Telecommunications Network Assignment II - Desklib

This article provides solutions to Telecommunications Network Assignment II. It covers topics like link transmission rate, node separation, receiver buffer size, packet delay, and more. It also includes information on the required buffer size and average packet delay introduced by the router. The article is relevant to students studying ELEC3500 Telecommunications Network in S2, 2018 at any college or university.

ELEC3500 TELECOMMUNICATIONS NETWORK
S2, 2018
ASSIGNMENT II
STUDENT NAME
STUDENT ID NUMBER
INSTITUTIONAL AFFILIATION
LOCATION (STATE, COUNTRY)
DATE OF SUBMISSION

PART A
Question 1
Link transmission rate --2.048mbps
Node separation --200km
Receiver buffer size --1.00mB
Information frame size --2500 bytes
Receiver informatino processing rate --10kbps
Xon/Xoff control packet size --80 bytes
(a) Value of P in a lossless communication while offering full utilization of the receiver
buffer.
(b) Calculate the time interval between the Xoff and Xon with signals transmitted by the
receiver when the buffer occupancy level goes down to 70% or below.
Solution
Every N downstream departure,
DD ( t 1 ) =DD ( t 0 ) + N
CC =RTT∗Rpk + DD ( t−tf ) −UD (t)
Question 2
1 frame length is used on the link is 400 bytes
20 byte header
ACK frame length is 20 bytes
2000 bytes 11 bits are corrupted due to the channel error
Solution
The transmission efficiency is given as,
α=
(1− S0
Sf )
1+
( Sa
Sf )+ 2 (t¿ +tproc ) r
Sf
s0=sa=200

For the stop and wait efficiency,
s0 =sa=t proc=0
At a 50% transmission delay of the ACK,
r =4 kbps , t¿=20 ms
The ACK frame size with a header/CRC overhead = 1250 bytes
W s ≥ (2 t¿ )∗r
Sf
=11bits for 1 frame
The emission delay for one frame,
¿ 1250 x 8
1 mbps =10 ms

The automatic repeat request and the positive acknowledgement with retransmission,
d= d 1
2
3 c
= 20 km
2e8
d∗2+ safet yconstant
T transmit = L
R = 200 kbppkt
10∗¿ 9 bps =2 μs
Delay=ttransmit + T¿+ Tqueue
The utilization shows the fraction of time the sender is busy sending,
U sender= ( L
R )
RTT + L
R
= 0.008
30.008 =0.0027

Question 3
8000 frames/sec
16 time slots/frame
8 input traffic sources
Output link data rate 6.144mbps
Data activity factor, α=0.7
Link distance =100km
Input packet arrival (inter arrival time of 2ms)
Solution
For a statistical time division multiplexer,
K= M
IR

α < K <1
λ=αIR
T s= 1
M
The average arrival rate is the total potential input with the fraction of time that each source is
transmitting. The service time is the time it takes to transmit one bit,
ρ=λ T s= αIR
M = α
K = λ
M
Question 4
Statistical multiplexer 20 different input links
Exponential packet interval arrival 10ms average value for each input
Output link operating at 10mbps
Input packets sizes; average value of 100 bytes
Multiplexer has a maximum buffer length 500 bytes
Solution

The average delayed message,

Using the binomial closed form
Question 5
LAN- four terminals connected to a hub operating at 100mbps.
Frame length = 2000 byte
Link is 125m long
LAN uses a fixed jamming interval of 2.4 μ sec.
Solution
The LAN delay determines the arrival rate and its packet processing speed using the queuing
delay. It increases with time and as the length of the queue is unbounded, the latter packets
experience a very high delay,

Question 6
Solution
In hub-polling, the monitoring channel is provided to indicate the appropriate station that it
started transmitting. It is the go-ahead is transmitted directly from one station to another. The
length of average cycle time,
T c=M [ Nm X
R +ω ]
From the input values,
Nm =λ T c
T c= Mω
1− MλX
R

The throughput is given as,
S= MλX
R
It is given as,
T c= Mω
1−S
The random arrival occurs in intervals of T c ( 1− ρ )
On average the delay is,
Tc ( 1−ρ )
2
W1= Mω ( 1−ρ )
2 ( 1−Mρ )
For total delay,

PART B
Calculate the traffic and switch parameters
(i) Calculate the aggregation factor used to generate the 1280 byte payload used to
generate the VoIP packets.
(ii) Total load generated at the input of the queue by the incoming data and voice traffic
arriving from all terminals. Calculate the traffic load in bits/sec

(iii) Find out the required buffer size (in packets) that will keep the average packet loss
below 1.5%. Write an iterative code to calculate the buffer size.
(iv) What is the average packet delay introduced by the router
(v) Traffic load increases by 10 % compared to the average traffic level as calculated in
(i). Calculate the packet loss probability, average queue size and packet delay.
(vi) Calculate the average throughput (in packets/sec) and goodput (in bits/sec) of the
router.
SOLUTION
tota lpacket ¿¿=( laye r 2header ) +( IP/ UDP/RT Pheader )+ ¿¿¿
tota lpacket ¿¿=18+ 40+ 1280¿ ¿
tota l packet ¿¿=1338bytes ¿¿
tota lpacket ¿¿=1338bytes∗8 bits per byte=10,704 bits ¿¿
Voice payload size,
¿ ( 160 bytes default voice payload )∗8 bits per byte
¿ 1280 bits
Packe t persecond
= code cbit rate
voic e payload¿ ¿¿ ¿
¿ 64
1280 =50 pps
Bandwidt hper call
=¿
¿ 1744∗50=87.2 kb/s
To obtain the maximum throughput of the network,
¿ 10 ( p−h )
p Mbps
The latency of the network is the time required to get the first byte of data across a network. For
an entire packet to be transmitted even for 1 byte, the latency of the network is the time to
transfer 1 packet. The latency of the network is,
¿ 4.(1+ 2e-4 p)ms

The bandwidth delay product measures the required buffer that is based on its average round trip
time and link capacity. The bandwidth delay product to determine the buffer size state size,
B=RTT xC
RTT- average round trip time
C-capacity of the limit
For small buffers and the router backbone design in the order of magnitude is given as,
B=RTT x C
√ N
To obtain the buffer size,
B= ( μ . RTT ) 2
32 n3
n−number of TCP connections
The network and window size of tiny buffers can have buffer size is:
B=O ( log W ) … ..W −window ¿ ¿
For a packet loss of below 1.5%, using the Poisson process to determine the probability,
Pr (X >m+ n∨x >m)=Pr ( X > n )
The average inter-arrival time is
¿ 1
λ
Composed of a collection of { N ( t ) :t ≥ 0 } random variables where the N ( t ) is the number of
events time t. N ( t ),
f ( n; λ )= λn e− λ
n !

P [ ( N ( t+ τ )−N ( t ) )=k ]= e−λτ ( λτ )k
k ! k=0,1,2… .
¿
e−( 5
60 )(16)
( ( 5
60 )16 )5
16 !
¿ 0.17347 % packet loss
For a stable system, the utilization factor is given as,
p= λ
μ p<1( stable system)
To determine the average packet delay including the transmission plus queuing,
T = 1
μ +Q
Following the little’s law,
N= λT ∧N q= λQ
P {n packets }=p n (1− p)
T = N
λ = p
λ ( 1− p )= 1
μ− λ
The average time in the queue according to Pollaczek-kinchine formulae,

¿
λ ( σ2+ 1
μ2 )
2 ( 1− p )
The average delay is given as,
¿ 1
μ +
λ (σ2 + 1
μ2 )
2 ( 1− p )
These two equations determine the steady state occupancy of the probability and as the σ 2
increases, the delay further increases.
REFERENCES
H. Han, S. Shakkottai, C. V. Hallot, R. Srikant and D. Towsley, “Multipath TCP: A joint
congestion control and routing scheme to explore path diversity in the internet,” IEEE/ACM
Transactions on networking, vol. 14, pp.1260-1271, Dec. 2006.