What’s up mga idol? Nung nakaraan, pinag-usapan natin ang basic about sa IP address. You can check again the Part I here and Part II here. Ngayon naman, itutuloy natin ang ating free ccna lessons at today ang pag-uusapan naman natin ay about subnetting. Ito ay isa sa pinaka-importanteng skill na kelangang matutunan ng isang aspiring CCNA.
Maraming tanong sa CCNA exam ang tungkol sa subnetting at syempre ito rin ay magagamit natin sa “real world” kapag tayo ay nag-tatrabaho na bilang isang network administrator. Alright, simulan na natin.
What is subnetting?
Kung matatandaan n’yo, sa Part II ng ating IP addressing lesson ay pinag-usapan natin ang IP address classes. Kasama ng mga IP address classes na ito ay ang kanilang mga default subnet mask. Ito din ay tinatawag din na classful addressing or pagsunod or paggamit ng mga default subnet mask. Katulad ng nakikita n’yo sa sample image natin sa ibaba. At gaya nga ng sabi natin, ang subnet mask ang nagtatakda ng network portion at host portion sa isang IPV4 ip address.
In real world or sa mga real networking scenario, kadalasan na hindi ginagamit ang mga default subnet mask or classful addressing. Ang kadalasang ginagamit ay ang classless addressing. Ito ay ang pag-gamit ng customized na mga subnet mask para sa mga IP addresses. Ito rin ang tinatawag na subnetted networks.
Ang subnetting ay ang proseso ng pag-customize ng default subnet mask para i-extend ang network portions ng isang IP address. Ito ay ginagawa sa pamamagitan ng pag-hiram or “pag-steal” ng mga bits galing sa host portions. Sa pamamagitan ng subnetting, maari nating ma-accomodate ang required number of needed network/s for a given IP address.
For example, kailangan nating mag-create ng 3 new networks para sa ating existing network na 192.168.1.0/24. Ibig sabihin, sa 192.168.1.0/24 din manggaling ang ating new networks. Magagawa natin ito sa pamamagitan ng subnetting. Kailangan lang nating “manghiram or mag-steal ng bits” galing sa host portions para ma-accomodate ang hinihinging number ng new networks. Let’s see how that works.
How to subnet. An easy and fast subnetting tutorial.
Ngayon mga idol, ito ang isa sa pina-importanteng skill na dapat nating matutunan bilang CCNA. Hindi lamang para sa CCNA exam kundi para na rin sa real world scenario. Malaki ang maitutulong nito para maipasa mo ang CCNA certification exam at maintindihan kung paano ito ginagamit sa real networking world.
Ang subnetting tutorial na ituturo ko sa inyo ay ang pinaka-madali at pinaka-mabilis na paraan ng subnetting na aking natutunan(atleast for me). Sa subnetting questions sa CCNA exam, oras ang kalaban mo so kinakailangan mo na mabilis magawa or ma-solve ang mga subnetting questions. Maximum of 1 minute per questions! Although maraming subnetting tutorial ang available sa internet, mostly ay English at super technical ng mga explanations kaya hindi agad natin ma-adapt or matutunan lalo na ng mga beginners. Kung problema mo ang subnetting at nahihirapan ka, ito na ang sagot. Padadaliin at pabibilisin natin sa maliwanag na paraan upang ma-solve ang iyong problema. Power mga kapatid! Power! Lol. Haha. Let’s go.
Sa given example natin kanina, naatasan tayong mag-create ng 3 new networks sa ating existing network na 192.168.1.0/24. Para madali at mabilis natin itong makuha, kailangan nating i-identify ang sumusunod.
- No. of networks needed
- No. of bits “stolen or borrowed“
- New subnet mask
Yung number of networks needed is ofcourse given na sa ating questions which is 3. Ang kailangan natin alamin is yung no. of bits needed to steal or borrow sa host portion para ma-accomodate ang hinihinging number of networks. Dito papasok ang pag-gamit ng ating mga fingers or tinatawag ding finger subnetting. Let’s continue.
Para naman makuha natin ang number of bits needed to accomodate the given networks(3 in our example), kelangan lang natin gamitin ang ating mga daliri sa pagbilang ng bits starting from 2 then duplicating or doubling itself(2, 4, 8, 16, 32, 64, 128 and so on if needed) hanggang sa ma-accomodate na natin ung hinihinging bilang ng network.
To do this, tingnan mo lang yung kanang kamay mo na nakatalikod sayo(yung parang nanunumpa) gaya ng nasa larawan sa taas. Then starting from hinliliit, magbilang ka starting from 2 then gaya ng sinabi ko kanina, doblehin or duplicate mo lang yung value hanggang makuha mo yung needed number of networks.
So sa ating example, 3 networks ang needed natin. Magbilang tayo from hinliliit starting from 2 hanggnang ma-abot or ma-accomodate na natin yung 3 na bilang ng networks na kelangan. Therefore sa value pa lang ng pangalawang daliri(4) pasok na yung 3 networks na hinahanap natin. So dalawang bits lang ang kailangan.
So ang number of bits stolen or borrowed natin is 2. Balikan natin yung mga kelangan.
Given IP/Network: 192.168.1.0/24
- No. of networks needed = 3
- No. of bits = 2
- New subnet mask = ?
Para naman makuha natin ang new or bagong subnet mask, napaka-simple lang. Ito ang secret formula.
Sa ating example, NSM = 24 + 2. So ang new subnet mask natin is 26 or /26 or in long format is 255.255.255.192. Therefore we can conclude na para makapag-create tayo ng 3 new networks from 192.168.1.0/24 our new subnetted network is 192.168.1.0/26.
Pano nakuha yung 255.255.255.192? Sa unang given network natin, given na yung /24 diba, so ibig sabihin yung first 3 octets ay 255 na or 3 groups of eight 1s (11111111.11111111.11111111.?). Ang kailangan lang natin is yung last octet kung saan tayo nanghiram ng bits.
Before tayo manghiram ng bits, ang bits ng last octet ay eight 0s(00000000) kasi nga siya ay para sa host. After natin manghiram ito ay magiging 11000000 na. Ibig sabihin yung dalawang bits na hiniram natin is ginawa nating 1. Kagaya ng nabanggit natin sa mga unang lessons, sa subnet mask ang 1s ay para sa network portion at ang 0s ay para sa host portions in binary. At ang value ng 11000000 in decimal is 192. Make sense mga idol?
Para mas mabilis n’yo makuha, gumamit din tayo ng ating mga fingers.
Makikita n’yo sa ating image sa taas ang walo sa ating mga daliri(hindi kasama ang hinlalaki). Ito ay katumbas din ng 8 bits sa bawat octet ng isang IPV4 subnet mask(eight 0s or 1s – eight na daliri). At since subnet mask ang focus natin dito, dun tayo sa green numbers focus. Kapag subnet mask ang usapan, ang tingin natin or pagbilang natin is ALWAYS FROM LEFT TO RIGHT kagaya ng pinag-usapan natin sa IP Addressing Part I and Part II. Kung ilang bits ang hiniram natin or ginamit natin kanina(or bit 0 na ginawa nating bit 1), kukunin lang natin ang value noon FROM LEFT TO RIGHT using the green numbers.
Sa example natin kanina, nakuha natin ang 192 sa subnet mask na 255.255.255.192 kasi 2 bits ang hiniram or ginamit natin(11000000), and from LEFT TO RIGHT gamit ang values ng finger natin (128, 192, 224, 240, 248, 252, 254, 255). 192 ang value ng pangalawang finger(2 bits).
Pano naman ito naging /26 in slash format? Sa given meron na tayong /24 which is tatlong 255.255.255 or 24 na 1s(11111111.11111111.11111111). Idinagdag lang natin yung dalawang hiniram or ginamit natin from host portions. /24 + 2 = /26 or (11111111.11111111.11111111.11000000).
Madali lang diba? Kung hindi agad nakuha mga idol, break muna. Basahin at balikan ulit kapag medyo unwind na. 🙂 Kung meron ka naman ng existing way of subnetting at mas mabilis at bihasa ka na doon, mas ok na dun kana mag-focus. Sabi ko nga, maraming paraan basta kelangan nagagawa mo ng mabilis.
Madali lang di ba mga idol? Let me summarize.
- Ilang networks ang kailangan
- Ilang bits ang nagamit mo(bilang ng daliri) para ma-accomodate mo yung bilang ng networks na kailangan(RIGHT TO LEFT sa ating example)
- New Subnet Mask(NSM) = Old Subnet Mask(OSM) + No. of bits borrowed
Let’s see more examples.
A.) 201.50.65.0/24, 10 networks >>> 4 bits | NSM = /28 or 255.255.255.240 | New subnetted network = 201.50.65.0/28
Explanation: Sa pagkuha ng bits, ang value ng fourth finger natin is 16(2,4,8,16) | Sa pagkuha ng subnet mask, ang value ng 4th finger is 240(128,192,224,240) | NSM = OSM + Bits or /24 + 4 = /28 (255.255.255.240)
B.) 160.40.0.0/16, 1000 networks >>> 10 bits | NSM = /26 or 255.255.255.192 | New subnetted network = 160.40.0.0/26
Explanation: Sa pagkuha ng bits, ang value ng 10th finger natin is 1024(2,4,8,16,32,64,128,256,512,1024) | Sa pagkuha ng subnet mask, ang value ng 2nd finger is 192(128,192) Dito nasa 4th octet kana kasi ung first 3 octets ay puro 1s na or 24 na 1s| NSM = OSM + Bits or /16 + 10 = /26(255.255.255.192)
C.) 100.0.0.0/8, 2000 networks >>> 11 bits | NSM = /19 or 255.255.224.0 | New subnetted network = 100.0.0.0/19
Explanation: Sa pagkuha ng bits, ang value ng 11th finger natin is 2048(2,4,8,16,32,64,128,256,512,1024,2048) | Sa pagkuha ng subnet mask, ang value ng 3rd finger is 224(128,192,224) Dito nasa 3rd octet kana kasi ung first 2 octets ay puro 1s na or 16 na 1s| NSM = OSM + Bits or /8 + 11 = /19(255.255.224.0)
In case na yung number of needed networks ay matataas na, basta ituloy lang natin ung rule na duplicate or idouble yung value at tandaan kung ilang daliri na ang nagamit natin. So sa value na 1000 nakagamit tayo ng 10 daliri na ang value eh 1024 kaya pasok ito. Pati na din sa 2000 network na may 11 bits(or 11 daliri) na may value na 2048. And it goes on and on.
Wag din kayong malilito sa pag-kuha ng bits para sa needed number of networks at sa pag-kuha ng bits para sa subnet mask. Sa pag-kuha ng no. of bits para sa no. of needed networks, always start at 2 then doubling or duplicate the value hanggang sa ma-accomodate na nung value yung needed no. of networks then kunin mo kung ilang daliri ang nagamit(10 bits or daliri sa 1000 networks). It doesn’t matter where you start! Sa sample ko lang ginamit na sa kanan kasi dito ako nasanay. 🙂 Basta make sure the rule is applied.
Sa shortcut ng pag-kuha naman ng subnet mask decimal value, yung value naman ng fingers natin from LEFT TO RIGHT starting from 128. Kada-eight 1s 255 na yun so next octet kana, then ganun ulit, value ng fingers from LEFT TO RIGHT starting from 128.
Sa sample natin na 160.40.0.0/16, 1000 networks needed, meron na tayong 16 1s(11111111.11111111.00000000.00000000) given and since 10 bits ang kelangan para ma-accomodate yung 1000 networks, hiniram or ginamit natin yung 10 0s galing sa host portions kaya naging (11111111.11111111.11111111.11000000) or 255.255.255.192 or /26(or 26 consecutive 1s). Nakita n’yo na ang last octet na lang ang ni-compute natin kasi nga kada eight 1s is 255 na. 🙂
Alright mga idol, hanggang dito na lang muna itong subnetting tutorial natin. Para hindi ma-overload lalo na yung mga beginners. Sa part II itutuloy natin ito. Pag-uusapan natin kung pano naman kukunin yung number of networks, number of hosts, increment, network range, valid host range, network address or broadcast address of a certain network at iba pa. Mas mahalaga ito kasi ito ang karaniwang kasama sa CCNA exam. Sasamahan din natin ng mga sample problems or questions para makapag-practice tayo.
Sana naintindihan at kahit papano ay may natutunan kayo sa lesson na ito. Paki-share na rin sa iba para mas marami pa tayong matulungan. Don’t forget to like our FB page and subscribe para sa mga updates. To our success, cheers!
Update: Check Subnetting tutorial Part II here.
mark says
October 5, 2015 at 9:37 PMsir my tnung po ako ex. 192.168.1.0 /24 tpos 4networks ung needed san po sya 2bits(under po kay 4) or 3 bits(under po kay 8)?? slmat po in advance
Carlos says
April 28, 2016 at 5:53 PMsa 8 bits sir kasi pag sa 2 bits ka mag kukulang ng usable host.
Network – 192.168.1.0
FH – 192.168.1.1
LH – 192.168.1.7
SM – 255.255.255.248
Next Network – 192.168.1.8
Carlos says
April 28, 2016 at 5:58 PMNetwork – 192.168.1.0
FH – 192.168.1.1
LH – 192.168.1.6
BA – 192.168.1.7
SM – 255.255.255.248
Next Network – 192.168.1.8
medyu mali yung sa taas nakalimutan yung Broabcast Address. hehe
jaybie says
May 25, 2016 at 2:46 AMCarlos misleading ka kaibigan.
RAFING says
August 25, 2016 at 11:13 PM3 bits ang klangan i borrows , pra 32 ang range niya
192.168.1.0
192.168.1.32
192.168.1.64
192.128.1.128 =4 networks
pag 2 bits lang ang range niya is 64
192.168.1.0
192.168.1.64
192.168.1.128 =3 networks which is kulang sa requiremnt na 4 networks dapat
try ko lang 😛 sinagot
RAFING says
August 25, 2016 at 11:51 PMay mali pala sagot ko ahahaha sorry
Sonny Delos Santos says
June 2, 2017 at 6:51 PMTama dapat sa 3bits sya kasi kulang sa 2bits pano yung Network and Broadcast
mark says
October 6, 2015 at 11:00 PMsir ask ko lang apnu pag create a 4 networks existing network na 192.168.1.0/24… two bits parin un sir? na kayang i accomodate ung 4 networks
Billy says
October 11, 2015 at 7:57 AMYes Mark! Kindly use the same procedure in the sample at makukuha m ang sagot jan. Actually they have the same answer. 🙂
Bong says
March 24, 2016 at 2:14 AMPwede mag tanung? Yung pagkuha ng bits or /8,16,24,32 ang gets ko kung paano makuha. Pero yung sa subnet mask ay still nalilito parin ako. Kung paano naging 192 ang subnet mask kapag 1000 ang needed na network pati na rin yung 2000. Hoping for a much clearer explanation sorry if im too slow hehe thanks
Billy says
March 26, 2016 at 6:45 AMBalik-balikan mo lang idol makukuha mo din yan. Practice makes perfect! 🙂
leejay says
August 4, 2016 at 10:00 PM160.40.0.0/16 – 16 bits is 255.255.0.0
160.40.0.0/24 – 24 bits is 255.255.255.0 you need 10 bits, 8bits was coverd by the 3rd octet
160.40.0.0/26 – 26 bits is 255.255.255.192 the last 2 bits coverd by the 4th octet but you only need 2bits
1bit 2bit 3bit 4bit 5bit 6bit 7bit 8bit
128 64 32 16 8 4 2 1 = 255
128 64 = 192
frank says
January 24, 2018 at 2:49 PMleejay, naliwanagan at nakuha ko na ung sa 160.40.0.0 /26 kung pano siya naging 255.255.255.192. salamat sa explanation mo.
RCT says
November 21, 2018 at 2:04 PMyung 255.255.255. kasi is kinocover na sya ng as /24 then since nangailangan ka ng 2 bits ifollow lang yung procedure from left – right ang value ng second finger dun is 192. So kung by combining the them, the result is 255.255.255.192, based on my understanding
nhielo says
March 30, 2016 at 9:48 PMsir billy ask lng po
ilang networks po ang pwede kada octet
octet 1 octet 2 octet 3
salamat sir
Billy says
March 31, 2016 at 3:52 AMWhat do you mean? In IPv4, we use 4 octets and each octet consists of 32 bits. The decimal value of each octet is from 0-255.
Mheljun Enanoria says
April 11, 2016 at 10:44 AMsir ask ko lang diba po ang finger is
128,64,32,16,8,4,2,1
pero bat pag nakuha kayo umpisa sa (2,4,8,16,32,64,128,256,512,1024,2048)
hindi sa 1
Mheljun Enanoria says
April 11, 2016 at 10:56 AMgets ko na hahahha master inulit ko na lang naintindihan ko na salamat boss sa tutorial POWER! hahaha lol
Billy says
April 14, 2016 at 5:01 PMNice idol! 🙂
Marck Angelo M. Obrador says
April 17, 2016 at 2:18 PMThank you dito Sir, Daming pwedeng matutunan
Billy says
April 19, 2016 at 2:42 AMWelcome idol! Keep learning. 🙂
JOHN RAYMOND S. PEREZ says
August 3, 2016 at 2:15 PMPwede po ba ito gamitin sa VLSM sir? Thanks po hehe
Dell says
May 17, 2019 at 7:10 PMpwedeng pwedeng gamitin yan sa VLSM, syempre may konting changes pero yung concept is parehong pareho.
Jam says
August 31, 2016 at 6:10 PMHi po, ask ko lang po what if lumagpas sa 32 bits,
ex 210.203.25.192/29 144 networks , base po sa naintindihan ko ang sagot dito ay
210.203.25.192/36, sa isang subnet mask po ay meron po bang 32 bit lang or (11111111.11111111.11111111.11111111)=32 bit e ang kailangan ko is 36 bit. Salamat po.
Billy says
September 3, 2016 at 4:05 AMNdi ko gets ung question mo Jam. Can you elaborate? 😀 The given is already subnetted, it’s a class C address with default subnet mask of /24. Try to read Part II here. Thanks!
Jam says
September 5, 2016 at 4:33 PMSalamt Sir, Gets ko na. 🙂
asbcit says
January 9, 2017 at 8:40 PMidol anu yung classless at classful? thank you
fibrella says
May 17, 2019 at 7:22 PMClassful is yung mga default # of bits per class.
Example:
Class A ==> 8 bits or /8
Class B ==> 16 bits or /16
Class C ==> 24 bits or /24
Classless is yung HINDI na default, haha pano yun? Learn about VLSM and CIDR.
basta pag hindi default, classless na yun
Ruck says
February 26, 2017 at 1:02 PMIdol Billy,
San ka nag based sa pagamit ng 2,4,8,16 so on so forth?
128, 192, 244 so on and so forth?
Ang pagkakaalam ko kpag cnsabi n base2 start ka sa 1,2,4,8,16… so on and so forth.
Thanks,
Ruck
Ruck says
February 26, 2017 at 1:16 PMNa gets ko pla yang nasa taas idol. Pero bkit nag start k pagbilang ng 2 sa pag kuha ng 3 network required? Pwede nman start sa 1,2 then nakuha n ntin ung required n 3 networks (1+2=3) at pag kakaalam ko we must always start at 1?
Salamat idol in advance sa pag sagot.
Regard,
Ruck
Billy says
March 11, 2017 at 12:48 PMThat’s what I learned before kaya yan na ginamit ko, wala nman ako nkkitang issue idol. Thanks!
Jhun Dela Cruz says
March 18, 2017 at 2:04 PMNetwork ID: 177 173 0 0
Subnets Required: 31
boss pa help nga pano makukuha to yung mask nakukuha ko pero iba nalilito pa sorry
Subnet Mask:
1st Available Host
Address of Subnet 1:
Max # of hosts/subnet:
Billy says
March 20, 2017 at 1:43 PMIt’s all there idol, try to re-read if ndi mo kaagad makuha.
1st available host – the Ip address after the network address
Address of subnet 1 – Unang subnet or network including na mask
Max # of hosts per subnet – Kung ilan daw IP on that network or subnet(ex. /24 = 256 IP pero 254 usable hosts IP)
Hope it helps. Thanks!
Joey says
May 12, 2017 at 3:44 AMDito ko lang naintindahan tong subnetting sa dami dami ng video na napanood ko. maraming salamat talaga idol #road2CCNA
Billy says
May 12, 2017 at 11:13 AMThanks idol! Goodluck!
jayson says
May 30, 2017 at 5:15 PMSir billy kung nakagawa na po tayo ng new subnetted networks, panu po cia iapply sa realworld. 192.168.1.0/26 ang new subnetted network natin, pde po ba malaman ang list ng new networks natin at mga host list na pde dun sa bagong subnetted networks natin.
and 2nd question ko po kung magkikita ba old network at new subnetted network maraming salamat idol.
Billy says
June 11, 2017 at 8:39 AMJason, you put the network addresses in the dhcp server. Let say 192.168.1.0/24. Then most of the time(in real world), ni-rereserve yung 1st, 2nd and 3rd addresses para sa ip ng devices kasi kadalasan ini-static sila. Then ung iba auotmatic na, so automatic sila mkka obtain ng ip from the dhcp server. Given na it was configured correctly. 🙂
lene says
June 1, 2017 at 5:08 PMhi sir billy, dumudugo na po ilong ko 🙂
tanong ko lang po ang counting po ba from LEFT to RIGHT mag sta start sa 128, 64,32, 16, 8, 4, 2, 1 ? diba po mag sisimula sa 1, 2, 4, 8, 16, 32, 64, 128? like s a example nyo po sa itaas 192.168.1.0/24
no. of networks needed = 3
no. of bits “stolen or borrowed = 2
new subnet mask = ?
New Subnet Mask = Old Subnet Mask + No. of bits borrowed
24 + 2 = 26 so ang subnet mask is 255.255.255.3 ? 11111111 . 11111111 . 11111111 . 11000000 kung magsisimula ang counting sa 1 – 128, pag sa 128-1 eh 192 ang sagot.
sana masagot nyo po tanong ko 🙂 salamat po 🙂 more power po sa blog nyo
#GOD BLESS po
Billy says
June 11, 2017 at 8:37 AMHello Lene, I don’t get your question. Sorry. It’s all laid out na in the article. Just try to re-read na lang, or if you have specific questions, email me. 🙂
frank says
January 24, 2018 at 2:31 PMLene, ang counting is left to right (128,192,224,240,248,252,254,255) pag Subnetmask ang usapan. Pero when getting a bits right to left(1,2,4,8,16,32,64,128..and soon). Tama po ba Sir Billy?
RCT says
November 21, 2018 at 2:07 PMHindi ata magstart sa 1 kapag bits.
arnold says
July 27, 2017 at 7:41 AMSIr I LOVE YOU SO MUCH ON THIS!!!! kelan tayo pde mag kita ililibre kita pnadali mo buhay ko !!! <3!!
Billy says
July 28, 2017 at 12:55 AMWheee! i-reserved mo mna yan idol. Hehe. God bless!
frank says
January 24, 2018 at 1:54 PMIdol Billy, Sa Example mo na 160.40.0.0/16, 1000 networks >> 10 borrowed bits nakuha ko ung 160.40.0.0 /26 pero pag dating sa subnet ang question ko is, pano nakuha ung 255.255.255.192? kung ang 8 bits ay may value na (128,192,224,240,248,252,254 at 255).
frank says
January 24, 2018 at 3:07 PMnakuha ko na Sir Billy.. Kung bakit siya naging 255.255.255.192..
Christian Pogi says
January 26, 2018 at 4:59 AMSir paki explain naman po pano nakuha ung B and C doon sa “Let’s see more example”.
Letter A lang po kasi na gets ko.
Mask Prefix Subnets Hosts
255.0.0.0 (/8) 1 network with 16,777,214 hosts
255.128.0.0 (/9) 2 subnets with 8,388,606 hosts each
255.192.0.0 (/10) 4 subnets with 4,194,302 hosts each
255.224.0.0 (/11) 8 subnets with 2,097,150 hosts each
255.240.0.0 (/12) 16 subnets with 1,048,574 hosts each
255.248.0.0 (/13) 32 subnets with 524,286 hosts each
255.252.0.0 (/14) 64 subnets with 262,142 hosts each
255.254.0.0 (/15) 128 subnets with 131,070 hosts each
255.255.0.0 (/16) 256 subnets with 65,534 hosts each
255.255.128.0 (/17) 512 subnets with 32,766 hosts each
255.255.192.0 (/18) 1,024 subnets with 16,384 hosts each
255.255.224.0 (/19) 2,048 subnets with 8,190 hosts each
255.255.240.0 (/20) 4,096 subnets with 4,094 hosts each
255.255.248.0 (/21) 8,192 subnets with 2,046 hosts each
255.255.252.0 (/22) 16,384 subnets with 1,022 hosts each
255.255.254.0 (/23) 32,768 subnets with 510 hosts each
255.255.255.0 (/24) 65,536 subnets with 254 hosts each
255.255.255.128 (/25) 131,072 subnets with 126 hosts each
255.255.255.192 (/26) 262,144 subnets with 62 hosts each
255.255.255.224 (/27) 524,288 subnets with 30 hosts each
255.255.255.240 (/28) 1,048,576 subnets with 14 hosts each
255.255.255.248 (/29) 2,097,152 subnets with 6 hosts each
255.255.255.252 (/30) 4,194,304 subnets with 2 hosts each
Available Class A subnet masks
mk says
January 26, 2018 at 5:10 AMoo nga sir nalito din ako sa B at C..
Christian Pogi says
January 27, 2018 at 1:17 AMBalak ko sana bumili ng book, kaso inaantay ko pa reply nya pra ma verify ko..
Daniella says
May 17, 2019 at 8:20 PMB. 160.40.0.0/16, 1000 networks
From the finger method, we only use 8 fingers correct? (don’t include your thumb)
|128| |64| |32| |16| |8| |4| |2| |1|
# of bits: 8 7 6 5 4 3 2 1
Normally, kapag ang network na given ay equal or below sa 128, napakadali natin malaman kung ilang bits diba? eh pano kung lampas? Gaya nung sa problem? 1000 networks daw? naloko na…
Meron na tayong values ng 1-8 bits diba? so pano pag pataas?
Ganito yan…
|256| |512| |1024| |2048| |4096| and so on… x2 ka lang ng x2 habang tumataas ang bits
# of Bits: 9 10 11 12 13 and so on…
Pano nakuha yung 256? edi 128×2, eh yung 512? edi 256×2, eh yung 1024? edi 512×2, and so on….
So proceed tayo…
NSM = OSM + bits
= 16 (given sa problem) + 10
= 26 or /26 (CIDR notation)
oh bakit naging 10 yung bits? Ilan bang networks ang required? 1000 diba? Yung 1000 ba mas madami sa 1024? NO, yung 1000 ba mas madami sa 512? YES…ilang bits yung 512? 10 diba? pak na pak! 😛
eh pano yung long format na 255.255.255.192? pano nakuha yun?
Ganito yan…
A Subnet Mask consists of four (4) octets diba? And each octets consists of 8 bits. 8 bits x 4 octets = 32 bits in total.
Also, Each octet has a subnet value ranging from 0-255
And in every 8 bits, ang value nyan ay 255
Example: 255.255.255.255, ilang bits to?
8 + 8 + 8 + 8 = 32 BITS
Eh sa example B, nakalagay eh 255.255.255.192
Balikan natin yung mahiwagang FINGER! 😛
Binary Value :|128| |64| |32| |16| |8| |4| |2| |1|
Binary Sum: |128| |192| |224| |240| |248| |252| |254| |255|
# of bits: 8 7 6 5 4 3 2 1
Ang NSM natin or CIDR ay /26 diba? ilang 8 meron sa 26? 3 diba? 8×3=24, o may sukli, ilan ang sukli? 2 tama?
Dyan ngayon papasok yung BINARY SUM.
Magsisimula kang magbilang from left to right. Pano yun?
Binary Sum: |128| |192| |224| |240| |248| |252| |254| |255|
1 2 3 4 5 6 7 8
Oh game?
ilang 8 meron sa /26? 3, ang sukli ay 2 diba?
oh bilang tayo
Anong value ng 2 sa binary sum? diba 192?
oh sige buuin na natin yung Long Format
remember na in every 8 bits, ang value nyan ay 255
Ilang 8 nga ulet? 3
So…
255.255.255.192
8 + 8 + 8 + 2 = 26 or /26
Oh, Letter C tayo: 100.0.0.0/8, 2000 Networks
2000 networks is under how many bits? 11 diba? Pano nangyari? yung 2000 ba mas madami sa 2048? NO, eh sa 1024? YES. Ilang bits ang meron sa 1024? diba 11? makes sense?
Oh move on…
NSM = OSM + bits
= 8 (from the problem) + 11
= 19 (CIDR)
Pano yung long format?
Ilang 8 nga ulit meron sa 19? 2 diba? oh, 8×2=16
Oh, may sukli na 3, oh bilang tayo dun sa binary sum sa taas, anong binary sum ng 3? diba 224?
Oh, buuhin na natin…
255.255.224.0
8 + 8 + 3 = 19 or /19
oh teka muna, baka magtaka ka, bakit 0 yung dulo?
diba ang /19 ay may 2 multiples of 8 and remainder o sukling 3? Ngayon, pag naubos mo na yung multiples of 8 at remainder, logically ganito lang dapat ang itsura nyan…
255.255.224 tama ba? 3 octet lang kumbaga diba?
eh kaso diba nga na ang subnet mask at ip address ay may 4 octets. Ex. 111.111.111.111
hindi naman pwedeng tatlong (3) octet lang mga bes. kulang yun..di pwede yun.
So para makumpleto ang 4 octets, yung 255.255.224 ay dadagdagan natin ng 0
So magiging 255.255.224.0
malinaw?
Eh pano kunwari ang result ng subnet mo is ganito, 255.252
Oh dalawa na lang? edi ganun pa din, kumpletuhin mo lang. lagyan mo ng 0
magiging 255.252.0.0 na sya.
ganun din kung isa na lang..kunwari eh, 255 lang nakuha mo, edi kumpletuhin mo ulit. Gawin mong 255.0.0.0
Super linaw na ba?
Kevin says
February 1, 2018 at 7:50 AMOo nga noh. Nakakalito nga, paano ba nakuha rin yun?
Billy says
February 17, 2018 at 9:07 PMSee the explanation na lang idol. I think na-explain ko naman ng ayos, yung iba gets naman. Hindi ko rin lagi monitor ‘tong comments, you can send an email if you have specific questions. Salamats!
Shiela says
February 6, 2018 at 3:36 AMNa stuck ako dito sa example na yun.
Digong says
May 17, 2019 at 8:27 PMBasahin mo yung comment ni Daniella sa taas. Yung reply nya kay Christian Pogi. hehe
Kyle says
February 8, 2018 at 8:29 AMWala pa po bang update dito? Dito rin ako na stuck e. Pero nabasa ko na ung iba.
Billy says
February 17, 2018 at 9:05 PMKindly see the explanation idol, it’s all there. Then you can also read other comments(like Leejay did). I can’t always check the comments, better to email me if you have specific questions. Salamats!
Kendrick says
February 9, 2018 at 1:16 AM@Chris, pareho tyo ng concern, mag cocomment sna ko ng ganyan kaso nakita ko comment mo. PM mo ko sir if makita mo to, gusto ko malaman yung totoong sagot talaga.
email nyo po ako
kendrickcruz@yahoo.com
Lyn says
March 3, 2018 at 4:42 AMSir ask ko lng paano kung walang given na networks needed? paano kung 192.168.1.0/24 tas hahanapin mo yung subnet mask and network? paano po pag ganun? ganun po kase magbigay prof namin
Billy says
March 5, 2018 at 11:14 PMHindi ko gets Lyn. :/
Mercy says
May 17, 2019 at 8:34 PMHi Lyn, from the given address, makukuha mo na agad yung long Format subnet mask.
192.168.1.0/24
Remember that this is a classful address.
A Class C address.
And a Class C address has 24 bits tama po ba? Kaya ko nasabing Classful yung example mo kasi pasok sa dun sa banga ni Class C classful address.
so para makuha mo yung long format, alamin mo lang ilang multiples of 8 meron sa /24
diba 3?
8 + 8 + 8 = 24 or /24
Remember that each octet contains 8 bits and in every 8 bits has a value of 255 tama?
So game, 255.255.255.0
8 + 8 + 8 = 24 or /24
bakit may 0 sa dulo? kasi di naman pwedeng 255.255.255 lang, remember na 4 anf octets natin, at di pwedeng tatlo lang, eh pano yun kulang nga diba? edi pag ganyang case, lagyan mo lang ng 0
So, 255.255.255.0
yan po yun, hope it helps.
slash says
October 9, 2018 at 9:16 AMsir ask ko lang. pano ko po malalaman yung broadcast address at network address?
Alegria says
May 17, 2019 at 8:38 PMA network address is the first address on the network range while the Broadcast address is the last address.
melvic says
December 10, 2018 at 1:50 AMIto natutunan ko dito , baka mkatulong din to sa iba 🙂
2048 1024 512 256 128 64 32 16 8 4 2 1
To Count bits borrowed = start at “2” RIGHT TO LEFT
ex. 160.40.0.0 /16 1000 networks
Given = 1000
Bits borrowed = 10 (start at “2” RIGHT TO LEFT) : 1024
/16 (given ) + 10 (bits) = /26
_____________________________________
128 64 32 16 8 4 2 1 = 1 octet or 8 bits
To get subnet mask = start at “1” LEFT TO RIGHT
8(11111111) + 8(11111111) + 8(11111111) + 2(11000000) = 26 or /26
/26 = 11111111.11111111.11111111.11000000 = 255.255.255.(start at “1” LEFT TO RIGHT)
128 + 64 = 192 | 255.255.255.192
Hope this will help 🙂
Thanks idol
MayAnn says
March 5, 2019 at 5:55 PMfavorite ko po ito….
Clinton Costales says
May 20, 2019 at 3:10 PMHi sir billy, ask ko lang bakit sa 2 nagstart ung bilnag ng bits ? bakit hindi sa 1 ?
thanks Idol