Bit Binary Value

I had a hard time understanding the subnetting at first. But thanks to a method from Mastering Computers combined with various other sites, books, charts, and braindumps it’s a breeze now.

I have laid out the below chart in hopes that it will help someone else pass the TCP/IP and IIS 4.0 exam and more importantly, help someone subnet their company’s IP address.

SUBNET CHART:

Bit Binary Value

Bits

Mask

Binary Pos. from Right

"C" Class
Hosts

Binary Pos. from Right

"B" Class
Hosts

Binary Pos. from Right

"A" Class
Hosts

Subnets
(Max Usable)

Bits²

*1

2^h-2=

2^h-2=

2^h-2=

*2

00000001

1

255

0

-

8

254

16

65,534

254

00000010

2

254

1

-

9

510

17

131,070

126

00000100

4

252

2

2

10

1,022

18

262,142

62

00001000

8

248

3

6

11

2,046

19

524,286

30

00010000

16

240

4

14

12

4,094

20

1,048,574

14

00100000

32

224

5

30

13

8,190

21

2,097,150

6

01000000

64

192

6

62

14

16,382

22

4,194,302

2

10000000

128

-

7

126

15

32,766

23

8,388,606

-

Layout your Bits first
Layout your Mask – See Rule 1
Layout Class "C" Hosts – for class "C" only = Bits minus 2 or 2^h-2; h = Binary position from right
Layout Subnets – See Rule 2

*Rule1 = (Mask + Bits to the right) Notice the Pattern – m 128+ b 64=192, m 192+b 32=224, m 224+ b 16=240, m 240+ b 8=248

*Rule2 = Carry Class C #14 across and reverse the order of Class Hosts Numbers (Ex. 6, 2 down & 30, 62, 126, 254 up)

X.X.X.X

1^st Binary Digit

Octet = 8 bits

Minus 2 Block Chart
The first and last address blocks are not useable

Class A – N.H.H.H

0

1 – 126

Block Size 64

0-63, 64-127, 128-191, ~~192-255~~

Loop Back Address – 127.0.0.1

127

Class B – N.N.H.H

10

128 – 191

Block Size 32

0-31, 32-63, 64-95, 96-127, 128-159, 160-191, 192-223, ~~224-255~~

Class C – N.N.N.H

110

192 – 223

Block Size 16

0-15, 16-31, 32-47, 48-63, 64-79, 80-95, 96-111, 112-127, 128-143, 144-159, 160, 175, 176-191, 192-207, 208-223, 224-239, ~~240-255~~

Multicasting, Research

224, 225

Minus 2 Block Chart: 1) The top and bottom ranges are not usable. 2) In the valid ranges, the first and last IP addresses are not usable. Example: checking the validity of a class IP address 198.204.35.14 with a mask of 255.255.255.224. If you forget the invalid ranges, you will incorrectly think that there’s nothing wrong with the address and subnet mask combination after the ANDing process.

EXAM QUESTIONS:

As far as the actual test questions, the TCP/IP 4.0 site (MCSE dated 18.5.98) could not have been more accurate, as far as, what transcender questions were similar on the test.

Thanks BRAINDUMPS

Bit Binary Value	Bits	Mask	Binary Pos. from Right	"C" Class Hosts	Binary Pos. from Right	"B" Class Hosts	Binary Pos. from Right	"A" Class Hosts	Subnets (Max Usable)
	Bits²	*1	2^h-2=		2^h-2=		2^h-2=		*2

00000001	1	255	0	-	8	254	16	65,534	254
00000010	2	254	1	-	9	510	17	131,070	126
00000100	4	252	2	2	10	1,022	18	262,142	62
00001000	8	248	3	6	11	2,046	19	524,286	30
00010000	16	240	4	14	12	4,094	20	1,048,574	14
00100000	32	224	5	30	13	8,190	21	2,097,150	6
01000000	64	192	6	62	14	16,382	22	4,194,302	2
10000000	128	-	7	126	15	32,766	23	8,388,606	-

X.X.X.X	1^st Binary Digit	Octet = 8 bits	Minus 2 Block Chart The first and last address blocks are not useable
Class A – N.H.H.H	0	1 – 126	Block Size 64	0-63, 64-127, 128-191, ~~192-255~~
Loop Back Address – 127.0.0.1		127
Class B – N.N.H.H	10	128 – 191	Block Size 32	0-31, 32-63, 64-95, 96-127, 128-159, 160-191, 192-223, ~~224-255~~
Class C – N.N.N.H	110	192 – 223	Block Size 16	0-15, 16-31, 32-47, 48-63, 64-79, 80-95, 96-111, 112-127, 128-143, 144-159, 160, 175, 176-191, 192-207, 208-223, 224-239, ~~240-255~~
Multicasting, Research		224, 225