QUESTION IMAGE
Question
complete subnetting information below:
class c ip address 192.168.10.0
class c default mask 1111 1111.1111 1111.1111 1111.0000 0000
new subnet mask in binary: 1111 1111.1111 1111.1111 1111.1100 0000
item 14 (2 points) highlight the right - most 1 in the binary mask.
item 15 (2 points) the decimal value of that bit is the block size. block size =
item 16 (2 points) new subnet mask in decimal:
Item 14:
The new subnet mask in binary is \(1111\ 1111.1111\ 1111.1111\ 1111.1100\ 0000\). The right - most \(1\) is in the \(3\)rd position from the left in the fourth octet.
Item 15:
The formula for the value of a bit in a binary number is \(2^{n}\), where \(n\) is the position of the bit (starting from \(0\) on the right). For the bit in the \(3\)rd position (counting from the right in the fourth octet), \(n = 3\). So, \(2^{3}=8\).
Item 16:
- First octet: \(1111\ 1111\) in binary. Using the formula \(\sum_{i = 0}^{7}b_{i}\times2^{i}\), where \(b_{i}\) is the \(i\)th bit. \(\sum_{i = 0}^{7}1\times2^{i}=2^{8}-1 = 255\)
- Second octet: \(1111\ 1111\) in binary. Using the formula \(\sum_{i = 0}^{7}b_{i}\times2^{i}\), where \(b_{i}\) is the \(i\)th bit. \(\sum_{i = 0}^{7}1\times2^{i}=2^{8}-1 = 255\)
- Third octet: \(1111\ 1111\) in binary. Using the formula \(\sum_{i = 0}^{7}b_{i}\times2^{i}\), where \(b_{i}\) is the \(i\)th bit. \(\sum_{i = 0}^{7}1\times2^{i}=2^{8}-1 = 255\)
- Fourth octet: \(1100\ 0000\) in binary. Using the formula \(\sum_{i = 0}^{7}b_{i}\times2^{i}\), where \(b_{6}=1,b_{7}=1\) and \(b_{i}=0\) for \(i = 0,1,2,3,4,5\). \(2^{6}+2^{7}=64 + 128=192\)
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- Item 14: The right - most \(1\) is in the \(3\)rd position from the left in the fourth octet (binary: \(1111\ 1111.1111\ 1111.1111\ 1111.1100\ 0000\), fourth octet \(1100\ 0000\), right - most \(1\) is the second \(1\) from the left in the fourth octet).
- Item 15: \(8\)
- Item 16: \(255.255.255.192\)