ch9

Question 1

M/G/1 PS for single class: formula mean (conditional)
sojourn time, slow-down factor
Understand the formulas, interpret slowdown factor 1/(1-ρ)

Answer 1

Each user in the system gets equal part of the capacity
With k users this is 1/k
Mean Sojourn time: The mean time that an arbitrary job spends in the system (models response time of the system)
Expected sojourn time = B / (1 - p)
Expected conditional sojourn time = x / (1 - p) where p is the load and x is a specific job
Speed slows down as more users enter the system. slowdown factor = 1/(1-p)

Question 2

M/G/1 PS for multi-class: product-form solution

Know what product-form formula means + relation to HSDPA

Answer 2

Product-form solution, gives the joined probability distribution of the number of jobs (of the different classes) in the system

Mean service time in a further away ring is larger
This is used in HSDPA for each of the circles of the onion model
The capacity of a specific circle/ring is equally divided using the processor sharing model
Where each ring is a separate class in the multi-class model

Question 3

Little’s formula

Know what it is, and what it can be used for

Answer 3

When a customer leaves the system.
The number of customers that are in the system when that customer leaves, is: Lambda * average time you spent in system

When a customer leaves the queue and enters the system
The number of customers that are in the queue is: lambda * waiting time in the queue

Little’s formula: E[Lq] = lambda * E[W]
Where Lq = number of jobs in queue
W = time spent waiting in queue

Using Markov chains we can calculate Lq, lambda is known, hence we can calculate the expected waiting time in the queue

Question 4

M/M/c (“Erlang-c” model)

Know what it is and understand Markov chain analysis

Answer 4

c parallel servers
Load per server: p = lambda * beta / c = lambda / (c * mu)

Departure rate is number of jobs * mu,
until you reach c jobs, then the departure rate stays c * mu

Question 5

Bus paradox, length biasing

Understand the idea

Answer 5

On specific arrivals:
Say on average a bus comes twice every hour
Arrival exactly every 30 minutes
Mean waiting time is 15 minutes

Uncertain arrivals:
Interarrival time either 50 minutes or 10 minutes
Mean interarrival time = 0x5 * 50 + 0x5 * 10 = 30 minutes (the same)
Mean waiting time is 5/6 * 25 + 1/6 * 5 = 130/6 = 21.67 minutes

Since there is a larger chance that you enter in the bigger interarrival time, and the average waiting time in that interval is larger,
the average waiting time overall is larger.
The waiting times do not cancel each other out (between the smaller and bigger interarrival times)

Question 6

M/G/1 formula for E[W]

Understand that variability in service time increases E[W]

Answer 6

Since there is a larger chance that you enter in the bigger interarrival time, and the average waiting time in that interval is larger,
the average waiting time overall is larger.
The waiting times do not cancel each other out (between the smaller and bigger interarrival times)

High variability means a long waiting time
This means there is no insensitivity to the service time
Because the waiting time increases due to other users taking longer to exit the system

“Old lady in front of counter, may cause longer waiting times in the queue”

Question 7

M/G/1 with priorities (preemptive and non-
preemptive)
Understand difference preemptive and non-preemptive

Answer 7

Preemptive resume:
If a low priority packet is in service, and a high priority packet come in:
The low priority packet is interrupted
Strict priority!

Non-preemptive resume:
High priority packets are served first, but low priority packets are not interrupted when a high priority packet comes in during sending