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Two computers $C1$ and $C2$ are configured as follows. $C1$ has IP address $203.197.2.53$ and netmask $255.255.128.0$. $C2$ has IP address $203.197.75.201$ and netmask $255.255.192.0$. Which one of the following statements is true?

1. $C1$ and $C2$ both assume they are on the same network
2. $C2$ assumes $C1$ is on same network, but $C1$ assumes $C2$ is on a different network
3. $C1$ assumes $C2$ is on same network, but $C2$ assumes $C1$ is on a different network
4. $C1$ and $C2$ both assume they are on different networks.
edited | 3.1k views
+12
$Remark:$
A host knows its own IP address and the subnet mask, when a host wants to send a packet to another host, it checks if the destination host is in the same network, if the destination host is in the same network, Source Host directly sends the packet to it else it sends the packet to the router.

Subnetmask for C1 is $255.255.128.0.$ So, it finds the network ID as:

$203.197.2.53$ AND $255.255.128.0 = 203.197.0.0$
$203.197.75.201$ AND $255.255.128.0 = 203.197.0.0$

Both same.

Bow subnetmask for $C2$ is $255.255.192.0.$ So, the respective network IDs are:

$203.197.2.53$ AND $255.255.192.0 = 203.197.0.0$
$203.197.75.201$ AND $255.255.192.0 = 203.197.64.0$

Both not same. So, option C.

edited
0
Two hosts in the same network have different netmask then is one belongs to   another's subnetwork?I mean does C2 belong to  a subnet of the network where C1 belongs?Then what is the reason of this scenario?
+1

They are classless.
Subnetmask for C1 is 255.255.128.0. So, we can network ID reserved first 17 bits as it it classless.

Similarly, for C2  255.255.192.0. network ID reserved first 18 bits.

for C1:  203.197.2.53 AND 255.255.128.0 = 203.197.0.0 /17
for C2:  203.197.75.201 AND 255.255.192.0 = 203.197.65.0/18

as C2 has reserved 1 extra bit for subnet ID,

C1 assume the C2 is inside its network.

so,

C1 assumes C2 is on same network, but C2 assumes C1 is on a different network
Option C
ans d)
+3
no.. it's (C)
+3
i agree its option c.
0
Can anyone explain why is it 'c' ?

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