need a little cisco help (basic)

mango123

Senior member
Sep 1, 2012
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Looking for a fellow Cisco student or ccna/p to answer a couple basic subnetting questions for a project, please be familiar with cisco jargon...
Would be easiest to do over real time chat / irc please respond.
 
Last edited:

ch33zw1z

Lifer
Nov 4, 2004
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You'll have better luck posting the basic subnetting questions with an idea of what you understand and don't...
 

mango123

Senior member
Sep 1, 2012
214
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Review the existing network, and upgrade it by adding an 1841 router and standalone 2960 switch to
supplement and offload the existing Linksys WRT300N. The new switch will support connections from
wired clients on one subnet. The existing Linksys will support wireless clients on another subnet.
Configure the 1841 as a DHCP server for the wired network, and the Linksys to support wireless users.
The wired and wireless client traffic from each subnet is routed through the new 1841 customer router.
RIPv2 is to be used between the 1841 and the ISP, and the encapsulation on the WAN link between them
is PPP. The customer router must use a static address. The ISP router serial interface IP address it must
communicate with is: ___________________________________
If your local network is connected to the ISP as AnyCompany1, the IP address of the ISP serial 0/0/0
interface is 10.100.1.5 /22.

Part B – Develop the Subnet Scheme
The customer has been assigned an IP address and subnet mask ____________________________
If the local network customer is AnyCompany1, use 192.168.111.0 /24.
If the local network customer is AnyCompany2, use 192.168.222.0 /24.
Develop a subnet scheme with this address that allows the customer network to support two subnets of
up to 30 clients each, and allow for growth to as many as six subnets in the future.
The first subnet is for the wired clients. The second subnet is used to assign an IP address to the Linksys
external Internet interface. The internal wireless network clients use the default IP addressing (network
192.168.1.0 /24) assigned by the Linksys. The Linksys uses NAT/PAT to convert internal wireless client
addresses to the external address. The internal wireless clients do not require a subnet from the base
address.
Step 1: Determine the number of hosts and subnets.
a. The largest subnet must be able to support 30 hosts. To support that many hosts, the number of host
bits required is ______________. --- this is 5, correct?
b. What is the minimum number of subnets required for the new network design that also allows for
future growth? _________ well they want 6, but u get 8 with /27 so 8 right?
c. How many host ID bits are reserved for the subnet ID to allow for this number of subnets with each
subnet having 30 hosts? _________ what do they mean by this?
d. What is the maximum possible number of subnets with this scheme? ___________ agaiun what do they mean?
Step 2: Calculate the custom subnet mask.
Now that the number of subnet ID bits is known, the subnet mask can be calculated. A class C network
has a default subnet mask of 24 bits, or 255.255.255.0.
The custom subnet mask for this network will be _____._____._____._____, or /_____.
Step 3: Identify subnet and host IP addresses.
Now that the subnet mask is identified, the network addressing scheme can be created. The addressing
scheme includes the subnet numbers, the subnet broadcast address, and the range of IP addresses
assignable to hosts.
Complete the table showing all the possible subnets,......



ok so understand they are asking for a /27.

my question lies in the wording of the bits and "minimum number" of subnets req'd
also in how to do the address of actual interfaces, specifically the WNDR linsys device WAN link going from there to the customer router which is an 1841...
 
Last edited:

Danimal1209

Senior member
Nov 9, 2011
355
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0
a. Yes, it is 5. if you do 2^host bits = amount of hosts for that subnet. For example, 2^5 = 32 hosts.

b. Correct.

c. You need to do a little research on host bits and network bits.

d. Maximum number of subnets. Lets say, like in this scenario, you have a /27.
So, each subnet is 32 addresses, correct? In a /27 the first three octets do not change, only the last octect. So you have 255.255.255.0 - 255.255.255.31 for each subnet up until 255.255.255.255. The easiest way to determine how many subnets for a /24 or higher (class C) would be to take the number of hosts allowed per subnet and divide it by 256. So, 32/256 = 8.
 
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