IP Addressing Help plzzzz
HossamElbanna
Registered Users Posts: 3 ■□□□□□□□□□
in CCNA & CCENT
Hi Guys,Currently i study Networking and i am doing pretty good.But i am always getting stuck in Subneting CCNA . this example on packet tracer given IP address of 192.15.75.0 for Subnet A and 192.15.75.9 Subnet B Design a basic IP addressing scheme that satisfies the following requirements Subnet A has 59 Host Subnet B has 25 Host and its just ask me to write all the subnet Details in this tables of A and B
Specification Result
Number of bits in the subnet ....... IP subnet mask (binary) ....... IP subnet mask (decimal) ....... Number of usable hosts in subnet ....... IP Subnet (Base subnetwork address) ....... First IP Host address ....... Last IP Host address ...... i will appreciate if someone could expline to me or show me how i can do it so i can carry on more examples. Thank you
Specification Result
Number of bits in the subnet ....... IP subnet mask (binary) ....... IP subnet mask (decimal) ....... Number of usable hosts in subnet ....... IP Subnet (Base subnetwork address) ....... First IP Host address ....... Last IP Host address ...... i will appreciate if someone could expline to me or show me how i can do it so i can carry on more examples. Thank you
Comments
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Deathmage
Banned Posts: 2,496
click on this link and use this workbook. it will help you understand subnetting.
Let me google that for you -
Priston
Member Posts: 999 ■■■■□□□□□□
What do you have so far?
Also do you mean "192.15.75.0 for Subnet A and Subnet B" or "192.15.75.0 for Subnet A and 192.15.75.9 Subnet B"A.A.S. in Networking Technologies
A+, Network+, CCNA -
HossamElbanna
Registered Users Posts: 3 ■□□□□□□□□□
What do you have so far?
Also do you mean "192.15.75.0 for Subnet A and Subnet B" or "192.15.75.0 for Subnet A and 192.15.75.9 Subnet B"
Thanks for your reply.
Yes Subnet A 192.15.75.0 and Subnet B 192.15.75.9 i just need to see how its done according to the table blow
Number of bits in the subnet .......
IP subnet mask (binary) .......
IP subnet mask (decimal) .......
Number of usable hosts in subnet .......
IP Subnet (Base subnetwork address) .......
First IP Host address .......
Last IP Host address ......
i will appreciate if someone could explain to me or show me how i can do it.Thank you -
samwinchester
Member Posts: 28 ■□□□□□□□□□
HossamElbanna wrote: »Thanks for your reply.
Yes Subnet A 192.15.75.0 and Subnet B 192.15.75.9 i just need to see how its done according to the table blow
Number of bits in the subnet .......
IP subnet mask (binary) .......
IP subnet mask (decimal) .......
Number of usable hosts in subnet .......
IP Subnet (Base subnetwork address) .......
First IP Host address .......
Last IP Host address ......
i will appreciate if someone could explain to me or show me how i can do it.Thank you
Ohkk. Let's just take Subnet A for example. Now you have to remember some stuff. And practice. Otherwise there is no way it will come to you.
You need 59 hosts in Subnet A. Now go back to the simplest most common table you should know.
128 64 32 16 8 4 2 1 <--- Start with 128 and keep it halving by 2 until u reach one. U shud know this by heart.
So, here we have number of hosts. We need 59 and the closest is 64 in the line above. See now we don't wanna go over too much and leave too much space. If we choose 128 that will waste the hosts, where as 32 doesn't satisfy the need for 59 hosts. Agreed? good.
So, now number of usable hosts will be 64 minus 2 = 62. Why? Because the first address is the network and last is the broadcast. Will mention this in a bit.
now look at the table below which I love writing down for easy easy subnetting.
128
64
32
16
8
4
2
1
17
18
19
20
21
22
23
24
25
26
27
28
29
30
31
32
128
192
224
240
248
252
254
255
So, Since we are using Class C subnet --> 255.255.255.0
Look under 64, what do u see? 18? But that will come in class B. Why? Because add up the bits now if u don't get it. 4 octets right? 0.0.0.0 each has 8 in it.
so 8 + 8 which is 8.8.0.0 is class B. 8 + 8 = 16 and 18 comes in the 3rd octet get it? if not read again, really simple stuff.
Ohkk so, since class C, we look under 18 which takes us to 26. Now this 26 is the /26 you see in subnet masking. so how will u get to 26??
8 + 8 + 8 = 24.
24 + 2 is 26. So you know 24 is 255.255.255.0 right??
So now what is 26 ?? Well, look at the table, under 26 --->> 192.
That makes the subnet mast 255.255.255.192 -->> ur base subnet address.
Now, network address and broadcast address. Remember this: 1st address is network address and last is the broadcast address.
Let's go to the subnet A:
192.15.75.0 ---->> now at the place of 0 you put 64.
ohk so here's how it goes.
192.15.75.0
192.15.75.64
192.15.75.128
You keep adding 64. Now these are all different networks.
Take 192.15.75.0 for example, since this is the first address it's the network address.
What's the broadcast address?
Look at the next network and minus one from the last octet.
so Broadcast address will be 192.15.75.63 get it??
all the addresses inbetween are the usable hosts. so 192.15.75.1, 192.15.75.2.....192.15.75.62
first usbale host::: 192.15.75.1
last usable host:::: 192.15.75.62
Pheww. Finished my CCENT a while ago. So much fun to teach subnetting.
Hope this helps!!Achieved CCNA R & S. Next : MCSA 2012
"My Fault Is That I Don't Realize How Great I Really Am." - Muhammad Ali -
HossamElbanna
Registered Users Posts: 3 ■□□□□□□□□□
samwinchester wrote: »Ohkk. Let's just take Subnet A for example. Now you have to remember some stuff. And practice. Otherwise there is no way it will come to you.
You need 59 hosts in Subnet A. Now go back to the simplest most common table you should know.
128 64 32 16 8 4 2 1 <--- Start with 128 and keep it halving by 2 until u reach one. U shud know this by heart.
So, here we have number of hosts. We need 59 and the closest is 64 in the line above. See now we don't wanna go over too much and leave too much space. If we choose 128 that will waste the hosts, where as 32 doesn't satisfy the need for 59 hosts. Agreed? good.
So, now number of usable hosts will be 64 minus 2 = 62. Why? Because the first address is the network and last is the broadcast. Will mention this in a bit.
now look at the table below which I love writing down for easy easy subnetting.
128
64
32
16
8
4
2
1
17
18
19
20
21
22
23
24
25
26
27
28
29
30
31
32
128
192
224
240
248
252
254
255
So, Since we are using Class C subnet --> 255.255.255.0
Look under 64, what do u see? 18? But that will come in class B. Why? Because add up the bits now if u don't get it. 4 octets right? 0.0.0.0 each has 8 in it.
so 8 + 8 which is 8.8.0.0 is class B. 8 + 8 = 16 and 18 comes in the 3rd octet get it? if not read again, really simple stuff.
Ohkk so, since class C, we look under 18 which takes us to 26. Now this 26 is the /26 you see in subnet masking. so how will u get to 26??
8 + 8 + 8 = 24.
24 + 2 is 26. So you know 24 is 255.255.255.0 right??
So now what is 26 ?? Well, look at the table, under 26 --->> 192.
That makes the subnet mast 255.255.255.192 -->> ur base subnet address.
Now, network address and broadcast address. Remember this: 1st address is network address and last is the broadcast address.
Let's go to the subnet A:
192.15.75.0 ---->> now at the place of 0 you put 64.
ohk so here's how it goes.
192.15.75.0
192.15.75.64
192.15.75.128
You keep adding 64. Now these are all different networks.
Take 192.15.75.0 for example, since this is the first address it's the network address.
What's the broadcast address?
Look at the next network and minus one from the last octet.
so Broadcast address will be 192.15.75.63 get it??
all the addresses inbetween are the usable hosts. so 192.15.75.1, 192.15.75.2.....192.15.75.62
first usbale host::: 192.15.75.1
last usable host:::: 192.15.75.62
Pheww. Finished my CCENT a while ago. So much fun to teach subnetting.
Hope this helps!!
Thump up. Thanks a lot this is a fantastic explanation.