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Computer Networks - Queues

A buffer is filled over a single input channel and emptied by a single channel with a capacity of 64 kbps. Measurements are taken in the steady state for this system with the following results:
Average packet waiting time in the buffer = 0.05 second
Average number of packets in residence = 1 packet
Average packet length = 1000 bits
The distributions of the arrival and service processes are unknown and cannot be assumed to be exponential!
What are the average arrival rate \lambda in units of packets/second and the average number of packets w waiting to be serviced in the buffer?
The text in bold basically tells me that this can't be modeled as an M/M/1 so I have to use Little's Law and the Conservation of Time and Messages formulas. How do I solve this using them? Any help would be great!
Formulas:
Conservation of Time and Messages

1 Answer

2 votes

Answer:

the average arrival rate \lambda in units of packets/second is 15.24 kbps

the average number of packets w waiting to be serviced in the buffer is 762 bits

Step-by-step explanation:

Given that:

A single channel with a capacity of 64 kbps.

Average packet waiting time
T_w in the buffer = 0.05 second

Average number of packets in residence = 1 packet

Average packet length r = 1000 bits

What are the average arrival rate \lambda in units of packets/second and the average number of packets w waiting to be serviced in the buffer?

The Conservation of Time and Messages ;

E(R) = E(W) + ρ

r = w + ρ

Using Little law ;

r = λ × T_r

w = λ × T_w

r / λ = w / λ + ρ / λ

T_r =T_w + 1 / μ

T_r = T_w +T_s

where ;

ρ = utilisation fraction of time facility

r = mean number of item in the system waiting to be served

w = mean number of packet waiting to be served

λ = mean number of arrival per second

T_r =mean time an item spent in the system

T_w = mean waiting time

μ = traffic intensity

T_s = mean service time for each arrival

the average arrival rate \lambda in units of packets/second; we have the following.

First let's determine the serving time T_s

the serving time T_s
= (1000)/(64*1000)

= 0.015625

now; the mean time an item spent in the system T_r = T_w +T_s

where;

T_w = 0.05 (i.e the average packet waiting time)

T_s = 0.015625

T_r = 0.05 + 0.015625

T_r = 0.065625

However; the mean number of arrival per second λ is;

r = λ × T_r

λ = r / T_r

λ = 1000 / 0.065625

λ = 15238.09524 bps

λ ≅ 15.24 kbps

Thus; the average arrival rate \lambda in units of packets/second is 15.24 kbps

b) Determine the average number of packets w waiting to be serviced in the buffer.

mean number of packets w waiting to be served is calculated using the formula

w = λ × T_w

where;

T_w = 0.05

w = 15238.09524 × 0.05

w = 761.904762

w ≅ 762 bits

Thus; the average number of packets w waiting to be serviced in the buffer is 762 bits

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