CS610 TODAY FINAL TERM PAPERS STARTING DATE 15/02/20 TO 26-02-20
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CS610 current Final term papers Dated 15-02-2020
mcqs past papers se repeat hoye hen
If a data exceeds MTU, which technique is best used to transfer data? (5 Marks)
Find the number of useable host addresses in the block if one of the subnet addresses is 22.214.171.124/20. (3 Marks)
Different Multicast Routing Methodologies have their different mechanisms of forwarding the datagram according to situation. You are required to write the names of these multicast routing methodologies and suggest the suitable choice of these methodologies for the following cases:
When the group includes few members.
When the members are scattered. (5 Marks)
Enlist 3 error message and 2 informational message by ICMP?(5 Marks)
Reason for different network technologies. (5 Marks)
NAT (Network Address Translation) and its implementation. (5 Marks)
Which multicast protocol designed for use within organization? Write its complete name. (3 Marks)
Identify the Ethernet frame regarding given fields. CRC, Frame type, flag, preamble, sequence no. (3 Mark)
How TCP is Reliable transfer protocol, prove with its function. (5 Marks)
How administrator updates the routing table in routing.
Answer the Question regarding Static routing and dynamic routing
CS610 Current Finalterm Paper Dated 17-02-2020
Delay ki types
GP and EGP b/w
ICMP ki types
CS610 Current Finalterm Paper Dated 22-02-2020
In order to understand 255.255.254.0 subneting, we first look at the usual 255.255.255.0 subnet mask.
If you use 255.255.255.0 (on binary each 255 is 11111111 and each 0 is 00000000), it means that you decided to use the last 8 bits of the address for your devices. 8 binary bits gives you 256 decimal, so you can have 256 addresses for the devices on your network ranging from 192.168.0.0 to 192.168.0.255. In real life you can’t use the first and the last so you are in this case limited to 254 addresses.
Now, if you change the network mask to 255.255.254.0, then you will have 9 bits for your devices (254 is 11111110). 9 binary bits translated to decimal gives you 512, so you can have 512 addresses for your devices ranging from 192.168.0.0 to 192.168.1.255. Just like before you can’t use the first and the last so you are in this case limited to 510 addresses.