### Discussion :: Networking Basics - Networking Basics (Q.No.7)

Guru said: (Feb 1, 2012) | |

I didn't understand, please. Can anyone explain clearly. |

Shahnawaz said: (Feb 7, 2012) | |

I didn't understand. Can anyone explain me clearly please? |

Dinesh Rathinam said: (Feb 11, 2012) | |

Actually the value given is wrong. the value should be 192.168.168.128 then only we can get the correct value. 192-128=64 so that we can get 64 values.that valid host range value should be between 129 to 191. i think correct answer is B. does anybody refuse it....? |

Rohini said: (Jun 7, 2012) | |

Why the value should be between 129 and 191. It should be between 128-191. Also why the broadcast address is 191? |

Sandeepyadav said: (Jun 12, 2012) | |

This value is right not wrong. First of all we need to be the see the address and now. We take broadcast address 256 then in this we subtract the value of last gourd subnet mask (192). Then we add the same bit of subtracting then we find the 128 also and then we re add the value (64) then we find out the 192 this will be the broadcast address of segment. |

Lavanya S S said: (Jul 6, 2012) | |

Here the subnet mask is 255.255.255.192 means only the last value is changing so subtract it from 255, you will get 64 means out of 255 IP's 64 is already used only 192 IP's are available. 256/4=64 means 0-63, 64-127, 128-191, 192-255 all 4 segments contains 64 IP's, out of which one segment is already used (lets not worry about it). Now check the IP given which is 192. 168. 168. 188 which falls in the range 128-191 so here the valid host range will be from 129-190 because first one is always network IP, the last one is always the broadcast IP. |

Tushar said: (Oct 23, 2012) | |

Subnet masks 255.255.255.192 means that it has 6 bits to represent hosts. which means in the last octet: first two bits represent network bits and other 6 host bits. Now you are looking to fit 192.168.168.188. Out of two bits you can have : either 10 or 11( 11 is not possible it will go beyond 192) so network bit is 10 , which makes it 128.( 255.255.255.128) Now , possible values for hosts (for 6 bits) will start with: all zeroes to all 11s( but zero and all ones are meant for this network and broadcast) Hence the possible values are: from 1 to 62 Which will be: 129 to 190 |

Richa said: (Oct 31, 2012) | |

sbnet mask=255.255.255.{192} equiv to [11111111.11111111.11111111.11]00 0000 in binary. Now the binary values in [] represents the n/w portion i.e. first 26 bits is allocated for n/w portion and remaining 6 bits for the host portion. Now take a look into the IP address.. 192.168.168.188 where the 1st 26 bits represent d N/W portion. So the binary representation of IP address is [11000000.10101000.10101000.10]111100,the bits in [] again representing the N/W portion. There are four imp. things dat can be concluded from an IP address given with subnet mask,namely, -> N/W ADDRESS Represent d IP address in binary form and convert d HOST portion into ZEROS,and leave d N/W portion as it is. for eg. 11000000.10101000.10101000.10"111100" (192.168.168.188) is changed to 11000000.10101000.10101000.10"000000"(192.168.168.128;represents d n/w address). NOTICE THE CONVERSIONS REPRESENTED BY ".." ->1ST USABLE ADDRESS The next IP address immediately after the N/W address is d 1st usable address. For eg. 192.168.168.129 is the 1st usable address in this case. ->LAST USABLE ADDRESS Note that d host portion is of 6 bits i.e.2^6(=64) hosts can be connected. So d host portion can be divided in 64 blocks each i.e 0-63,64-127,128-191..and so on. Here in this case,since the n/w add. started from .128 so it must end on .191. So the LAST USABLE IP ADDRESS IS THE IP ADDRESS WHICH COMES IMMEDIATELY BEFORE THE BROADCAST ADDRESS. So the last usable IP address is 192.168.168.190. ->BROADCAST ADDRESS It refers to d last IP address of the block. In this eg, the broadcast address is 192.168.168.191 since d block is 128-191 as seen in the previous step. I hope u must have understood clearly. |

Chandan Kumar said: (Jun 22, 2013) | |

192.168.168.188 255.255.255.192 mask 26. And now have to find first address, Applying AND operation with first address. 192.168.168.188 8 8 8 2 188 = 10111100. And last 2 bit of mask, 11. Hence first address, 192.168.168.128. And last address, Adding complement of mask into first address. 00000000 00000000 00000000 0011111111. (63). Since 1 address 192.168.168.128. + 0 0 0 63. 192.168.168.191. |

Vini said: (Sep 20, 2013) | |

I don't know that how we can find the IP address so please explain in details? |

Ritu said: (Feb 5, 2014) | |

Can anybody explain me this question in more elaborate way from the scratch? |

Lakshmi said: (Aug 31, 2014) | |

First of all you have to find the base number for that the formula:-. 256-mask = base number will be used. Now that will be 256-192 = 64. Now to find the number of valid subnet's do the following. Subnet 1: 64. Subnet 2: 64+64 = 128. Subnet 3: 128+64 = 192 (not valid as it comes upto subnet 's mask address). So 2 valid subnets. Next find the broad cast address that would be 127 and 191. So first and last valid address of the subnet would be. 65, 129 (first valid address). And 126, 190 (last valid address). So the range would be 129 to 190. Hence the answer. :). |

Teddy said: (Jul 27, 2015) | |

@Richa. Where did you get that 191? As you said n/w add starts from 128 and end to 191 how? |

Ravi Sankar said: (Oct 19, 2015) | |

To find valid host range, we need subnet mask. Here subnet mask 255.255.255.192 (11111111.11111111.11111111.1100000). So for 192 (128+64+0+0+0+0+0+0). Here magic number is 64. 0-63 (192.168.168.0 (n/w address) & 192.168.168.63 (broadcast address). So we shouldn't use as IP address in between we can use these IP addresses). 64-127. 128-191. 192-255. Our question where 192.168.168.188 address lies in these range? Answer is 192.168.168.129-192.168.168.190. I hope you understood well. Thank you. Ravi sankar. |

Sumit Sain said: (Feb 8, 2016) | |

I really don't understand these answer correctly. Please explain it in brief. |

Sanjay Singhaniya said: (Mar 15, 2016) | |

Please explain other very simple method. |

Frank said: (May 9, 2016) | |

@Ravi Sankar. Why you call 64 as magic number and why not 192? Please help me. |

Harish said: (Jul 21, 2016) | |

64 - 2 = 62 then b is the answer. |

Shyamnarayan2003@Gmail.Com said: (Aug 3, 2016) | |

Hi, @All. Subtract 192 from 256 = 64, Means, the subnetting has been done by dividing into four parts, 64 * 4 = 256. Now see the option. 64 * 2 = 128, And 64 * 3 = 192. So IP add should be in between. 192.168.168.129 to 192.168.168.191. |

Punith said: (Sep 26, 2016) | |

@Shyamnarayan. Your answer is perfect with the simple solution. |

Robel said: (Oct 3, 2016) | |

128 is called a network address and 191 is called broadcast address. Thanks it is easy to understand. |

Vivek Singh Rathore said: (Jan 16, 2017) | |

Not getting the solution, please explain me in detail. |

Sumit said: (Mar 5, 2017) | |

The correct answer will be 192.168.168.128 - 191. |

Iglooroy said: (Mar 12, 2017) | |

The answer is right they said "usable" ip address, so the network is 192.168.168.128 to .191, 128 is network address and 191 is broadcast address. |

Richard said: (Mar 23, 2017) | |

256-192 = 64 so increment by 64. Networ - Broadcast 64 - 127 128 - 191 > so the usable host address range is 127 - 190 (answer is A). 192 - 255. |

Darth Vader said: (Apr 3, 2017) | |

Everybody is confused because desimal binary conversation, here is the way to better understand this: subnet is all 1s but the last 6 digits determines hosts. You cant do anything for the fist 24 bits because they are fixed (255.255.255) you have play with the first 2 bits of the last octet, they can be: 00xxxxxx 01xxxxxx 10xxxxxx 11xxxxxx In other word, you can create 4 different networks out of these bits, now convert them to decimal: The first one's max value will be 63 (if all x were 1) which is not even close to 188. Second posibility or network can be max 127 not close to 188. Third one is good, it can be max 191 (don't forget you can't change the 0 on the second digit from left, it is belong to network portion not host, so max=1011111y. Here you cannot make y=1 because when it is 1 with other bits, it becomes network address. 11xxxxxx's min address is 192 so it will never contain 188 although other bits were 0. |

Dr@Cul said: (May 9, 2017) | |

It's simple guys. Given IP address : 192.168.168.188. Given subnet mask address:- 255.255.255.192. To find the N/W ID step1: Convert both addresses into binary form. step2: Perform AND operation b/w them. step3: Result will be your network ID i.e(192.168.168.128). step4: So range of the N/W starts from 192.168.168.129 (starting host ID). step5: The last id of the N/w is 192.168.168.191 (which is broadcast domain). step6: So the last id of the N/W will be 192.168.168.190. Thank you. |

Madhu said: (Jul 3, 2017) | |

How you get 192.168.168.191 as broadcast id? |

Astha said: (Jul 13, 2017) | |

Hello, all. The answer is obtained as follows: IP: 192.168.168.188 Subnet: 255.255.255.192 192 in binary : 1 1 0 0 0 0 0 0 188 in binary: 1 0 1 0 1 1 0 0 192 AND 188: 1 1 0 0 0 0 0 0 1 0 1 0 1 1 0 0 Ans: 1 0 0 0 0 0 0 0 In 192 first 2 digits are 1 therefore in Ans 1 0 will be part of subnet id. While remaining 6 will give host address; First Host addr: 1 0 0 0 0 0 0 1 = 129. Last host addr: 1 0 1 1 1 1 1 0 = 190. And Broad cast addr will be : 1 0 1 1 1 1 1 1. |

Hitanshi said: (Jul 13, 2017) | |

Hi, I am not getting this, Please explain it in detail. |

Varun said: (Aug 23, 2017) | |

Thank you @Lavanya S S. |

Srinu said: (Sep 18, 2017) | |

Why other options are in correct, anyone can explain it briefly? |

Jitendra said: (Nov 12, 2017) | |

Here are the answers using 255.255.255.192: How many subnet bits are used in this mask? Answer: 2 2^2-2=2 subnets. How many host bits are available per subnet? Answer: 6 2^6-2=62 hosts per subnet. What are the subnet addresses?Answer: 256-192=64 (the first subnet)64+64=128 (the second subnet)64+128=192. However, although 192 is the subnet mask value, it's not a valid subnet. The valid subnets are 64 and 128. What is the broadcast address of each subnet?Answer: 64 is the first subnet and 128 is the second subnet. The broadcast address is always the number before the next subnet. The broadcast address of the 64 subnet is 127. The broadcast address of the 128 subnet is 191. What is the valid host range of each subnet? Answer: The valid hosts are the numbers between the subnet number and the mask. For the 64 subnet, the valid host range is 64-126. For the 128 subnet, the valid host range is 129-190. |

Shami said: (Nov 28, 2017) | |

I can't understand. Please anyone help me. |

Mehulkumar Patel said: (Dec 14, 2017) | |

I think answer should be B. Because 2nd subnet should be 128-192. I. E 128 is network address and 192 is broadcast address. Hence valid host range should be 129-191. So answer is B. |

Rashidalamsiwani said: (Feb 28, 2018) | |

Subnet should be C 192.168.168.128-190. |

Sivakrishna said: (May 15, 2018) | |

Thank you @Jitendra. |

Daniel Ajo said: (Jun 24, 2018) | |

Where did 156-192 come from? |

Evens Ridore said: (Aug 3, 2018) | |

IP address is 192.168.168.188 .188 in binary is (1 0 1 1 1 1 0 0) =.188. Subnet mask is 255.255.255.192 .192 in binary is (1 1 0 0 0 0 0 0) =.192. as .255 is all ones, last octet of first network ID will be (1 0 0 0 0 0 0 0 )=.128. with 256 - 192 = 64, N=(64) which is our magic number. N-2 = 62 hosts/network. first N/W ID 192.168.168.128, 1st Host ID 192.168.168.129. last Host ID 192.168.168.190. Broadcast 192.168.168.191. with (128+64=192), 2nd N/W ID will be 192.168.168. 192. |

Baefred said: (Aug 8, 2018) | |

This is the trick for this one. The given is 192.168.188.188 with a subnet of 255.255.255.192. This subnet is /26. You should know the value of the increment of each subnet mask. In this case, it's 64. Since it's in the fourth octet, we can now find the CLOSEST number which can be divided by the increment value. Only below the range of the given IP. In this case, 128 is the CLOSEST number to the given 188 that CAN BE DIVIDED by the INCREMENT VALUE which is 64. So we know that the IP will be 192.168.188.188 /26. Since the increment value is 64 which is also the number of IP addresses, we will just ADD 64 to 128. Which will give us the last IP 192.168.188.192. So, what do we have now? First IP: 192.168.188.128 which is the NETWORK ADDRESS. Last IP: 192.168.188.192 which is the BROADCAST ADDRESS. These 2 IPs cannot be used. So the usable IP range will be 192.168.188.129 to 192.168.188.191 Always remember that the first and last IP of the IP range is the Network and Broadcast addresses, respectively. And they are cannot be used. I hope this will be a clear solution. |

Jayanti Moger said: (Sep 18, 2018) | |

The subnet mask is 26, but the actual number of bits is 32 so 32-26 =6 bits for the one subnet. So 2^6 is 64 one subnet will have 64 addresses. So, 0-63,64-127, 128-191 and 192-255 is the 4 subnets, in that 128 lies in 128-191 subnet. |

Ali Hasan said: (Jan 26, 2020) | |

Answer (A) is right. Because valid host means usable ip. The CIDR is; Subnet=192.168.168.128 First Usable IP= 192.168.168.129 Last usable IP is= 192.168.168.190 Broadcast IP is= 192.168.168.191. so, answer is correct. |

Karthikeyan said: (Feb 7, 2021) | |

The answer is correct. Subnet Mask = 255.255.255.192 where. CIDR /26 means, 11111111.11111111.11111111.11000000. The default subnet mask is 255.255.255.0, So, We can only change the last byte. Subnet = 2^2 = 4 Total Subnets. Valid IP Addresses = 2^6 = 64 Valid IP's. Valid Host Ranges = 2^6 - 2 = 62 Host addresses. Subnet Number ----- First network adrs ----- Valid Host Range ----- Broadcast adrs 1. 192.168.168.0 ----- 192.168.168.1 ----- 192.168.168.62 ----- 192.168.168.63 2. 192.168.168.64 ----- 192.168.168.65 - ----- 192.168.168.126 ----- 192.168.168.127 3. 92.168.168.128 ----- 192.168.168.129 - ----- 192.168.168.190 ----- 192.168.168.191( 192.168.168.188 is here) 4.192.168.168.192 ----- 192.168.168.191 ----- 192.168.168.254 ----- 192.168.168.255. Hence, Ans is correct. |

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