Networking - Subnetting - Discussion

Discussion :: 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.

[A]. 1 only
[B]. 2 only
[C]. 3 and 4 only
[D]. 1 and 5 only
[E]. 2 and 5 only

Answer: Option D

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).

Nathi said: (Jul 23, 2013)  
Please explain why would the two hosts be in different sub nets as they both have the 255.255.255.240 and also wouldn't they belong in the same IP range?. Please explain?.

Shivani Jain said: (Oct 29, 2013)  
As AND operation of IP address and Subnet mask gives us the Network address. In this case network address is different with given subnet address.

But when subnet mask (255.255.255.0) is used same network address will obtained and hence both will be on same network.

Alooma said: (Dec 26, 2013)  
But looking at the mask address they are both on the same mask address 255.255.255.240, please explain.

C.R.Nishanth said: (Mar 11, 2014)  
The Increment (or) Block size is 16, which means the IP Address range will be 192.168.1.0 to 192.168.15.255 if you can see the IP Addresses 192.168.1.20 & 192.168.1.201 falls in between the IP Address. Hence, the IP Address 192.168.1.20 & 192.168.1.201 having mask 255.255.255.240 can communicate.

Can anyone explain clearly why the Answer shows the Masks are different. Please Help?

Daisy said: (Apr 22, 2014)  
The series is as follows:

192.168.1.0, valid host(1 to 14).
192.168.1.16, valid host(17 to 30).
192.168.1.32, valid host.
.
.
.
.
192.168.1.192 (193 to 206).
192.168.1.208.
192.168.1.224 ().
192.168.1.240.. hence the given IP address of both host falls in different subnets.

Derek Mercado said: (Jun 1, 2014)  
I too am confused as it seems they both have the same subnet mask (x.x.x.240) while i agrees the they are different subnets eg .20 is in subnet .16 and .201 is in subnet .192 they clearly share the same subnet mask x.x.x.240 because it states it in the example and it states in the answer that the hosts have different subnet masks, maybe I am not understanding it correctly.

Mokonnin Alemu said: (Jun 19, 2014)  
First, if you have two hosts directly connected, 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.

Yasser said: (Aug 7, 2014)  
The .244 in mask means that there are only 4 allowable host bits thus can't afford to make address .201 for right host if all 4 bits where ones.

Tabrez said: (Jan 21, 2015)  
Please explain why?

Peerzada Shabir said: (Mar 12, 2015)  
We cannot use subnet mask 255.255.255.240 with both IP addresses as this subnet mask will take them in two different subnets(networks).

The binary equivalent of 240 is 11110000. i.e 16 is the increment.

The possible networks are:

192.168.1.0 to 192.168.1.15.
192.168.1.16 to 192.168.1.31------>192.168.1.20 belongs to here.
.
.
.
.
.
192.168.1.192 to 192.168.1.207----->192.168.1.201 belongs here.
192.168.1.208 to 192.168.1.223.
192.168.1.224.

So we can clearly see that if we use subnet mask 255.255.255.240 then communication will not be possible as both will be on separate networks. So we have to use 255.255.255.0 subnet in both so that they belong to same network.

Jay said: (Jun 27, 2016)  
Please give a little explanation about crossover cable theory.

Ehifameh Egbiremolen said: (Oct 25, 2016)  
D answer is perfect. You check again. It is a private IP address.

Shaoor said: (Nov 18, 2016)  
Suppose a company ABC wants to develop two LANs for its different departments in the same building. For LAN network the main factors to consider are reliability, security, and speed. The company has also no cost constraints for the development of LANs.

After LAN development, this company also needs to connect these two LANs with each other and with its remote offices located in different cities.

1. Identify the suitable medium that will fulfill the requirements of the company in order to develop LANs.

2. Which type of address will be associated when data will travel from one user to another user within same LAN?
VLAN. if two people attempt to send information at the same time,

3. If the user of one LAN wants to communicate, with a remote user located in another city network then which type of address will be used?

4. Which type of networking device will be used to transfer data from one of the LANs to a remote user located in the network of another city?

Please tell me the solution.

Sudheer Gupta said: (Dec 11, 2016)  
In this mask, Only 14 host address will be used.

20 comes between 16 to 31.

But 201 is not between 16 to 31.

So it is another subnet. Hence communication is possible. And cause of cable as will.

Ehifameh said: (Dec 25, 2016)  
It was kind of confusing. But crossover cable should be used and IP add should be in a class C. There won't be any problem.

Ehifameh said: (Dec 25, 2016)  
It was kind of confusing. But crossover cable should be used and IP add should be in a class C. There won't be any problem.

Marcus said: (Jan 6, 2017)  
If 255.255.255.240 is used how they are on different networks. Someone show me how( 255.255.255.0) works?

Marcus said: (Jan 6, 2017)  
If someone could find the subnet of these addresses using /24 CIDR.

Saravanan said: (Jan 11, 2017)  
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

Darth Vader said: (Apr 3, 2017)  
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.

Manjumudhuval said: (Jun 1, 2017)  
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.

Makuupaul said: (Jul 28, 2017)  
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.

Chiranjee said: (Jun 15, 2018)  
Please elaborate working process between system to system with cross cable connectivity.

Avinash said: (Nov 5, 2018)  
Thanks for explaining this.

Karanjit Singh said: (Nov 13, 2018)  
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.

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