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Ip Addressing Host Bits Minimized

davidex95
Level 1
Level 1

Hi, i have one problem with Ip addressing,i didnt understand well how set the masks for mine subnets.

 

I have to use one ip address with A Class 10.0.62.0 for 4 department with a different number of hosts.

 

1° 11 addresses

2° 42

3° 22

4° 36

I have to do that with the same size for each subnet one time, and with minimized size the second time. I have problem with the second task.

The first time i used as size 10.0.62.0/26.

 

The router has 10.0.62.1/26 with the first switch. The devices have 10.0.62.2/26 to 10.0.62.62/26.

With the second switch (2° Dept.) 10.0.62.65/26 and the devices 10.0.62.66/26 to 10.0.62.126

3th 10.0.62.129/26 on the router and devices 10.0.62.130/26 to 10.0.62.190/26

4th 10.0.62.193/26; 10.0.62.194/26 to 10.0.62.253/26

In this too the request is to have the minimum number of hosts bits.

 

I'm not sure if i did well but anyway the network work correctly.

 

On the second i tried as follow:

 

1st router has 10.0.62.1/28 with the first switch. The device have 10.0.62.2/28 to 10.0.62.62/28.

and i continued following this line, so 2° router with mask /26; 3rd /27; and 4th/26 with the Ip address starting on the first free number after the previuos department as on the first case.

But in this case i dont have connetivity between the departments, the error usual is to not find the ip address on the subnet so ask at the router but dont find notthing.

 

For sure im doing some mistakes with the masks but i dont know how solve this and what is the problem. Someone can hel me?

 

Ps. On the first case when i try to ping another device give me than the request is failed but if i try again work correctly. This is normal or happen because i did some mistake on the addressing?

 

I add one pic of the progect:ù

 


image.png

 

1 Accepted Solution

Accepted Solutions

Martin L
VIP
VIP

 

this exercise is about VLSM. Think about block of subnets: 128, 64, 32, 16, 8, 4, 2. The best technique is to start subnetting Or allocation of IPs based on number of hosts requested.  which department vlan requesting the most hosts, going towards the less.  in your example, 42, then 36, 22, 11 last. 

This way you sure will be able to fulfill the request to have the minimum number of hosts bits. aka no wasting bits. 

you gave all departments block of 64, 62 hosts per each. this works but you wasting bits for all but 2 vlans. 

2nd try, giving /28, 2-/26; 3rd /27; and 4th/26 - may work but usually will not work. you started with block of 16, then 64 which overlaps block of 16 (depending where u start).

Anyway, you should start with block of 64, aka /26, for dept #2, then another /26 for 36 hosts for #4, then /27 - 30 hosts for #3, and finally /28. 

starting with network id of 10.0.62.0/26, range from 10.0.62.1/26 thru 10.0.62.62/26. then another 64 block starting at 10.0.62.65/26 thru 10.0.62.126/26; then you can start at 10.0.62.128 to allocate block 32, 30 IPs, aka /27, the range will be 10.0.62.127 thru 10.0.62.158/27. you will end up at network id 10.0.62.160.  To assign last block of 16 for 11 IPs, that is /28. Range is 10.0.62.161 - 10.0.62.174, B-cast 10.0.62.175. 

 

answer (in order) should be: 

2° 42 - 10.0.62.0/26 - giving 62 IPs

4° 36 - 10.0.62.64/26 - 62 IPs

3° 22 - 10.0.62.128/27 - 30 IPs

1° 11 - 10.0.62.160/28 - 16 IPs

 

 

Regards, ML
**Please Rate All Helpful Responses **

View solution in original post

4 Replies 4

Martin L
VIP
VIP

 

this exercise is about VLSM. Think about block of subnets: 128, 64, 32, 16, 8, 4, 2. The best technique is to start subnetting Or allocation of IPs based on number of hosts requested.  which department vlan requesting the most hosts, going towards the less.  in your example, 42, then 36, 22, 11 last. 

This way you sure will be able to fulfill the request to have the minimum number of hosts bits. aka no wasting bits. 

you gave all departments block of 64, 62 hosts per each. this works but you wasting bits for all but 2 vlans. 

2nd try, giving /28, 2-/26; 3rd /27; and 4th/26 - may work but usually will not work. you started with block of 16, then 64 which overlaps block of 16 (depending where u start).

Anyway, you should start with block of 64, aka /26, for dept #2, then another /26 for 36 hosts for #4, then /27 - 30 hosts for #3, and finally /28. 

starting with network id of 10.0.62.0/26, range from 10.0.62.1/26 thru 10.0.62.62/26. then another 64 block starting at 10.0.62.65/26 thru 10.0.62.126/26; then you can start at 10.0.62.128 to allocate block 32, 30 IPs, aka /27, the range will be 10.0.62.127 thru 10.0.62.158/27. you will end up at network id 10.0.62.160.  To assign last block of 16 for 11 IPs, that is /28. Range is 10.0.62.161 - 10.0.62.174, B-cast 10.0.62.175. 

 

answer (in order) should be: 

2° 42 - 10.0.62.0/26 - giving 62 IPs

4° 36 - 10.0.62.64/26 - 62 IPs

3° 22 - 10.0.62.128/27 - 30 IPs

1° 11 - 10.0.62.160/28 - 16 IPs

 

 

Regards, ML
**Please Rate All Helpful Responses **


let me know if that makes sense and if it works. u can attach your PT file as zip here.
btw. 1st ping usually fails due to no cache in ARP table so ICMP sends ARP first. then, consequently pings will work because MAC to IP is already cache in ARP ( until time out).  so , This is normal.

I ordered everyone as you suggested, now work perfectly thanks a lot!! I didnt expect than would have been so simple.

glad to help