Introduction to closed-loop queuing theory
In the introduction to queuing theory , the author introduces the basic content of queuing theory in detail. In this paper, the application flow is from outside the system, and its intensity (or density) does not depend on the system itself, nor does it depend on the state of the system. In this article, another queuing theory will be discussed, the intensity of the application flow is related to the state of the system, so it is called a closed-loop queuing system .
1. System scenario
Imagine the following scenario. In a factory, the debugger watches over the nnn machine tools. Each machine tool may break down and stop working at any time, requiring repair by the debugger, and the intensity (density) of the failure isλ \lambdalambda . If the debugger is free at this time, start maintenance, maintenance time
t ˉ mt = 1 μ \bar t_{mt} = \frac{1}{\mu}tˉmt=m1And if the debugger is not idle (in a busy state) when the machine tool fails, the faulty machine tool will join the waiting queue until the debugger comes to repair.
For this scenario, we are generally interested in the following three indicators:
- The probability that the debugger is idle;
- The probability of a queuing queue appearing;
- The average number of machine tools awaiting repair.
In this scenario, the flow of requests comes from the machine tools themselves, which are limited in number and issue requests based on their own status (normal/faulty). Therefore, the strength of the overall application flow naturally depends on how many machine tools are linked to maintenance (whether they are being repaired or waiting for repairs).
Different from the introduction to queuing theory , in the closed-loop queuing system, the number of application streams is limited (the number of machine tools is limited). When the number of application flows is too large, the state of the queuing system itself will hardly affect the state of the application flow, so it can be considered that the two are irrelevant.
2. Mathematical description
In this scenario, according to nnThe number of faults in n
machine tools can define different states of the system: S 0 S_0S0– All machines are working normally, no machine failures;
S 1 S_1S1– There is a machine tool failure, the debugger repairs this machine tool, and other machine tools work normally;
S 2 S_2S2– There are 2 machine failures, the debugger repairs one, and the other waits, while the other machines work normally;
⋮ \vdots⋮
S n S_n Sn– all nnAll n machine tools are faulty, the debugger repairs one, and the rest( n − 1 ) (n-1)(n−1 ) Waiting in line.
The state diagram is shown in the figure.
From state S 0 S_0S0to S 1 S_1S1In the process, initially all machine tools are working, so the intensity (density) of the process is n λ n\lambdanλ . Similarly, from stateS 1 S_1S1to S 2 S_2S2In the process, initially ( n − 1 ) (n-1)(n−1 ) One machine tool works and one machine tool fails, so the intensity (density) of the process is( n − 1 ) λ (n-1) \lambda(n−1 ) λ . The process of transforming from low occupancy to high occupancy can be deduced by analogy.
And since there is only one debugger, only one machine tool can be repaired at a time, so the strength is the same when changing from high occupancy to low occupancy every time, which is μ \mum .
Here the different states S i S_i are givenSiFunctional equations:
p 1 = n λ µ p 0 p 2 = n ( n − 1 ) λ 2 µ 2 p 0 ⋮ pn = n ( n − 1 ) ( n − 2 ) ⋯ 2 ⋅ 1 ⋅ λ n µ np 0 p 0 = 1 1 + n ( λ / µ ) + n ( n − 1 ) ( λ / µ ) 2 + ⋯ + n ( n − 1 ) ⋯ 2 ⋅ 1 ⋅ ( λ / µ ) n p_1 = \frac { n \lambda }{\mu} p_0\\p_2 = \frac{n(n-1) \lambda^2}{\mu^2}p_0 \\\vdots\\p_n = \frac{n(n-1) 1) (n-2) \cdots 2 \cdot 1 \cdot \lambda ^n}{\mu^n} p_0 \\p_0 = \frac{1}{ 1 + n \left( \lambda / \mu \right ) + n (n-1) \left( \lambda / \mu \right)^2 + \cdots + n (n-1) \cdots 2 \cdot1 \cdot \left( \lambda / \mu \right)^ n }p1=mn.mp0p2=m2n(n−1 ) l2p0⋮pn=mnn(n−1)(n−2)⋯2⋅1⋅lnp0p0=1+n( l / m )+n(n−1)( l / m )2+⋯+n(n−1)⋯2⋅1⋅( l / m )n1任
λ / μ = ρ \lambda / \mu = \rhol / m=ρThen the above formula can be written as
p 0 = 1 1 + n ρ + n ( n − 1 ) ρ 2 + ⋯ + n ( n − 1 ) ⋯ 2 ⋅ 1 ⋅ ρ n p_0 = \frac{1}{ 1 + n \rho + n (n-1) \rho^2 + \cdots + n (n-1) \cdots 2 \cdot1 \cdot \rho^n }p0=1+nρ+n(n−1 ) p2+⋯+n(n−1)⋯2⋅1⋅rn1p 1 = n ρ p 0 p_1 = n \rho p_0p1=nρp0 p 2 = n ( n − 1 ) ρ 2 p 0 ⋮ p n = n ( n − 1 ) ⋯ 2 ⋅ 1 ⋅ ρ n p 0 (1) p_2 = n (n-1) \rho^2 p_0 \\ \vdots \\ p_n = n (n-1) \cdots 2 \cdot 1 \cdot \rho^n p_0 \tag{1} p2=n(n−1 ) p2p0⋮pn=n(n−1)⋯2⋅1⋅rnp0In formula ( 1 ) , the strength of entering each square is multiplication, and the strength of going out of each square is division.
For this system, absolute passability refers to the average number of failures per unit time . The probability when there is no failure is p 0 p_0p0, then obviously, the probability of being faulty (and also the probability that the debugger is not idle) is
P = 1 − p 0 P = 1 - p_0P=1−p0If the debugger is busy, the number of machine tools he can repair per unit time is μ \muμ , so the absolute passability is
A = ( 1 − p 0 ) μ A = \left( 1 - p_0 \right) \muA=(1−p0)μ and the probability that the debugger is idle is the probability that no fault occursp 0 p_0p0。
The average number of faulty machine tools can be calculated with the following expectation
w ˉ = 1 ⋅ p 1 + 2 ⋅ p 2 + ⋯ + n ⋅ pn = ∑ i = 1 ni ⋅ pi \bar w = 1 \cdot p_1 + 2 \cdot p_2 + \cdots + n \cdot p_n = \sum_{i=1} ^ni \cdot p_iwˉ=1⋅p1+2⋅p2+⋯+n⋅pn=i=1∑ni⋅piThis value can also be calculated by absolute passability. We know that the failure intensity of each machine tool is λ \lambdaλ ; meanwhile, on average the system will work normally with( n − w ˉ ) \left( n - \bar w \right)(n−wˉ )machine tools; thus the number of failures of these normal machine tools should be( n − w ˉ ) λ \left( n - \bar w \right) \lambda(n−wˉ)λ ; according to the definition of absolute passability, there is
( n − w ˉ ) λ = ( 1 − p 0 ) μ \left( n - \bar w \right) \lambda = \left( 1 - p_0 \right) \mu(n−wˉ)l=(1−p0)The µ functional value
w ˉ = n − µ λ ( 1 − p 0 ) = n − 1 − p 0 ρ \bar w = n - \frac{\mu}{\lambda} \left( 1 - p_0 \right); = n - \frac{1-p_0}{\rho}wˉ=n−lm(1−p0)=n−r1−p0Next, calculate the number of machine tools waiting to be repaired in the waiting queue. Let the number of all machine tools related to "repair" be WWW , including the number RRwaiting in the queueR and the number of Ω being repaired\OmegaΩ :
W = R + Ω W = R + \OmegaW=R+Ω Obviously, since there is only one debugger,Ω \OmegaΩ can only take 1 or 0. Correspondingly, its expectation is
Ω ˉ = 0 ⋅ p 0 + 1 ⋅ ( 1 − p 0 ) = 1 − p 0 \bar \Omega = 0 \cdot p_0 + 1 \cdot \left(1 - p_0 \right) = 1 - p_0Ohˉ=0⋅p0+1⋅(1−p0)=1−p0Then the average number of machine tools waiting to be repaired in the queue is
R ˉ = w ˉ − Ω ˉ = n − 1 − p 0 ρ − ( 1 − p 0 ) = n − ( 1 − p 0 ) ( 1 + 1 ρ ) \bar R = \bar w - \bar \Omega = n - \frac{1 - p_0}{\rho} - \left(1 - p_0 \right) = n - \left(1 - p_0 \right) \left ( 1 + \frac{1}{\rho} \right)Rˉ=wˉ−Ohˉ=n−r1−p0−(1−p0)=n−(1−p0)(1+r1) Finally, let’s look at indicators related to production capacity. Let the production capacity of a machine tool belll , then the capacity loss caused by failure per unit time can be calculated as
L = w ˉ l L = \bar wlL=wˉl