Question

An organization requires a range of IP addresses to assign one to each of its 1500 computers. The organization has approached an Internet Service Provider (ISP) for this task. The ISP uses CIDR and serves the requests from the available IP address space 202.61.0.0/17. The ISP wants to assign an address space to the organization which will minimize the number of routing entries in the ISP's router using route aggregation. Which of the following address spaces are potential candidates form which the ISP can allot any one to the organization ?
I. 202.61.84.0/21
II. 202. 61.104.0/21
III. 202.61.64.0/21
IV. 202.61.144.0/21

• A. I and II only
• B. II and III only
• C. III and IV only
• D. I and IV only

Answer 1

The correct option is B II and III only

$\Rightarrow 11111111111111111000000000000000$
$\Rightarrow \mathrm{255.255.128.0}$
$1500\simeq 2048=2^{11}=2^3\times 2^8$
$=2^{32-21}=2^{11}=/21$
$/17\Rightarrow \mathrm{202.61.0.0}/17$
$\begin{array}{c}0.0\Rightarrow \begin{array}{c}0\underset{–}{0000}\underset{–}{0000.00000000}\\ 0\underset{–}{0000}\underset{–}{111.11111111}\end{array}\}(0.0-7.255)\end{array}$
$\begin{array}{c}\begin{array}{c}0\underset{–}{0001}\underset{–}{000.00000000}\\ 0\underset{–}{0001}\underset{–}{111.11111111}\end{array}\}(8.0-15.255)\end{array}$
$\begin{array}{c}\begin{array}{c}0\underset{–}{0010}\underset{–}{000.00000000}\\ 0\underset{–}{0010}\underset{–}{111.11111111}\end{array}\}(16.0-23.255)\end{array}$
Sequence is 0,8, 16, 24,32,40, 48, 56, 64, 72, 80, 96, 104, 112, 120.
Hence 64 and 104 are present in sequence so it is the possible IP addresses.
So option (d) is correct.