Question

start with the default mask. in the fourth octet,
from the right to the left, keep zeros until you
have enough host addresses. the rest of the bits
will be changed to 1s. for 30 host addresses, we
need 5 zeros. 2^5 = 32 make the fourth octet
1110 0000.
item 12 (4 points) new subnet mask in binary
(write out all 32 bits; 8 bits in each box below):
11111111 11111111
11111111 11100000
question 13 3 pts
the right - most 1 in the binary mask is the subnet
boundary.
item 13 (3 points) the decimal value of the
subnet boundary (right most 1 bit) is the block
size. block size =

Explanation:

Step1: Identify the right - most 1 in the fourth octet

The fourth octet of the subnet mask is \(11100000\). The right - most 1 is in the \(2^{5}\) position.

Step2: Calculate the block size

The formula for block size based on the position of the right - most 1 (say \(n\) is the position from the right, starting from 0) is \(2^{n}\). Here, for the bit pattern \(11100000\), the right - most 1 is the 5th bit from the right (since \(11100000 = 1\times2^{7}+1\times2^{6}+1\times2^{5}+0\times2^{4}+0\times2^{3}+0\times2^{2}+0\times2^{1}+0\times2^{0}\)). So, using the formula \(2^{n}\), with \(n = 5\), we have \(2^{5}=32\).

Answer:

\(32\)