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41 votes
41 votes

A company has a class $C$ network address of $204.204.204.0$. It wishes to have three subnets, one with $100$ hosts and two with $50$ hosts each. Which one of the following options represents a feasible set of subnet address/subnet mask pairs?

  1. $204.204.204.128/255.255.255.192$
    $204.204.204.0/255.255.255.128$
    $204.204.204.64/255.255.255.128$
  2. $204.204.204.0/255.255.255.192$
    $204.204.204.192/255.255.255.128$
    $204.204.204.64/255.255.255.128$
  3. $204.204.204.128/255.255.255.128$
    $204.204.204.192/255.255.255.192$
    $204.204.204.224/255.255.255.192$
  4. $204.204.204.128/255.255.255.128$
    $204.204.204.64/255.255.255.192$
    $204.204.204.0/255.255.255.192$
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6 Answers

Best answer
39 votes
39 votes

Answer is D.

MSB in last $8$ bits helps us to get two subnets

  • $10000000 \to $ subnet$1$
  • $00000000 \to $ subnet$2$

subnet$2$ is divided into $2$ more subnets using $7$th bit

  • $00000000 \to $ subnet$2(0)$
  • $01000000 \to$ subnet$2(1)$
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51 votes
51 votes

the company has a class C network address of 204.204.204.0

The above given is network id.

So , host bits in above Id= 8.

Now total addresses in this network can be 2^8=256.

Now these 256 available addresses have to be divided into three subnets with one with 100 hosts and two with 50 hosts each. Since, in subnetting number of hosts can be allocated only in powers of 2, So our division should look like below

So, it is clear that one should have subnet mask of 255.255.255.128 and other two should have subnet mask of 255.255.255.192

Considering this, we can eliminate options (a) and (b).

Now, options (c) and (d) have to be inspected closely.

204.204.204.128/255.255.255.128

matches in both options, so don't need to consider it because the subnet address is given as

204.204.204.128/255.255.255.128

Last octet has 7 host bits, hence we can allocate 128 addresses using this.

Look at this value of option (c)

204.204.204.224/255.255.255.192

subnet address last octet= 224=1110 0000. Host bits=5. So we cannot allocate 64 hosts using this subnet address. So this is wrong.

Option (c) eliminated.

If we look at last two values of option (d)

204.204.204.64/255.255.255.192 
204.204.204.0/255.255.255.192

Considering last octets, 64= 0100 0000 Since mask is 192(first two bits from the left are subnet bits in bold) This is subnet 1

0-0000 0000 and this is subnet 0.

Both can give 64 hosts as they have host bits = 6.

So answer is (d)

27 votes
27 votes

Here we will check options one by one, although the explanation seems to me long but once you will get how to check then yo will able to handle ;

considering option (A)

1) 11001100.11001100.11001100.10000000/11111111.11111111.11111111.11000000 Network bits-26 (as 26 bits are 1 in the subnet mask) so sub-network id = 204.204.204.128 .  6 bits ‘0’ in subnet mask,i.e. 64(i.e. 2^6) -2 = 62 hosts.

2) 11001100.11001100.11001100.00000000/11111111.11111111.11111111.10000000 Network bits-25 (as 25 bits are 1 in the subnet mask) so sub-network id = 204.204.204.0 .  7 bits ‘0’ in subnet mask,i.e. 128(i.e.2^7) -2 =126 hosts.

3) 11001100.11001100.11001100.01000000/11111111.11111111.11111111.10000000 Network bits-25 (as 25 bits are 1 in the subnet mask) so sub-network id = 204.204.204.0 .  7 bits ‘0’ in subnet mask,i.e. 128(i.e.2^7) -2 = 126 hosts .

Though the networks are divided into subnets containing 62,126,126 hosts which satisfy our minimum criteria of 50,50,100. But, two subnets have same network ID thus,they are not different.Thus,this division is only of two subnets but we need 3 ,so A option is incorrect.

consider option (B)

1) 11001100.11001100.11001100.00000000/11111111.11111111.11111111.11000000 Network bits-26 (as 26 bits are 1 in the subnet mask) so sub-network id = 204.204.204.0 .  6 bits ‘0’ in subnet mask,i.e. 64(i.e. 2^6)-2=62 .

2) 11001100.11001100.11001100.11000000/11111111.11111111.11111111.10000000 Network bits-25 (as 25 bits are 1 in the subnet mask) so sub-network id = 204.204.204.128 . 7 bits ‘0’ in subnet mask,i.e. 128(i.e.2^7)-2=126 hosts 

3) 11001100.11001100.11001100.01000000/11111111.11111111.11111111.10000000 Network bits-25 (as 25 bits are 1 in the subnet mask) so sub-network id = 204.204.204.0 . 7 bits ‘0’ in subnet mask,i.e. 128(i.e.2^7)-2=126 hosts .

Though the networks are divided into subnets containing 62,126,126 hosts which satisfy our minimum criteria of 50,50,100. But, two subnets have same network ID thus,they are not different.Thus,this division is only of two subnets.hence incorrect

consider option (C)

1) 11001100.11001100.11001100.10000000/11111111.11111111.11111111.10000000 Network bits-25 (as 25 bits are 1 in the subnet mask) so sub-network id == 204.204.204.128 . 7 bits ‘0’ in subnet mask,i.e. 128(i.e. 2^7) -2 =126 hosts.

2) 11001100.11001100.11001100.11000000/11111111.11111111.11111111.11000000 Network bits-26 (as 25 bits are 1 in the subnet mask) so sub-network id == 204.204.204.192  . 6 bits ‘0’ in subnet mask,i.e. 64(i.e.2^6)-2=62 hosts.

3) 11001100.11001100.11001100.11110000/11111111.11111111.11111111.11000000 Network bits-26 (as 26 bits are 1 in the subnet mask) so sub-network id == 204.204.204.192 .  6 bits ‘0’ in subnet mask,i.e. 64(i.e.2^6) -2=62 hosts.

Though the networks are divided into subnets containing 62,62,126 hosts which satisfy our minimum criteria of 50,50,100. But, two subnets have same network ID thus,they are not different.Thus,this division is only of two subnets. hence incorrect

consider option (D)

1) 11001100.11001100.11001100.10000000/11111111.11111111.11111111.10000000 –> Network bits-25 (as 25 bits are 1 in the subnet mask) so sub-network id == 204.204.204.128 .   7 bits ‘0’ in subnet mask,i.e. 128(i.e. 2^7)-2=126 hosts.

2) 11001100.11001100.11001100.01000000/11111111.11111111.11111111.11000000 –> Network bits-26 (as 25 bits are 1 in the subnet mask) so sub-network id == 204.204.204.64. 6 bits ‘0’ in subnet mask,i.e. 64(i.e.2^6)-2=62 hosts.

3) 11001100.11001100.11001100.00000000/11111111.11111111.11111111.11000000 –> Network bits-26 (as 26 bits are 1 in the subnet mask) so sub-network id == 204.204.204.0 .  6 bits ‘0’ in subnet mask,i.e. 64(i.e.2^6)-2=62 host bits. This satisfies the minimum criteria of 50,50 and 100 hosts and all subnet IDs are different .

THUS , option (D) is correct. satisfying the requirements.  

13 votes
13 votes

Hope this helps.....................................................

So, Option D is the answer.

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