Local and Metropolitan area Network Problem-1 Problem-2 Problem-3 Problem-4 Problem-5 Problem-6 Problem-7 Problem-8 Problem-9 Problem-10

Local and Metropolitan area Network
Problem 1:
Solution : Vulnerability Period when time increases from  to 2 = 4
Explanation :
The total period of time when collision may occur for a packet is called vulnerable
Period.
In Pure Aloha Scheme if the packet transmission time is T , then vulnerable period is 2T .
How ?
Consider a packet scheduled to transmitted at some time t . This packet will be transmitted
successfully only of no other packet is scheduled for transmission at interval (t-T, t+T) .
This period of 2T is called vulnerable period .
So for Pure Aloha when transmission time is 2T vulnerable period will be 4T .
Problem 2:
Solution 2 :
As in this case both 802.11a/g has same physical link data rate of 54Mbps , 802.11 g should be
selected for upgrading the existing wireless infrastructure .
Following are the reasons why 802.11 g should be prefe
ed :
a) Cost Effective : Operating Frequency of 802.11g is 2.4 GHz and that of 802.11a is
5 GHz . As range is inversely proportional to Frequency , so 802.11g has larger range .
So while deploying 802.11a would require more access points and hence more
investment compared to 801.11g. So selecting 801.11g for upgrading will provide a cost
effective solution .
) Interoperability : 802.11g has same frequency as cu
ent infrastructure 802.11b so
existing clients can use same interface card that operates at 2.4 GHz. While 802.11a
would create interoperability problem as existing clients may not have interface card that
operates at 5GHz.
c) Provide Better Connectivity : 802.11a operate at high frequency so signals will have
high wavelengths , So signals will have problems penetrating office walls and cubicles
leading to poor connectivity in campus . While 802.11g has lower frequency hence
providing better connectivity .
Problem 3 :
Solution 3:
Node Can Transmit Successfully
to ….
Can Receive successfully
from …
Node-3 None None
Node-4 5 None
Node-5 None 4
Node-6 None None
Node-7 None None
Node-8 None None
It is given that a transmission is going on from node 1 to node 2. If we consider transmission
etween node 3 and 8 , it will collide with ongoing transmission at node 2 causing interference
which we would not want to happen .Similarly transmission between 7 and 4 would collide with
ongoing transmission at node 1 . The only transmission which is possible is from node 4 to 5 . So
node 4 can transmit successfully to node 5 and node 5 can receive successfully from node 4 .
Problem 4:
Solution 4 :
4(a) Throughput Without Using RTS/CTS transaction :
Total data in one Frame = Mac Layer Data Payload = 1452 bytes
Mac Header = 28 Bytes
Total Frame Size = 1452 + 28 = 1480 bytes
Time required for transmitting Frame (Data Time) = Total frame size / Transmission Rate
...