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+15 votes
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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. $204.204.204.128/255.255.255.128$
    $204.204.204.64/255.255.255.192$
    $204.204.204.0/255.255.255.192$
asked in Computer Networks by Boss (19.1k points)
edited by | 2.2k views
+2

$Remark:$
Subnet Address: First IP in the subnet. 
Subnet Mask: all the net bits + subnet bits will be 1 and host bits will be 0.
DBA: Last IP in the subnet. 

0

DBA: all the net bits + subnet bits will be 1 and host bits will be 0.

Are you sure?

0
@Manu

DBA: all the n/w + subnet bits will be as it is, and host bits will be all 1's..
0
m sorry @anu, i wanted to write subnet mask, correcting it
0
no need to say sorry. i just want to make sure m not study wrong things.:)
0

DBA: keep network id and subnet id bits same, make all host bits as 1s

5 Answers

+25 votes
Best answer

Answer is D.

MSB in last 8 bits helps us to get two subnets

10000000 $\to $ subnet1

00000000 $\to $ subnet2

subnet2 is divided into 2 more subnets using 7th bit

00000000 $\to $ subnet2(0)

01000000 $\to$ subnet2(1)

answered by Junior (939 points)
edited by
+1
Here subnetmask(SM) of subnet1 is 255.255.255.128 because we do all 1's in NID and SID and here only one bit is chosen for subnet id
In. subnet2 we choose two bit for SID so here the SM becomes 255.255.255.192 because two 1's in fourth octet
Also for subnet 3 SM becomes 255.255.255.192
0

well i have a doubt what if options were given like this which one you would  select

[A]   204.204.204.128/255.255.255.128
       204.204.204.192/255.255.255.192
       204.204.204.0/255.255.255.192

[B]  204.204.204.128/255.255.255.128
      204.204.204.192/255.255.255.192
      204.204.204.64/255.255.255.192

[C]  204.204.204.128/255.255.255.128
      204.204.204.64/255.255.255.192
      204.204.204.0/255.255.255.192

0
A little more explanation was needed .....
+9 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.  

answered by Junior (783 points)
+9 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)

answered by Boss (11k points)
+6 votes

MSB in last 8 bits is in use to get subnet since it is class c ip

10000000 / 128(mask)subnet id bit(10) --->subnet1 ( since need 100 host so we assign max 128 here so 28 gone waste).
01000000/192( mask) subnet id bit(01)--->subnet2 ( since need 50 host so we assign max 64 here so 14 gone waste). here we can go for option c b/c we get coolision of ip address since subnet 1 also have 11000000 and subnet 2 also . so here D is answer
00000000 /192( mask)subnet id bit(00)--->subnet2(0)  ( since need 50 host so we assign max 64 here so 14 gone waste).

D is answer.

answered by Veteran (60.4k points)
0
In your first case, 1000000 or 128 the subnet ID would be just 1 and not 10. If it would have been 10 then we would have not got 128 addresses or hosts. It would have been just 64 hosts.
+1
yupp..me too have the same issue..instead of 10 the subnet id would be just 1 ...!!
+1 vote

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

So, Option D is the answer.

answered by Junior (559 points)
edited by
Answer:

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