08-26-2018 12:11 PM
Im doing a practice test which asks the following:
You have been given an IP address block of 192.168.3.0 /24. You must provide for existing networks as well as future growth. Subnet assignments are:
• 1st subnet, existing student LAN, up to 28 hosts (Fa0/0 on R2-Central, connected to Fa0/24 on S1-Central)
• 2nd subnet, future student LAN, up to 28 hosts (not yet implemented)
• 3rd subnet, existing ISP LAN, up to 14 hosts (Fa0/0 on R1-ISP)
• 4th subnet, future ISP LAN, up to 7 hosts (not yet implemented)
• 5th subnet, existing WAN, point-to-point link (S0/0/0 on R1-ISP and S0/0/0 on R2-Central)
My first thought was to change the /24 into a /27 which gives you 8 subnets and 32 hosts per subnet. is my thinking right? or if its not can someone point me in the right direction?
Solved! Go to Solution.
08-26-2018 01:36 PM
Hi,
One of the point in subnets is to conserve IPs (ofcourse if you have Class A or B that may change the game). Subnet based on what needed per subnet but not based on total number of subnets. For example in your case you need 1st and 2nd subnet /27 to cover 28 hosts but 5th is only 2 IPs (point to point) - so /30. That way you left with few IPs for future growth as well. I hope this put you in right direction. If you still need assistance, post here or check online tutorials on IP Subnet procedures.
hth
MS
08-26-2018 01:36 PM
Hi,
One of the point in subnets is to conserve IPs (ofcourse if you have Class A or B that may change the game). Subnet based on what needed per subnet but not based on total number of subnets. For example in your case you need 1st and 2nd subnet /27 to cover 28 hosts but 5th is only 2 IPs (point to point) - so /30. That way you left with few IPs for future growth as well. I hope this put you in right direction. If you still need assistance, post here or check online tutorials on IP Subnet procedures.
hth
MS
08-26-2018 07:55 PM
That answers my question perfectly. thanks
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