School of Computer Science and Engineering
Digital Assignment – I, Mar 2021
B-Tech Computer Science and Engineering
Course Code: CSE 1004
Course Name: Network & Communication
Slot: D2
Name: Ramadugula Naga Sai Venkat
Registration number: 19BDS0017
1. An XYZ software limited located at tech park have
implemented a star topology in their network branch. A
centralized switch has installed and configured to connect the
whole systems that exist in the environment. In the training
department cell, all systems are connected in closed-loop
format. A network admin wants to connect these two networks
so that data can be shared among various divisions. Describe
the following questions with the necessary illustration a. In this
case, what network topology would be more appropriate? b.
Create a suitable topology for the scenario described above. c.
For this case, what kind of network would you recommend?
Ans:
The 2 branches one with star topology and the other with the
closed format can be connected in a single network with the
help of a centralized switch using star topology. It costs less as
only 2 extra cables are required and the connection is less
complex so easy to establish.
As it has ring and star topology both it is a hybrid network.
It looks as below
2. List the devices around you that are not currently part of a
computer network?
1) Air conditioner
2) Tube light
3) Coffee machine
4) Micro oven
5) Freezer
6) Television
7) Washing machine
8) Ceiling fan
9) Dustbin
3. Could you design various types of networks to link these
devices? List and describe the "new" computer networks that
you were able to build.
⮚ Air conditioner could be linked to the PC or laptop in this way,
whenever PC is switched on Air conditioner will be switched on
too unless the temperature at that instant is less than 24 degrees
Celsius
⮚ Tube light could be linked to PC and television if any one of them
is switched on after 6pm then proper lighting is required so tube
light will be switched on
⮚ Coffee machine could be linked to television in this way whenever
the television is switched on after 6pm on weekdays it will make
coffee automatically
⮚ Micro oven could be linked to television in the same way as coffee
machine it makes popcorn instead if oven is not empty ⮚ Freezer and
washing machine could be connected to mobile phone so that they
can be controlled from a remote places ⮚ Ceiling fan could be
connected to mobile phone in this way if the distance between
phone and ceiling fan is more than a certain number it means the
person should be outside the room so fan will be switched of
otherwise it will be switched
⮚ Similar to ceiling fan the lid of dustbin can be opened if the
distance from the smart phone is less than certain limit otherwise
it will be closed
A simple star topology of above network connected with a hub
4. We transmit messages directly between two servers 6,000
kilometers apart via a geostationary satellite positioned exactly
between them at a distance of 10,000 kilometers from Earth.
This network receives data at a rate of 100 Mb/s. i) Calculate
the propagation delay if data travels at speed of 2.3*108 m/s.
ii) Determine how many bits were in transit during the
propagation delay. iii) Calculate the time it takes to send 10
bytes of data.
(i)
t propagation = distance/speed
So form the above diagram we can see that
Distance(AC) = 2 x sqrt[(10000)^2 + (3000)^2]
= 2 x 10440.30
= 20880.61 Km
= 20880610 m
Therefore,
t propagation = 20880610 m / (2.3 x 10^8) m/s
= 0.09078 sec
Therefore,
t propagation = 90.78 ms
(ii)
Number of bits transmitted = transmission rate x t propagation
= 100 x 10^6 x 0.09078
Number of bits = 9.07 M bits
(iii)
Time taken to spend 10 bytes in transmission = (10 bytes x 8)/ (100 x
10^6)
= ( 8 x 10^-7 sec)
5. Interpretation Go back N ARQ (Sliding Window
Configuration) for the following Scenario. a) In the case of m=3,
Calculate the Sender and Receiver Window Sizes. b) Assume
that the sender wants to transmit 25 frames. c) Assume that
Frames 8 and 12 of the transmission are unavailable. (Frames
16 and 18 have no acknowledgement).
Part a)
The window size for sending packets is 3, that is three packets
can be transferred at one time before expecting any
acknowledgement of the first one as m is given 3 in this case
and receiver window size is always 1. These frames are referred
to with the sequence numbers to distinguish between them.
Part b)
If the sender wants to send 25 data packets then all the packets
are sent in the window of three frames that is collectively three
frames can be sent to the receiver.
If the receiver acknowledges frame no. 1 that it has been
received by the receiver, then the sender will send the next
packet that is frame number 4. But the window size will remain
three only. No more than 3 packets can be sent.
If the receiver sends the acknowledgement of frame 2 then the
sender sends the next frame that is frame number 5 and so on.
This process will be repeated till the window reaches packet
number 25.
Part c)
As frames 8 and 12 are not present in the whole set of the 25
frames then, the window will consider the next frame to
transmit to the receiver. In the case of frame number 8, it
would be the 9th frame number and in the case of the 12th
frame, the frame transferred would be the 13th one.
When there isn't any acknowledgement for frame number 16
to the sender then, the sender will go back to the previous
frame numbers and again retransmit the frames to the sender.
In this case, the previous window frame contains the numbers
15, 19 and 17.
Similarly in the case of frame number 18. The sender doesn't
get the acknowledgement for it then, the sender's window will
go back to the previous frame numbers which contain 17, 18
and 19 frames and retransmit to the receiver again.
Explanation:
Go-Back -N ARQ follows the principle of pipelining protocol
which can send multiple frames even before receiving the
acknowledgement of the first frame.
In this, the frames are labelled with a sequential number as this
approach sends the multiple frames at a single point of time to
make difference between them.
The number of packets sent to the receiver depends on the
window size of the sender.
If the acknowledgement is not sent for any of the frames then,
the current window containing all the frames are retransmitted
to the receiver.
Part a)
In the given scenario m = 3, so the sender's window size is 3
that is it can transfer the frames in one go without waiting for
the first frame's acknowledgement. And the receiver's window
size is always 1 as the receiver receives only one frame at a
time and sends an acknowledgement to the sender after which
the sender sends the next frame.
Part b)
As the sender wants to send the 25 frames to the receiver and
the window size of the sender is 3 so at one time only 3 packets
can be transferred.
Initially, the first three will be transmitted and then the receiver
sends the acknowledgement to the sender which gives the
sender the confirmation of sending the next packet.
Now, the sender will send the next packet that is frame number
3, now the window will have the frame numbers 1, 2 and 3.
After getting the confirmation for the delivery of frame number
1, the sender will send frame number 4. A window will have the
frame numbers 2, 3 and 4.
The process will be continued till the frame number reaches 25.
Part c)
Step1:
As the sender's window size is 3, so initially the first three
packets that are 0,1 and 2 are transferred to the receiver. After
receiving the acknowledgement of frame number 0, it will send
the other frame.
Step2:
Now, after receiving the acknowledgement of the first frame,
the window will have the updated frame that is 1,2 and 3. Now,
the sender will receive the acknowledgement of frame number
1.
Step3:
After receiving the acknowledgement of frame number 1, the
window will be updated with another frame that is the 4th one.
It will have frame numbers 2, 3 and 4.
Step4:
After receiving the acknowledgement of frame number 2, the
sender will send frame number 5 now.
Step5:
After receiving the acknowledgement of frame number 3, the
sender will send the packet with sequence number 6.
Step6:
After receiving the acknowledgement of frame number 4, the
sender will send frame number 7 along with the previous two
that are 5 and 6.
Step7:
After receiving the acknowledgement of frame number 5, the
sender will send the packet with sequence number 9 as the 8th
frame is not available. So, the window has frames 6,7 and 9.
Step8:
After receiving the acknowledgement of frame number 6, the
sender will send frame number 10 along with the previous two
that are 7 and 9.
Step9:
After receiving the acknowledgement of frame number 7, the
sender will send frame number 11 along with the previous two
that are 9 and 10.
Step10:
After receiving the acknowledgement of frame number 9, the
sender will send frame number 13 as the 12th frame is not
available. So, the window has frames 10,11 and 13.
Step11:
After receiving the acknowledgement of frame number 10, the
sender will send frame number 14 along with the previous two
that are 11 and 13.
Step12:
After receiving the acknowledgement of frame number 11, the
sender will send frame number 15 along with the previous two
that are 13 and 14.
Step13:
After getting the acknowledgement of frame number 13, the
sender will send frame number 14, 15 and 16.
Step14:
After getting the acknowledgement of frame number 14, the
sender will send frame number 15, 16 and 17.
Step15:
After getting the acknowledgement of frame number 15, the
sender will send frame number 16, 17 and 18.
Step16:
Now, the acknowledgement of the 16th frame has not been
received by the sender, so the sender will go back to 15 which
is the initial packet of the current window and will retransmit
all the frames.
Step17:
After receiving the acknowledgement of the 15th frame again,
the sender will send the next three frames that are 16, 17 and
18 will be transferred to the receiver.
Step18:
Now the receiver will again send the acknowledgement of
frame 16th to the sender and then the next three frames are
transferred that are 17, 18 and 19.
Step19:
After getting the acknowledgement of frame number 17, the
sender will send frame number 18, 19 and 20.
Step20:
Now, the acknowledgement of the 18th frame has not been
received by the sender, so the sender will go back to 17 which
is the initial packet of the current window and will retransmit
all the frames.
Step21:
After receiving the acknowledgement of the 17th frame again,
the sender will send the next three frames that are 18, 19 and
20 will be transferred to the receiver.
Step22:
Now the receiver will again send the acknowledgement of
frame 18th to the sender and then the next three frames are
transferred that are 19, 20 and 21. And next frames are also
transferred to the receiver as in the previous process like by
getting the acknowledgement of the first frame of the window
and sending the next frame by keeping the count of the
sender's window 3.