GATE CSE 2004 | Question: 55

Question

The routing table of a router  is shown below:
$$\begin{array}{|l|l|l|} \hline \textbf {Destination} & \textbf {Subnet Mask} & \textbf{Interface}  \\\hline \text {128.75.43.0} &  \text{255.255.255.0} & \text{Eth$0$} \\\hline\text {128.75.43.0} &  \text{255.255.255.128} & \text{Eth$1$} \\\hline\text {192.12.17.5} &  \text{255.255.255.255} & \text{Eth$3$}\\\hline \text {Default} &  \text{} & \text{Eth$2$}\\\hline\end{array}$$
On which interface will the router forward packets addressed to destinations $128.75.43.16$ and $192.12.17.10$ respectively?

• A. Eth$1$ and Eth$2$
• B. Eth$0$ and Eth$2$
• C. Eth$0$ and Eth$3$
• D. Eth$1$ and Eth$3$

Comments

No two destination have same IP address but two IP address are in same Network.

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Two destination always have different IP address but more than one IP address can belong to same Network.

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GO to the Root of Concept : Popular Misconceptions & what is Supernetting | Longest Match | Explained CLEARLY | With NOTES

Answer 1

The answer must be A.

(Using $\wedge$ to denote bitwise AND)

For 1$^{st}$packet,

$(128.75.43.16) \wedge (255.255.255.0) = (128.75.43.0)$ since $\{16 \wedge 0 = 0\}$, as well as 

$(128.75.43.16) \wedge (255.255.255.128) = (128.75.43.0)$ since $\{16 \wedge 128 = 0\}$.

Now, since both these subnet masks are producing the same Network ID, hence The one with greater number of ones will be selected, and the packet will be forwarded there. Hence packet $1$ will be forwarded to Eth$1$.

For $2^{nd}$ packet,

$(192.12.17.10)$ when ANDed with each of the subnet masks does not match with any of the network ID, since:

$(192.12.17.10) \wedge (255.255.255.0) = (192.12.17.0)$ {Does not match with any of the network addresses}

$(192.12.17.10) \wedge (255.255.255.128) = (192.12.17.0)$ {Does not match with any of the network addresses}

$(192.12.17.10) \wedge (255.255.255.255) = (192.12.17.10)$ {Does not match with any of the network addresses}

Hence, default interface must be selected for packet $2$, i.e., Interface Eth$2.$

Comments

@JAINchiNMay Why longest prefix match of subnet mask are always preferred over shorter one?

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A network is divided in the many subnets,

But now the entries of the original network and new subnets both are present in routing tables
and the subnet mask of the new subnet will be longer. so it is more meaningful to send the data to subnet
Longest Prefix Matching in Routers - GeeksforGeeks here you can understand it better.

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@Nisha Bharti Longest prefix match subnet mask is preferred because in Routing table Network ID are stored in decreasing order of subnet mask , also can understand the network mask having less number of host is more specific than the network have more number of host.

Answer 2

How router makes decisions ?

Router check the destination address by decapsulating packet and perform bitwise AND with subnet mask of all interfaces and if the resulting n/w address matches with corresponding interface n/w address then router send packet to this interface , in case of tie , router uses Longest prefix match

interface eth0 :- 128.75.43.0/24 (128.75.43.0 to 128.75.43.255)

interface eth1 :- 128.75.43.0/25 (128.75.43.0 to 128.75.43.127)

for 128.75.43.16 we can see there is a tie as it falls within the n/w addresses for eth0 and eth1...which route would the router choose? It depends on the prefix length, or the number of bits set in the subnet mask. Longer prefixes are always preferred over shorter ones when forwarding a packet.

https://en.wikipedia.org/wiki/Longest_prefix_match

https://stackoverflow.com/questions/9335504/network-longest-prefix-matching

Comments

"Longer prefixes are always preferred over shorter ones when forwarding a packet." why so ?

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read about RIPv2

and packet forwarding

Answer 3

Start with the maximum mask:

128.75.43.16 while doing masking(bit wise AND operation) with

255.255.255.128 will give 128.75.43.0 .

we always takes the maximum mask so ans would be eth1

Answer 4

It'll be through eth1. If bit wise OR of packet address and subnet mask matches with more than one destination address then we go with the one with longer run of 1s.