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An organization has a class $C$ network and wants to form subnets for four departments, with hosts as follows.
$$\begin{array}{|c|c|}  \hline Department & \text{No. of hosts required} \\ \hline W & 72 \\ \hline X & 35 \\ \hline Y & 20 \\ \hline Z & 18 \\ \hline   \end{array}$$
What is the possible subnet masks for the departments $W,X,Y\:  and \: Z$
 

  1. $W:255.255.255.223, \: X:255.255.255.192,\: Y\&Z :255.255.255.128$
  2. $W: 255.255.255.128,\: X :255.255.255.192,\: Y\&Z : 255.255.255.224$
  3. $W: 255.255.255.192,\: X :255.255.255.128,\: Y\&Z : 255.255.255.224$
  4. $W: 255.255.255.223,\: X \& Y\& Z : 255.255.255.223$
asked in Computer Networks by Boss (17.2k points)
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+2 votes

Class C, 24 bits reserved for Network Portion and 8 Bits reserved for Host portion.

W requires 72 hosts, But we allot always in the power of 2, ===> 128 we can allot

So, fix 0$^{th}$ bit of Last octate from MSB as 0, it will indicate allocated ip address for W.

Subnet Mask = Network Portion + Subnet Portion = 24 + 1 = 25 Bits==> 255.255.255.128/25

Note :- Prefix for W = 0

 

X requires 35 hosts, But we allot always in the power of 2, ===> 64 we can allot but not 32 due to satisfy our requirement

Already we fixed 0$^{th}$ bit of Last octate from MSB as 0 for W

So, we have to differentiate X from W, fix 0$^{th}$ bit as 1, and 1$^{st}$ Bit as 0, it will indicate allocated ip address for X

Subnet Mask = Network Portion + Subnet Portion = 24 + 2 = 26 Bits==> 255.255.255.192/26

Note :- Prefix for X = 10

 

Y requires 20 hosts, But we allot always in the power of 2, ===> 32 we can allot

Already we fixed 0$^{th}$ bit and 1$^{st}$  of Last octate from MSB.

So, we have to differentiate Y from X , fix 1$^{st}$ bit as 1, 2$^{nd}$ bit as 0, it will indicate allocated ip address for Y

Subnet Mask = Network Portion + Subnet Portion = 24 + 3 = 27 Bits==> 255.255.255.224/27

Note :- Prefix for Y = 110

 

Z requires 18 hosts, But we allot always in the power of 2, ===> 32 we can allot

Already we fixed 0$^{th}$ bit and 1$^{st}$  of Last octate from MSB.

So, we have to differentiate Z from Y , fix 1$^{st}$ bit as 1, 2$^{nd}$ bit as 1, it will indicate allocated ip address for Z

Subnet Mask = Network Portion + Subnet Portion = 24 + 3 = 27 Bits==> 255.255.255.224/27

Note :- Prefix for W = 111

answered by Veteran (60.2k points)
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