ccna workbook 6

8/8/2019 Ccna Workbook 6

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CSCO 120

Workbook 6:

Intermediate Subnetting

Student Name: Jacob Thomas

CSCO 120 D01 3/20/2010


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CSCO 120: CCNA Networ Fun amenta s Wor oo 6


pg. 2

Worksheet 1: Review from Workbook 5

Problem 1: You have been given the following IP address space 215.32.254.0 /24. You have ten Local Area Networks

(LANs) that you need to allocate the addresses to.

Step 1 How many host bits? 8

Step 2 How many host bits need to be borrowed? 4

Step 3 Do you have enough host bits? Yes

Step 4 What will be the size of the subnets? 24 = 16

Step 5 How many host IP addresses will each subnet have? 24

-2 = 14

Subnet

Number

1st

, 2nd

, and 3rd

octets (Network

Bits)

4th

Octet (Host Bits) IP Address in Dotted Decimal Notation

1

11010111.00100000.11111110 0 0 0 0 0 0 0 0 215.32.254.0 Subnet ID

11010111.00100000.11111110 0 0 0 0 0 0 0 1 215.32.254.1 First host IP address

11010111.00100000.11111110 0 0 0 0 1 1 1 0 215.32.254.14 Last host IP address

11010111.00100000.11111110 0 0 0 0 1 1 1 1 215.32.254.15 Broadcast IP address

2

11010111.00100000.11111110 0 0 0 1 0 0 0 0 215.32.254.16 Subnet ID




3

11010111.00100000.11111110 0 0 1 0 0 0 0 0 215.32.254.32 Subnet ID




4

11010111.00100000.11111110 0 0 1 1 0 0 0 0 215.32.254.48 Subnet ID




5

11010111.00100000.11111110 0 1 0 0 0 0 0 0 215.32.254.64 Subnet ID




6

11010111.00100000.11111110 0 1 0 1 0 0 0 0 215.32.254.80 Subnet ID




7

11010111.00100000.11111110 0 1 1 0 0 0 0 0 215.32.254.96 Subnet ID




8

11010111.00100000.11111110 0 1 1 1 0 0 0 0 215.32.254.112 Subnet ID

11010111.00100000.11111110 0 1 1 1 0 0 0 1 215.32.254. 113 First host IP address



Use the table on the next page to complete this worksheet if you need to.


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pg. 3

Worksheet 1: Continued

Problem 1: Continued

Subnet

Number

1st

, 2nd

, and 3rd

octets (Network

Bits)

4th

Octet (Host Bits) IP Address in Dotted Decimal Notation

9

11010111.00100000.11111110 1 0 0 0 0 0 0 0 215.32.254.128 Subnet ID




10

11010111.00100000.11111110 1 0 0 1 0 0 0 0 215.32.254.144 Subnet ID




11

11010111.00100000.11111110 1 0 1 0 0 0 0 0 215.32.254.160 Subnet ID




12

11010111.00100000.11111110 1 0 1 1 0 0 0 0 215.32.254.176 Subnet ID




13

11010111.00100000.11111110 1 1 0 0 0 0 0 0 215.32.254.192 Subnet ID




14

11010111.00100000.11111110 1 1 0 1 0 0 0 0 215.32.254.208 Subnet ID


11010111.00100000.11111110 1 1 0 1 1 1 1 0 215.32.254.222 Last host IP address11010111.00100000.11111110 1 1 0 1 1 1 1 1 215.32.254.223 Broadcast IP address

15

11010111.00100000.11111110 1 1 1 0 0 0 0 0 215.32.254.224 Subnet ID




16

11010111.00100000.11111110 1 1 1 1 0 0 0 0 215.32.254.240 Subnet ID





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pg. 4

Worksheet 2: Examining subnetting from a Dotted Decimal perspective.

Instructions: Using the results from worksheet 1 in the workbook, complete the table below.

Subnet

# Subnet ID First Host IP Last Host IP Broadcast IP

1 215.32.254.0 215.32.254.1 215.32.254.14 215.32.254.15

2 215.32.254.16 215.32.254.17 215.32.254.30 215.32.254.31

3 215.32.254.32 215.32.254.33 215.32.254.46 215.32.254.47

4 215.32.254.48 215.32.254.49 215.32.254.62 215.32.254.63

5 215.32.254.64 215.32.254.65 215.32.254.78 215.32.254.79

6 215.32.254.80 215.32.254.81 215.32.254.94 215.32.254.95

7 215.32.254.96 215.32.254.97 215.32.254.110 215.32.254.111

8 215.32.254.112 215.32.254.113 215.32.254.126 215.32.254.127

9 215.32.254.128 215.32.254.129 215.32.254.142 215.32.254.143

10 215.32.254.144 215.32.254.145 215.32.254.158 215.32.254.159

11 215.32.254.160 215.32.254.161 215.32.254.174 215.32.254.175

12 215.32.254.176 215.32.254.177 215.32.254.190 215.32.254.191

13 215.32.254.192 215.32.254.193 215.32.254.206 215.32.254.207

14 215.32.254.208 215.32.254.209 215.32.254.222 215.32.254.223

15 215.32.254.224 215.32.254.225 215.32.254.238 215.32.254.239

16 215.32.254.240 215.32.254.241 215.32.254.254 215.32.254.255

1) What is the new subnet mask that will have to be used for these subnets? (list both the prefix notation andthe dotted decimal notation). 255.255.255.240

2) Convert the new subnet mask into binary and place it into the table below.11111111.11111111.11111111.11110000

3) Locate the rightmost 1 in the subnet mask. Circle the power of 2 above the rightmost 1 in the subnet mask.4) What is the value of the power of 2 you circled? 24


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pg. 5

Worksheet 3: Analysis and conclusions from Worksheet 2.

1) In which octet does the Subnet ID value change? 4th2) What is the difference in value between the Subnet 1 ID and the Subnet 2 ID? .163) What is the difference in value between the Subnet 2 ID and the Subnet 3 ID? .164)

What is the difference in value between the Subnet 3 ID and the Subnet 4 ID? .16

5) What is the difference in value between the Subnet 10 ID and the Subnet 11 ID? .166) What is the difference in value between the Subnet 15 ID and the Subnet 16 ID? .167) What is the difference in value between the Subnet 1 ID and the First Host in Subnet 1? .18) In which octet does the First Host IP value change? 4th9) What is the difference in value between the First Host IP in subnet 1 and the first host IP in subnet 2? .1610)What is the difference in value between the First Host IP in subnet 2 and the first host IP in subnet 3? .1611)What is the difference in value between the First Host IP in subnet 10 and the first host IP in subnet 11? .1612)In which octet does the Last Host IP value change? 4th13)What is the difference in value between the last Host IP in subnet 1 and the last host IP in subnet 2? .2914)What is the difference in value between the last Host IP in subnet 2 and the last host IP in subnet 3? .1615)What is the difference in value between the last Host IP in subnet 12 and the last host IP in subnet 13? .1616)What is the difference between the Last Host IP address for Subnet 1 and the Broadcast IP address for Subnet 1?

.1

17)In which octet does the Broadcast IP value change? 4th18)What is the difference in value between the broadcast IP in subnet 1 and subnet 2? .1619)What is the difference in value between the broadcast IP in subnet 2 and subnet 3? .1620)What is the difference in value between the broadcast IP in subnet 7 and subnet 8? .121)What is the difference in value between the Broadcast IP address in Subnet 1 and the Subnet ID in Subnet 2? .122)Are the Subnet IDs odd or even? even23)Are the First Host IPs odd or even? odd24)Are the Last Host IPs odd or even? even25)Are the Broadcast IPs odd or even? odd26)What is your conclusion? Is there a pattern? If so describe it. If not, why? Yes all the subnet IDs, first host IPs,

last host IPs, and Broadcast IPs are 16 away from the next Subnets, subnet, first, last, and broadcast. And all the

subnets and last hosts are even and the first host and broadcast are odd.


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pg. 6

Worksheet 4: Subnetting problems in Dotted Decimal Practice.

Instructions: Complete the problems using the method you learned in the preceding worksheets. If you have to, you

can check your work in binary.

Problem 1: You have been given the IP address space 192.168.0.0 /24. Subnet the space so that you create equal sized

subnets that can accommodate 62 hosts. (Note, you may not need all of the rows in the table below.)

1) This approaches subnetting from the host perspective. So ask yourself the following questions:a. How many bits do you have in the IP address space? 8b. How many bits do you need for 62 hosts in each subnet? 2c. How many bits will you have left over for subnetting? 6d. What is the new subnet in binary? 11111111.11111111.11111111.11000000e. What number is above the rightmost bit in the new subnet mask? 64f. In what octet is that number located? 4th

Subnet

# Subnet ID First Host IP Last Host IP Broadcast IP1 192.168.0.0 192.168.0.1 192.168.0.62 192.168.0.63

2 192.168.0.64 192.168.0.65 192.168.0.126 192.168.0.127

3 192.168.0.128 192.168.0.129 192.168.0.190 192.168.0.191

4 192.168.0.192 192.168.0.193 192.168.0.254 192.168.0.255

5

6

7

8

9

10

11

12

13

14

15

16


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pg. 7




Subnet


1 200.5.1.0 200.5.1.1 200.5.1.30 200.5.1.31

2 200.5.1.32 200.5.1.33 200.5.1.62 200.5.1.63

3 200.5.1.64 200.5.1.65 200.5.1.94 200.5.1.95

4 200.5.1.96 200.5.1.97 200.5.1.126 200.5.1.127

5 200.5.1.128 200.5.1.129 200.5.1.158 200.5.1.159

6 200.5.1.160 200.5.1.161 200.5.1.190 200.5.1.191

7 200.5.1.192 200.5.1.193 200.5.1.222 200.5.1.223

8 200.5.1.224 200.5.1.225 200.5.1.254 200.5.1.255

9

10

11

12

13

14

15

16


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pg. 8




Subnet


1 210.21.13.0 210.21.13.1 210.21.13.14 210.21.13.15

2 210.21.13.16 210.21.13.17 210.21.13.30 210.21.13.31

3 210.21.13.32 210.21.13.33 210.21.13.46 210.21.13.47

4 210.21.13.48 210.21.13.49 210.21.13.62 210.21.13.63

5 210.21.13.64 210.21.13.65 210.21.13.78 210.21.13.79

6 210.21.13.80 210.21.13.81 210.21.13.94 210.21.13.95

7 210.21.13.96 210.21.13.97 210.21.13.110 210.21.13.111

8 210.21.13.112 210.21.13.113 210.21.13.126 210.21.13.127

9 210.21.13.128 210.21.13.129 210.21.13.142 210.21.13.143

10 210.21.13.144 210.21.13.145 210.21.13.158 210.21.13.159

11 210.21.13.160 210.21.13.161 210.21.13.174 210.21.13.175

12 210.21.13.176 210.21.13.177 210.21.13.190 210.21.13.191

13 210.21.13.192 210.21.13.193 210.21.13.206 210.21.13.207

14 210.21.13.208 210.21.13.209 210.21.13.222 210.21.13.223

15 210.21.13.224 210.21.13.225 210.21.13.238 210.21.13.239

16 210.21.13.240 210.21.13.241 210.21.13.254 210.21.13.255


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pg. 9



subnets that can accommodate the needs to 6 LANS. (Note, you may not need all of the rows in the table below.)

1) This approaches subnetting from the perspective of networks needed. So ask yourself the following questions:a. How many bits do you have in the IP address space? 16b. How many bits do you need for creating enough subnets for your needs? 3c. How many bits will you have left over for host addresses? 13d. What is the new subnet in binary? 11111111.11111111.11000000.00000000e. What number is above the rightmost bit in the new subnet mask? 64f. In what octet is that number located? 3rd

Subnet


1 150.58.0.0 150.58.0.1 150.58.31.254 150.58.31.255

2 150.58.32.0 150.58.32.1 150.58.63.254 150.58.63.255

3 150.58.64.0 150.58.64.1 150.58.95.254 150.58.95.255

4 150.58.96.0 150.58.96.1 150.58.127.254 150.58.127.255

5 150.58.128.0 150.58.128.1 150.58.159.254 150.58.159.255

6 150.58.160.0 150.58.160.1 150.58.191.254 150.58.191.255

7 150.58.192.0 150.58.192.1 150.58.223.254 150.58.223.255

8 150.58.224.0 150.58.224.1 150.58.255.254 150.58.255.255

9

10

11

12

13

14

15

16


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pg. 10




Subnet


1 171.17.0.0 171.17.1.1 171.17.14.254 171.17.15.255

2 171.17.16.0 171.17.17.1 171.17.30.254 171.17.31.255

3 171.17.32.0 171.17.33.1 171.17.46.254 171.17.47.255

4 171.17.48.0 171.17.49.1 171.17.62.254 171.17.63.255

5 171.17.64.0 171.17.65.1 171.17.78.254 171.17.79.255

6 171.17.80.0 171.17.81.1 171.17.94.254 171.17.95.255

7 171.17.96.0 171.17.97.1 171.17.110.254 171.17.111.255

8 171.17.112.0 171.17.113.1 171.17.126.254 171.17.127.255

9 171.17.128.0 171.17.129.1 171.17.142.254 171.17.143.255

10 171.17.144.0 171.17.145.1 171.17.158.254 171.17.159.255

11 171.17.160.0 171.17.161.1 171.17.174.254 171.17.175.255

12 171.17.176.0 171.17.177.1 171.17.190.254 171.17.191.255

13 171.17.192.0 171.17.193.1 171.17.206.254 171.17.207.255

14 171.17.208.0 171.17.209.1 171.17.222.254 171.17.223.255

15 171.17.224.0 171.17.225.1 171.17.238.254 171.17.239.255

16 171.17.240.0 171.17.241.1 171.17.254.254 171.17.255.255


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pg. 11




1) This approaches subnetting from the perspective of networks needed. So ask yourself the following questions:a. How many bits do you have in the IP address space? 24b. How many bits do you need for creating enough subnets for your needs? 3c. How many bits will you have left over for host addresses? 21d. What is the new subnet in binary? 11111111.11100000.00000000.00000000e. What number is above the rightmost bit in the new subnet mask? 32f. In what octet is that number located? 2nd

Subnet


1 10.0.0.0 10.0.0.1 10.31.254.254 10.31.255.255

2 10.32.0.0 10.32.0.1 10.63.254.254 10.63.255.255

3 10.64.0.0 10.64.0.1 10.65.254.254 10.65.255.255

4 10.96.0.0 10.96.0.1 10.127.254.254 10.127.255.255

5 10.128.0.0 10.128.0.1 10.159.254.254 10.159.255.255

6 10.160.0.0 10.160.0.1 10.191.254.254 10.191.255.255

7 10.192.0.0 10.192.0.1 10.223.254.254 10.223.255.255

8 10.224.0.0 10.224.0.1 10.255.254.254 10.255.255.255

9

10

11

12

13

14

15

16


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pg. 12




Subnet


1 10.0.0.0 10.0.0.1 10.15.254.254 10.15.255.255

2 10.16.0.0 10.16.0.1 10.31.254.254 10.31.255.255

3 10.32.0.0 10.32.0.1 10.47.254.254 10.47.255.255

4 10.48.0.0 10.48.0.1 10.63.254.254 10.63.255.255

5 10.64.0.0 10.64.0.1 10.79.254.254 10.79.255.255

6 10.80.0.0 10.80.0.1 10.95.254.254 10.95.255.255

7 10.96.0.0 10.96.0.1 10.111.254.254 10.111.255.255

8 10.112.0.0 10.112.0.1 10.127.254.254 10.127.255.255

9 10.128.0.0 10.128.0.1 10.143.254.254 10.143.255.255

10 10.144.0.0 10.144.0.1 10.159.254.254 10.159.255.255

11 10.160.0.0 10.160.0.1 10.175.254.254 10.175.255.255

12 10.176.0.0 10.176.0.1 10.191.254.254 10.191.255.255

13 10.192.0.0 10.192.0.1 10.207.254.254 10.207.255.255

14 10.208.0.0 10.208.0.1 10.223.254.254 10.223.255.255

15 10.224.0.0 10.224.0.1 10.239.254.254 10.239.255.255

16 10.240.0.0 10.240.0.1 10.254.254.254 10.254.255.255


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pg. 13

Worksheet 5: Creating Subnets the Right Size

Instructions: For the problems below, complete the steps as they are outlined.

Example 1: You have been given the 192.168.0.0 /24 address space. You need to create a subnetting scheme to meet

the needs of the hosts on each of the LANs as specified below.

LAN 1 68 hostsLAN 2 40 Hosts

LAN 3 25 Hosts

LAN 4 10 hosts

WAN 1 2 hosts

Step 1 Complete the table below:

Network

Host

Addresses

Needed

Power of 2 that

will equal or

exceed the

number of hostaddresses

Starting IP Address

(Subnet ID)

Ending IP Address

(Broadcast IP)

LAN 1 68 128 192.168.0.0 192.168.0.127

LAN 2 40 64 192.168.0.128 192.168.0.191

LAN 3 25 32 192.168.0.192 192.168.0.223

LAN 4 10 16 192.168.0.224 192.168.0.239

WAN 1 2 4 192.168.0.240 192.168.0.243

145 244


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pg. 14


Problem 1: You have been given the 192.168.0.0 /24 address space. You need to create a subnetting scheme to meet


LAN 1 90 hosts

LAN 2 55 Hosts

LAN 3 22 HostsLAN 4 13 hosts

WAN 1 2 hosts

WAN 2 2 hosts

WAN 3 2 hosts

Network

Host

Addresses

Needed

Power of 2 that

will equal or

exceed the

number of host

addresses

Starting IP Address

(Subnet ID)

Ending IP Address

(Broadcast IP)

LAN 1 90 128 192.168.0.0 192.168.0.127

LAN 2 55 64 192.168.0.128 192.168.0.191

LAN 3 22 32 192.168.0.192 192.168.0.223

LAN 4 13 16 192.168.0.224 192.168.0.239

WAN 1 2 4 192.168.0.240 192.168.0.243

WAN 2 2 4 192.168.0.244 192.168.0.247

WAN 3 2 4 192.168.0.248 192.168.0.251

186 252


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pg. 15


Problem 2: You have been given the 192.168.0.0 /20 address space. You need to create a subnetting scheme to meet


LAN 1 90 hosts

LAN 2 55 HostsLAN 3 22 Hosts

LAN 4 13 hosts

WAN 1 2 hosts

WAN 2 2 hosts

WAN 3 2 hosts

Network

Host

Addresses

Needed

Power of 2 that

will equal or

exceed the

number of host

addresses

Starting IP Address

(Subnet ID)

Ending IP Address

(Broadcast IP)

LAN 1 90 128 192.168.0.0 192.168.0.127

LAN 2 55 64 192.168.0.128 192.168.0.191

LAN 3 22 32 192.168.0.192 192.168.0.223

LAN 4 13 16 192.168.0.224 192.168.0.239

WAN 1 2 4 192.168.0.240 192.168.0.243

WAN 2 2 4 192.168.0.244 192.168.0.247

WAN 3 2 4 192.168.0.248 192.168.0.251

186 252


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pg. 17


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ccna workbook 6

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