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Consider the following routing table at an IP router:
$$\begin{array}{|l|l|l|} \hline \textbf {Network No} & \textbf {Net Mask} & \textbf{Next Hop} \\\hline \text {128.96.170.0} & \text{255.255.254.0} & \text{Interface 0} \\\hline\text {128.96.168.0} & \text{255.255.254.0} & \text{Interface 1} \\\hline\text {128.96.166.0} & \text{255.255.254.0} & \text{R2}\\\hline \text {128.96.164.0} & \text{255.255.252.0} & \text{R3}\\\hline \text {0.0.0.0} & \text{Default} & \text{R4}\\\hline \end{array}$$
For each IP address in Group I Identify the correct choice of the next hop from Group II using the entries from the routing table above.
$$\begin{array}{l|ll} \hline \textbf {Group I} & \textbf {Group II} \\\hline \text {i) 128.96.171.92} & \text{a) Interface 0} \\ \hline \text {ii) 128.96.167.151} & \text{b) Interface 1} \\\hline \text {iii) 128.96.163.151} & \text{c) R2}\\ \hline \text {iv) 128.96.164.121} & \text{d) R3}\\ \hline \text {} & \text{e) R4}\\\hline \end{array}$$

1. i-a, ii-c, iii-e, iv-d
2. i-a, ii-d, iii-b, iv-e
3. i-b, ii-c, iii-d, iv-e
4. i-b, ii-c, iii-e, iv-d

iii matches with none. so it goes to default router.
edited

Taking the 1st IP Address: 128.96.171.92

Bitwise AND between $128.96.171.92$ and $255.255.254.0$ we get the subnet ID as follows:$$\begin{array} {rrrr}255&255&11111110&0 \\ 128&96&10101011&92 \\ \hline \textbf{128}&\textbf{96}&\textbf{10101010}&\textbf{0}\end{array}$$

∴ Subnet ID = 128.96.170.0
∴ 128.96.171.92 will forward to interface 0

Taking the 2nd IP Address: 128.96.167.151

Bitwise AND between $128.96.167.151$ and $255.255.254.0$ we get, $$\begin{array} {rrrr}255&255&11111110&0 \\ 128&96&10100111&151 \\ \hline \textbf{128}&\textbf{96}&\textbf{10100110}&\textbf{0}\end{array}$$

∴ Subnet ID = 128.96.166.0
∴ 128.96.167.151 will forward to interface R2

Taking the 3rd IP Address: 128.96.163.151

Bitwise AND between $128.96.167.151$ and $255.255.254.0$ we get, $$\begin{array} {rrrr}255&255&11111110&0 \\ 128&96&10100011&151 \\ \hline \textbf{128}&\textbf{96}&\textbf{10100010}&\textbf{0}\end{array}$$

∴ Subnet ID = 128.96.162.0 (Doesn’t match with any given interface)

Now, Bitwise AND between $128.96.167.151$ and $255.255.252.0$ we get, $$\begin{array} {rrrr}255&255&11111100&0 \\ 128&96&10100011&151 \\ \hline \textbf{128}&\textbf{96}&\textbf{10100000}&\textbf{0}\end{array}$$

∴ Subnet ID = 128.96.160.0 (Doesn’t match with any given interface)
∴ 128.96.163.151 will forward to default interface R4

Taking the last IP Address: 128.96.164.121

Bitwise AND between $128.96.164.121$ and $255.255.254.0$ we get, $$\begin{array} {rrrr}255&255&11111110&0 \\ 128&96&10100100&121 \\ \hline \textbf{128}&\textbf{96}&\textbf{10100100}&\textbf{0}\end{array}$$

∴ Subnet ID = 128.96.164.0
∴ 128.96.167.151 will forward to interface R3

∴ Option (A) is the correct answer.

Option $A$ is correct . Do the $AND$ operation of group $1$ with net mask you will get the answer.

Can u explain this what to check with result of  AND operation to get next hop ???

The next hop is decided according to the longest prefix matching obtained by AND operation, hope this helps

So now we can eliminate the options easily.

i -> a

ii -> c

Option - A

i) 128.96.171.92  --- > First addr 128.96.170.0         a) Interface 0

ii) 128.96.167.151 ---> First  addr 128.96.166.0        c) R2

iii) 128.96.163.151--> first addr matches none e) R4

iv) 128.96.164.121---> first addr  128.96.164.0       d) R3

Consider the following routing table at an IP router:

Network No                      Net Mask            Next Hop

128.96.170.0/23 255.255.254.0     Interface 0

128.96.168.0/23 255.255.254.0     Interface 1

128.96.166.0/23 255.255.254.0     R2

128.96.164.0/22 255.255.252.0     R3

0.0.0.0   Default R4

128.96.171.92

To get First Net addr make the last 32-N bits of the addr to zero.[N is prefix]

First prefix to check is 23 --> 32-23=9

Make the last 9 bits of the addr to zero. The last 9 bits of address are in the last two octets.

171.92 -  - - - - -> 170.0   hence first netaddress is 128.96.170.0 matches first entry.