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Nairaland Forum / Science/Technology / Computers / Help On Subnetwork (3337 Views)
Help On Subnetwork by software(m): 7:40pm On Jul 24, 2007 |
Hi Guys, i need some help on this. thanks The major network IP number is 8.0.0.0/23: 1. How many host bits are there before subnetting? _____ 2. How many host bits are there after subnetting? _____ 3. How many host bits were borrowed to subnet? _____ 4. How many usable hosts per subnet are there? _____ 5. What is the subnet mask written in dotted decimal? _______._____._____._____ 6. What is the subnet IP number of subnet 800? _____._____._____._____ 7. What is the broadcast IP number of subnet 800? _____._____._____._____ 8. What are the usable host IP numbers of subnet 800? _____._____._____._____ through _____._____._____._____ The major network IP number is 191.48.0.0/25: 9. How many host bits are there before subnetting? _____ 10. How many host bits are there after subnetting? _____ 11. How many host bits were borrowed to subnet? _____ 12. How many usable subnets are there? _____ 13. What is the subnet mask written in dotted decimal? _______._____._____._____ 14. What is the subnet IP number of subnet 500? _____._____._____._____ 15. What is the broadcast IP number of subnet 500? _____._____._____._____ 16. What are the usable host IP numbers of subnet 500? _____._____._____._____ through _____._____.______.______ |
Re: Help On Subnetwork by nbeet2000(m): 7:51pm On Jul 24, 2007 |
What sort of help you need?? |
Re: Help On Subnetwork by software(m): 8:04pm On Jul 24, 2007 |
The major network IP number is 8.0.0.0/23: 1. How many host bits are there before subnetting? _____ 2. How many host bits are there after subnetting? _____ 3. How many host bits were borrowed to subnet? _____ 4. How many usable hosts per subnet are there? _____ 5. What is the subnet mask written in dotted decimal? _______._____._____._____ 6. What is the subnet IP number of subnet 800? _____._____._____._____ 7. What is the broadcast IP number of subnet 800? _____._____._____._____ 8. What are the usable host IP numbers of subnet 800? _____._____._____._____ through _____._____._____._____ The major network IP number is 191.48.0.0/25: 9. How many host bits are there before subnetting? _____ 10. How many host bits are there after subnetting? _____ 11. How many host bits were borrowed to subnet? _____ 12. How many usable subnets are there? _____ 13. What is the subnet mask written in dotted decimal? _______._____._____._____ 14. What is the subnet IP number of subnet 500? _____._____._____._____ 15. What is the broadcast IP number of subnet 500? _____._____._____._____ 16. What are the usable host IP numbers of subnet 500? _____._____._____._____ through _____._____.______.______ |
Re: Help On Subnetwork by Maleeq(m): 9:36pm On Jul 24, 2007 |
@software Are you trying to get your assignment done for you? |
Re: Help On Subnetwork by software(m): 10:25pm On Jul 24, 2007 |
Maleeq: Nope, its not an assignment, Just preparing for an Exam and this is an aspect of it i dont understand, Its just a past question, Am actually studying about it. Just that i need them answered for me so i could learn faster, |
Re: Help On Subnetwork by Maleeq(m): 3:28pm On Jul 25, 2007 |
Okay, since you boldly placed the CCNA title on your signature, I shall skip some basics and get down to the real deal. Network ID = 8.0.0.0 Mask= /23 = 255.255.254.0 Note: 1. 24bits. Explanation: class A addresses use a default mask of 255.0.0.0, which leaves 22bits for subnetting since you must leave 2bits for host addressing. 2. we are subnetting using the /23 mask, thus we have 11111111.11111111.11111110.00000000 (Count the bits that are off) = 9bits N.B: x = number of off bits = 9 y = number of on bits = 23 - 8(from the default class A mask) = 15 3. Compare with the default mask; /8 = 11111111.00000000.00000000.00000000 /23= 11111111.11111111.11111110.00000000 Borrowed bits = 15bits 4. Usable host per subnet: 2^x - 2 2^9 - 2 = 510 5. /23 = 255.255.254.0 in dotted decimal 6. Network ID for Subnet 800: To get the network multiplier factor, we have 256 - 254 = 2 Thus, we av our networks as; # Network Range Broadcast 1 8.0.2.0 8.0.2.1 - 8.0.3.254 8.0.3.255 2 8.0.4.0 8.0.4.1 - 8.0.5.254 8.0.5.255 3 8.0.6.0 8.0.6.1 - 8.0.7.254 8.0.7.255 . . . . 32766 8.255.252.0 8.255.252.1 - 8.255.253.254 8.255.253.255 Extrapolating to # 800, 800 in binary = 11 0010 0000 We now pick the /23 in binary for comparison /23 = 11111111.11111111.11111110.00000000 write out, from right to left, the 800 in binary starting from under the 1st '1' in the /23 mask i.e 11111111.11111111.11111110.00000000 XXXXXXXX.XXXXX110.0100000X.XXXXXXXX Perform a binary to decimal conversion: (treat the x as a '0') XXXXX110 = 6 0100000X = 64 thus, the network ID for subnet 800 is 8.6.64.0 7. The broadcast address for subnet 800 is thus: 8.6.65.255 (Follow the logic at the start of the solution to question 6) 8. Usable address for hosts; 8.6.64.1 - 8.6.65.254 The same applies for the other network in the second question. Just follow the same procedures. I will leave that one out for you to practice on. Good luck in your career advancement drive. |
Re: Help On Subnetwork by software(m): 7:44pm On Jul 25, 2007 |
thanks so much Maleek I really must thank your for the great support, It was well understood. I have been able to solve the second part myself, God bless , Thank you NL |
Re: Help On Subnetwork by Maleeq(m): 6:05am On Jul 26, 2007 |
@software No wahala. Take care |
Re: Help On Subnetwork by software(m): 7:19am On Jul 26, 2007 |
Maleeq: thanks bro |
Re: Help On Subnetwork by software(m): 8:34pm On Jul 29, 2007 |
hi Maleeq The Second part of the questions u assisted me with the other day, My results came out, and i failed the second part of it. But i got the first part right, Pls i would like you to kindly still throw more light on it for me, It would greatly be apprciated, Thanks The major network IP number is 191.48.0.0/25: 9. How many host bits are there before subnetting? _____16bits 10. How many host bits are there after subnetting? _____ 11. How many host bits were borrowed to subnet? _____ 12. How many usable subnets are there? _____ 13. What is the subnet mask written in dotted decimal? _______._____._____._____ 14. What is the subnet IP number of subnet 500? _____._____._____._____ 15. What is the broadcast IP number of subnet 500? _____._____._____._____ 16. What are the usable host IP numbers of subnet 500? _____._____._____._____ through _____._____.______.______ |
Re: Help On Subnetwork by software(m): 8:46pm On Jul 29, 2007 |
Attached to this post, is a MSWord Doc, Please kindly hell me go through, Anyhelp would be greatly appreciated, I need explanations also. THanks Thanks in advance
|
Re: Help On Subnetwork by Maleeq(m): 1:13pm On Jul 31, 2007 |
Here the solution to the second one. I'll get the final exam's solution to you later. Network ID = 191.48.0.0/25 8.0.0.0 Mask= /25 = 255.255.255.128 Note: 1. 16bits. Explanation: class B addresses use a default mask of 255.255.0.0 (/16), which leaves 14bits for subnetting since you must leave 2bits for host addressing. 2. we are subnetting using the /25 mask, thus we have 11111111.11111111.11111111.10000000 (Count the bits that are off) = 7bits N.B: x = number of off bits = 7 y = number of on bits = 25 - 16(from the default class A mask) = 9 3. Compare with the default mask; /16 = 11111111.11111111.00000000.00000000 /25 = 11111111.11111111.11111111.10000000 Borrowed bits = 9bits 4. Usable host per subnet: 2^x - 2 2^7 - 2 = 126 5. /25 = 255.255.255.128 in dotted decimal # ID Range Broadcast 510 191.48.255.0 191.48.255.1 - 191.48.255.126 191.48.255.127 6. Network ID for Subnet 800: To get the network multiplier factor, we have 256 - 128 = 128 Thus, we av our networks as; # Network Range Broadcast 1 191.48.0.128 191.48.0.129 - 191.48.0.254 191.48.0.255 2 191.48.1.128 191.48.1.129 - 191.48.1.254 191.48.1.255 3 191.48.2.0 191.48.2.1 - 191.48.2.126 191.48.2.127 . . . . 510 191.48.255.0 191.48.255.1 - 191.48.255.126 191.48.255.127 Extrapolating to # 500, 500 in binary = 1 1111 0100 We now pick the /25 in binary for comparison /25 = 11111111.11111111.11111111.10000000 write out, from right to left, the 500 in binary starting from under the 1st '1' in the /25 mask i.e 11111111.11111111.11111111.10000000 XXXXXXXX.XXXXXXXX.11111010.0XXXXXXX Perform a binary to decimal conversion: (treat the x as a '0') 11111010 = 250 0XXXXXXX = 0 thus, the network ID for subnet 500 is 191.48.250.0 7. The broadcast address for subnet 800 is thus: 191.48.250.127 (Follow the logic at the start of the solution to question 6) 8. Usable address for hosts; 191.48.250.1 - 191.48.250.126 |
Re: Help On Subnetwork by soldier4gd: 5:23pm On Dec 11, 2007 |
Maleeq: What advice did you give the user called "software" in reference to the attachment he posted on July 29, 2007 (Help On Subnetwork)? I have a similar problem in which I am trying to understand. The attachment was concerning VLSM. |
Re: Help On Subnetwork by Maleeq(m): 11:38pm On Dec 11, 2007 |
@soldier4gd Men, its been a long time. I cant remember the exact solution I offered him (software). I shall attempt is once more for you. Here we go: 1- I couldn't make out how to use, thats if the solution requires we use the table with values provided in them. So, I'll just take it as though the table was not provided. 2- We have a total of 120 + 50 + 25 + 2 + 2 = 197Hosts that we must accommodate in our addressing scheme. . .thus a minimum of 8bits for subnetting. Base address = /24. Starting with the network segment that has the most host(Houston, 120Hosts): Network Address: 220.108.38.0 For 120Hosts, we need 7bits for subnetting: 2^7 - 2 = 126 possible hosts 7Bits = 1000 0000 Thus for Houston: Address: 220.108.38.0/25 This runs from 220.108.38.1 - 220.138.38.127 The next available block would be 220.138.38.128 block For 50Hosts(Waco segment), we need at least 6 bits 2^6 - 2 = 62 possible hosts 6Bits = 1100 0000 Thus for Waco: Address: 220.108.38.128/26 This runs from 220.108.38.129 - 220.108.38.191 Next block would be 220.138.38.192 For the 25Host(Corpus Cristi Segment), we nee at least 5bits 2^5 -2 = 30 possible hosts 5Bits = 1110 0000 Thus for Corpus Cristi Address: 220.108.38.192/27 This runs from 220.108.38.193 - 220.108.38.223 Next block would be 220.108.38.224 For the WAN links, we need only 2 address. Thus,2bits mask is required 2^2 - 2 = 2 possible hosts 2Bits = 1111 1100 WAN #1: Address: 220.108.38.224/30 This spans 220.108.38.225 - 220.108.38.227 WAN #2: Address: 220.108.38.228/30 Thats all! There's some standard chart used that simplifies VLSM/CIDR, but I dont a copy of the chart right now. This is the best I can offer right now. If you need any more clarification, dont hesitate to ask me. Cheers |
Re: Help On Subnetwork by soldier4gd: 2:33pm On Dec 15, 2007 |
Did you mean to put 220.138.38.128 or is it suppose to be 220.108.38.128? Also, are the ranges considered to the but subnet ranges? |
Re: Help On Subnetwork by soldier4gd: 2:34pm On Dec 15, 2007 |
Did you mean to put 220.138.38.128 or is it suppose to be 220.108.38.128? Also, are the ranges considered to be subnet ranges? |
Re: Help On Subnetwork by Maleeq(m): 10:44pm On Dec 15, 2007 |
C'mon, you should know that 220.138.38.x was a typographical error. The network address we were assigned is 220.108.38.0 And yes, they are subnet ranges. |
Re: Help On Subnetwork by gabby74: 1:57am On Apr 20, 2009 |
The major network IP number is 12.0.0.0/23: 1. How many host bits are there before subnetting? _____ 2. How many host bits are there after subnetting? _____ 3. How many host bits were borrowed to subnet? _____ 4. How many usable hosts per subnet are there? _____ 5. How many usable subnets are there? _____ 6. What is the subnet mask written in dotted decimal? _______._____._____._____ 7. What is the subnet mask written in hex-decimal? 8. What is the subnet IP number of subnet 731? _____._____._____._____ 9. What is the broadcast IP number of subnet 731? _____._____._____._____ 10. What are the usable host IP numbers of subnet 731? |
Re: Help On Subnetwork by Maleeq(m): 10:43am On Apr 21, 2009 |
@gabby74 This problem you have stated here is actually about the same as the initial poster's problem. Both addresses are in the CLASS A scheme, thus no much difference. Follow the explanation here closely and you will see the solution easily. Attempt it, and paste your solution here for correction wherever necessary. Believe me, you'd learn faster that way. |
Re: Help On Subnetwork by gabby74: 11:35am On Apr 21, 2009 |
ok maleek i will do that because i really want to learn |
Re: Help On Subnetwork by gabby74: 2:03am On Apr 22, 2009 |
Ok, Maleek, I have tried my best so do help me out. this is something new to me to your help would be very much appreciated. The major network IP number is 12.0.0.0/23: 1. How many host bits are there before subnetting? __24___ 2. How many host bits are there after subnetting? __15___ 3. How many host bits were borrowed to subnet? __15___ 4. How many usable hosts per subnet are there? ___510__ 5. How many usable subnets are there? ____ 6. What is the subnet mask written in dotted decimal? __255____._255____.__254___.___0__ 7. What is the subnet mask written in hex-decimal? 8. What is the subnet IP number of subnet 731? _____._____._____._____ 9. What is the broadcast IP number of subnet 731? _____._____._____._____ 10. What are the usable host IP numbers of subnet 731? _____._____._____._____ through _____._____._____._____ Don’t forget to define starting subnet indices (you may assume Subnet [0] The major network IP number is 188.49.0.0/25: 11. How many host bits are there before subnetting? __14___ 12. How many host bits are there after subnetting? _7____ 13. How many host bits were borrowed to subnet? __9___ 14. How many usable hosts per subnet are there? ___126__ 15. How many usable subnets are there? _____ 16. What is the subnet mask written in dotted decimal? _______._____._____._____ 17. What is the subnet mask written in hex-decimal? 18. What is the subnet IP number of subnet 241? _____._____._____._____ 19. What is the broadcast IP number of subnet 241? _____._____._____._____ 20. What are the usable host IP numbers of subnet 241? _____._____._____._____ through _____._____.______.______ |
Re: Help On Subnetwork by Maleeq(m): 11:03am On Apr 23, 2009 |
Note (Key): 0 = OFF Bit 1 = ON Bit x = Number of OFF bits y = Number of ON bits Alright, the given IP is 12.0.0.0/23 From inspection, the following can be deduced: Network ID = 12.0.0.0 Mask= /23 = 255.255.254.0 (/23 = 11111111.11111111.11111110.00000000 Convert the bytes into decimals and you'll get 255.255.254.0) 1. 24bits. Explanation: Class A addresses use a default mask of 255.0.0.0 . Count the number of bits in the 3bytes (x.0.0.0) we have 24bits (1byte = 8bits) 2. We are subnetting using the /23 mask, thus writing /23 out in binary we get: 11111111.11111111.11111110.00000000 (Count the bits that are off) = 9bits Thus, we have 9bits left after subnetting. N.B: x = number of OFF bits = 9 y = number of ON bits = 23 - 8(from the default class A mask) = 15 3. To get the number of borrowed bits, we compare with the default mask; /8 = 11111111.00000000.00000000.00000000 /23= 11111111.11111111.11111110.00000000 Now, do a bit wise comparison between the 2-rows. Count the number of bits that are not in pairs (i.e 0/0 or 1/1) This is equivalent to ( 23 - 8 ) = 15. The 8 is from the default CLASS A mask Borrowed bits = 15bits 4. Usable host per subnet: Note that in subnetting, we MUST leave atleast 2bits for host addressing. Why? You should know that the first IP is the default gateway and the last IP is the broadcast for any subnet (Ideal situation). So, the formula to get the number of useable IP is 2^x -2. 2^x - 2 2^9 - 2 = 510 (How did I get the value of x as 9? Look at the solution to question 2 above) 5. How many usable subnets are there? To get this, we use the formula 2^x : 2^x 2^9 = 512 usable subnets 6. What is the subnet mask written in dotted decimal? To get this, take the mask /23 and write out '1' 23 times, grouping with dots in every bytes i.e /23 = 11111111.11111111.11111110.00000000 Now convert each byte to from binary to decimal /23 = 255.255.254.0 in dotted decimal 7. What is the subnet mask written in hex-decimal? Similarly as #6 above, I expect u have a knowledge of hexadecimal numbers (0-9, A-F). If not, then do a research on BINARY numbers /23 = 11111111.11111111.11111110.00000000 = FF.FF.FE.00 8. Network ID for Subnet 731: To get the network multiplier factor, we take the last byte in the mask (/23 = 11111111.11111111.11111110.00000000 = 255.255.254.0) that is not all OFF bits. This happens to be 254 and then subtract this value from the maximum decimal value of a byte (256) ie. 256 - 254 = 2 Thus, we have our networks as, starting from the base address; # Network Range Broadcast 1 12.0.2.0 12.0.2.1 - 12.0.3.254 12.0.3.255 2 12.0.4.0 12.0.4.1 - 12.0.5.254 12.0.5.255 3 12.0.6.0 12.0.6.1 - 12.0.7.254 12.0.7.255 . . . . 32766 12.255.252.0 12.255.252.1 - 12.255.253.254 12.255.253.255 Extrapolating to # 731, 731 in binary = 10 1101 1011 We now pick the /23 in binary form for comparison: /23 = 11111111.11111111.11111110.00000000 write out, from right to left, the 731 in binary starting under the 1st (rightmost) ON bit ('1') in the /23 mask i.e 11111111.11111111.11111110.00000000 XXXXXXXX.XXXXX101.1011011X.XXXXXXXX Perform a binary to decimal conversion: (treat the x as a '0') XXXXX101 = 5 0100000X = 182 thus, the network ID for subnet 731 is 12.5.182.0 9. The broadcast address for subnet 731 is thus: 12.5.183.255 (Follow the logic at the start of the solution to question 10. Usable address for hosts; 12.5.183.1 - 12.5.183.254 If you have any questions about any step, feel free to ask for further clarification. I left the solution to the second one for you to attempt in full. Cheers |
Re: Help On Subnetwork by gabby74: 1:20pm On Apr 23, 2009 |
ok
|
Re: Help On Subnetwork by opensource(m): 5:28pm On Apr 23, 2009 |
@ Maleeq are u a teacher , i really envy your patience and tolerance |
Re: Help On Subnetwork by Maleeq(m): 6:10pm On Apr 23, 2009 |
@opensource |
Re: Help On Subnetwork by Nobody: 11:16pm On Apr 24, 2009 |
Na to commoth biro and paper remain! |
Re: Help On Subnetwork by Dune: 9:25pm On Apr 24, 2010 |
Maleeq, thanks for the tutorial! The way you worked it makes total sense and I must have visited a dozen websites before finding your post. THe other websites gave some of the answers, but never EXPLAINED it. Thanks again for the help! |
Re: Help On Subnetwork by Maleeq(m): 2:24pm On May 08, 2010 |
u welcome bro! |
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