Eindhoven University of Technology BACHELOR. Queueing models involving cross-selling. Kisters, Thijs. Award date: Link to publication

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1 Eindhoven University of Technology BACHELOR Queueing models involving cross-selling Kisters, Thijs Award date: 2017 Link to publication Disclaimer This document contains a student thesis bachelor's or master's, as authored by a student at Eindhoven University of Technology. Student theses are made available in the TU/e repository upon obtaining the required degree. The grade received is not published on the document as presented in the repository. The required complexity or quality of research of student theses may vary by program, and the required minimum study period may vary in duration. General rights Copyright and moral rights for the publications made accessible in the public portal are retained by the authors and/or other copyright owners and it is a condition of accessing publications that users recognise and abide by the legal requirements associated with these rights. Users may download and print one copy of any publication from the public portal for the purpose of private study or research. You may not further distribute the material or use it for any profit-making activity or commercial gain

2 Bachelor Final Project 2WH40 Queueing Models Involving Cross-Selling Author: Thijs Kisters Supervisor: Onno Boxma July 11, 2017

3 Contents 1 Introduction 2 2 Literature study Paper of Armony et al. [1] Model description Probability generating functions Matrix geometric method Paper of Boxma et al. [2] Model with customer types involving cross-selling Model description Balance equations Finding the probability generating functions Results obtained from PGFs M/M/1 queue with service time depending on waiting time Approach Applying level crossing to the model Conclusion and recommendations 19 Department of Mathematics and Computer Science 1

4 1 Introduction Cross-selling is a phenomenon in marketing which involves companies trying to sell their customers additional things after they have already bought something. In this report, we will look at several models involving the concept of cross-selling. We will not look at companies as a whole here, but for example at cashiers in a store trying to sell extra things to the customers that they have just served. We will see that it might be beneficial to only apply cross-selling when the length of the queue is smaller than some fixed value. In section 2 we will study some papers in which models are analysed which involve cross-selling. Next, in section 3 we analyse a model ourselves which is based on one of the models from section 2 and therefore also involves cross-selling. After this, in section 4 we look at a model which has some elements of cross-selling in it, but differs from the models that were discussed before, since we now look at the waiting time instead of the queue length. Finally, in section 5 we draw some conclusions about what was discussed in this report and give some recommendations for further research. 2 Literature study In this section we analyse two papers, which discuss models that involve cross-selling in some way, to hopefully get a better mathematical understanding on the concept of cross-selling. 2.1 Paper of Armony et al. [1] Model description Armony et al. [1] introduce the following model. They consider a multiserver queueing system where N parallel servers are present and where at each server the arrival process is a Poisson process with parameter λ. Every customer can encounter two potential phases. First we have phase 1, which lasts an exponentially distributed time with mean 1 µ. After completing this phase, a customer will either become a cross-selling candidate with probability p or he will completely leave the system with probability q = 1 p. If the customer is picked to be offered a cross-selling deal, he enters phase 2. In this phase he then spends again an exponentially distributed time, but this time with mean 1. Next they denote the number of customers at time t in phase 1 by L 1 t and the number of customers at time t in phase 2, which is the cross-selling phase, by L 2 t. Furthermore, let L q t be the queue length at time t and Lt = L q t L 1 t L 2 t the total number of customers in the system at time t. They then prove that it is sufficient to describe the whole system with the two-dimensional state-space L 2, L. After this, they introduce a threshold T 1, which denotes whether a customer that completed phase 1 and is chosen to be a cross-selling candidate actually enters phase 2. Namely, if the total number of people is less than or equal to this threshold, meaning L T, then the cross-selling candidate will enter the cross-selling phase. If L > T, meaning the total number of customers exceeds the threshold T, then the cross-selling candidate will not enter phase 2 and will simply leave the system. This T may be greater than the number of servers, which means that cross-selling will be initiated if L q T N. In a similar fashion, if T is less than or equal to N then we initiate cross-selling if the number of idle servers is greater than N T. The continuous time Markov chain L 2 t, Lt using this threshold is then further analysed Probability generating functions Armony et al. [1] then proceed to derive the steady-state distribution of the two-dimensional process mentioned above. They define this steady-state distribution by P jm = lim PL 2 t = j, Lt = m. Also, they assume t λ < Nµ, since this is apparently the stability condition for this system. To derive the steady-state distribution, they first find the balance equations for the steady-state probabilities. They divide these equations in the cases j = 0, 1 j N 1 and j = N. We will omit these equations as they are quite lengthy, but we will give the equations which are the result of multiplying these balance equations for P jm by z m and then summing them over m. Before they do this they first introduce the conditional probability generating function PGF of the number of customers in the system. This PGF is given for each j = 0, 1, 2,... N by G j z = P jm z m. m=j By multiplying the equations with z m and then summing them over m we get for j = 0: Department of Mathematics and Computer Science 2

5 N mp 0m z m1 qµ N 1 µ λz Nµ1 zg 0 z G 1 z = N T T mp 0m z m Nqµ P 0m z m Nµ P 0m z m. m=1 m=n1 1 For j = 1,..., N 1 we get: λz N jµ1 z jzg j z j 1G j1 z = N 1 T N mp jm z m1 pµ m j 1P j 1,m z m1 N j 1pµ p j 1,m z m1 N 1 µ m=j N 1 qµ m=j m j 1P j,m1 z m1 N jqµ m=j T 1 m=n mn P j,m1 z m1 N jµ T P jm z m. m=j 2 Finally, for j = N we have: λ1 z NG N z = pµ T m=n P N 1,m z m. 3 Next, Armony et al. [1] try to solve this system of equations for each G j z by first writing it as Az Gz = bz, where Gz = G 0 z, G 1 z,..., G N z T, which are the PGF s that we want to find and α 0 z α 1 z 2 0. Az = 0 0 α 2 z , αn 1 z N α N z where α j z = λz N jµ1 z jz for j = 0, 1,..., N 1 and α N z = λ1 z N. For the elements of bz we have: N 1 b 0 z = µ N mp 0m z m1 qµ For j = 1,..., N 1 the elements are N mp 0m z m Nqµ m=1 T m=n1 P 0m z m Nµ T P 0m z m. 5 N 1 N 1 b j z = µ N mp jm z m1 pµ m j 1P j 1,m z m1 N j 1pµ m=j N 1 qµ m=j m j 1P j,m1 z m1 N jqµ m=j T 1 m=n T m=n P j,m1 z m1 N jµ p j 1,m z m1 T P jm z m, m=j 6 and finally T b N z = pµ P N 1,m z m. 7 This means that the elements of bz are precisely the right hand sides of 1, 2 and 3 respectively. m=n To now solve Az Gz = bz, where Az, Gz and bz are as above, Armony et al. [1] use Cramer s rule, which gives G j z = Ajz Az. Here A is the determinant of the matrix A and A jz is the matrix which can be found by replacing the j-th column of Az by bz. In bz there are many unknown probabilities P jm to be precise 1 2 N 12T 2 N of them, which are Department of Mathematics and Computer Science 3

6 thus also contained in A j z. To find all of these probabilities Armony et al. [1] build a system of equations to solve them. However, just using the balance equations that we omitted from this summary does not give enough equations as we also need 1 2 N 12T 2 N equations to solve that many unknowns, which is why they use a few methods to find the remaining equations. As these methods are not really relevant for us, we will not go into very many details of how to find these remaining equations. Since G j z is a PDF defined for all 0 z 1, they note that each root of Az is a root of A j z in that interval. Furthermore they note that Az = N α j z. They find that for j = 1,..., N 1 each α j z is a j=0 quadratic polynomial with only one of the roots in 0, 1, call this root z j,1, while α 0 z and α N z have no roots in 0, 1. By then writing A j z j,1 = 0 for j = 1, 2,..., N 1 they find N 1 more equations in terms of the unknown probabilities P jm. The last equation is found by applying so called diagonal cuts to seperate {0, m, 1, m 1,..., N, m N} from the states {0, m 1, 1, m 2,..., N, m N 1}. We will not elaborate on this method any further, but eventually this yields the last desired equation. After having found enough equations, Armony et al. [1] are able to solve the linear system, which gives us all the P jm that were needed to find bz as well as A j z. Now we found A j z, we can calculate G j z = Ajz Az. Next Armony et al. [1] find the marginal probability of having j customers in the cross-selling, namely P j. They do this by first substituting z = 1 in 1 which gives for j = 1 the expression P 1 = b01. Also, by substituting z = 1 in equation 2 we get for j = 2 the expression P 2 = b11 P1 2 = 1 2 b 01 b 1 1. Doing this for all 1 j N we achieve P j = 1 j 1 j b i 1. As the sum of the probabilities is 1 we furthermore get P 0 = 1 N P j. j=1 i=0 Apparently, by summing the equations 1, 2 and 3 over all j and then substituting z = 1, the left hand side becomes zero, which means that the right hand also has to be zero. This results in N b i 1 = 0. We use this to write the above expression for P j as follows: P j = 1 j N b i 1 for j = 1,..., N. After this, Armony et al. [1] also derive the PGFs G j z for j = 0, 1,... N. They do not use the formula which was achieved from Cramer s rule, but they instead use the system Az Gz = bz. Namely from 4 and this system we find i=j G N z = b Nz α N z, 8 G j z = 1 α j z b jz j 1G j1 z, j = 0, 1,..., N 1. 9 Using the recursion found in 9 with starting point 8 we then find the PGFs for j = 0, 1,... N: N j G N j z = 1 N i! N b N i z α i z N j! j i α k z. 10 Here they define N k=n1 = 1. i=n j i=0 k=n i1 Finally, Armony et al. [1] also compute the mean queue length of the system. For this, they first take the marginal probability P m of having m customers in the system, meaning P m = PL = m. We can calculate this P m for m = 0, 1,..., N 1 by the formula P m = m P jm and for m N we have P m = N P jm. Armony et al. [1] also note that the mean number of customers which are in cross-selling is given by E[L 2 ] = N N jp j, where P j = 1 j b i 1 for j = 1,..., N, as was seen above. j=1 Next they define E[L j ] = i=j m=j j=0 mp jm for j = 0, 1,..., N. They then state that E[L j ] = d dz G jz z=1, which is quite obvious. By then summing over all j we get E[L], the mean number of customers in the system: E[L] = N E[L j ] = N d dz G jz z=1 = mp m. Armony et al. [1] then calculates the derivatives of G j z j=0 j=0 i=0 j=0 Department of Mathematics and Computer Science 4

7 explicitly, but we will not do that here, as it does not add much to the idea on how to find E[L]. Finally, they also give a formula for the mean number of waiting customers in the system, which is given by E[L q ] = E[L] N N N mp m Matrix geometric method After using PGFs to find the steady-state distribution P jm, Armony et al. [1] also use a different approach to analyse the model given in section Namely, they first construct a quasi birth and death QBD process, after which they use the matrix geometric method to further solve this. The matrix geometric method is another way to analyse the M/M/1 queue or birth and death processes, using matrices. Since we will not use this method to solve any further problems in this report, we will only give a brief overview of this method without going into very much detail. First they find the infinitesimal generator of the earlier mentioned QBD, which is very large and will therefore be left out. Next they describe, like in the method with the PDFs, how the probabilities P jm can be found, which is by solving a system of linear equations. Using these probabilities the mean queue length can again be found. Department of Mathematics and Computer Science 5

8 2.2 Paper of Boxma et al. [2] The paper of Boxma et al. [2] is not on cross-selling, but the authors apply the main results of the paper to some special cases of which some lie close to the concept of cross-selling. Therefore, it is worthwhile to discuss the paper in this report. Boxma et al. [2] introduce the following model. Let {X n, n = 0, 1,... } be a one-dimensional Markov chain, where X 0 = 0, { maxx n 1 n, 0 if X n 1 > a, X n = maxx n 1 η n m, 0 if X n 1 = m, m = 0,..., a. 11 Here a is a nonnegative integer and { n } and {ηn m }, m = 0,..., a are independent sequences of independent integer-valued random variables, which are identically distributed within each sequence. Next, Boxma et al. [2] try to find the limiting distribution of this model given by p k = lim PX n = k, k = n 0, 1,.... Since they also assume that E[ 1 ] < 0 and that the Markov chain is irreducible and aperiodic, this limiting distribution exists. Next, they derive expressions for PX n = 0 and PX n = k, k = 1, 2,.... We will not give these equations here, but we will give the result of multiplying these equations by z k and then summing them over k. First we introduce an expression for the probability generating function for PX n = k, namely P n z = z k PX n = k, where z 1. By multiplying the equations with z k and summing them over k we get P n z = a m=a1 PX n 1 = mpη m n m < 0 12 z k PX n 1 = mp n m < 0 a k=0 z k k=0 m=a1 PX n 1 = mpη m n = k m PX n 1 = mp n = k m Next, they introduce the following notation for any D {..., 1, 0, 1,... } and for an arbitrary function of the form k= a k z k : Using 13 to rewrite 12 gives [ k= k=0 a k z k] D = a k z k 13 k D P n z = [P n 1 zfz] [0, 14 a PX n 1 = m z k {Pη n m = k m P n = k m} a k=0 PX n 1 = m{pη m n m < 0 P n m < 0} PX n 1 = mp n m < 0. Here fz is the probability generating function of the distribution of n. Department of Mathematics and Computer Science 6

9 Now we let n and we introduce P z = lim n P nz, z 1, 15 lz = [P zfz], 1], z 1, q m z = Pη m 1 m < 0 P 1 m < 0 z m [g m z fz] [ m,, m = 0, 1,..., a, z 1, where g m z is the probability generating function of the distribution of η m n. Applying 15 to 14 gives for z = 1 P z1 fz = a p m q m z l1 lz. 16 Boxma et al. [2] then find the standard Wiener-Hopf factorization of the factor 1 fz, so where 1 fz = R zr z, 17 R z = 1 E[z χ ν < ], z 1, 18 R z = 1 E[z χ ], z Here they define ν as the first weak ascending ladder index, so ν = min{n 1 : 1 n 0} and they define ν as the first descending ladder index, so ν = min{n 1 : 1 n < 0}. Also, χ = 1 ν, χ = 1 ν. Since apparently both sides of 16 vanish for z = 1 and since R 1 = 0, we define R z = R z 1 z and get from 16 that q mz P zr z = a l1 lz p m h m z 1 z R z, 20 where h m z = 1 z R. Boxma et al. [2] then give some arguments why P zr z z S0,, R z 1 S, 1 and R z S, 1, where Ss, t denotes the set of all convergent power series of the form t b k z k, z = 1, where s, t are arbitrary integers with s t. We will not go in further detail about these k=s arguments. They also state that ψz := l1 lz 1 z R S, 0, but we will not further elaborate on why this z is true. We then write c := ψ and apply Liouville s theorem to 20 to get which can be rewritten to get a P zr z = p m [h m z] [0, c, z 1, 21 a p m [h m z],1] l1 lz 1 z R c = 0, z z Boxma et al. [2] prove that 22 can be deduced from 21 and that summing up 21 and 22 gives 20, meaning that 20 and 21 are equivalent. However, we will not give this proof here, since it is not really relevant to this report. After this proof, they state that the main result of their paper is basically 21, namely for z 1 P z = a p m R 1 z[h m z] [0, cr 1 z. 23 Department of Mathematics and Computer Science 7

10 They also show how the values for p m, m = 0,..., a can be obtained, since these are still unknown, but we will again not include this in this report. Next, they dedicate a section of their paper to the asymptotics of the steady-state probabilities p k for k. Since this analysis is not relevant for our research, we will omit this section from the summary of this paper. However, what we are interested in is the next section, which is devoted to some applications of the results that were found earlier. We will mention two of these applications, since these somewhat resemble cross-selling models. The first application that is somewhat interesting to us is the case where a = 0, fz = B λ1 z z and g 0 z = B λ1 z. Here B λ1 z is the generating function of the number of Poisson V λ1 z V λ z1 V λ events with rate λ that take place during a length of time which is distributed according to B with mean E[B] and Laplace-Stieltjes Transform LST B. Namely, if we take these particular choices we get a model that apparently corresponds to an M/G/1 queue with multiple server vacations, where the vacation times have LST V. Server vacations can be interpreted as some extra service which is given to customers if the queue length is smaller than or equal to some value in this case this value is 0. We also see this phenomenon when looking at cross-selling, where people get an extra service if the queue length is smaller than some value, which is why this case is interesting to us. Boxma et al. [2] deduce a formula for P z by using the aforementioned chosen parameters which is given by P z = 1 ρ1 zb λ1 z B λ1 z z 1 V λ1 z. 24 λe[v ]1 z Next, we look at another application given in the paper, which also has similarities to cross-selling and therefore has our interest. In this application we take fz = B a λ1 z z, g 0 z = Bb λ1 z and g mz = B b λ1 z z, m = 1,..., a. This apparently corresponds to an M/G/1 queue with service time distribution B a with LST Ba if the last departing customer leaves more than a customers behind and service time distribution B b with LST Bb otherwise. P z will be the PDF of the queue length distribution, right after departures events. Boxma et al. [2] note that, like in the ordinary M/G/1 queue, the queue length distribution after a departure event is the same as before an arrival event and that it is also the same as the steady-state queue length distribution. This can be proven using an up- and downcrossing approach and PASTA Poisson Arrivals See Time Averages, but we will not go into any further detail about this. This result implies that this model basically represents the concept of cross-selling, since there we also change the service time distribution depending on the current queue length. This model is somewhat different to the model described by Armony et al. [1], since they take two exponentially distributed service times, whereas here we consider the more general case. Also, in their model they account for the number of customers that are in cross-selling, as well as the total number of customers in the system, which is not the case here. Boxma et al. [2] use the choices of the parameters mentioned above to derive an expression for P z for this particular model which is given by 1 z P z = p Baλ1 0 B z z aλ B aλ1 z zbb λ1 z 1 z a p m z m B aλ1 z Bb λ1 z c z m=1 They then again describe how the unknown constants p 0,..., p a and c can be determined in this case, but we will omit this from this report. Department of Mathematics and Computer Science 8

11 3 Model with customer types involving cross-selling In this section we will look at a different model with cross-selling, which has some similarities to the model that Armony et al. [1] used, but does not involve the number of people that are currently in the cross-selling phase. 3.1 Model description We will first describe the model we use. We consider a single server queuing system, where the arrival process is a Poisson process with parameter λ. At the arrival of a new customer, it is determined of which type the customer will be. With probability p the customer will be of type 1, which means that he will be a cross-selling customer. This results in him lasting an exponentially distributed time with mean 1 µ in the regular service phase, after which he enters the cross-selling phase, which also lasts an exponentially distributed time, but now with mean 1. After the cross-selling phase, the customer leaves the system. Once he enters this cross-selling phase, he will become a customer of type 2. With probability 1 p the customer will be of type 0, meaning that he is a regular customer. This means he will only enter the regular service phase, which lasts an exponentially distributed time with mean 1 µ, after which he leaves the system. We will describe this system with the two-dimensional state space X w, S. Here X w is the total number of customers in the system on an arbitrary moment and can thus take values in {0, 1, 2,... }. S is the type of the customer that is currently being served, meaning that S can take the values 0, 1 or 2 as seen above. 3.2 Balance equations We would like to find the steady-state distribution of this two-dimensional process X w, S. We will define this steady-state distribution as q n,j := PX w = n, S = j. Using this definition we can find several balance equations for the steady-state probabilities. For n = 0 we have: λq 0 = µq 1,0 q 1,2, 26 where q 0 = PX w = 0, as we cannot speak of a type of customer in service if there are no customers in the system. For n = 1 we have: j = 0 : λ µq 1,0 = 1 pλq 0 1 pµq 2,0 1 pq 2,2 27 j = 1 : λ µq 1,1 = pλq 0 pµq 2,0 pq 2,2 28 For n 2 we have j = 2 : λ q 1,2 = µq 1,1. 29 j = 0 : λ µq n,0 = λq n 1,0 1 pµq n1,0 1 pq n1,2 30 j = 1 : λ µq n,1 = λq n 1,1 pµq n1,0 pq n1,2 31 j = 2 : λ q n,2 = λq n 1,2 µq n, Finding the probability generating functions We will now do the same procedure as Armony et al. [1] did in their paper. Namely, we will multiply the equations 27, 28, 29, 30, 31 and 32 by z n on both sides and then sum them over n. However, before we do this, first define the probability generating function PGF of the q n,j, which is given by G j z = q n,j z n, for j = 0, 1, 2. Now we multiply the equation 30 by z n and then sum it over n, together with 27, to obtain the following Department of Mathematics and Computer Science 9

12 equation for j = 0: so hence λ µ q n,0 z n = λ1 pq 0 z λ q n 1,0 z n 1 pµ q n1,0 z n 1 p q n1,2 z n, 33 n=2 λ µg 0 z = λ1 pq 0 z λzg 0 z 1 pµ G 0 z q 1,0 z z 1 p G 2 z q 1,2 z, 34 z λz1 z µz 1 pµg 0 z 1 pg 2 z = λ1 pq 0 z 2 1 pq 1,0 zµ 1 pq 1,2 z. 35 To obtain 34 from 33, we shifted some of the indices and then used the formula G j z = q n,j z n. To then get 35, we collected all of the similar terms and also multiplied by z. We can further simplify equation 35 by using equation 26, which gives us: λz1 z µz 1 pµg 0 z 1 pg 2 z = λ1 pzz 1q Now we do the same thing for equation 31, together with 28, which gives the following equation for j = 1: so hence λ µ q n,1 z n = λpq 0 z λ q n 1,1 z n pµ q n1,0 z n p q n1,2 z n, 37 n=2 λ µg 1 z = λpq 0 z λzg 1 z pµ z G 0z q 1,0 z p z G 2z q 1,2 z, 38 λz1 z µzg 1 z pµg 0 z pg 2 z = λpq 0 z 2 pµq 1,0 z pq 1,2 z. 39 To obtain 38 from 37, we again shifted some of the indices and we again used the formula G j z = q n,j z n. To obtain 39, we collected all of the similar terms and also multiplied by z. We can again further simplify equation 39 by using equation 26, which gives us: λz1 z µzg 1 z pµg 0 z pg 2 z = λpzz 1q Lastly, to obtain the equation for j = 2, we multiply 32 by z n and together with 29 this gives: yielding and finally λ q n,2 z n = λ q n 1,2 z n µ q n,1 z n, 41 n=2 λ G 2 z = λzg 2 z µg 1 z, 42 λ1 z G 2 z µg 1 z = To obtain 41 from 42, we shifted some indices and used the formula G j z = q n,j z n. To then get 43, we collected all of the similar terms. Now we will solve the system of equations formed by the results of equations 36, 40 and 43 for G 0 z, G 1 z and G 2 z. We can rewrite equation 43 as: λ1 z G 1 z = G 2 z. 44 µ Department of Mathematics and Computer Science 10

13 Similarly, we can rewrite equation 36 as: G 0 z = 1 p λz1 z µz 1 pµ G λ1 pzz 1q 0 2z λz1 z µz 1 pµ. 45 Now we will substitute equations 45 and 44 into equation 40 to get: We can simplify 46 to get: λz1 z µzλ1 z p1 pµ G 2 z µ λz1 z µz 1 pµ G 2z pµλ1 pzz 1q 0 λz1 z µz 1 pµ pg 2z = λpzz 1q G 2 z = λµpzz 1q 0λz1 z µz 1 pµ λpµ 2 1 pzz 1q 0, 47 Az where Az = λz1 zµzλ1 zλz1 zµz 1 pµ p1 pµ 2 pµλz1 zµz 1 pµ. Equation 47 can again be simplified to obtain: G 2 z = λµpzq 0 µ λz λλz 1z µ1 z p. 48 Since we now have an expressions for G 2 z, we can use this together with equations 44 and 45 to find expression for G 1 z and G 0 z respectively. Substituting 48 in 44 gives: which can be simplified to G 1 z = λ1 z µ G 1 z = Similarly, we can substitute 48 in 45 to obtain: λµpzq 0 µ λz λλz 1z µ1 z p, 49 λpzq 0 λ1 z µ λz λλz 1z µ1 z p. 50 G 0 z = 1 p λz1 z µz 1 pµ λµpzq 0 µ λz λλz 1z µ1 z p λ1 pzz 1q 0 λz1 z µz 1 pµ, 51 which can be simplified to G 0 z = λ1 pzq 0 λ1 z µ λz λλz 1z µ1 z p. 52 We have now found expressions for G 0 z, G 1 z and G 2 z in equations 52, 50 and 48 respectively. However, there is still an unknown parameter in these equations, namely q 0. Since we know the sum of all q n,j should be equal to 1, we know that q 0 G 0 1 G 1 1 G 2 1 = 1. Using this we get which can be solved for q 0 to get λq 0 µp µ λ λµp = 1 q 0, 53 q 0 = 1 λ µ λp. 54 This probability q 0 can be interpreted as the fraction of the time that the server is not busy. This fraction is usually denoted as 1 ρ, where ρ is the fraction of the time that the server is occupied. Since q 0 = 1 λ µ λp = 1 ρ, we get that ρ = λ µ λp. Also, we know that the stability condition of an M/G/1 queue is given by ρ < 1, which in this case results in λ µ λp < 1. Armony et al. [1, p. 14] further support this result, where in this case N = 1. Department of Mathematics and Computer Science 11

14 Substituting 54 into 52, 50 and 48 gives G 0 z = G 1 z = G 2 z = λ1 pzλ1 z 1 λ µ λp µ λz λλz 1z µ1 z p, 55 λpzλ1 z 1 λ µ λp µ λz λλz 1z µ1 z p, 56 λµpz1 λ µ λp µ λz λλz 1z µ1 z p. 57 Since G 0 z, G 1 z and G 2 z are probability generating functions, their power series have to converge absolutely at least for z 1, i.e. be analytical on the domain z 1. To prove that this is indeed true, we prove that the poles of G 0 z G 1 z and G 2 z lie outside the unit circle. Let gz be the denominator of G 0 z, G 1 z and G 2 z they all have the same denominator. Rearranging the terms of gz gives gz = λ 2 z 2 λ λ 2 λµz µ λµ1 p. 58 Since gz is quadratic in z, we know that it will have 2 roots. We prove that these roots are real and are greater than 1. First we check what value gz has for z = 1. g1 = λ 2 λ λ 2 λµ µ λµ1 p = µ λ λµp 59 > λ λµp λ λµp = 0. To get the inequality in 59, we used the stability condition λ µ λp see that g1 > 0. In addition to this, we also see that lim z ± λµp < 1 and rewrote it to µ > λ. So we gz =, since if z goes to plus or minus infinity then gz will behave as λ 2 z 2, which goes to infinity if z goes to plus or minus infinity. Next, we check the value of gz in z = µ λ. µ g = λ 2 µ λ λ 2 λ λ 2 λµ µ µ λµ1 p λ = µ 2 µ λµ µ 2 µ λµ1 p 60 = λµp < 0. To get the inequality in 60, we use that λ,, µ and p are greater than 0. We see that µ λ > 1, since by the stability condition we have λ µ < 1 λp µ λ > 1 = 1 λp λp > = 1. Since g1 > 0 and g µ λ < 0, we know that g must have a real root in the interval 1, µ λ. Also, since we stated earlier that lim gz =, g must also have a real root in the interval µ z ± λ,. g will only have two roots since it is quadratic in z and we found that these both are real and lie outside the unit circle. This proves that G 0 z, G 1 z and G 2 z are indeed analytical on the domain z 1. Now that we proved that G 0 z, G 1 z and G 2 z are indeed probability generating functions, we can try to rewrite 55, 56 and 57 so that we can deduce specific expressions for the probabilities q n,j, for n = 1, 2,... and j = 0, 1, 2. Let z 1 and z 2 be the roots of gz given in 58. We proved that both of these roots lie outside the unit circle. Using the quadratic formula on gz we find that z 1 = λ µ λ µ 2 4µ λ1 p 2λ z 2 = λ µ λ µ 2 4µ λ1 p. 62 2λ 61 Department of Mathematics and Computer Science 12

15 This means that gz = λ 2 z z 1 z z 2. We can now use partial fraction decomposition on G 0 z, G 1 z and G 2 z to rewrite them as follows. For G 0 z we get that G 0 z = 1 p1 λ µ λp Bz 1 Az 2 1 λ 2 z z 1 z 1 z 2 Bz 2 Az 2 2 λ 2 z z 2 z 1 z 2, 63 where A = λ 2 1 p1 λ µ λp and B = λ1 pλ 1 λ µ λp. This can be rewritten to G 0 z = 1 p 1 λ µ λp 1 p 1 λ µ λp λz 1 1 λz 1 z 2 Since 1 1 x = n=0 1 p 1 λ µ λp λz 1 z 2 x n for x 1, we get that G 0 z = 1 p 1 λ µ λp 1 p 1 λ µ λp λz 1 z 2 λ1 z 2 1 p λ1 z z z 2. 1 λ µ λp λz 1 z 2 n z. n=0 z 2 λz z z 1 64 n z 65 This expression will only converge for z z 1, since z 1 z 2 by the way we defined z 1 and z 2. We see that the first term in 65 cancels the n = 0 terms of both sums together, meaning that 65 reduces to G 0 z = 1 p 1 p 1 λ µ λp λz 1 z 2 1 λ µ λp λz 1 z 2 λz 1 1 λ1 z 2 which according to the definition of G j z implies that for n = 1, 2,... q n,0 = 1 p 1 λ µ λp λz 1 z 2 λz 1 1 n 1 1 p z 1 n=0 z 1 n z 66 z 1 n z, 1 λ µ λp z 2 λz 1 z 2 λ1 z 2 n z 2 We can use a similar approach to rewrite G 1 z and G 2 z to get q n,1 and q n,2 respectively. For G 1 z we get that p 1 λ µ λp λz 1 1 n z G 1 z = 68 meaning that for n = 1, 2,... p 1 λ µ λp q n,1 = λz 1 z 2 p 1 λ µ λp λz 1 1 λz 1 z 2 λ1 z 2 λz 1 z 2 n 1 p z 1 z 1 n z, 1 λ µ λp z 2 λ1 z 2 λz 1 z 2 n z 2 For G 2 z we get that G 2 z = µp 1 λ µ λp z 1 µp λz z 1 z 1 z 2 z 2 1 λ µ λp λz z 2 z 1 z 2, 70 Department of Mathematics and Computer Science 13

16 which eventually yields and thus for n = 1, 2,... µp 1 λ µ λp n z n z G 2 z =, 71 λz 1 z 2 µp 1 λ µ λp q n,2 = λz 1 z 2 z 2 z 1 n n z 2 z Results obtained from PGFs Now, using the formulas for G 0 z, G 1 z and G 2 z found in 55, 56 and 57, we can calculate several properties of the system. For example, from Armony et al. [1, p. 9], we know that the mean total number of customers in the system E[L] can be found by using the formula E[L] = E[L 0 ] E[L 1 ] E[L 2 ], where E[L j ] = d dz G jz z=1. Using this formula on the expressions from 55, 56 and 57 we get E[L] = d dz G 0z z=1 d dz G 1z z=1 d dz G 2z z=1 = λ1 p2 λpλ λ µ λµp = λpλ µ λµp1 p 2. λ µ λµp λp2 λpλ λ µ λµp λpλ2 µ λµ1 p λ µ λµp 73 It might be nice to compare the formula of the mean queue length given in 73 to some of the results from Armony et al. [1], in particular their result for the mean queue length. To be able to compare these two results, we will choose N = 1 in the paper of Armony et al. [1]. In 8 and 9 formulas are given for G N z and G j z j = 0, 1,..., N 1 respectively. Plugging in N = 1 gives us the expressions for G 1 z and G 0 z. To find the complete expression for these PGFs, we need to derive the unknown probabilities P jm with j = 0, 1 and m = j, j 1,..., T. This gives us 2T 1 unknown variables. As described in section 2.1.2, we can find these unknowns by solving a system of linear equations of size 2T 1. We will again not include these equations and we will also omit how we solved these for the P jm, as the expressions become very large even for small values of T. We will only include the formula for E[L] for T = 1, since for T 2 the expressions for E[L] become quite large and the formula for T = 1 should still be sufficient to compare the mean queue lengths of both models. By using that E[L] = N d dz G jz z=1 we find for T = 1 that j=0 E[L] = λ 3 µ λ 2 pµµ λ pλλ 2 2λµ 2µ 2 2 pλ 2 µ 2 λµ1 2p µµ λ λpλ µ1 p λ pµ Before we compare the expressions in 73 and 74, we will give a global overview of some of the differences and similarities of the two models. In the model of Armony et al. [1] customers can only enter a cross-selling phase if the total number of customers in the system is less than or equal to T, whereas in the model described in section 3.1 the total number of customers in the system does not influence the fact if a customer can become a cross-selling candidate or not. Furthermore we see that the stability condition of both models differ a bit. Namely, for the model of Armony et al. [1] the stability condition is given by λ < µ, for N = 1, while the stability condition of the other model is given by λ µ λp < 1. We can see from these conditions that the model from section 3.1 is a lot more sensitive for a change in the values, then the other one, as the stability condition depends on. This will also reflect in the mean queue lengths, which we will see later on. Now we look at the expressions in 73 and 74 itself to see if we can predict the behaviour of both formulas. If tends to 0, then both of the mean queue lengths will go to infinity, because of the factor in the denominator. Also, if tends to infinity, then both formulas reduce to λ µ, which is the mean queue length of the standard 1 λ µ M/M/1 queue, which we will mention again a bit later on. Letting λ go to 0 will in both cases result in a mean queue length of 0, because of the factor λ in the numerator. However if λ and µ are very close together then the mean queue length in 74 grows very large, if λ < µ which as we know is the stability condition for N = 1, while for the expression in 73 this is not necessarily the case. Namely, for this model the stability condition is given by λ µ λp < 1 as mentioned before, meaning that λ and µ can not lie too close together, since this would make the system unstable. Department of Mathematics and Computer Science 14

17 To confirm some of these beliefs and to further investigate both expressions, we will analyse them numerically by filling in values for the parameters for which both systems are stable. For convenience we call the mean queue length for the model of Armony et al. [1] E[L a ] so the formula in 74 and the mean queue length of the other model we call E[L b ] so the formula in 73. First we take the parameters λ = 1, µ = 2, p = 0.5 and we let vary. This gives the following table: values of E[L a ] E[L b ] Table 1: λ = 1, p = 0.5, µ = 2, varying Interestingly, if we increase, we see that E[L a ] increases, but only by a little bit, while E[L b ] decreases quite a lot. This is most likely the result of the fact that the stability condition of the first model does not depend on, as mentioned before. Therefore changing will not have as much impact on the mean queue length as it does for the second model. Also, since we only consider T = 1, we know that the customer in service will only enter the cross-selling phase if the queue is empty when he finishes his regular service phase. This will not happen very often, resulting in also not affecting the mean queue length very much. Both mean queue lengths seem to converge to 1, which is indeed true if we further increase. It is not surprising that both mean queue lengths converge to the same value as increases, because this means both systems will turn into standard M/M/1 queues, with mean queue length equal to λ µ, like we predicted before. This would 1 λ µ give with our choices of the parameters indeed a mean queue length of 1. Next we choose the parameters λ = 1, = 2, p = 0.4 and let µ vary, which gives us the following table: values of µ E[L a ] E[L b ] Table 2: λ = 1, p = 0.4, = 2, varying µ In both cases we see that increasing µ results in a lower mean queue length. This is a logical result, because increasing µ means that the server will serve the customer faster, resulting in a shorter queue. Now we will vary λ and take µ = 3, p = 0.5 and = 3 to get the following table: values of λ E[L a ] E[L b ] Table 3: µ = 3, p = 0.5, = 3, varying λ Decreasing λ seems to lower the mean queue length in both cases. This is a reasonable result, as lowering λ means less customers will arrive at the server on average, resulting in a shorter queue. In conclusion we can say that both systems appear to act the same and also as expected, except when we change, because then the systems act differently. This may however be caused by us only considering the case T = 1 for the model of Armony et al. [1]. If we were to investigate higher values of T, we could get different results, but since the expressions for E[L] become a bit too large, we will not consider those. Department of Mathematics and Computer Science 15

18 4 M/M/1 queue with service time depending on waiting time In this section we will consider a standard M/M/1 queue, but with a slight change. For every arriving customer we will check what his waiting time will be. If his waiting time is smaller than some constant T, then this customer will have an exponentially distributed service time with mean 1 µ. However if his waiting time will be larger than or equal to T, then his service time will be exponentially distributed with mean 1 γ. Here, we assume that 1 γ > 1 µ. To approach this problem, we will use something called level crossing, which will be explained in the following section. 4.1 Approach Since we want to analyse the distribution of the waiting time of arriving customers, we will note the following. According to PASTA Poisson Arrivals See Time Averages we know that the waiting time is in distribution equal to that of the steady-state workload, i.e. the amount of work, V. We define fx as the density of this steady-state workload. We can find the distribution of this steady-state workload by using the level crossing method, which is explained thoroughly in [3] and therefore we will not go into very much detail about this. This method is based on the idea that the long run average number of downcrossings of some level x > 0 is fx and that this should be equal to the long run average number of upcrossings of level x. By a downcrossing we mean crossing level x from above and by an upcrossing we mean crossing it from below. We will not go into any more details about this concept. Using this results in the following integral equation for the ordinary M/G/1 queue, which can be found in [3]: fx = λ x 0 1 Gx wfwdw λpv = 01 Gx, x > Here Gx is the cumulative distribution function of the service time distribution. The left hand side of 75 represents the rate to cross level x > 0 from above and the right hand side of 75 represents the rate to cross level x from below. The first term of the right hand side represents the jumps from every w 0, x and the second term represents jumps from Applying level crossing to the model Now we will apply this method to the model described above. Since the service time distribution depends on the fact if the waiting time so also the amount of work is less than T or greater than T, we will consider two cases while using the level crossing method. First the case x < T. Since the amount of work will always be less than T in this case, we will not see any change in the service time distribution and can therefore use 75. We consider an M/M/1 queue, so the service times are exponentially distributed with mean 1 µ, since x < T, meaning Gx = 1 e µx. This gives fx = λ x 0 e µx w fwdw λpv = 0e µx. 76 We will now solve 76 for fx. Differentiating 76 on both sides with respect to x yields f x = λµ x 0 e µx w fwdw λfx λµpv = 0e µx. 77 From 76 we find that λ x 0 e µx w fwdw = fx λpv = 0e µx. Substituting this into 77 gives f x = µfx λpv = 0e µx λfx λµpv = 0e µx, 78 and so f x = λ µfx. 79 This first order differential equation can easily be solved to get the general solution fx = ce λ µx, 80 Department of Mathematics and Computer Science 16

19 where the constant c has to be determined. We can do this by calculating f0 in both 76 and 80 and then equating the results. This gives c = λpv = 0, so for x < T we get the final result fx = λpv = 0e λ µx. 81 Next, we look at the case where x T. Here the service time distribution can change depending on how much waiting time an arriving customer has and therefore we have to take this into account when finding an equation similar to 75. For x T we find the following equation: T fx = λ e µx w fwdw λ 0 x T e γx w fwdw λpv = 0e µx. 82 Here we can see that the first term of the right hand side represents the jumps from any level w 0, T, where the exponential service time distribution will have mean 1 µ ; the second term represents jumps from any level w T, x, where the exponential service time distribution will have mean 1 γ ; the last term represents jumps from level 0. We will again solve 82 for fx. Differentiating 82 with respect to x gives T f x = λµ e µx w fwdw λγ 0 x T e γx w fwdw λfx λµpv = 0e µx. 83 Writing 82 as λ T 0 e µx w fwdw = fx λ x T e γx w fwdw λpv = 0e µx and substituting this into 83 yields x f x = µ fx λ e γx w fwdw λpv = 0e µx and so λγ x Differentiating 85 with respect to x gives T T e γx w fwdw λfx λµpv = 0e µx, 84 f x = λ µfx µ γλ f x = λ µf x γµ γλ x T x T e γx w fwdw. 85 e γx w fwdw λµ γfx. 86 We can rewrite 85 to µ γλ x T e γx w fwdw = f x λ µfx and by substituting this into 86 we get yielding f x = λ µf x γf x λ µfx λµ γfx, 87 f x = λ µ γf x µλ γfx is a second order differential equation, which has characteristic polynomial r 2 λ µ γr µλ γ = 0. This polynomial has two unique solutions, namely r 1 = λ γ and r 2 = µ. This results in the general solution of 88 being given by fx = c 1 e λ γx c 2 e µx. 89 The only thing that is left now is to determine the constants c 1 and c 2. We know that it must hold that fxdx PV = 0 = 1. Since we know that fx is defined as in 81 for 0 x < T and for x T as in 89, we get T 0 λpv = 0e λ µx dx T c 1 e λ γx c 2 e µx dx PV = 0 = For the integrals in 90 to be finite we need that λ < γ and µ < 0. It is obvious that µ > 0. The stability condition of the system is λe[b] < 1, where B is the service time. This service time distribution has either mean 1 µ or 1 γ, but for the stability condition we will only have to use 1 γ, since the service time with mean 1 µ Department of Mathematics and Computer Science 17

20 only occurs for amounts of work that are lower than T. Thus we see that λ γ integrals in 90 will be finite. Solving these integrals gives λpv = 0 e λ µt 1 λ µ and so < 1, so λ < γ. Therefore, the c 1 λ γ eλ γt c 2 µ e µt PV = 0 = 1, 91 µ c 2 = c 1 λ γ eλµ γt µe µt µe µt PV = 0 1 λ e λ µt λ µ Next, we find a second expression involving c 1 and c 2. Evaluating 85 in x = T gives the relation f T = λ µft. 93 Differentiating 89 with respect to x, evaluating the result in x = T and substituting this in 93, along with 89 evaluated in x = T then gives We can rewrite 94 as c 1 λ γe λ γt c 2 µe µt = λ µc 1 e λ γt c 2 e µt. 94 c 1 µ γe λ γt c 2 λe µt = We can now substitute 92 into 95 and solve for c 1 to get λ γ c 1 = λµ γ 2 e λ γt 1 PV = 0 1 λ e λ µt γλ µ λ µ Then, substituting 96 into 92 yields an expression for c 2, namely γ 2 γλ µ λµ c 2 = µ γ 2 e µt 1 PV = 0 γλ µ 1 λ λ µ e λ µt Therefore, we find the complete solution of 82, which is given by 89 with c 1 and c 2 given by 96 and 97 respectively. We still need to determine what the value of PV = 0 is, so the probability that there is no work. To do this, we find two expressions for ft, equate them and then solve for PV = 0. First we substitute x = T into 81 to get ft = λpv = 0e λ µt. 98 The expression in 81 only holds for x < T, but we can still evaluate it in x = T, since we know fx has to be continuous in x = T so lim fx = ft. x T Next we substitute x = T into 89 along with the expressions for c 1 and c 2 in 96 and 97 respectively to get ft = 1 PV = 0 1 λ λ µ e λ µt 1 Combining 98 and 99 and solving for PV = 0 eventually gives PV = 0 = µ λµ γ. 99 µλ µγ λ. 100 µ 2 λ γ λ 2 γ µe λ µt An interesting case to look at is if we take T = 0. This would mean that one of the integrals in 82 vanishes and this equation reduces to fx = λ x 0 e γx w fwdw λpv = 0e µx. 101 This implies that every customer that arrives will get an exponential service time with mean 1 γ except if the system is empty, because in that case the customer would still get an exponential service time with mean 1 µ. This kind of models, where customers that encounter an empty system have a different service time, is studied extensively in [4] and therefore we will not go into any further detail about this here. We could also take the limit for T, meaning that every customer will have an exponential service time with mean 1 µ, turning the model into a standard M/M/1 queue. This would also mean that for all x 81 holds, where PV = 0 is given by 100, which for T reduces to PV = 0 = 1 λ µ. This is the known probability that the server is not busy in the standard M/M/1 queue, which is in this case equivalent to the amount of work being equal to 0. Department of Mathematics and Computer Science 18

21 5 Conclusion and recommendations In this report we analysed different kinds of models which all had some similarities to cross-selling. The main result from section 3 were the probability generating functions, which gave us the mean queue length and in section 4 we found the probability density function for the steady state workload for different values of x. However, due to time constraints we could not analyse everything that we wanted to. For example, we only compared the results from section 3 to one particular case of the model discussed in Armony et al. [1]. It would be interesting to see for higher values of T how much both models still differ from each other. Also, we could turn the M/M/1 queue of section 4 into an M/G/1 and see if we can still get nice closed-form expressions for fx. Department of Mathematics and Computer Science 19

22 References [1] M. Armony, E. Perel, N. Perel, and U. Yechiali, Exact Analysis for Multiserver Queueing Systems with Cross Selling, pp. 1 25, [2] O. Boxma and V. Lotov, On a Class of One-Dimensional Random Walks, [3] P. Brill, Level Crossing Methods in Stochastic Models, Springer, New York, [4] P. Welch, On a Generalized M/G/1 Queuing Process in Which the First Customer of Each Busy Period Receives Exceptional Service, Operations Research, vol. 12, no. 5, pp , Department of Mathematics and Computer Science 20