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Interface $A - 255.255.255.128$

Interface $B -255.255.255.192$

Interface $C -255.255.255.224$

 

1) which one of the following IP addresses belong to interface $B$ ?

  1. $200.200.200.130$ & $200.200.200.157$                       
  2. $200.200.200.200$ & $200.200.200.205$
  3. $200.200.200.30$ & $200.200.200.130$
  4. $200.200.200.20$ & $200.200.200.200$


 

2) Based on the subnet masks above , what could be the maximum number of possible subnets in the network $200.200.200$ ?

  1. $2$
  2. $4$
  3. $8$                     
  4. $3$
asked in Computer Networks by Loyal (7.9k points)
edited by | 160 views
0
I am getting 1. b and 2. a

correct or not??
+1
no its 1.a ) and 2 . c)

1) for interface B only a option has same network other are belongs from different

2) if it says 200.200.200 are network id part and remaining are subnet bits so  224 which has 3 1,s in last octent so it will be 2^3= 8 ... rt ?
+1

What is the source of the question. I think the question is itself incomplete because whenever an incoming packet arrives at the router, then the router performs the Logical AND operation between the subnet mask and the destination address provided in the IP header of the packet, the result of the AND operation is subnet id provided that the network uses the SUBNET Technology. And every interface has its own subnet id associated with the interface (which is missing in the question) based on the subnet mask only we can't decide where this packet will be forwarded.

If the subnet ID associated with the interface B is $200.200.200.128$ then option (A) is correct. And if the subnet ID associated with the interface B is $200.200.200.192$ then option (B) is the correct answer.

As per the subnet mask, not ID provided in the question to the two bits are for subnet and 6 bits are for host.

taking option (A):- when $200.200.200.128$ subnet ID is assigned to interface B

$200.200.200.130$ expanding the last octet it is 1000 0010 means 2nd subnet.

$200.200.200.157$ expanding the last octet it is 1001 1101 means 2nd subnet, Now both belong the same subnet.

taking option (B):- when $200.200.200.192$ subnet ID is assigned to interface B

$200.200.200.200$ expanding the last octet it is 1100 1000 means 3rd subnet.

$200.200.200.205$ expanding the last octet it is 1100 1101 means 3rd subnet, Now both belong the same subnet.

Hence to forward the packet we should have subnet ID associated with each interface is the necessary condition.

0
yes its not complete
0
@Jason , can you plz explain what do we mean when we say we have a subnet mask associated with an interface , we have an interface between two networks and subnet mask is associated with a network so how come we are saying that an interface has a subnet mask , plz explain this precisely .

1 Answer

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$(1)$ According to the given data, the maximum range of addresses possible in each interfaces are:

 Interface $A\Rightarrow 255.255.255.128$ to $255.255.255.191$

Interface $B\Rightarrow 255.255.255.192$ to $255.255.255.223$

Interface $C\Rightarrow 255.255.255.224$ to $255.255.255.255$

Clearly, the pair of IP addresses which belong to Interface $B$ is $200.200.200.200$ & $200.200.200.205$,

which is option $(B)$

 

$(2)$  

Interface $A \Rightarrow 255.255.255.128 = 255.255.255.100\space 0\space 0000$

Interface $B \Rightarrow 255.255.255.192 = 255.255.255.110\space 0\space 0000$

Interface $C \Rightarrow 255.255.255.224 = 255.255.255.111 \space 0\space 0000$

Looking at the subnet masks, we can infer that first 3 of the host bits are used for subnetting. So maximum possible number of subnets in this configuration is $2^3 = 8.$

So, the answer is $(C)$
answered by Active (1.3k points)

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