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This is a question i got and i just dont under stand how to do it.

An organization is assigned the network number 172.18.0.0/16, and it must create a set of subnets that supports up to 90 hosts on each subnet.

How do you Define subnet mask for this organization

What is the IP number for the Subnet number 421 explicitly in binary and decimal notation

What is the explicitly IP address of the host number 68 on the Subnet number 421


What is the broadcast address for the Subnet number 421

Any guideience would be great cause im stressing out :(

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Last Post by Casper429
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Where are you coming up with 421? A /16 subnet is 255.255.0.0. The ip would be 172.18.0.whatever

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Yes That the part I mstly didn't understand
I assumed that If the subnet number was 421 what would the ip address be

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First off what are you trying to do. I'm assuming connecting an Internet router? Your ip should be 172.18.0. Whatever they gave you. You subnetmask should be 255.255.0.0 Your gateway should be 172.18.0. Whatever the isp assigned.

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Here it goes :_)

Bits from the left to right
8 7 6 5 4 3 2 1 8 7 6 5 4 3 2 1th bit
65535, 32768, 16384, 8192, 4096, 2048, 1024, 512, 256, 128, 64, 32, 16, 8, 4, 2

We will use VLSM to create such a beautiful class B address with 512 networks,
and 126 host in each one (128 - 2, one-for network address, one-for broadcast
is reserved)

172.18.0.0/16

11111111.11111111.00000000.00000000 network bits
00000000.00000000.11111111.11111111 host bits

Using VLSM we will borrow host bits to create subnets with 90 host (126 actually)

0 1 1 1 1 1 1 1 < last octet (4th)
7 6 5 4 3 2 1th bit
128, 64, 32, 16, 8, 4, 2

7 bits will give as more than enough space for hosts. We need 90, we must have at least 126.

1 1 1 1 1 1 1 1 10000000 3rd and 4th octet
8 7 6 5 4 3 2 1 8th bit of the 4th octet
512, 256, 128, 64, 32, 16, 8, 4, 2


Last bit from 4th octet will go to the 3rd octet giving au 512 networks to run for.

172.18.0.0/25
172.18.0.0 255.255.255.128
So now we have 512 subnets with 126 host each.

172.18.0.0/25 network 1th
172.18.0.1-126 hosts
172.18.0.127 broadcast

172.18.0.128/25 network 2nd
172.18.0.129-254 hosts
172.18.0.255 broadcast

172.18.1.0/25 network 3rd
172.18.1.1-126 hosts
172.18.1.127 broadcast

172.18.1.128/25 network 4th
172.18.1.129-254 hosts
172.18.1.255 broadcast

172.18.2.0/25 network 5th
172.18.2.1-126 hosts
172.18.2.127 broadcast

and so on
172.18.2.128/25
172.18.3.0/25
172.18.3.128/25
172.18.4.0/25

172.18.210.0/25 network 420th
172.18.210.1-126 hosts
172.18.210.127 broadcast

172.18.210.128/25 network 421 The One :_)
172.18.210.129-254 hosts
172.18.210.255 broadcast


172.18.210.129 129 first host + 67 = 196

172.18.210.196/25 This is 68th host on the 421th subnet of 172.18.0.0/25
network made by Variable Length Subnet Masking (VLSM) from 172.18.0.0/16
address giving us 512 networks with 126 host each. (Remember that first host
address is reserved for network, and last for broadcast.)


172.18.210.196/25
172.18.210.196 255.255.255.128

host
10101100.00010010.11010010.11000100

subnet mask
11111111.11111111.11111111.10000000

[End of File]

Edited by Windgator: n/a

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Thanks for that but can you describe it by simple way, please???
I need to find the question solution of the close time.

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Windgators answer is accurate..
but i have a question on the last part..
How did you come up with the 42st subnet??
Did u use any binary conversion?? (Or is it just by dividing the number of subnets )
If you can please explain this..
Thanks

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