# Calculating subnet hosts

Discussion in 'MCSE' started by BagelBoy, Nov 4, 2005.

1. ### BagelBoyGuest

everything up to the part about "the remaining 5 bits identify the
hosts on that subnet". Here is the question and answer:

Your company has been assigned a network ID of 197.132.240.0. Your
network has 5 subnets. The max number of hosts per subnet is 25. Which
subnet mask should you use for the network?
A. 255.255.255.192
B. 255.255.255.224
C. 255.355.255.240
D. 255.255.255.248

Correct answer is B. A network ID of 197.132.240.0 is a class C
network. A class C network falls in the range of 192.0.0.0 to
223.255.255.0. The high-order bits of a class C network ID are always
set to 1 1 0. The next 21 bits (completing the first three octets)
identify the remaining portion of a class C network ID. In this case,
the first three octets of the network ID are 197.132.240, which
translates to a binary value of 11000101 10000100 11110000. The default

subnet mask of a class C network is 255.255.255.0 (binary 11111111
11111111 11111111 00000000), indicating that the first three octets
define the network ID. The remaining octect can be used to define which

bits should be used to identify the network's subnetting structure and
which should be used to identify the host on those subnet. For a class
C network, a subnet mask of 255.255.255.224 (binary 11111111 11111111
11111111 11100000) indicates that the first three octets and the first
3 bits of the last octet identify the subnetted network ID. The
remining 5 bits identify the hosts on that subnet. Because 3 bits are
used to define the subnetting structure of this class C network, the
network will support up to 8 subnets and up to 30 hosts per subnet.

I understand everything up to this:
Because 3 bits are used to define the subnetting structure of this
class C network, the network will suport up to 8 subnets and up to 30
hosts per subnet.

Can someone please elaborate on this for me?

This is what I understand about calcing subnets so far:
Binary is a Base2 system VS. Decimal is a Base10 system.
This means that all binary numbers can be expressed as 2n.
Which means each 8-bit field matches this chart:
128 64 32 16 8 4 2 1
27 26 25 24 23 22 21 20
Use this chart to convert number between decimal and
binary.
Example:
What is the binary value of 153?
128 64 32 16 8 4 2 1
27 26 25 24 23 22 21 20
1 0 0 1 1 0 0 1
Or 128+16+8+1=153

Classes:
Class A addresses have the highest level bit (27) turned
off. So the address must be less than 128.
Class A range = 0-127

Class B addresses have the highest level bit (27) turned
on and the 2nd highest-level bit (26) turned off. So the
addresses will differ from Class A by the first 64 bytes.
Class B range = 128-191 (127+64=191)

Class C addresses have the highest level bit (27) and the
2nd highest-level bit (26) turned on and the 3rd highest-
level bit (25) turned off. So the addresses will differ
from Class B by the first 32 bytes.
Class C range = 192-223 (191+32=223)

# Of networks and hosts:
Points to remember
- The 0. and 127. networks are used for loopback and
must be subtracted from the Class A networks.
- Each 8-bit field has a maximum value of 255, but
to avoid the possibility of all 1's or all 0's subtract 2
when computing the number of networks and hosts.

Class A
Net: 0-127= 128-2= 126 networks
Hosts: 253x253x253=16 million
Class B
Net: 128-191=64 (different first bytes) x253=16,000
networks
Hosts: 253x253=64,000 hosts
Class C
Net: 32 (different first bytes) x253x253=2 million
Hosts: 253 hosts

It's this last part to calc the hosts that's baffling me. Any help to
understand this would be greatly appreciated.

BB

BagelBoy, Nov 4, 2005

2. ### Frisbee®Guest

Frisbee®, Nov 4, 2005

3. ### KurtGuest

The "remaining number of bits" is a terrible way of saying "the in the mask
bits that are zeros". Subnetting works from left to right. So working with
Answer A. 255.255.255.192, the last octet looks like:

11000000 there are 2 "set" bits (1's) meaning that the corresponding bits in
the IP address are part of the network. The 6 bits of an IP address that
correspod to the 6 bits of the mask that are "cleared" (zeros) determine the
unique host. So 2^6 = 64 (minus 2), for a total of 62 usable hosts.
Answer B. 255.255.255.224 has 5 "zero" bits, 2^5 =32, so you have 30 usable
hosts. Since each successive answer has one more network bit, it will have
1/2 as many hosts. So Answer C works out to 16, and D is 8 (always minus 2

....kurt

Kurt, Nov 4, 2005
4. ### kpgGuest

That is calssic.

kp "I understand everything except..." g

kpg, Nov 4, 2005
5. ### BagelBoyGuest

Thanks all! I really appreciate the advice.

BagelBoy, Nov 4, 2005