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?
- A crossover cable should be used in place of the straight-through cable.
- A rollover cable should be used in place of the straight-through cable.
- The subnet masks should be set to 255.255.255.192.
- A default gateway needs to be set on each host.
- The subnet masks should be set to 255.255.255.0.
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:
4 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.
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.
(3)
Karanjit Singh said:
7 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.
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.
(2)
Manjumudhuval said:
8 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.
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.
(2)
C.R.Nishanth said:
1 decade ago
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?
Can anyone explain clearly why the Answer shows the Masks are different. Please Help?
(1)
Sudheer gupta said:
9 years ago
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.
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.
Avinash said:
7 years ago
Thanks for explaining this.
Chiranjee said:
7 years ago
Please elaborate working process between system to system with cross cable connectivity.
Makuupaul said:
8 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.
Darth vader said:
8 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.
Never forget computers think in binary.
Saravanan said:
9 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
Class C Network occupies 24 networks bits & 8 hosts bits
So network bits full 1's.
11111111.11111111.11111111.00000000
255 . 255 . 255 . 0
Post your comments here:
Quick links
Quantitative Aptitude
Verbal (English)
Reasoning
Programming
Interview
Placement Papers