Variable Independiente

In this subsection, we study a system with two branches and three service stations as shown in Figure 3.5. Stations 2 and 3 are in series and parallel to station 1 as a whole. A job after being served at station 0 will join station 1 with probabilityp1 or station 2 with probability

p2. This queueing system is motivated by the nurse staffing problem studied in Yankovic

and Green [35]. We consider this closed queueing system as a hospital ward, where a patient seeks admission (station 3), then stays at a bed (station 0), then requests nursing service (station 1) and returns back to his/her bed after receiving service (this process may repeat for several times), and at last seeks discharge (station 2). We assume that a new patient

comes to the ward immediately after a patient is discharged. We defineXi to be the random

variable denoting the i.i.d. service time at station i for i = 1,2,3. We provide a partial

Figure 3.5: A closed queueing system with an automated station and three service stations.

characterization of the optimal policy that maximizes the long-run average throughput of the system in Theorem 3.3.3.

Theorem 3.3.3. If X1[X3]≤lr X3[X1], then there exists an optimal policy that gives the highest priority to station 1[3].

Theorem 3.3.3 partically characterizes the optimal policy: the faster station between the two stations which directly connect to the entry of the automated station should be prioritized, if the service times at these two stations are ordered according to likelihood ratio ordering.

Proof of Theorem 3.3.3. We show that ifX1≤lrX3, then there exists an optimal policy

that gives the highest priority to station 1. The proof that an optimal policy gives the highest priority to station 3 if X3 ≤lrX1 is similar.

Suppose policyπ is a policy under which there exists at least one decision epoch where the highest priority is not given to station 1. Specifically, let ε be the first time policy π

does not give priority to station 1 even if there is a job in station 1. Suppose {j1, . . . , jm}

(j1, . . . , jm∈ {2,3}) gives the sequence of stations that the server visits after timeεbefore

it visits station 1 under policyπ. We will next construct a new policyγ which serves station 1 right before it serves the last job at station jm.

Let τ be the time under policy π that the server starts to work on a job at station 2 with service time S2 right before it moves to station 1. Then after completing serving this

job, the server immediately switches to station 1 to serve a job there with service time S1. We construct the new policy γ as follows: γ followsπ during [0, τ) and then serves a job at station 1 with service time S₁0, switches to station 2, and serves a job with service time S₂0. We directly couple the service times of all jobs taken into service during [0,∞), which yields S₁0 = S1 and S20 = S2. Let S0 be the service time at station 0 for the job entering

station 0 at timeτ+S2+S1under policyπ and for the job entering station 0 at timeτ+S₁0

under policy γ. γ followsπ during [τ +S1+S2,∞). This is possible because same jobs or

more are available to policyγ compared to policyπ. The system states for the two policies will become identical after time τ +S2+S1 +S0. The problem can be analyzed in three

intervals as follows: D₀γ(t) =          Dπ₀(t), 0≤t < τ+S₁0 +S0, Dπ₀(t) + 1, τ +S₁0 +S0 ≤t < τ+S2+S1+S0, Dπ₀(t), t≥τ+S2+S1+S0. Therefore, T H₀γ(t)−T H₀π(t) = lim inf t→∞ 1 t[D γ 0(t)−D0π(t)]≥0, for all t≥0. Case 2. (jm = 3)

Let τ be the time under policy π that the server starts to work on a job at station 3 with service time S3 before it moves to station 1. Then after completing serving this job,

the server immediately switches to station 1 to serve a job there with service time S1. We construct a new policyγ as follows: γ followsπduring [0, τ) and then serves a job at station 1 with service time S₁0, switches to station 3 and serves a job with service timeS₃0.

We directly couple the service times of all jobs taken into service during [0, τ) and we cross coupleS1,S3,S₁0,S₃0 as in the proof of Theorem 3.3.2 by first generating the minimum and maximum of S1 and S3, namely m and M, respectively, conditioning on their values

and using these values in both policies. We need to consider three couplings: i. S1 =m,S3=M,S10 =M,S30 =m,

iii. S1 =m,S3=M,S₁0 =m,S₃0 =M,

all of which yieldS₁0 ≤S3 (andS1 ≤S₃0). In the first two cases all arrival times to station 0 for policiesπ and γ are identical. The system states for the two policies are identical after time τ +m+M. Hence, policy γ can follow policy π thereafter. By directly coupling the service times of all jobs taken into service after τ+m+M, we find that the throughput of the system is the same for policy π and γ.

In the third case, letS0 be the service time at station 0 for the job entering station 0 at

time τ +S3 under policyπ and for the job entering station 0 at time τ +S₁0 under policy

γ. Consider the following two sub-cases:

1. 0 < S0 ≤m. We directly couple the service times of all jobs taken into service after τ +m+M. The system states for the two policies will become identical after time

τ +m+M. Hence, policy γ can follow policy π after τ +m+M.

2. S0 > m. We directly couple the service times of all jobs taken into service after τ +m+M. The system states for the two policies will become identical after time

τ +M +S0. However, policy γ can follow π after τ +m+M because same jobs or

more are available to policy γ compared to policy π.

In both sub-cases, the problem can be analyzed in three intervals as below.

Dγ₀(t) =          D₀π(t), 0≤t < τ+m+S0, D₀π(t) + 1, τ+m+S0 ≤t < τ+M+S0, D₀π(t), t≥τ +M +S0. Therefore, T H₀γ−T H₀π = lim inf t→∞ 1 t[D γ 0(t)−D π 0(t)]≥0, for all t≥0.

We have shown that a job at station 1 should be served ahead of a job at station 3 if jobs are available at both stations.

If we follow the same argument iteratively for the two cases, we can show that a job at station 1 should be served ahead of the sequence of jobs {j₁, . . . , jm−1}. In other words, the

optimal policy which maximizes the throughput of the system gives the highest priority to station 1.

Furthermore, we provide a corollary to Theorem 3.3.3 whenX3≤lrX1.

Corollary 3.3.1. Suppose that X3 ≤lrX1. IfX1 ≤lr X2+X3, then there exists an optimal policy that gives priority to station 1 over station 2. If X1 ≥_lr X2+X3, then there exists an optimal policy which gives priority to station 2 over station 1.

Proof. SinceX3 ≤lr X1, Theorem 3.3.3 tells that we need to only consider policies that give

the highest priority to station 3 whenever a job visits that station. This implies that we need to only consider policies that serve stations 2 and 3 sequentially, i.e., we can think of stations 2 and 3 as single station with service timeX2+X3. Then, the result follows from Theorem 3.3.2.

Corollary 3.3.1 gives conditions under which two priority policies are optimal: 1) If

X3 ≤lr X1 ≤lr X2+X3, then the priority order is station 3, station 1 and station 2; 2) If

X3 ≤lr X1 andX2+X3 ≤lr X1, then the priority order is station 3, station 2 and station 1.