Help Please - Assignment

[aelar]

ICANN address of 193.24.221.0

Prove by means of calculation that the new subnet is capable of seven(7) subnets and atleast 10 nodes.

Confused! help plz

 

[Drake2007]

Goto binary and you're missing the netmask.

 

[ponder]

...and you're missing the netmask.

It's Class C so assume /24 if not specified?

CIDR notation	  Network Mask	     Sub-network   Available Hosts/network	Total usable hosts
/24	                  255.255.255.0	        1	            256                               254
/25	                  255.255.255.128	2	            128                               126
/26	                  255.255.255.192	4	            64	                                 62
/27	                  255.255.255.224	8	            32	                                 30
/28	                  255.255.255.240	16	            16	                                 14
/29	                  255.255.255.248	32	            8	                                 6
/30	                  255.255.255.252	64	            4	                                 2

You will however have to do the binary calculations in order to prove this, the following should get you going:
http://www.cisco.com/en/US/tech/tk365/technologies_tech_note09186a00800a67f5.shtml
http://www.tcpipguide.com/free/t_IPVariableLengthSubnetMaskingVLSM.htm
http://searchnetworking.techtarget.com/generic/0,295582,sid7_gci1154229,00.html
http://conft.org/Cisco_Ip-Addressing-And-Subnetting-For-New-Users.pdf
http://www.docstoc.com/docs/2164762/Barrys-CIDRVLSM-Quick-Reference-Guide

If I can give you some advice then it would be to learn this very well and be able to do it in your head after some practice.

 

[Drake2007]

It's Class C so assume /24 if not specified?

Class C would have to be assumed yes, though the question infers that it's a new subnet meaning it was working fine until someone broke it

 

[ponder]

...though the question infers that it's a new subnet meaning it was working fine until someone broke it

Ah, reading it again clarifies things, "the new subnet"

 

[zvakarehwa]

i have an address 193.24.221.0/24 i need to have 7 subnets and accomodate at least 10 pcs.
what will be number of hosts before modification?
what will be the modified subnet ?
first network ID, first node, last node, broadcast address for all subnets created?

 

[syntax]

i have an address 193.24.221.0/24 i need to have 7 subnets and accomodate at least 10 pcs.
what will be number of hosts before modification?
what will be the modified subnet ?
first network ID, first node, last node, broadcast address for all subnets created?

I have two issues with you asking this.

1. There are ip subnet calculators on the internet, which if you had googled you could have got the answer quite quickly. So, you didnt even google.

2. If you used the calculator, and didnt understand the answer, you should have asked why the answer is the way it is.
For the record, there are 2 answers, to be lazy, i would pick the biggest one = less subnets

 

[staunchy20]

I have two issues with you asking this.

1. There are ip subnet calculators on the internet, which if you had googled you could have got the answer quite quickly. So, you didnt even google.

2. If you used the calculator, and didnt understand the answer, you should have asked why the answer is the way it is.
For the record, there are 2 answers, to be lazy, i would pick the biggest one = less subnets

Well said Syntax.

People no longer wants to think for themselves and google for help, they expect to be spoon fed everything.

 

[czc]

I have two issues with you asking this.

1. There are ip subnet calculators on the internet, which if you had googled you could have got the answer quite quickly. So, you didnt even google.

2. If you used the calculator, and didnt understand the answer, you should have asked why the answer is the way it is.
For the record, there are 2 answers, to be lazy, i would pick the biggest one = less subnets

Alternatively this mybb page was ranked higher in the google ranking than anything else. I've see that happen.