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44 votes
44 votes

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?

  1. Eth$1$ and Eth$2$
  2. Eth$0$ and Eth$2$
  3. Eth$0$ and Eth$3$
  4. Eth$1$ and Eth$3$
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Does Eth3 even have any valid hosts? There are $2^{32}$ subnets on that network, and $2^0 = 1$ host per subnet. However, this one host is the subnet ID itself isn't it (not considering classful addressing)?
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please explain how 2 destination have same IPA
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5 Answers

60 votes
60 votes
Best answer

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$2.$

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network id and subnet mask for default entry are 0.0.0.0 and 0.0.0.0 for default entry.

if the IP doesn’t matches with any entry of routing table then, the packet should be forward to the default interface
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@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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11 votes
11 votes

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

2 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

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1 vote
1 vote
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
0 votes
0 votes
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.
Answer:

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