Queueing Flashcards
population arrival
infinite generation following a poisson distribution: random but with a fixed mean rate. interarrival times follow an exponential time
queue
where the population wait before being served
discipline, how someone from the queue is selected, random, first (usual way), …
queue length is number waiting
servers
serve the population, can be in parallel (then completely served by it) and can in sequence (going from one server to another). here usually its 1 single sequence system
the time stayed at a server is the service time, usually exponential distribution. here independent and identically distributed
state of system
n population in system
steady state
at first system affected by initial state: transient condition
then with sufficient time it reaches steady state
rate diagram
the states are the number of people in the queue. the more people can be in the system & queue the longer you make it
arrow in are the numbers entering, arrow out are the numbers leaving
the servers should be at the beginning, 1 node 1 server and how many can handle, 2nd node 2nd server and how many can handle together with 2 and so on, the last nodes are the waiting nodes (with the same arrows as the last server)
computing stationary distribution from the rates
rate in = rate out for each state
=> rate * pi where arrow comes from
sum rates = 1
there will always be one redundant equation that you can remove (can cancel terms from each other by comparing equations)
try to solve equations by reducing terms and having only 1 remaining unknown pi for all (the same one for each equation). eg: pi_1= 3/2 pi_0 & pi_2 = 1/2 pi_0
Put all fraction over the same denominator X. The remaining pi will be X/X, eg pi_0= 2/2 . Sum numerator for all pi to find new denominator which make sum of fraction = 1, eg 3+1+2 = 6 => pi_0=2/6,pi_1=3/6,pi_2=1/6.
Now put all fractions found in pi=(pi_1,pi_2,…), eg: (2/6,3/,6,1/6)
getting probas(compute time in each state then compute stationary with them) from rates
transform the rates such that the sum of transition from each node = 1
time spent in a node: unit of time/ how many leave(go to next and previous)
take the probas and multiply by expected time at node
how many expected people in the system
sum_i pi_i * i
how long will it take to reach y from x
compute the probas (transitions from x to i. Do fractions, customers leaving to i / total customers leaving from x)
compute how much time spent at each node (full queue time / people leaving)
do Ex equations ( proba of going to state i from x * E_i), instead of 1 + E do time spent at node + E
solve for Ex and have Ey = 0