### Discussion :: Subnetting - Subnetting (Q.No.5)

Jaffar Shaikh said: (Oct 15, 2012) | |

How 255.255.255.224 provides 8 subnets ,each with 30 hosts? please tell me answer. |

Safdar Dogar said: (Jun 21, 2013) | |

Yes I agree but nobody tell me how we create subnetting? |

Mokonnin A Lemu said: (Jun 19, 2014) | |

How 255.255.255.224 provides 8 subnets ,each with 30 hosts? how could it be? |

Vinay said: (Jul 1, 2014) | |

Network bits = 3. Host bits = 5. Formula: To find number of subnet = 2^n(n = no>of borrowed network bits). To find no.of host = 2^h-2(h = no>of borrowed host bits). |

Tabe Rudith Ayuk Ashu said: (Oct 10, 2014) | |

Is there a possibility of using only the subnets and host values given in the question to get the subnet mask rather than testing the answers one after the other to get a match? |

Hagos said: (Mar 17, 2015) | |

255.255.255.224 in binary would be written as: 11111111.11111111.11111111.11100000. To find number of host: Starting from the octet consisting both 1 and 0. Count number of zeros to the end of the subnet mask. The last octet consist of 5 zeros therefore. 2^h-2 (h = number of zeros) = 2^5-2 = 32 -2 = 30. To find number of sub nets: 2^n (n = number of ones) 2^3 = 8. |

Roopa said: (Apr 22, 2015) | |

Host bits required = 16. By formula, 2^h-2 = 16. => 2^h = 18. Best suited value here for h is 5. By this sub net we get is 1111 1111.1111 1111.1111 1111.1110 0000. 255.255.255.224. |

Pratik said: (Jul 8, 2015) | |

224 = 11100000 in binary. So (2^5)-2 = 32-2 = 30. So 30 hosts. |

Mack said: (Aug 22, 2015) | |

Please tell me if I'm wrong, but another way to think of it is: 128 64 32 16 8 4 2 1. 1 1 1 0 0 0 0 0. 32 is the smallest bit used so. 32 - 1 (domain) - 1 (server) = 30. Honestly, I think I remember seeing it this way in a tutorial, so if I'm wrong please correct and/or email me. |

Mohamed said: (Sep 26, 2015) | |

@Vinay and @Hagos explained well Thanks for your valuable inputs, I keep expect from your side. |

Zewdu said: (Dec 3, 2015) | |

Matching and work out equation place including. |

Peeyush said: (Jan 20, 2016) | |

As we require 16 Host in each subnet. We will first check Binary Table. 128 64 32 16 8 4 2 1. If we take 16 we will not get 16 usable IP we will only get 14 usable IP. 2 ID will be taken for Subnet id and Broadcast ID. So will take 32. Now if we check mask it will be created like below: 11111111.11111111.11111111.11100000 By converting this into decimal it will then become: 255.255.255.224 This is the right answer. |

Fariha said: (Feb 16, 2016) | |

Suppose you are given the following class C IP address: 192.100.1.1. You need to perform Subnetting in such a way that you get 3 (three) subnets such that each subnet contains 62 (sixty two) valid hosts. You need to specify the network and broadcast address of each subnet and the valid host range in each of the subnet. Please solve it as early as possible. |

Angel said: (Feb 16, 2016) | |

Suppose you are given the following class C IP address: 192.100.1.1. You need to perform Subnetting in such a way that you get 3(three) subnets such that each subnet contains 62 (sixty two) valid hosts. You need to specify the network and broadcast address of each subnet and the valid host range in each of the subnet. solve it and give answer |

Gopi Kanaparthi said: (Mar 15, 2016) | |

Class c default subnet mask \24 like 255.255.255.0 Example: Converting bit 1, n=2^1=2. Converting bit 2, n=2^2=4. Converting bit 3, n=2^3=8. So first converting bit 1 is two subnets, second one is 4 subnets and third one is 8 Subnets support. We want 5 subnets so by using 3 converting bits. Default network bits+converting bits, N+n = 24+3 = 27. By using subnetmask 255.255.255.224 |

S .Hemanathan said: (Jul 20, 2016) | |

class c 192.168.2.0 Subnets = 5. Hosts = 16. 128 64 32 16 8 4 2 1 1 1 1 0 0 0 0 0. and 2^5 - 2=30 host 2^3 = 8 subnets. So, 255.255.255.224. |

Siri said: (Nov 4, 2016) | |

Prove if a host on a network has the address 172.16.45.14/30, the broadcast address of the subnet is 172.16.45.15. |

Shahnawaz said: (Nov 23, 2016) | |

Thank you @Peeyush. |

Jaganpj said: (Apr 4, 2017) | |

@Roopa Why the best value for h is 5? Isn't it would be 9 because h=18/9. |

Chaithanya Katari said: (Jun 13, 2017) | |

225.255.255.192=/26 so 24+2(2N we need to borrow) formula 2^2=4 subnets. for hosts 256-192=64 hosts. 255.255.255.224=/27 so 24+3(3n we need to borrow)formula 2^n=2^3=8 sub. for hosts 256-224=32 hosts. 255.255.255.240=/28 so 24+4(4N we need to borrow) formula 2^N=16 subnets. for hosts 256-240=16 hosts. |

Manikandan said: (Jul 7, 2017) | |

I need formula 2^16=32 value of network. So I taken 128.64.32. value mention after another 0.0.0.0.0 calculate 128+64+32=224. Ans 255.255.255.224. |

Czar said: (Mar 5, 2019) | |

5 subnets 16 hosts First, you have to meet the number of the host which is 16. By borrowing 4 bits we will meet 16 hosts but there will be no room for the network and broadcast address so we will use 5 bits that has 32 hosts. Using the 2^5 -2 will give us 30 which satisfy our required number of hosts. Secondly, we will use the formula an actual number of network = 32 (by default) minus no. of borrowed bits so 32-5 equals 27. Our new subnet mask is /27 which is equivalent to 255.255.255.224. |

Pollob Basak said: (Jun 24, 2021) | |

4 subnets, each with at least 20 hosts. Which subnet mask? Please explain about it. |

Jayden said: (Sep 24, 2021) | |

16 hosts < 32, 32 = 2^5, so it takes 5 bits, and each subnet has 30 hosts. 32 " 5 = 27, it means the subnet-mask has /27 bits. 5 subnets < 8, 8 = 2^3, therefore, 27 " 3 = 24 bits, = 255.255.255.0. 8 subnets x 32 IPs " 2 = 254, the answer should be 255.255.255.0. For example class c IP, 192.168.1.1/24 with subnet-mask 255.255.255.224. |

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