Networking question and solution(3)

Q1 How many 8-bit characters can be
transmitted per second over a
9600 baud serial communication
link using asynchronous mode of
transmission with one start bit,
eight data bits, two stop bits and
one parity bit?

 

So, transmission rate here 
=9600bps
An eight bit data (which is a char) requires 
 start bit, 
 stop bits and 
 parity bit 
=12bit

So, number of characters transmitted per second 
=9600/12=800bps.
.



        

Q2 . Determine the maximum length of
the cable (in km) for transmitting
data at a rate of 500 Mbps in an
Ethernet LAN with frames of size
10000 bits. Assume the signal speed
in the cable to be 200000 km/s.

Transmission time>=propogayion time
Transmission time=framesize/bandwidth
Propogation time=2*distance/speed
10000/500*10^6=2*x/2*10^8
x=2km

Q3 . In the IPv4 addressing format, the
number of networks allowed under
Class C addresses is

In class c address
Net I'd =24bit
Host I'd =8bit
3 bit for class representation 110
21 bit make 2^21 network.

Q4 .A link has a transmission speed of
106 bits/sec. It uses data packets of
size 1000 bytes each. Assume that
the acknowledgement has negligible
transmission delay and that its
propagation delay is the same as the
data propagation delay. Also assume
that the processing delays at the
nodes are negligible. The efficiency
of the stop-and-wait protocol in this
setup is exactly 25%. The value of
the one-way propagation delay
(in milliseconds) is 

Transmission time Tt=1000*8/10^6=8/10^3 sec=8 ms

efficiency =25%=1/4=Tt/(Tt+2*Tp)

8/(8+2*Tp)=1/4

Tp=12

so propagation Delay is 12 ms

  

Q5 Consider an IP packet with a
length of 4500 bytes that includes
a 20-byte IPv4 header and a
40-byte TCP header. The packet is
forwarded to an IPv4 router that
supports a Maximum Transmission
Unit (MTU) of 600 bytes. Assume
that the length of the IP header in
all the outgoing fragments of this
packet is 20 bytes. Assume that
the fragmentation offset value
stored in the first fragment is 0.
The fragmentation offset value
stored in the third fragment is

MTU=600B
MTU payload=600-20=580B
But payload should be multiple of 8 so number nearest to 580 and multiple of 8 is 576
So MTU payload=576B
Size of offset=576/8=72
fragmentation offset value stored in first fragment =0
fragmentation offset value stored in second fragment=72
fragmentation offset value stored in third fragment=144