Which of the following matching is true concerning the Protocol Data Unit (PDU) and its corresponding OSI layer location?
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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
Explanation:
Given that:
A single channel with a capacity of 64 kbps.
Average packet waiting time 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
= 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