Questions and Answers:
1. Advertise only odd network
210.8.40.0
210.8.41.0
210.8.42.0
210.8.43.0
even 40,42
40 0010 1000
42 0010 1010
0000 0010
odd 41,43
41 0010 1001
43 0010 1011
0000 0010
answer:
acl odd 210.8.41.0 0.0.2.x
acl even 210.8.40.0 0.0.2.x
x = 0 = route
x = 255 = host
dont care = 1
stings of 32 bits
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2. pick only even /24 from 202.78.0.0 / 16
make as few line as possible, permit no more no less.
202.78.00000000.0
202.78.00000010.0
202.78.x.0
0 0000 0000
2 0000 0010
4 0000 0100
6 0000 0110
8 0000 1000
10 0000 1010
1111 1110 = mask = 254
answer : 202.78.0.0 0.0.254.0
even: acl permit 202.78.0.0 0.0.254.0 .0 on last octed means a route!
odd : acl permit 202.78.1.0 0.0.254.0 .0 on last octed means a route!
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3. Group / Summarize: is a few lines a possible
172.16.31.0
172.16.32.0
172.16.33.0
172.16.34.0
answer:
31.0
32.0 /22
deny 35.0
acl deny 172.16.35.0 0.0.0.0
acl permit 172.16.31.0 0.0.0.0
acl permit 172.16.32.0 0.0.3.0
31 0001 1111
32 0010 0000
33 0010 0001
34 0010 0010
35 0010 0011 <<—-not present!
0000 0011 Mask = (3) , 2 bit position difference, so 2 to the 2 is 4, thats 32-35.
Tip: You can over permit but make sure you deny the over permitted route.
—————–
4. Group / Summarize: is a few lines a possible
172.16.31.0
172.16.32.0
172.16.33.0
172.16.34.0
172.16.35.0
172.16.36.0
172.16.37.0
answers:
acl permit 172.16.31.0 0.0.0.0
acl permit 172.16.32.0 0.0.3.0
acl permit 172.16.36.0 0.0.1.0
or
acl deny 172.16.38.0 0.0.1.0
acl permit 172.16.31.0 0.0.0.0
acl permit 172.16.32.0 0.0.7.0
Tip: Depends on how you group the routing according the position, will result in same 3 line answer.
——————–
5. Summarize in as few lines a possible.
168.192.3.0 /24
168.192.14.0 /24
168.208.11.0 /24
168.208.14.0 /24
168.208.3.0 /24
168.192.11.0 /24
—
answer, group first.
168.192.3.0 /24
168.192.11.0 /24
168.208.3.0 /24
168.208.11.0 /24
168.192.14.0 /24
168.208.14.0 /24
then,
192 1100 0000
208 1101 0000
0001 0000 mask = 16
3 0000 0011
11 0000 1011
0000 1000 mask = 8
14 0000 1110
answer:
acl permit 168.192.3.0 0.16.8.0
acl permit 168.192.14.0 0.16.0.0
———
5. summarize in as few lines as possible
207.49.164.0 /24
208.49.164.0 /24
205.49.165.0 /24
207.49.165.0 /24
192.49.164.0 /24
answer, group!
192.49.164.0 /24
208.49.164.0 /24
192 1100 0000
208 1101 0000
0001 0000 mask 16
207.49.164.0 /24 <-stand alone.
207.49.165.0 /24
205.49.165.0 /24
205 1100 1101
207 1100 1111
0000 0010 mask 2
answer:
acl permit 192.49.164.0 16.0.0.0
acl permit 205.49.165.0 2.0.0.0
acl permit 207.49.164.0 0.0.0.0
or
192.49.164.0 /24
208.49.164.0 /24
192 1100 0000
208 1101 0000
0001 0000 mask 16
207.49.164.0 /24
207.49.165.0 /24
164 1010 0100
165 1010 0101
0000 0001 mask 1
205.49.165.0 /24 <-stand alone.
so, another answer is:
acl permit 192.49.164.0 16.0.0.0
acl permit 207.49.164.0 0.0.1.0
acl permit 205.49.165.0 0.0.0.0
————–
7. Create an ACLin as few line as possible
182.17.73.0
182.17.77.0
182.81.73.0
182.81.77.0
190.17.73.0
190.81.73.0
190.17.77.0
190.81.77.0
190 1011 1110
182 1011 0110
0000 1000 mask 8
17 0001 0001
81 0101 0001
0100 0000 mask 64
73 0100 1001
77 0100 1101
0000 0100 mask = 4
In every octet there is a differnce, so we can put this on a
single entry!
answer:
acl permit 182.17.73.0 8.64.4.0
checking tip: put the middle network say 190.81.73.0 and
mask, the router should put the lowest IP address (
182.17.73.0) to the running config. This will ensure you that
you have computed correctly the mask.
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I can’t help it, I have a feeling this should come out and be part of the exam. Therefore this should be a must for all CCIE candidate to master!