ferdon.susan

!"#$% '()*+%, -./-01 223 !455

Module 4, Assignment 3-2

Lab 3.4 Basic Subnetting Question 3.4.1:

A company has applied for and received a Class C network address of 197.15.22.0. The physicalnetwork is to be divided into 4 subnets, which will be interconnected by routers. At least 25 hosts will be needed per subnet. A Class C custom subnet mask needs to be used and a router is needed between the subnets to route packets from one subnet to another. Determine the number of bits that need to be borrowed from the host portion of the network address and the number of bits that will be left for host addresses.

Note: There will be 8 possible subnets, of which 6 can be used.

---------------------------------------------------------------------------WORKSPACE:

(Class C) NNNNNNNN NNNNNNNN NNNNNNNN SSSHHHHH

255 255 255 224197 15 22 0197 15 22 0

256 – 224 = 32 [magic number]

N = 24S = 3 (2^3 – 2 = 6 usable subnets, 4 needed)H = 5 (2^5 – 2 = 30 usable hosts, 25 needed)---------------------------------------------------------------------------

Fill in the following table and answer the following questions:

Subnet No. Subnet BitsBorrowedBinary Value

Subnet BitsDecimal andSubnet No.

Host Bits Possible Binary Values (Range) (5 Bits)

Subnet/HostDecimalRange

Use?

0 Subnet 000 0 (197.15.22.0) 00000 - 11111 0 -31 No (zero subnet)

1st Subnet

001 32

197.15.22.32

00000 - 11111 32 - 63 YES

2nd Subnet

010 64

197.15.22.64

00000 - 11111 64 - 95 YES

3rd Subnet

011 96

197.15.22.96

00000 - 11111 96 - 127 YES

4th Subnet

100 128197.15.22.128

00000 - 11111 128 - 159 YES

5th Subnet

101 160(197.15.22.160)

00000 - 11111 160 - 191 YES

6th Subnet

110 192

197.15.22.192

00000 - 11111 192 - 223 YES

7th Subnet

111 224

(197.15.22.224)

00000 - 11111 224 - 255 No (broadcast)

!"#$% '()*+%, -./-01 223 !455

Use the table just developed to help answer the following questions:

1. Which octet(s) represent the network portion of a Class C IP address? The first three

2. Which octet(s) represent the host portion of a Class C IP address? The last one

3. What is the binary equivalent of the Class C network address in the scenario? 197.15.22.0

Decimal network address: 197. 15. 22. 0

Binary network address: 11000001. 00001111. 00010110. 00000000

4. How many high-order bits were borrowed from the host bits in the fourth octet? 3

5. What subnet mask must be used? Show the subnet mask in decimal and binary.

Decimal subnet mask: 255.255.255.224 Binary subnet mask: 11111111.11111111.11111111.11100000

6. What is the maximum number of subnets that can be created with this subnet mask? 8

7. What is the maximum number of useable subnets that can be created with this mask? 6

8. How many bits were left in the fourth octet for host IDs? 5

9. How many hosts per subnet can be defined with this subnet mask? 2^5 – 2 = 30 usable subnets

10. What is the maximum number of hosts that can be defined for all subnets with this scenario? Assume the lowest and highest subnet numbers and the lowest and highest host ID on each subnet cannot be used.

6 (# usable subnets) x 30 (# usable hosts/subnet) = 180

11. Is 197.15.22.63 a valid host IP address with this scenario? Why or why not?

No. It’s the broadcast address for the 1st subnet (197.15.22.32)

12. Is 197.15.22.160 a valid host IP address with this scenario? Why or why not?

No. It is the address for the 5th subnet.

13. Host A has an IP address of 197.15.22.126. Host B has an IP address of 197.15.22.129.

Are these hosts on the same subnet? Why?

No. Host A (126) is in the 3rd

subnet (96-127) and Host B (129) is in the 4th subnet (128-

159).

!"#$% '()*+%, -./-01 223 !455

Lab 3.5 Subnetting a Class A Network

Question 3.5.1:

-----------------------------------------------WORKSPACE:

Address Network bits Subnet bits Host bits # of subnetsin network

# if hostsper subnet

10.0.0.0 / 24 8 (Class A) 16 8 2^16 =65,536

2^8 -2 = 254

Subnet 16 Address: 10.0.16.0Subnet 16 host range: 10.0.16.1 – 10.0.16.254Subnet 16 broadcast: 10.0.16.255Last subnet: 10.255.254.0-----------------------------------------------

Given a Class A network address of 10.0.0.0 / 24 answer the following questions:

1. How many bits were borrowed from the host portion of this address? 16

2. What is the subnet mask for this network?

Dotted decimal? 255.255.255.000

Binary? 11111111.11111111.11111111.00000000

3. How many usable subnetworks are there? 2^16 – 2 = 65,534 (can’t use first or last)

4. How many usable hosts are there per subnet? 2^8 – 2 = 254

5. What is the host range for usable subnet sixteen? 10.0.16.1 – 10.0.16.254

6. What is the network address for usable subnet sixteen? 10.0.16.0

7. What is the broadcast address for usable subnet sixteen? 10.0.16.255

8. What is the broadcast address for the last usable subnet? 10.255.254.255

9. What is the broadcast address for the major network? 10.255.255.255

Lab 3.6 Subnetting a Class B Network

Background / Preparation

This is a written lab and is to be performed without the aid of an electronic calculator.

ABC Manufacturing has acquired a Class B address, 172.16.0.0. The company needs to create a subnetting scheme to provide the following:

! 36 subnets with at least 100 hosts



It is not necessary to supply an address for the WAN connection since it is supplied by the Internet service provider.

!"#$% '()*+%, -./-01 223 !455

Question 3.6.1:

Given this Class B network address and these requirements answer the following questions

1. How many subnets are needed for this network? 70

2. What is the minimum number of bits that can be borrowed? 7N = 16 (Class B) H = 9 (2^8 -2 = 254 … too few hosts)



Binary? 11111111.11111111.11111110.00000000

4. Slash format /23

5. How many usable subnetworks are there? 2^7 – 2 = 126 (formula: 2^s – 2)

6. How many usable hosts are there per subnet? 2^9 – 2 = 510 (formula: 2^h – 2)

NNNNNNNN NNNNNNNN SSSSSSSH HHHHHHHH

255 255 254 0172 16 0 0172 16 0 0

256-254 = 2 [magic number]

Question 3.6.2:

Complete the following chart listing the first three subnets and the last 4 subnets.

Subnetwork # Subnetwork ID Host Range Broadcast ID

0 172.16.0.0 ZERO SUBNET

1 172.16.2.0 172.16.2.1 –172.16.3.254

172.16.3.255

2 172.16.4.0 172.16.4.1 –172.16.5.254

172.16.5.255

3 172.16.6.0 172.16.6.1 –172.16.7.254

172.16.7.255

… … … …123 172.16.246.0 172.16.246.1 –

172.16.247.254172.16.247.255

124 172.16.248.0 172.16.248.1 –172.16.249.254

172.16.249.255

125 172.16.250.0 172.16.250.1 –172.16.251.254

172.16.251.255

126 17.16.252.0 172.16.252.1 –172.16.253.254

172.16.253.255

127 17.16.254.0 172.16.254.1 –172.16.255.254

172.16.255.255network broadcast

128 172.16.256.0 INVALID

!"#$% '()*+%, -./-01 223 !455

1. What is the host range for subnet two? 172.16.4.1 – 172.16.5.254

2. What is the broadcast address for the 126th subnet? 172.16.253.255


Lab 3.7 Subnetting a Class C Network

Background / Preparation

This is a written exercise and is to be performed without the aid of an electronic calculator.

The Classical Academy has acquired a Class C address, 192.168.1.0. The academy needs to create subnets to provide low level security and broadcast control on the LAN. It is not necessary to supply an address for the WAN connection. It is supplied by the Internet service provider. The LAN consists of the following, each of which will require its own subnet:

• Classroom #1 28 nodes

• Classroom #2 22 nodes

• Computer lab 30 nodes

• Instructors 12 nodes

• Administration 8 nodes

Question 3.7.1:

Given this Class C network address and these requirements answer the following questions

1. How many subnets are needed for this network? 5



Binary? 11111111.11111111.11111111.11100000

3. Slash format /27

4. How many usable hosts are there per subnet? 30 (2^5 – 2)

NNNNNNNN NNNNNNNN NNNNNNNN SSSHHHHH

11111111 11111111 11111111 11100000

255 255 255 224192 168 1 0

256 – 224 = 32 [magic number]

N = 24S = 3 (2^3 = 8 subnets, 5 needed)H = 5 (2^5 – 2 = 30, 30 hosts per subnet needed)

!"#$% '()*+%, -./-01 223 !455

Question 3.7.2:

Complete the following chart

Subnetwork # Subnetwork IP Host Range Broadcast ID

0 192.168.1.0 192.168.1.1 -

192.168.1.30 192.168.1.31

1 192.168.1.32 192.168.1.33 -

192.168.1.62192.168.1.63

2 192.168.1.64 192.168.1.65 -

192.168.1.94 192.168.1.95

3 192.168.1.96 192.168.1.97 -

192.168.1.126 192.168.1.127

4 192.168.1.128 192.168.1.129 -

192.168.1.158 192.168.1.159

5 192.168.1.160 192.168.1.161 -

192.168.1.190 192.168.1.191

6 192.168.1.192 192.168.1.193 -

192.168.1.222 192.168.1.223

7 192.168.1.224 192.168.1.225 -

192.168.1.254 192.168.1.255(network broadcast)

192.168.1.256 INVALID

1. What is the host range for subnet six? 192.168.1.193 - 192.168.1.222

2. What is the broadcast address for the 3rd subnet? 192.168.1.127


ferdon.susan

Documents