Repair systems with exchangeable items and the longest queue mechanism
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DOI: 10.1007/s11134-012-9319-5
- Cite this article as:
- Ravid, R., Boxma, O.J. & Perry, D. Queueing Syst (2013) 73: 295. doi:10.1007/s11134-012-9319-5
Abstract
We consider a repair facility consisting of one repairman and two arrival streams of failed items, from bases 1 and 2. The arrival processes are independent Poisson processes, and the repair times are independent and identically exponentially distributed. The item types are exchangeable, and a failed item from base 1 could just as well be returned to base 2, and vice versa. The rule according to which backorders are satisfied by repaired items is the longest queue rule: At the completion of a service (repair), the repaired item is delivered to the base that has the largest number of failed items.
We point out a direct relation between our model and the classical longer queue model. We obtain simple expressions for several probabilities of interest, and show how all two-dimensional queue length probabilities may be obtained. Finally, we derive the sojourn time distributions.
Keywords
Repair system Longest queue Queue lengths Sojourn timeMathematics Subject Classification
60K25 90B221 Introduction
In this paper, we consider a repair facility consisting of one repairman and two arrival streams of failed items, from bases 1 and 2. The arrival processes are independent Poisson processes with rates λ _{1} and λ _{2}. The repair times are independent and identically distributed, with exp(μ) distribution regardless of the type of failed item. The item types are exchangeable, and a failed item from base 1 could just as well be returned to base 2, and vice versa. The rule according to which backorders are satisfied by repaired items is the longest queue rule: At the completion of a service (repair), the repaired item is delivered to the base that has the largest number of failed items. In case of a tie, the item will be delivered to base 1 or base 2 with probability \(\frac{1}{2}\).
We are interested in key performance measures of this repair facility, such as the (joint) queue length distribution of failed items of both types, and their sojourn time distribution (the time between arrival and departure of a failed item). In the literature, several studies have appeared about the so-called longest queue system. That is a queueing system with one server and (typically) two queues with customers of two different types; the server choosing a customer from the longest queue upon service completion. Cohen [2] has studied the case of two customer types with Poisson arrival streams with rates λ _{1} and λ _{2}, having service time distributions B _{1}(⋅) and B _{2}(⋅). If the server has completed a service, then the next customer to be served is the one at the head of the longest queue if the queue lengths are not equal; if both queues have equal length, then the next customer in service is of type i with some probability α _{ i }. He determines the generating function of the joint steady-state queue length distribution right after service completions, by solving a boundary value problem of Riemann–Hilbert type.
Zheng and Zipkin [9] consider the completely symmetric exponential case (λ _{1}=λ _{2}; \(\alpha_{1} = \alpha_{2} = \frac{1}{2}\); B _{1}(x)=B _{2}(x)=1−e^{−μx }). They calculate the steady-state distribution of the difference between the two queue lengths, and they provide a recursive scheme for the calculation of the joint queue length distribution and the marginal distributions. They also briefly consider the case that λ _{1}≠λ _{2}. Flatto [5] also considers the symmetric exponential case. He allows preemption, and derives an expression for the probability generating function of the joint queue length distribution. He uses this expression to derive asymptotic results.
Van Houtum et al. [7] also focus on the completely symmetric exponential model. They consider two variants: a longest queue system with threshold rejection of customers and one with threshold addition of customers. They show that these systems can be analyzed in detail using matrix-geometric methods, and that this provides lower and upper bounds for the longest queue system.
The repair facility with exchangeable items, that is the subject of our paper, has already been studied by Daryanani and Miller [4]. Using taboo sets and taboo probabilities, they derive various relations between the steady-state queue length probabilities; however, they do not solve those equations.
Remark 1
While the above described classical longer queue model is closely related to the model studied in [4] and the present paper, there are significant differences. To demonstrate these, let us consider the classical system and our repair system, receiving exactly the same input and having exactly the same service times. Suppose both systems start empty, and then a type-1 item arrives. The server starts serving. During the service, there are no type-1 arrivals and three type-2 arrivals. In the classical system, the type-1 customer leaves and the server starts serving the first of the three type-2 customers. The state of the classical system now is (0,3): one type-2 customer is just entering service and the other two type-2 customers are waiting.
In our repair system, the items are exchangeable. The repaired item is assigned to the first type-2 customer, even though it was brought as a type-1 item. Hence, the state of the system right after the repair is (1,2): there is still one waiting type-1 customer and there are two waiting type-2 customers.
Motivation
The longer queue model is related to the join-the-shortest-queue model: both models feature a mechanism that tends to equate the queue lengths. The longer queue model is a very natural one, but it has received much less attention than the join-the-shortest-queue model. We believe our paper yields valuable new insight into the longer queue model and a variant of it.
Contributions
Our main contributions are: (i) We point out a direct relation between our model and the classical longer queue model of Cohen [2]; (ii) we obtain simple expressions for several probabilities of interest, and we show how all two-dimensional probabilities may be obtained; (iii) we derive the sojourn time distributions—this performance measure was not studied in the papers mentioned above; and (iv) we present some methodological ideas which might be more broadly applicable; one example is the use of the “difference busy period.”
Organization of the paper
In Sect. 2, we first give the balance equations for the joint steady-state queue length distribution. We then study its generating function (GF), deriving various special results like the distribution of the difference of the two queue lengths and the probability that there are n _{1} customers of one type and none of the other type. Then we point out a direct relation between Cohen’s model and our model, which in principle allows us to use his results for obtaining the GF of the joint queue length distribution. However, we are interested in providing explicit results for the probabilities P(i,i) of having i customers of either type, i=1,2,…, which will also give us the marginal distributions explicitly. These probabilities are studied in Sect. 3; in that section, we also give an iterative method for obtaining all queue length probabilities. Finally, Sect. 4 is devoted to the determination of the sojourn time distribution of a customer of either type.
2 Queue lengths—a generating function approach
2.1 P(N _{1}+N _{2}=n)
2.2 The probabilities P(n _{1},0) and P(0,n _{2})
2.3 P(N _{1}−N _{2}=n|N _{1}>N _{2}) and P(N _{2}−N _{1}=n|N _{2}>N _{1})
So far, we have not yet tackled the general problem of finding \(E[z_{1}^{N_{1}}z_{2}^{N_{2}}]\); we only showed that various relevant performance measures have a geometric distribution. The natural approach to the general solution of (2) seems to be to translate the problem into a boundary value problem, like a Riemann–Hilbert problem (cf. [3]). That was also the approach chosen by Cohen [2] in his analysis of the two-dimensional queue length process right after departures in the case of Poisson arrivals and generally distributed service times; see also Flatto [5] for the case of exponential service times.
When we compared (2) with formula (1.7) of Cohen [2], we came to the conclusion that his formula for exp(μ) service times reduces to our formula (2). This is surprising in view of Remark 1, where it is explained that the exchangeability feature of our repair system leads to different queue length behavior in both models. Below we shall show that, despite that different behavior, the steady-state joint queue length distributions in both models are the same.
So, although Cohen [2] and we study different quantities (cf. Remark 1), the above reasoning shows that in the case of exp(μ) service times, his \((x_{n}^{(1)},x_{n}^{(2)})\) and our (N _{1}(t),N _{2}(t)) have the same limiting distribution. This is confirmed by the fact that Cohen’s formula (1.7) for the generating function, when taking exp(μ) service times, agrees with our formula (2).
For general service times, this reasoning fails because then successive service times do not generate a Poisson process, and PASTA cannot be applied.
3 Queue lengths—a probabilistic approach
In this section, we shall first determine the probabilities P(i,i), and then present a procedure to obtain all P(i,j).
3.1 Determination of P(i,i)
We use an argument from Markov renewal theory to derive an expression for (the generating function of) P(i,i).
Step 1: relate P(i,i) to the steady-state probabilities π _{ i } of an underlying Markov chain.
- (i)
An arrival from base 1 occurs first. As indicated in the previous section, one may argue that this arrival starts a busy period, call it B _{1}, with arrival rate λ _{1} and service rate λ _{2}+μ. Indeed, all the time until equality of the two queue lengths occurs again for the first time (at some level (i,i)), repaired items will be handed back to base 1 and not to base 2, since queue 1 is the longer queue. Notice that the stability condition λ _{1}+λ _{2}<μ implies that λ _{1}<λ _{2}+μ.
- (ii)
Similarly, if an arrival from base 2 occurs first when the queue length process is in state (k,k), then it takes a busy period B _{2} with arrival rate λ _{2} and service rate λ _{1}+μ until the system is back at some state (i,i) with equal queue lengths.
- (iii)
Finally, if a service completion occurs first, then with probability \(\frac{1}{2}\) the queue length process moves to (k,k−1), respectively, to (k−1,k), and again a busy period B _{1} respectively B _{2} occurs. We shall sometimes speak of a ‘difference busy period’.
We now determine the probability that the underlying Markov chain jumps from (k,k) to (i,i), in each of these three possible events.
Remark 2
3.2 Determination of P(i+1,i) and P(i,i+1)
In Sects. 2.1, 2.2, and 2.3, we successively derived simple geometric expressions for P(N _{1}+N _{2}=n), P(n _{1},0) and P(0,n _{2}), and for P(N _{1}−N _{2}=n|N _{1}>N _{2}) and P(N _{2}−N _{1}=n|N _{2}>N _{1}). In Sect. 3.1, we obtained a slightly more complicated expression for the (GF of) P(i,i). In the present subsection, we shall determine P(i+1,i) and P(i,i+1); that is an important building block for obtaining all P(i,j). Once more, there is a crucial role for the idea of having a “difference busy period” that starts at the moment the queue length vector leaves the state (i,i), and that lasts until equality is reached again. In the next subsection, we shall subsequently show how all P(i,j) can be determined once we know the above mentioned probabilities.
3.3 Determination of P(i,j)
Remark 3
An interesting feature of the present model is that, for general service times, it has only been solved via the boundary value method (cf. Cohen [2]). In almost all two-dimensional queueing problems which have been solved via the boundary value method, taking exponential service times does not simplify the problem to such an extent that one no longer needs to rely on that method. In the present problem, though, there seems to be so much structure that, in the exponential case, the P(i,j) have the nice form indicated in (30).
3.4 Numerical example
Queue length probabilities P(i,j) for the case λ _{1}=2, λ _{2}=1, μ=4
ij |
0 |
1 |
2 |
3 |
4 |
5 |
6 |
7 |
---|---|---|---|---|---|---|---|---|
0 |
0.250000 |
0.066432 |
0.010425 |
0.001636 |
0.000257 |
0.000040 |
0.000006 |
0.000001 |
1 |
0.121068 |
0.086662 |
0.030848 |
0.005411 |
0.000938 |
0.000161 |
0.000028 |
0.000005 |
2 |
0.043537 |
0.057328 |
0.043391 |
0.015993 |
0.002840 |
0.000502 |
0.000088 |
0.000015 |
3 |
0.015657 |
0.024413 |
0.030185 |
0.023189 |
0.008623 |
0.001531 |
0.000271 |
0.000048 |
4 |
0.005630 |
0.010145 |
0.013431 |
0.016414 |
0.012686 |
0.004734 |
0.000838 |
0.000149 |
5 |
0.002025 |
0.004139 |
0.005875 |
0.007430 |
0.009052 |
0.007018 |
0.002624 |
0.000464 |
6 |
0.000728 |
0.001665 |
0.002532 |
0.003326 |
0.004131 |
0.005028 |
0.003906 |
0.001462 |
7 |
0.000262 |
0.000662 |
0.001076 |
0.001473 |
0.001871 |
0.002305 |
0.002805 |
0.002182 |
Remark 4
In the case of equal arrival rates, the computation of the steady-state probabilities simplifies a bit. In particular, one now has P(i,j)=P(j,i). P(j,0)=P(0,j) follow as before from (8). The P(i,i) follow from (22). P(i,i+1)=P(i+1,i) follow using the relations (2λ+μ)P(i,i)=2λP(i−1,i)+2μP(i+1,i). Note that P(1,0)=P(0,1) were already determined. Finally. P(i,j)=P(j,i) follow for |i−j|≥2 as in Sect. 3.3.
4 Sojourn times
The main purpose of this section is to express the LST of the sojourn time distribution of a customer into the joint queue length distribution that was derived in the previous sections. We focus on a customer who has brought a failed item of type 1; by interchanging indices 1 and 2 (in particular, the arrival rates), we then also obtain the sojourn time LST for items of type 2.
4.1 The LST of the sojourn time distribution
Let X _{ k,j }:= sojourn time of a type-1 customer who increases the number of customers waiting in line 1 from j−1 to j, and whose arrival increases the difference between numbers of waiting customers in lines 1 and 2 from k−1 to k, k≥1, j≥1. We similarly define X _{ k,j } for k≤0; in that case, the difference between lines 1 and 2 again decreases from k−1 to k. Define \(\varPsi_{k,j}(\alpha) := E[\mathrm{e}^{-\alpha X_{k,j}}]\).
Case I: k≥0
Remark 5
To see that it indeed makes no difference for the sojourn time of the tagged customer whether the system is in state (j+m,j+m) or in state (j,j), suppose that a type-1 customer (just for convenience we refer to this customer as the red customer), arrives to find the system at state (j−1,j). That means that he finds j−1 type-1 customers in front of him in line and also j type-2 customers; however, it is not yet determined how many of them are served before him. The reason for that is that in principle, the sojourn time of the red customer depends on future arrivals of both types of customers. After the admittance of the red customer, the state of the system becomes (j,j), and the LST of the sojourn time of the red customer is Ψ _{0,j }(α). However, the latter dependence on future arrivals has a special regenerative property. For example, suppose that m type-1 customers and m type-2 customers were admitted to the system after the arrival of the red customer and before the service completion of the item being served. Then, by the memoryless property of the service, the residual sojourn time of the red customer (the time it takes from the arrival of the above 2mth customer until the red customer leaves the system), is stochastically equal to the sojourn time of the red customer. Thus, the LST of the above residual sojourn time is also Ψ _{0,j }(α).
In this way, we find the double GF of the sojourn time LST for a customer of type 1 who arrives to find j−1 customers waiting at Q _{1}, and whose arrival increases the difference between numbers of waiting customers in lines 1 and 2 from k−1 to k, for both k≥1 and k=0 (see G _{0}(z _{2};α)).
Case II: k<0
We thus obtain all conditional sojourn time LST’s. Multiplying by the probabilities P(j−1,j−k) as seen by an arriving customer (use PASTA to conclude that these arrival probabilities are exactly the probabilities which were calculated in Sects. 2 and 3) and summing yields the unconditional sojourn time LST.
4.2 Further results for the sojourn time distribution
One might use the results of the previous subsection to obtain mean conditional sojourn times E(X _{ k,j }). In the present subsection, we present an alternative approach for obtaining those means.
Let T _{ j } be the number of arrivals into base 1 minus the number of arrivals into base 2 that occur between the (j−1)th and the jth departure. Clearly, {T _{ j };j≥1} is a sequence of i.i.d. random variables; let T be the generic random variable of the sequence.
Lemma 1
Proof
4.3 Numerical example
Conditional expected sojourn time E _{ k,j } for the case λ _{1}=2, λ _{2}=1, μ=4
kj |
1 |
2 |
3 |
4 |
5 |
6 |
7 |
---|---|---|---|---|---|---|---|
0 |
0.379555 |
0.811411 |
1.224560 |
1.631373 |
2.035292 |
2.437686 |
2.839205 |
1 |
0.269332 |
0.627366 |
1.031134 |
1.434780 |
1.837252 |
2.238883 |
2.639965 |
2 |
0.252885 |
0.531077 |
0.873868 |
1.256923 |
1.649990 |
2.046662 |
2.444902 |
3 |
0.250430 |
0.506422 |
0.788239 |
1.119573 |
1.486777 |
1.869373 |
2.259407 |
4 |
0.250064 |
0.501170 |
0.759749 |
1.042309 |
1.364878 |
1.719494 |
2.092150 |
5 |
0.250009 |
0.500022 |
0.752156 |
1.011815 |
1.294536 |
1.609989 |
1.954075 |
6 |
0.250001 |
0.500000 |
0.750332 |
1.000703 |
1.263822 |
1.545459 |
1.854200 |
7 |
0.250000 |
0.500000 |
0.750039 |
1.000000 |
1.252476 |
1.515039 |
1.792684 |
One can see that for a constant j, as k increases, the expected sojourn time E _{ k,j }, tends to j/4 which is the expectation of the Erlang(j,4) random variable. This is not surprising, since as k gets larger, the probability that repaired items will be released only to base 1 during the customer’s sojourn time, is getting closer to 1.
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