Answer to H.W. : H.W.1 : Class C, 255.255.255.0, 40 Hosts Given so 101000 = 6 bits reserved from right to left, Subnet Generator = 64, Octet Position = 4, New Subnet Mask is 255.255.255.192 or /26, Number of Networks / Subnets = 2^2 = 4 and Usable Host IP Addresses = 2^6-2 H.W.2 : Class B, 255.255.0.0, 1000 Hosts Given so 1111101000 = 10 bits reserved from right to left, Subnet Generator = 4, Octet Position = 3, New Subnet Mask is 255.255.252.0 or /22, Number of Networks / Subnets = 2^6 and Usable Host IP Addresses = 2^10-2 H.W.3 : Class A, 255.0.0.0, 100 Hosts Given so 1100100 = 7 bits reserved from right to left, Subnet Generator = 128, Octet Position = 4, New Subnet Mask is 255.255.255.128 or /25, Number of Networks / Subnets = 2^17 and Usable Host IP Addresses = 2^7-2

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why is 2^17???

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Only two networks (Subnets) are possible in this problem and the octet position is 2. Please correct.

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IP Address: 201.1.1.0 New subnet mask: 255.255.255.192 Host Address range: 201.1.1.1 - 201.1.1.62 Host per subnet 62

Homework Answers✔✔✔✔ Answer 1: Given IP Add is 201.1.1.0, class C IP Address and default subnet mask is 255.255.255.0. Given number of hosts=40 which has 6 bits. So, Subnet generator will be 64 and Octet 4. New subnet mask will be 255.255.255.192 or /26. Number of subnets will be 2^2=4 and usable hosts in each subnet will be (2^6) - 2. Answer 2: Given IP Add is 170.15.0.0, class B IP Address and default subnet mask is 255.255.0.0. Given number of hosts=1000 (binary 1111101000) which has 10 bits. So, Subnet generator will be 4 and Octet 3. New subnet mask will be 255.255.252.0 or /22. Number of subnets will be 2^6=64 and usable hosts in each subnet will be (2^10) - 2. Answer 3: Given IP Add is 15.0.0.0, class A IP Address and default subnet mask is 255.0.0.0. Given number of hosts=100(binary 1100100) which has 7 bits. So, Subnet generator will be 128 and Octet 4. New subnet mask will be 255.255.255.128 or /25. Number of subnets will be 2^17 and usable hosts in each subnet will be (2^7) - 2.

How the network will be 2¹⁷ ? It will be 2²⁵ as there are 25 ones in binary notation????

can anyone tell how to solve these type of questions What is the maximum number of subnets and hosts per subnet for the 192.168.50.247 255.255.255.224 network?

what will be the last range of the question is it 10.255.255.0 to 10.255.255.127

Why would you define the new subnet mask as 255.255.255.128 (correct) but still modify the third octect? Yes initially the IP is Class A but once you limit to 100 hosts (as the question says) you're LOOKING for the subnet masks that adecuates, is not like you adecuate the subnest mask you found to the initial classification of the IP adress, like you are doing in the video. Am I missing something?

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2^25 subnets is not it?

Split the network 10.200.20.0/27 into 4 subnets and submit the broadcast address of the 2nd subnet as the answer. Can someone please explain to me why the answer is 10.200.20.15

Someone should Lindy verify this for me . The broadcast address of the fourth subnet is it correct . No body seem to find anything wrong with it it started with 128 and ended with 127 I mean the 4th octet is part

if the subnet mask is 255.255.255.128 then how can you have a value greater than 128 in the last octet? I am really confused about this