Networking - Subnetting - Discussion

Discussion Forum : Subnetting - Subnetting (Q.No. 3)
3.
A network administrator is connecting hosts A and B directly through their Ethernet interfaces, as shown in the illustration. Ping attempts between the hosts are unsuccessful. What can be done to provide connectivity between the hosts?
  1. A crossover cable should be used in place of the straight-through cable.
  2. A rollover cable should be used in place of the straight-through cable.
  3. The subnet masks should be set to 255.255.255.192.
  4. A default gateway needs to be set on each host.
  5. The subnet masks should be set to 255.255.255.0.
1 only
2 only
3 and 4 only
1 and 5 only
2 and 5 only
Answer: Option
Explanation:
First, if you have two hosts directly connected, as shown in the graphic, then you need a crossover cable. A straight-through cable won't work. Second, the hosts have different masks, which puts them in different subnets. The easy solution is just to set both masks to 255.255.255.0 (/24).
Discussion:
26 comments Page 1 of 3.

LUITEL M said:   3 years ago
Nowadays, the interface card automatically detects no matter if we use straight or crossover.

255.255.255.240 - useable IP range 192.168.1.16 - 192.168.1.30 - Wrong answer.
255.255.255.192 - usable IP range 192.168.1.1 - 192.168.1.62 - Wrong Answer.
255.255.255.0 - usable IP range from 192.168.1.1 to 192.168.1.254,

Hence the answer.
(1)

Karanjit Singh said:   6 years ago
Lets say the mask is x.x.x.240.

It means we have borrowed 4 network bit from fourth octet. If you know binary it will say you can have 16(2 power 4) subnets and 14(2 power 4 - 2(Network ID AND Broadcast address)) valid hosts in each subnet.

So having said that the given IPs will not fall in any of the subnet. So here is the problem with mask, which supposed to be /24, so that we can have 8 hosts bit(2 power 8-2= 254 Valid hosts). I hope it helps everybody.

Avinash said:   6 years ago
Thanks for explaining this.

Chiranjee said:   6 years ago
Please elaborate working process between system to system with cross cable connectivity.

Makuupaul said:   7 years ago
First of all, the 224 is a broadcast and falls at the very last bit of the 16th time. Meaning the subnet address available there and asked for cannot be used, given that 224 is not available. The subnet is 239 also because the 240 is not available on the range when working with turns of 16 subnets.

Manjumudhuval said:   7 years ago
2^4=16 subnets to be divided of each 16 valid IPs.

The only range starts 0-15, 16-31........176-191,192-207, 208-223,224-239, 240-255 valid IPs because The 207 comes under the same network.

If it's wrong any one make me understand.
(1)

Darth vader said:   7 years ago
They are in different networks, always make bitwise AND operation to find if the computers in the same network. One of them belong to 192.168.1.16 and the other one in 192.168.1.192.

Never forget computers think in binary.

Saravanan said:   7 years ago
Ip address 192.168.1.20 it is class C network.

Class C Network occupies 24 networks bits & 8 hosts bits
So network bits full 1's.

11111111.11111111.11111111.00000000
255 . 255 . 255 . 0

Marcus said:   7 years ago
If someone could find the subnet of these addresses using /24 CIDR.

Marcus said:   7 years ago
If 255.255.255.240 is used how they are on different networks. Someone show me how( 255.255.255.0) works?


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