = 43, g) P = 431, Omin = 152 h) P = 339, Omin = 30
E.2.22 Calculate the minimum number of stations kmin for the constant values p1 = 4, p2 = 9, p3 = 2, p4 = 13, p5 = 5, p6 = 9, p7 = 6 and for:
a) Cmax = 10, b) Cmax = 12, c) Cmax = 4, d) Cmax = 6 e) Cmax = 2, f) Cmax = 9
E.2.23 Calculate the minimum cycle time Cmin for the values from E.2.16 and E.2.21.
E.2.24 Let p1 + p2 + … + pn-1 + pn = 40. Find, independently of Cmin, Cmax, kmin and kmax, all positive values of C, such that for ALE = 1 the values of k are integer.
E.2.25 Draw a diagram (similar to figure 2.13) with the cycle time C on the horizontal axis and the number of stations k on the vertical axis. In this diagram, draw the maximum and minimum cycle time as well as the maximum and minimum number of stations for the following values: P = 210, Omin = 19, p1 = 5, p2 = 3, p3 = 2, p4 = 9, p5 = 8, p6 = 7, p7 = 1, p8 = 5, p9 = 6 and p10 = 4.
35
E.2.26 In the diagram from E.2.25, plot all integer points that lie within the rectangle given by Cmin, Cmax, kmin and kmax. Furthermore draw the following points: (7 │ 6), (12 │ 6), (10 │ 4), (10 │ 11).
Give concrete reasons, i.e. without using the formulas or the diagram, why the last four points are not feasible.
Excursus: Inversely proportional functions – equilateral hyperbolas
The curve discussed before belongs to the class of functions that are generally denoted by
( ) constantky f x kx
= = …
Such curves are called equilateral hyperbolas (→E.2.27). In this case we say: y is inversely proportional to x. When you consider equilateral hyperbolas with k > 0, then the following dependency between x and y holds: If x increases, then y decreases; if x decreases, then y increases. To what extent a change in x affects the parameter y is dependent on k.
( )
with 0
kf x yx
k
= =
>
( )
with 0
kf x yx
k
= =
<
y
x
y
x
Figure 2.14.a k is positive Figure 2.14.b k is negative
Equilateral hyperbolas have the following properties:
- The y-value at x = 0 is not defined. As the function with a positive k approaches the y-axis from the left, it converges to negative infinity (– ∞). As, in contrast, the function approaches the y-axis from the left, it converges to positive infinity (+ ∞). We say, the function has a jump discontinuity at x = 0.
- The function does not have a zero, i.e. there is no point where y = 0 holds. Indeed, the function approaches the y-axis as x converges to + ∞ or – ∞, but it never reaches the y-axis.
- The function is odd, i.e. it holds: ( ) ( )f x f− = − x .
This condition is satisfied here, since
36
( ) ( )k kf x fx x
− = = − = −−
x .
Odd functions are point symmetric about the point of origin (0 | 0). - The function is symmetric to the axis of reflection y = x. This means:
( ) ( )f x y f y x= ⇔ =
or
( )( )f f x x=
(→E.2.28).
Exercises E.2.27 Which of the following functions are equilateral hyperbolas, which are not? Explain your solutions.
2
1) ( )
1) ( )
3) ( )
4) ( )
a f x yx
b f x yx
c f x yx
d f x yx
= =
= =
= =
= = −
2
1
) ( )
2) ( )
29) ( )
) ( )
xe f x yx
xf f x yx
g f x yx
h f x y x−
= =
= =
−= =
= =
) ( )
) ( )2 3) ( )
4) ( ) 4
x
ei f x yx
j f x y exk f x y
x
l f x yx
−
= =
= =+
= =
= = −
E.2.28 Show that f(f(x)) = x holds for equilateral hyperbolas. For this, use the general formula of an equilateral hyperbola.
2. 6 Search for integer solutions
In the meantime, Oliver worked on the small diagram and marked all integer points (see figure 2.15).
37
1 2 3 4 5 6 7 8 9 10 11 12 13 14tzei
1
2
3
4
5
6
7
8
9
10
11
12
ionsanzahlk StatNumber of Stations k
Tak t TCycle time C
kmax
kmin
Cmin CmaxFigure 2.15
Oliver: Well, didn’t I do a good job? We’ve already said that k and C have to be integer values, so I’ve marked all integer points. The points show combinations of k and C where both values are integer. If the function goes through one of these points, then we would have found a solution that yields an assembly line efficiency of one.
Nadine: But what do we do if there’s no solution that lies on the curve, as it is the case in our example? Then we have to accept that ALE = 1 cannot be achieved. I’ve an idea: We should decrease the assembly line efficiency and find out what happens to our function then. Let’s just try ALE = 0,9.
1 2 1
1 2 1
0,9
53 58,80,90,9
n n
n n
p p p pALEC k
p p p pk kC C
−
−
+ + + += =
⋅+ + + +
⇒ ⋅ = ⇒ = =⋅ C
Nadine: You see? The constant in our function increased, since the ALE was reduced. So what happens to the function if the constant increases? Oliver, you’re good at drawing functions. Please construct a diagram that contains the new function.
38
1 2 3 4 5 6 7 8 9 10 11 12 13 14Taktzeit T
1
2
3
4
5
6
7
8
9
10
11
12
anzahlk StationsNumber of Stations k
ALE = 0,9
ALE = 1
Cycle time C
Figure 2.16
Selina: The new function moved up and to the right and is a bit more flat. Let’s try another value, e.g. ALE = 0.5. Maybe we can see the movement more clearly then.
53 53 1060,5 0,5
0,5ALE k k
C C= ⇒ ⋅ = ⇒ = =
⋅ C
1 2 3 4 5 6 7 8 9 10 11 12 13 14
1
2
3
4
5
6
7
8
9
10
11
k Stationsanzahl
12Number of Stations k
ALE = 0,5
ALE = 0,9
ALE = 1
Taktzeit TCycle time C
Figure 2.17
Sebastian: So the function varies with each new value for the assembly line efficiency. We should write this down in general:
39
1 2 1n np p pkALE C
− p+ + + +=
⋅.
Sebastian: The processing times bi are still constants, i.e. they’re fixed and don’t change in the course of our calculations. The ALE is a parameter in this function. This means, it can be chosen arbitrarily, but then is fixed for the calculations. The set of all functions that result from varying the assembly line efficiency is called a function family (→E.2.29).
Oliver: Stop it! You use all these technical terms.
Sebastian: Those are just the correct names. It’s good if we can use the right terms when we’re talking to our clients, isn’t it?
Nadine: You’re right. Let’s go on: We’re now looking for a function within this function family on which a solution to our problem can be found and that also has an as high as possible assembly line efficiency. But how can we get it?
Selina: First of all, I’d erase all points from the diagram that are out of the question. Those are for example the points that lie outside the bounds kmin, kmax, Cmin and Cmax.
Oliver: Good idea! We only mark the feasible solutions in our diagram. Those are the points that lie within the bounds. And since the ALE can never be greater than one, we can also omit the points below the curve.
1 2 3 4 5 6 7 8 9 10 11 12 13 14ktzeit
1
2
3
4
5
6
7
8
9
10
11
12
tationsanzahl
k SNumber of Stations k
ALE = 1
Ta TCycle time C
Figure 2.18 Feasible solutions
Oliver: The closer a point is to the curve, the better. In the diagram you can see that one point is very close to the curve: (9 | 6). The assembly line efficiency of (9 | 6) is 53/(9⋅6) = 53/54 ≈ 0,981.
Selina: Great! So we’ve already found the solution, right? Six stations with a cycle time of nine minutes and an assembly line efficiency of 0,981. And that’s a very good value, isn’t it?
Sebastian: I agree with that. The other assembly lines of the Car Corp. have an assembly line efficiency of around 0,75 – 0,85, Mr. Wiedner said.
40
Selina: Good, then let’s write down the production schedule. We’ve to assign the jobs to the stations, and in doing so we’ve to keep in mind that no station can work longer than nine minutes.
0 2 4 6 8
Station 1
Station 2
Station 3
Station 4
Station 5
Station 6
Station 7
Bearbeitungsdauer [Minuten]
10
Figure 2.19 Diagram: With a cycle time of nine minutes seven stations are required
Selina: Oh no, it’s not working out with six stations! If we don’t want to exceed nine minutes, we’ve to set up seven stations. Now we’ve found the best combination of cycle time and number of stations and cannot realize it.
Sebastian: I haven’t thought about that! Sure, we’ve ignored the lengths of the jobs.
Oliver: That’s a pity, but we’ve other points that are also very close to the curve ALE = 1. These points have an assembly line efficiency which is hardly worse. Maybe we can realize one of these combinations (→E.2.30)?
Nadine: Well, so if we have to check several points, I suggest that we use the computer. The best is if we write a program for the computer that calculates all feasible solutions with the relevant assembly line efficiency and then sorts the points by assembly line efficiency in descending order. According to this order we then can check whether the combination of cycle time and number of stations of these points can be realized.
Oliver: I’ll try to write an algorithm that gives us all feasible solutions.
So Oliver runs off to his computer and comes back a quarter of an hour later with the following algorithm.
41
Algorithm 2.1 Determination of the feasible points Input: processing times pi, minimum output Omin, time frame P Output: feasible points ListFP
Cmax := min
PO
If Cmax = 0 then Output: “The maximum production time per step is 0, i.e. the
problem is infeasible.“ Stop
kmin := 1 2 1
max
n np p p pC
− + + + +
kmax := n
Cmin := 1 2 1
1, ,max
max max , n nii n
p p p ppk
−
=
+ + + + …
i := Cmin j := kmin ListFP := empty set While j ≤ kmax do
While i ≤ Cmax do
If 1 2 1 1n np p p pi j
−+ + + +≤
⋅
Then add the following element to ListFP:
1 2 1, , n np p pi ji j
− + + + + ⋅
p
i := i + 1 End While i j := j + 1 i := Cmin
End While j If ListFP = ∅
Then Output: “There are no feasible points.“ Else Output: ListFP.
Oliver: I’m done! The algorithm gives us a list of feasible solutions. For each solution we get the cycle time, the number of stations and the assembly line efficiency.
42
Nadine: Could you please explain how the algorithm works?
Oliver: Of course, it’s not very complicated. First of all, the algorithm calculates all bounds that we’ve just worked out, this means Cmax, Cmin, etc. In the outer “while“-loop all feasible numbers of stations are examined. Then the inner “while“-loop generates the feasible cycle times for each number of stations. Afterwards time the assembly line efficiency is calculated for each pair of number of stations and cycle. If it is smaller than or equal to one, the algorithm saves this pair as a feasible solution.
Selina: So now we have to check the feasible solutions one after the other?
Sebastian: No way! We should also come up with an algorithm for this. How else can we sell our result to the Car Corp.? I don’t think that they want to calculate manually.
Oliver: Ok, I go right back to work.
Some time later, Oliver can provide an algorithm for checking the points.
43
Algorithm 2.2 Checking the feasible points Input: processing times bi, feasible points ListFP Output: feasible point with the best assembly line efficiency
1 2 1n np p pALEi j
−+ + +=
⋅p
Sort ListFP by the ALE in descending order (3rd element of the entries in the list) i := 0 While i < length(ListFP) do
i := i + 1 point := ListFP[i] CounterPS := 1 j := 0 While j < point[2] do
j := j + 1 sum := 0 ok := 1 While ok = 1 and CounterPS ≤ n do
If sum+bCounterP ≤ point[1] SThen sum := sum + b CounterPS
CounterPS := CounterPS + 1 Else ok := 0
End While ok If CounterPS = n + 1
Then Stop. Output: „Point is the best feasible solution.“
End While j End While i
(→E.2.31)
Oliver: I’m done! This algorithm is a bit more difficult. Do you want me to explain how it works?
Sebastian: Sure, go ahead!
Oliver: This algorithm uses three “while“-loops. The outer loop goes through the entries of ListFP, which are the feasible solutions. Each feasible solution is examined and saved as point. Point is a vector and consists of three elements: cycle time, number of stations, assembly line efficiency. The middle loop goes through the stations of the solution. Then in the innermost loop each station gets filled with jobs as long as the cycle time allows it. The parameter CounterPS shows the current job. Since at the beginning the feasible solutions were sorted by the assembly line efficiency the program can stop as soon as a solution is found where all jobs are scheduled.
Nadine: Clever. But if there were two solutions with the same assembly line efficiency, you’d only find one of them.
44
Oliver: That’s true. The best is if we ask the Car Corp., whether they want to find all equivalent solutions. It shouldn’t be too difficult to adjust the program.
Selina: That’s great! We simply let the computer do the work and we’ve got a solution.
Sebastian: Which solution do we get for the dates of the Car Corp.?
Oliver: I’ve already calculated it: The first solution from our list that can be realized is (9 | 7), a cycle time of nine minutes and seven stations.
Selina: I’ll draw the corresponding diagram:
0 2 8
Station 1
Station 2
Station 3
Station 4
Station 5
Station 6
Station 7
n]
10
Figure 2.20 Diagram for
Selina: Wait a minute, the diagrof nine minutes. A little while agseven stations.
Sebastian: But it’s just by chanc
Nadine: The assembly line effici= 63 minutes and seven cars ca
Sebastian: So this is the highesThus we’ve found an optimal schanged the technical data, e.g.
Selina: Wouldn’t this be a possmodify the technical data.
Oliver: First let’s see if we give athe other assembly lines as well
Nadine: And we could solve theach line we’d only need the coof efficiency and thus the econom
Selina: I still like blue the best, w
Nadine: Pardon?
Selina: Blue! For the car!
Nadine: ....
4 6
Bearbeitungsdauer [MinuteProcessing time [minutes]
seven stations with a cycle time of nine minutes
am looks exactly like the previous one! We’ve again a cycle time o, we’ve required six stations. That didn’t work out, it works with
e that the solution we’ve found first is the best.
ency is 84,13%. The whole production time of a sports car is 9 · 7 n be processed at then same time.
t assembly line efficiency that can be obtained for our problem. olution! We could only get a better assembly line efficiency if we split up or speed up the jobs.
ibility for another job? We could analyze where it’s profitable to
good impression with our result. It would be great if we analyzed .
e question, if a reorganization is profitable, mathematically. For sts of reorganization for one job, the remaining time, the increase
ies…
hat do you think?
45
Selina: Well, if we do that many jobs for the Car Corp., we’ve already worked hard enough to get the new car! Wouldn’t that be great? We need a company car!
Oliver: And on the weekend we simply share the car.
Sebastian: O.K., let’s get back to work then! Who’s going to write the field report?
Oliver: Sebastian, that’s very kind of you; it’s a great idea that you’re going to write the report!
Nadine: And please mention that we’ve many new ideas for improvement, the Car Corp. would be a good customer for us.
Selina: Bye Sebastian, and don’t work too long!
Sebastian: But, I didn’t mean it that way...
Selina: Exactly, neither did we. Of course we’ll write the report together, but right now we’re going to celebrate our success.
Exercises E.2.29 Draw the function family k(T) fort he values from E.2.17 and ALE {1; 0,8; 0,5}. Therefore choose 1 for the length of the x-axis and 600 for the length of the y-axis.
∈
E.2.30 Which of the following combinations can be realized if the values from E.2.17 are given?
a) (13 │ 4) b) (14 │ 4) c) (15 │ 3) d) (16 │ 3) e) (26 │ 2) f) (27 │ 2) g) (28 │ 2)
E.2.31 Use the algorithms 2.1 and 2.2 to obtain the best solution for an assembly line problem with the following dates:
p1 = 5, p2 = 6, p3 = 3, p4 = 7, p5 = 1, p6 = 5, p7 = 2, p8 = 3, Omin = 19, P = 170