Question

7. Which FTP command allows you to retrieve a file from within an FTP session?

39. Company X requests an Internet address from the ICANN. Company X will require 129 nodes to be connected to the network across 14 locations (separate physical networks).

1. What type of address will most likely be issued by the ICANN?

2. How will this address be subnetted to allow for the 14 physical networks to exist?

3. What is the total number of nodes that can be individually addressed at each location?

4. What will the subnet and host portion of the addresses look like?

40. For the following subnets answer the following. How many subnets can there be and how many hosts can be attached to each subnet.

Class B address with 4 bits as subnet, 12 bits representing hosts.

Class B address with 5 bits as subnet, 11 bits representing hosts.

Class B address with 6 bits as subnet, 10 bits representing hosts.

Class B address with 7 bits as subnet, 9 bits representing hosts.

Class C address with 3 bits as subnet, 5 bits representing hosts.

Answer 1

39.

Company X requests an Internet address from the ICANN. Company X will require 129 nodes to be connected to the network across 14 locations (separate physical networks).

ans. A. Type of address depends on number of address required = 129 * 14 = 2^11(approx)

2^11 address requred, class B = 2^16 IP address/Newtwork. So 2^11 falls near Class B

B. Subnetting of address depend on the location by location. If all location require same number of host then arrangements will be made accordingly.

14 network, 14 close to 2^4, so 4 bits selected.

16 bit = Network Id (NID)

16 bit = Host Id (HID)

Divide host id to Sub net id(4 bit) and Host Id (12 bit)

16 networks each hosting 2^12 network addresses.

C. HID = 12 bits

= 2^12 network addresses at each location

= 4096 nodes

D. Subnet = NID=.0000 0000

.0000(Subnet id) 0000(HID)

40.

Starting from the last question to the first for simplicity.

The default class C subnet mask is 255.255.255.0 , To make subnets, bits from the host portion of the mask is borrowed. We got only 8 bits as class C mask uses only the last octet for host addressing,

1.Class C address with 4 bits as subnet, 4 bits as representing hosts:

Mask = 255.255.255.240 (last octet as 1111000) Number of 1's indicate subnet bits and number of 0's indicate host bits.

Number of subnets/hosts = 2x - 2 (since two of these subnets are not usable) where x is the number of 1's for subnets and x is number of 0's for the hosts.

This gives 24 - 2 = 14 subnets and 24 - 2 = 14 hosts for each subnet.

2. Class C address with 3 bits as subnet, 5 bits representing hosts

The subnet mask is 255.255.255.224 as the last octet binary is 11100000 .

Number of subnets = 23 - 2 = 6

Number of hosts per subnet = 25 - 2= 30 hosts

3. Class B address with 7 bits as subnets and 9 bits representing hosts

For Class B, default subnet mask is 255.255.0.0, Class B has 16 bits for creating the subnets

Class B with 7 bits as subnets and 9 represeting hosts can be represented as 255.255.254.0 where binary for the last two octets is 11111110.00000000

Number of subnets = 27 - 2 = 126

Number of host for each subnet = 29 - 2 = 510

4. Class B address with 6 bits as subnets and 10 bits representing hosts

Binary for the last two octets : 11111100.00000000 , subnet mask is 255.255.252.0

Number of subnets = 26 - 2 = 62

Number of hosts for each subnet = 210 - 2 = 1022

5. Class B address with 5 as subnet and 11 represeting host bits

Binary for the last two octets is 11111000.00000000 , subnet mask 255.255.248.0

Number of subnets = 25 - 2 = 30

Number of host for each subnet = 211 - 2 = 2046

6. Class B with 4 bits as subnet and 12 bits representing hosts

Binary for the last two octets is 11110000.00000000 where as subnet mask is 255.255.240.0

Number of subnets = 24 - 2 = 14

Number of hosts for each subnet = 212 - 2 = 4094.