Network IP Addressing

Question

IP : 199. 166.15.119

SubNet Mask: 255. 255. 255. 240

Then Find the

(i) SubNet ID ?

(ii) SubNet No. ?

(iii) First host of the SubNet ?

(iv) Last host of the SubNet?

(v) third host of the first SubNet?

I found ambiguity in ans person to person .

Comments

????

---

Subnet ID = 199.166.15.112.

Subnet number = 16. But GATE still uses old representation hence 14.[follow Bikram sir's comment : https://gateoverflow.in/1265/gate2007-67-isro2016-72]

4 bits are used for subnetting, hence 0000,0001,0010...1111 are the subnets

First usable host of the first subnet = 199.166.15.0000 0001 = 199.166.15.1

Last usable host of the first subnet = 199.166.15.0000 1111 = 199.168.15.15

Last usable host of the whole subnet = 199.168.15.1111 1110 = 199.168.15.254

Third host of the 1st subnet = 199.168.15.0000 0011 = 199.166.15.3

---

@Ashwin Kulkarni the last octet of the subnet id is as you said 112 which is 0111 0000 right? Isn't it the 7th subnet id?

I read the comments from the link you shared. There is utter confusion with subnet numbers.

According to gate, subnet bits as 0000 and 1111 should not be considered right?

Please clear these doubts.

---

@minipanda,

Last octate of 199.166.15.112. is 0111 0000 So it will be 8th subnet. Because subnets are starting from 0000....1111

And hence given IP will belong to 8th subnet. This is the meaning.

And yes they haven't considered the subnets 0000 and 1111. But basically 16 subnets will be formed. And now a days all subnets are usable. so as per Bikram sir said go with the closest option.

---

Yes sorry...that should be 8. Thank you :)

---

@Ashwin Kulkarni 

Last usable host of the first subnet = 199.166.15.0000 1111 = 199.168.15.15

if we are talking about Last usable host of the first subnet then it should be 199.166.15.0000 1110 = 199.168.15.14

Please check once.

---

What is meant here by SubnetNo. ? - Is it no of possible subnets or something else (means 8 th subnet..)?

---

I think the question seems to be incomplete. It didn’t mention how many subnet bits should be there or number of bits of network ID. Without that we cannot have clear idea of number of subnets as there can be multiple.

Answer 1

IP : 199. 166.15.119 (0111 0111)

SubNet Mask: 255. 255. 255. 240 (1111 0000)

Subnet Id: ANDing of IP and Subnet Mask 199.166.15.119 AND 255.255.255.240 is 199.166.15.112

No. of Subnet = 199 (this is class C network) so number of 1's in fourth octect will give no of subnet that is 2^4 =16

(iii) First host of the SubNet

    199.166.15.0111 0001 tha is 199.166.15.113

(iv) Last host of the SubNet

199.166.15.0111 1110    (not 01110 1111) because it is for broadcasting

So ,  199.166.15.126 is answer .

(v) third host of the first SubNet

 199.166.15.0000 0011 here 0000 considered as first subnet and 0011 is considered as 3rd host as 4 bits are for host id and 4 bits are for network id.

So 199.166.15.3 will be the answer

Answer 2

This is "Ctrl + C & V" version of @Ashwin Kulkarni Answer.

I feel there should be a correction. (find as marked blue)

Subnet ID = 199.166.15.112.

4 bits are used for subnetting, hence 0000,0001,0010...1111 are the subnets

First usable host of the first subnet = 199.166.15.0000 0001 = 199.166.15.1

Last usable host of the first subnet = 199.166.15.0000 1111 = 199.168.15.15 should be 199.166.15.0000 1110 = 199.168.15.14

Last usable host of the whole subnet = 199.168.15.1111 1110 = 199.168.15.254

Third host of the 1st subnet = 199.168.15.0000 0011 = 199.166.15.3