Given the routing entries shown in the exhibit, which next-hop IP address will be used for the destination IP address 192.168.1.142? (Assume that only static routes are available for the destination)
A.
10.20.106.10
B.
10.20.14.131
C.
10.18.1.1
D.
10.20.14.130
D.
10.20.14.130
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Binary are 2^0, 2^1, 2^2, 3^2 1 , 2 , 4, 8 then total are 4 bite => /28 = 24+4
Destination IP address is 192.168.1.142 => the only ip address can cover destination is 192.168.1.128/28
rang is 192.168.1.128 – 192.168.1.143
D) is correct
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Can you explain more detailed? I cant seem to understand how you’ve come to the answer D….
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So…
If /28 = the magic no. thats 16
192.168.1.0 – 15
192.168.1.16 – 31
xxx.xxx.1.32 – 47
xxx.xxx.1.48 – 63
xxx.xxx.1.64 – 79
xxx.xxx.1.80 – 95
xxx.xxx.1.96 – 127
xxx.xxx.1.128 – 143
xxx.xxx.1.144 – 160 <— Are you looking at the network ID, thats 192.168.1.144?
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Sry, I meant if you were looking at the last usable host in subnet 8,
192.168.1.128 – 143
(-1 for the broadcast address that leaves us with last usable host, 192.168.1.142)
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