Electrical Engineering Network Problems Solutions Analysis

This document presents solutions to a series of network-related problems, focusing on concepts within electrical engineering. The solutions cover calculations for frame size, transmission delay, and propagation delay, addressing scenarios involving bandwidth and packet transmission. The document explores problems related to minimum frame size, probability of frame damage, and the application of Poisson's distribution. Furthermore, it includes detailed analysis of network throughput, considering factors such as bus length, transmission capacity, and packet size. The solutions also delve into the calculation of minimum window size, network utilization, and the impact of propagation delay on network performance. The final sections provide a comprehensive analysis of connection delay, packet transmission time, and the total delay in a multi-hop network, including calculations involving packet size, bit rate, and queuing time. The assignment also analyses the probability of frame damage under different conditions.

Tutorials
Problem 1.
Minimum framesize=Bandwith *delay(transmission)
But delay(transmission)=2*propagation delay.
Propagation delay therefore=(length wire/speed of signal)
=0.000005sec
=5μs
Delay(transmission)=2*2μs=10us
Frame size=20μs**5mbs=100bits
Problem 2
Assuming that the probability that the frame is sent successfully after transmission is equal to the
probability that frames are damages Q-1
Pq=P^(q-1-(q-1))=p^(q-1)(1-p)
The mean transmission number therefore is
The mean transmission therefore is:
¿ ∑
q−1
∞
αpq= 1
1− p
=E(x)=∑
q −1
∞
qpq
= ∑
q−i 0 r Pa
∞
❑
(i-p)∑
α−1
∞
α pq −1
= (1-p)[ 1
1− p + p
1− p ]
E(x)=( 1
1− p )

Problem 3
Using the poisons distribution for random packet data
P(x=x)=e^- yx
x ! Where x=0
Where e=2.718282
Problem 4
Stations=100
Distribution=Distance =4kms
Rate=5Mps
Date packet=1000 bits
Propagation time=5mbs/km
Distance between stations= 4000
100 40m
(i)
Maximum number of packets present=Bandwith *Delay
Delay=2*propagation
=2(length of the wire/speed of the signal)
=2(4*5)/2*0.8
=1.6 μs
Throughout =5mbps*1.6 μs =2bits/s=8bits per second

=8*1000=8000 bits/sec
(ii)if bus length=1km
Delay=2*(0.2 μ sec
km 0.2 μs)=2*0.2μs
=0.4μs
Throughout=5MbPs*0.4μs=2bits/sec
2*1000=2000bits/s
(iii).if the transmission capacity is doubled to 10mBps
Throughout = 10Mbs*1.64s=16 bits/sec
16*1000=16000bits/sec
(iv) if the packets are 10,000 bits long
Throughout=15Mbps *1.6=8 bits
= 8*1000
=8000 bits/sec.

Problem 5
(a) Propagation delay=length between/the speed of the signal
360000/300000
1.2s
(b). propagation delay is 1.2sec
Therefore RTT=2*propagation delay
= 2*1.25
= 2.5sec
Wmin= RTT ∗bandwith
datasize
2.5∗125000
1000 bytes
=300bytes
(c).Elapsed time=file transfer/speed(bps)
= 25MB/1Mbps
25mb=25*8
200Mb
Therefore=200Mb/1mbs
=200sec
Elapsed time=200Mb*10
=2000mbs.
d. With a repeat protocol, Minimum window size,

Utilization =W/1+2k and k=propagation/transamission
k=1.2s/t t=packet size/bit rate
t=80000b/8000bps=10s
k=1.2/10=0.12
Wmin= 2000/1+0.12=2000/1.12=1785mbs
Problem 6
a.
The connection delay per link is 3ms, total number of delay of 10000 characters is
therefore 30000ms
Transmission time for each call request is 100ms, having 10000 characters will call
1000000ms
Each character has 8bit, so 10000 characters will have 80000 bits
Time =80000bits/4800bits per sec=16.67sec=16670ms
The total time dalay therefore will be 16670+1000000+30000ms=1046670 ms
b. The packet transmission time is determines by then packet size/bit rate
From the above question, packet size =P+L
Bit rate =C
Therefore Transmission time=(P+L)/C
c. The total number of packets that can transmit message is determined by bandwidth and
propagation delay against packet length
Total packets therefore=C*D/P
d. From the figure, we have 3packets,the time to transmit one packet of data is (P+L)/C
For the 3 packets given, the transmission time therefore will be
(((P+L)/C)*3)+2Tx+Tq
e.
Total packet =C*D/P
Transmission time = ((P+L)/C)+2Tx+Tq

But Total delay =(((P+L)/C)+2Tx+Tq)/ C*D/P*N
f. Given the following values to calculate the total delay;
L=20
50 bytes of header, C=4800 bps, D=20 msec t
q = 300 msec queuing time at each node, and t
x = 20 msec packetization time (time between packets), N = 3 hops.
But Total delay =(((P+L)/C)+2Tx+Tq)/ C*D/P*N=
Problem 7
The probability that the frame is damaged in different situations of odd and even numbers are as
follows
Given that N is the number of characters contained in a single frame and n is the number of bits
in each frame
.n be odd
1-(1-PN )n
.n be even
1-(2-PN )n
N be odd
1-(1-Pn )N
N be even
1-(1-Pn )N