A
0 in each position indicates that the value for that position is not added to
the total. A 0 in every position yields a total of 0, as follows in Table 6-3.
Table 6-3
|
Binary
Positional Notation with Digit of 0
|
|
|
|
|
||||
Decimal Value
|
128
|
64
|
32
|
16
|
8
|
4
|
2
|
1
|
|
|
|
|
|
|
|
|
|
|
|
Binary Digit
|
|
0
|
0
|
0
|
0
|
0
|
0
|
0
|
0
|
|
|
|
|
|
|
|
|
|
|
Position Value
|
0
|
0
|
0
|
0
|
0
|
0
|
0
|
0
|
|
|
|
|
|
|
|
|
|
|
|
For
the conversion of a 32-bit IPv4 address, you identify the 4 bytes or octets
that make up this address. You then convert each of these four octets to
decimal. As an example, convert the IPv4 address
10101100000100000000010000010100 to the bytes of a dotted decimal address. The
conversion starts with the low-order bits,
in this case byte 00010100, and con-tinues up to the high-order byte.
How
To

Step
1. Divide the
32 bits into 4 octets, as follows:
10101100.00010000.00000100.00010100
Step
2. Convert the
low-order byte 00010100 first, as seen in Table 6-4.
Step
3. Convert the
next-highest byte, which is 00000100, as seen in Table 6-5.
Step
4. Continue
converting with the next byte, 00010000, as seen in Table 6-6.
Step
5. Convert the
highest byte, 10101100, as seen in Table 6-7.
Step
6. Write down the four
numbers with dots separating the octets, such as 172.16.4.20.
Table 6-4
|
Conversion of
Binary Number 00010100
|
|
|
|
|
||||
Decimal Value
|
128
|
64
|
32
|
16
|
8
|
4
|
2
|
1
|
|
|
|
|
|
|
|
|
|
|
|
Binary Digit
|
|
0
|
0
|
0
|
1
|
0
|
1
|
0
|
0
|
|
|
|
|
|
|
|
|
|
|
Position Value
|
0
|
0
|
0
|
16
|
0
|
4
|
0
|
0
|
|
|
|
|
|
|
|
|
|||
Total Value
|
|
|
|
0 + 0 + 0
+ 16 + 0 + 4 + 0 + 0 = 20
|
|
|
|||
|
|
|
|
|
|
||||
Table 6-5
|
Conversion of
Binary Number 00000100
|
|
|
|
|
||||
Decimal Value
|
128
|
64
|
32
|
16
|
8
|
4
|
2
|
1
|
|
|
|
|
|
|
|
|
|
|
|
Binary Digit
|
|
0
|
0
|
0
|
0
|
0
|
1
|
0
|
0
|
|
|
|
|
|
|
|
|
|
|
Position Value
|
0
|
0
|
0
|
0
|
0
|
4
|
0
|
0
|
|
|
|
|
|
|
|
|
|||
Total Value
|
|
|
0 + 0 + 0
+ 0 + 0 + 4 + 0 + 0 = 4
|
|
|
|
|||
|
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