102-500 Exam - Question 100

Question

Given the following routing table:How would an outgoing packet to the destination 192. 168. 2. 150 be handled?

Examice · Accepted Answer

To determine how an outgoing packet to the destination 192.168.2.150 would be handled, we need to inspect the routing table. The routing table lists routes based on their destination network, gateway, netmask, and the interface (Iface) they should use. In this case, the routing entry for the destination network 192.168.2.0/24 (which includes 192.168.2.150) has a gateway of 192.168.1.1 and is associated with the eth0 interface. Therefore, the packet would be passed to the router 192.168.1.1 on eth0.

dislexiccamel · Answer

A is the answer
0.0.0.0 is a route to any network
G flag indicates a gateway

il_biondo · Answer

D is correct
This routing table is about a router/gateway with ethernet (eth0) and wifi (wlan0) interface.
network 192.168.178.0 is connected via interface wlan0 to default gateway 0.0.0.0
network 192.168.1.0 is connected via interface eth0 to default gateway 0.0.0.0
network 192.168.2.0 is connected via interface eth0 to gateway 192.168.1.1 (on network 192.168.1.0)
-----> then outgoing packet to reach 192.168.2.150 on network 192.168.2.0 will passed to gateway 192.168.1.1  -------> answer D

Rebelpierre · Answer

Proved in a lab on virtual box with several vm's in different segments:
localhost:~ # route
Kernel IP routing table
Destination     Gateway         Genmask         Flags Metric Ref    Use Iface
default         192.168.1.254   0.0.0.0         UG    600    0        0 wlan0
10.10.20.0      192.168.56.201  255.255.255.0   UG    0      0        0 vboxnet1
172.17.0.0      0.0.0.0         255.255.0.0     U     0      0        0 docker0
172.18.0.0      0.0.0.0         255.255.0.0     U     0      0        0 br-2d142c106ac3
192.168.1.0     0.0.0.0         255.255.255.0   U     600    0        0 wlan0
192.168.49.0    0.0.0.0         255.255.255.0   U     0      0        0 br-38c8ba0fd0af
192.168.56.0    0.0.0.0         255.255.255.0   U     0      0        0 vboxnet1
192.168.57.0    0.0.0.0         255.255.255.0   U     0      0        0 vboxnet0
localhost:~ # traceroute 10.10.20.1
traceroute to 10.10.20.1 (10.10.20.1), 30 hops max, 60 byte packets
 1  192.168.56.201 (192.168.56.201)  4.347 ms  4.325 ms  4.308 ms
 2  10.10.20.1 (10.10.20.1)  5.540 ms  5.549 ms  5.529 ms
Destination IP 10.10.20.1 would represent the IP of option D

TT924 · Answer

D is correct, because route is defined

moepmann21 · Answer

D is correct because 192.168.2.0 route is "smaller" than 0.0.0.0 so the smaller route gets in first.

nepechon · Answer

man route 
Flags  Possible flags include
              U (route is up)
              H (target is a host)
              G (use gateway)
              R (reinstate route for dynamic routing)
              D (dynamically installed by daemon or redirect)
              M (modified from routing daemon or redirect)
              A (installed by addrconf)
              C (cache entry)
              !  (reject route)

rezart21 · Answer

Correct Answer is D. 192.168.2.0/24 is a static route and it is reached passing 192.168.1.1 via eth0. In routing table any static route (by default) is checked before default gateway .
Default gateway will be hited once no match criteria will be found in the routing table .

[Removed] · Answer

Correct Answer is D.

btomasz · Answer

D is Correct.

0.0.0.0/0 Matches the IP
192.168.1.0/24 no Match
192.168.2.0/24 Matches the IP -> this Route is used, becaus the Netmask on this Route is more precise, than that of the default Route.
192.168.178.0/24 no Match

102-500 Exam - Question 100

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