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CCNA - Route Summary question

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cisco222

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Jul 9, 2007
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The Ethernet networks connected to router R1 in the graphic have been summarized for router R2 as 192.1.144.0/20. Which of the following packet destination addresses will R2 forward to R1, according to this summary? (Choose two.)

A. 192.1.159.2

B. 192.1.160.11

C. 192.1.138.41

D. 192.1.151.254

E. 192.1.143.145

F. 192.1.1.144

Can someone explain to me how they have come to this answer? I understand how to summarize routes like 192.168.1.0,192.168.2.0 192.168.3.0 192.168.4.0 etc but i don't know to work out the type of question above?

Thanks
 
192.1.144.0/20 yields 192.1.144.0 - 192.1.159.255

Best regards.
 
Thanks.

How did you workout 192.1.159 255 is the a quick way i can work this out?

is it because 11111111.11111111.11110000.00000000 this = /20 do i need to increment by 16 in the 3rd octet to get my answer?

 
yes, consider this:
[tt]
IP address: 10110000 00000001 10010000 00000000
subnet mask:11111111 11111111 11110000 00000000
host bits start here -------------|

So, the range of host addresses is:
10110000 00000001 10010000 00000000
to:
10110000 00000001 10011111 11111111
[/tt]
which is: 192.1.159.255.

HTH
 
Quick clarification. "Yes" was meant the answer to if there's a quick way to figure this out.
 
yeah because changing to binary is always the quick way to do it...

/20 , this means bits were stolen from /16
/17 is 255.255.128.0
/18 is 255.255.192
/19 is 255.255.224
/20 is 255.255.240

256-240 = 16 block size.
this means 192.1.144.0 is the range from 192.1.144.0 - to - 19.1.(144+15).0
there is your range.

you have do some memorizing,
1 bit stolen is 128
2 bit stolen is 192
3 bits stolen is 224
4 bits stolen is 240
5 bits stolen is 248
6 bits stolen is 252
7 bits stolen is 254
8 bits stolen is 255


We must go always forward, not backward
always up, not down and always twirling twirling towards infinity.
 
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