Transcript VLSM Calculation

VLSM Calculation
Simplified Version
Problem
Provide VLSM based on the topology
requirements.
IP Block: 30.50.0.0/13
Network A – 1,000 hosts
Network B – 500 hosts
Network C – 250 hosts
Network D – 1,750 hosts
Network E – 750 hosts
Network F – 50 hosts
Solution
Step 1: Arrange the networks in descending
order based on its host requirements.
Network D – 1,750 hosts
Network A – 1,000 hosts
Network E – 750 hosts
Network B – 500 hosts
Network C – 250 hosts
Network F – 50 hosts
Solution
Step 2: Construct the VLSM Table
Host
Requirement
1,750
1000
750
500
250
50
250
30
2^n
How many
times the "n"
32-n value can be
subtracted to
8
Which
Octet
Remainder
2^N (where
N=value of
Remainder)
Network
Address
Solution
What’s in each column?
Host
Requirement
2^n
How many
times the "n"
32-n value can be
subtracted to
8
Which
Octet
Remainder
2^N (where
N=value of
Remainder)
Network
Address
1,750
1000
750
500
250
50
30
Host Requirements – This pertains to the
number of host for a particular network
Solution
What’s in each column?
Host
Requirement
2^n
1,750
1000
750
500
250
50
30
2^11
2^10
2^10
2^9
2^8
2^6
2^5
How many
times the "n"
32-n value can be
subtracted to
8
Which
Octet
Remainder
2^N (where
N=value of
Remainder)
Network
Address
2^n – This pertains to the power needed to satisfy
the host requirement using 2 as its base.
Note: 2^n ≥ host. Choose the least 2^n that can
satisfy our requirement!
What?
Note: 2^n ≥ host. Choose the least 2^n that
can satisfy our requirement!
Example: we need 10 hosts. Although 2^4
(16) and 2^5 (32) can satisfy the host
requirement, 2^4 is chosen due to
efficiency. (Less wasted IP address!)
Solution
What’s in each column?
Host
Requirement
2^n
1,750
1000
750
500
250
50
30
2^11
2^10
2^10
2^9
2^8
2^6
2^5
How many
times the "n"
32-n value can be
subtracted to
8
Which
Octet
Remainder
2^N (where
N=value of
Remainder)
21
22
22
23
24
26
27
32-n – This gives us our subnet mask for
that particular network.
Network
Address
Solution
What’s in each column?
Host
Requirement
2^n
1,750
1000
750
500
250
50
30
2^11
2^10
2^10
2^9
2^8
2^6
2^5
How many
times the "n"
32-n value can be
subtracted to
8
21
22
22
23
24
26
27
Which
Octet
Remainder
2^N (where
N=value of
Remainder)
Network
Address
1
1
1
1
1
0
0
How many times the "n" value can be subtracted
to 8 – This allows us to determine on which octet
should we add numbers starting from the fourth
octet. An increment of one tells us to go one
octet to the left.
What?
How many times the "n" value can be subtracted to 8 –
This allows us to determine on which octet should we
add numbers starting from the fourth octet. An increment
of one tells us to go one octet to the left.
Example: we need 10 hosts. 2^4 (16) will satisfy our
requirement. The n is then 4.
Now, how many times can we subtract 8 from 4? Note that
4-8 = -4 so the answer is 0. This means that we will
change the fourth octet for the next network.
What?
How many times the "n" value can be subtracted to 8 –
This allows us to determine on which octet should we
add numbers starting from the fourth octet. An increment
of one tells us to go one octet to the left.
Example: we need 1000 hosts. 2^10 (1024) will satisfy our
requirement. The n is then 10.
Now, how many times can we subtract 8 from 10? Note
that 10-8 = 2, then 2-8 = -6 so the answer is 1. This
means that we will change the third octet for the next
network.
Solution
What’s in each column?
Host
Requirement
2^n
1,750
1000
750
500
250
50
30
2^11
2^10
2^10
2^9
2^8
2^6
2^5
How many
times the "n"
32-n value can be
subtracted to
8
21
22
22
23
24
26
27
1
1
1
1
1
0
0
Which
Octet
3rd
3rd
3rd
3rd
3rd
4th
4th
Remainder
2^N (where
N=value of
Remainder)
Network
Address
Which octet – This gives us the octet that we
would change depending on the answer of the
previous column. 0 means the fourth octet, and
an increment of 1 tells us to change the octet to
the left. So, if it is 2, it means we should change
the second octet.
Solution
What’s in each column?
Host
Requirement
2^n
1,750
1000
750
500
250
50
30
2^11
2^10
2^10
2^9
2^8
2^6
2^5
How many
times the "n"
32-n value can be
subtracted to
8
21
22
22
23
24
26
27
1
1
1
1
1
0
0
Which
Octet
Remainder
3rd
3rd
3rd
3rd
3rd
4th
4th
3
2
2
1
0
6
5
2^N (where
N=value of
Remainder)
Network
Address
Remainder – This allows us to know the number that we
should add to the octet in consideration as it is equal to
2^remainder. Note that this is the remainder coming
from the subtraction of the n value to 8.
What?
Remainder – This allows us to know the number that we
should add to the octet in consideration as it is equal to
2^remainder. Note that this is the remainder coming
from the subtraction of the n value to 8.
Example: we need 10 hosts. 2^4 (16) will satisfy our
requirement. The n is then 4.
Now, how many times can we subtract 8 from 4? Note that
4-8 = -4. Since we can’t subtract 8 from 4, the remainder
is then 4.
What?
Remainder – This allows us to know the number that we
should add to the octet in consideration as it is equal to
2^remainder. Note that this is the remainder coming
from the subtraction of the n value to 8.
Example: we need 1000 hosts. 2^10 (1024) will satisfy our
requirement. The n is then 10.
Now, how many times can we subtract 8 from 10? Note
that 10-8 = 2, then 2-8 = -6. So, the remainder is 2.
Solution
What’s in each column?
Host
Requirement
2^n
1,750
1000
750
500
250
50
30
2^11
2^10
2^10
2^9
2^8
2^6
2^5
How many
times the "n"
32-n value can be
subtracted to
8
21
22
22
23
24
26
27
1
1
1
1
1
0
0
Which
Octet
Remainder
2^N (where
N=value of
Remainder)
3rd
3rd
3rd
3rd
3rd
4th
4th
3
2
2
1
0
6
5
8
4
4
2
1
64
32
Network
Address
2^N – This gives us the actual number to be added to
the octet in consideration. Take note that if the octet
reaches 256, simply make the current octet in
consideration 0 and add 1 on the octet to the right.
What?
2^N – This gives us the actual number to be added to the octet in
consideration. Take note that if the octet reaches 256, simply make
the current octet in consideration 0 and add 1 on the octet to the
right.
Example: Current network: 192.168.20.240 /28
How to solve for the next network?
n = 4 (32 – subnet mask = 32 – 28 = 4)
which octet? (4th)
how many to add? 2^4 = 16
Answer: 192.168.20.256? WRONG! (11111111 = 255)
Correct Answer: 192.168.21.0
Solution
What’s in each column?
Host
Requirement
2^n
1,750
1000
750
500
250
50
30
2^11
2^10
2^10
2^9
2^8
2^6
2^5
How many
times the "n"
32-n value can be
subtracted to
8
21
22
22
23
24
26
27
1
1
1
1
1
0
0
Which
Octet
3rd
3rd
3rd
3rd
3rd
4th
4th
Remainder
2^N (where
N=value of
Remainder)
Network
Address
3
2
2
1
0
6
5
8
4
4
2
1
64
32
50.30.0.0/21
50.30.8.0/22
50.30.12.0/22
50.30.16.0/23
50.30.18.0/24
50.30.19.0/26
50.30.19.64/27
Network Address – gives us the network address of the
current network in consideration. Take note that you start
from the IP block. Also, the subnet mask of the current
network affects the network address of the next
network, not the current one.
Solution
What’s in each column?
Host
Requirement
2^n
1,750
1000
750
500
250
50
30
2^11
2^10
2^10
2^9
2^8
2^6
2^5
How many
times the "n"
32-n value can be
subtracted to
8
21
22
22
23
24
26
27
1
1
1
1
1
0
0
Which
Octet
3rd
3rd
3rd
3rd
3rd
4th
4th
Remainder
2^N (where
N=value of
Remainder)
Network
Address
3
2
2
1
0
6
5
8
4
4
2
1
64
32
50.30.0.0/21
50.30.8.0/22
50.30.12.0/22
50.30.16.0/23
50.30.18.0/24
50.30.19.0/26
50.30.19.64/27
Note that IP Block subnet mask is insignificant!!
Try it yourself!
• Requirements: (IP Block = 192.168.0.0/11)
Network
Alpha
Amp
Anub'arak
Awts
B eta
C harlie
D ay
D elta
E cho
F oxtrot
K ing
L ightning
P rophet
R evenant
S and
V engeful
V iper
Y urnero
Hos t
10000
250
2
300
5000
2500
36
1000
500
450
40
10
50
10
7
70
5
150
Try it yourself!
• Answer:
Alpha
B eta
C harlie
Delta
E cho
F oxtrot
Awts
Amp
Y urnero
V engeful
Day
P rophet
K ing
S and
L ightning
R evenant
V iper
Anub'arak
10000
5000
2500
1000
500
450
300
250
150
70
36
50
40
7
10
10
5
2
2^n
14
13
12
10
9
9
9
8
8
7
6
6
6
4
4
4
3
2
S ubnet Mas k
18
19
20
22
23
23
23
24
24
25
26
26
26
28
28
28
29
30
Network addres s
192.168.0.0 /18
192.168.64.0/19
192.168.96.0/20
192.168.112.0/22
192.168.116.0/23
192.168.118.0/23
192.168.120.0/23
192.168.122.0/24
192.168.123.0/24
192.168.124.0/25
192.168.124.128/26
192.168.124.192/26
192.168.125.0/26
192.168.125.64/28
192.168.125.80/28
192.168.125.96/28
192.168.125.112/29
192.168.125.120/30