ccna workbook 6
TRANSCRIPT
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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
Intermediate Subnetting
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
11010111.00100000.11111110 0 0 0 1 0 0 0 1 215.32.254.17 First host IP address
11010111.00100000.11111110 0 0 0 1 1 1 1 0 215.32.254.30 Last host IP address
11010111.00100000.11111110 0 0 0 1 1 1 1 1 215.32.254.31 Broadcast IP address
3
11010111.00100000.11111110 0 0 1 0 0 0 0 0 215.32.254.32 Subnet ID
11010111.00100000.11111110 0 0 1 0 0 0 0 1 215.32.254.33 First host IP address
11010111.00100000.11111110 0 0 1 0 1 1 1 0 215.32.254.46 Last host IP address
11010111.00100000.11111110 0 0 1 0 1 1 1 1 215.32.254.47 Broadcast IP address
4
11010111.00100000.11111110 0 0 1 1 0 0 0 0 215.32.254.48 Subnet ID
11010111.00100000.11111110 0 0 1 1 0 0 0 1 215.32.254.49 First host IP address
11010111.00100000.11111110 0 0 1 1 1 1 1 0 215.32.254.62 Last host IP address
11010111.00100000.11111110 0 0 1 1 1 1 1 1 215.32.254.63 Broadcast IP address
5
11010111.00100000.11111110 0 1 0 0 0 0 0 0 215.32.254.64 Subnet ID
11010111.00100000.11111110 0 1 0 0 0 0 0 1 215.32.254.65 First host IP address
11010111.00100000.11111110 0 1 0 0 1 1 1 0 215.32.254.78 Last host IP address
11010111.00100000.11111110 0 1 0 0 1 1 1 1 215.32.254.79 Broadcast IP address
6
11010111.00100000.11111110 0 1 0 1 0 0 0 0 215.32.254.80 Subnet ID
11010111.00100000.11111110 0 1 0 1 0 0 0 1 215.32.254.81 First host IP address
11010111.00100000.11111110 0 1 0 1 1 1 1 0 215.32.254.94 Last host IP address
11010111.00100000.11111110 0 1 0 1 1 1 1 1 215.32.254.95 Broadcast IP address
7
11010111.00100000.11111110 0 1 1 0 0 0 0 0 215.32.254.96 Subnet ID
11010111.00100000.11111110 0 1 1 0 0 0 0 1 215.32.254.97 First host IP address
11010111.00100000.11111110 0 1 1 0 1 1 1 0 215.32.254.110 Last host IP address
11010111.00100000.11111110 0 1 1 0 1 1 1 1 215.32.254.111 Broadcast IP address
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
11010111.00100000.11111110 0 1 1 1 1 1 1 0 215.32.254.126 Last host IP address
11010111.00100000.11111110 0 1 1 1 1 1 1 1 215.32.254.127 Broadcast IP address
Use the table on the next page to complete this worksheet if you need to.
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Intermediate Subnetting
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
11010111.00100000.11111110 1 0 0 0 0 0 0 1 215.32.254.129 First host IP address
11010111.00100000.11111110 1 0 0 0 1 1 1 0 215.32.254.142 Last host IP address
11010111.00100000.11111110 1 0 0 0 1 1 1 1 215.32.254.143 Broadcast IP address
10
11010111.00100000.11111110 1 0 0 1 0 0 0 0 215.32.254.144 Subnet ID
11010111.00100000.11111110 1 0 0 1 0 0 0 1 215.32.254.145 First host IP address
11010111.00100000.11111110 1 0 0 1 1 1 1 0 215.32.254.158 Last host IP address
11010111.00100000.11111110 1 0 0 1 1 1 1 1 215.32.254.159 Broadcast IP address
11
11010111.00100000.11111110 1 0 1 0 0 0 0 0 215.32.254.160 Subnet ID
11010111.00100000.11111110 1 0 1 0 0 0 0 1 215.32.254.161 First host IP address
11010111.00100000.11111110 1 0 1 0 1 1 1 0 215.32.254.174 Last host IP address
11010111.00100000.11111110 1 0 1 0 1 1 1 1 215.32.254.175 Broadcast IP address
12
11010111.00100000.11111110 1 0 1 1 0 0 0 0 215.32.254.176 Subnet ID
11010111.00100000.11111110 1 0 1 1 0 0 0 1 215.32.254.177 First host IP address
11010111.00100000.11111110 1 0 1 1 1 1 1 0 215.32.254.190 Last host IP address
11010111.00100000.11111110 1 0 1 1 1 1 1 1 215.32.254.191 Broadcast IP address
13
11010111.00100000.11111110 1 1 0 0 0 0 0 0 215.32.254.192 Subnet ID
11010111.00100000.11111110 1 1 0 0 0 0 0 1 215.32.254.193 First host IP address
11010111.00100000.11111110 1 1 0 0 1 1 1 0 215.32.254.206 Last host IP address
11010111.00100000.11111110 1 1 0 0 1 1 1 1 215.32.254.207 Broadcast IP address
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 0 0 0 1 215.32.254.209 First host IP address
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
11010111.00100000.11111110 1 1 1 0 0 0 0 1 215.32.254.225 First host IP address
11010111.00100000.11111110 1 1 1 0 1 1 1 0 215.32.254.238 Last host IP address
11010111.00100000.11111110 1 1 1 0 1 1 1 1 215.32.254.239 Broadcast IP address
16
11010111.00100000.11111110 1 1 1 1 0 0 0 0 215.32.254.240 Subnet ID
11010111.00100000.11111110 1 1 1 1 0 0 0 1 215.32.254.241 First host IP address
11010111.00100000.11111110 1 1 1 1 1 1 1 0 215.32.254.254 Last host IP address
11010111.00100000.11111110 1 1 1 1 1 1 1 1 215.32.254.255 Broadcast IP address
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CSCO 120: CCNA Networ Fun amenta s Wor oo 6
Intermediate Subnetting
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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CSCO 120: CCNA Networ Fun amenta s Wor oo 6
Intermediate Subnetting
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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CSCO 120: CCNA Networ Fun amenta s Wor oo 6
Intermediate Subnetting
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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Intermediate Subnetting
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Worksheet 4: Continued
Problem 2: You have been given the IP address space 200.5.1.0 /24. Subnet the space so that you create equal sized
subnets that can accommodate 30 hosts. (Note, you may not need all of the rows in the table below.)
Subnet
# Subnet ID First Host IP Last Host IP Broadcast IP
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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Intermediate Subnetting
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Worksheet 4: Continued
Problem 3: You have been given the IP address space 210.21.13.0 /24. Subnet the space so that you create equal sized
subnets that can accommodate 14 hosts. (Note, you may not need all of the rows in the table below.)
Subnet
# Subnet ID First Host IP Last Host IP Broadcast IP
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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Intermediate Subnetting
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Worksheet 4: Continued
Problem 4: You have been given the IP address space 150.58.0.0 /16. Subnet the space so that you create equal sized
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
# Subnet ID First Host IP Last Host IP Broadcast IP
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
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Intermediate Subnetting
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Worksheet 4: Continued
Problem 5: You have been given the IP address space 171.17.16.0 /16. Subnet the space so that you create equal sized
subnets that can accommodate the needs to 13 LANS. (Note, you may not need all of the rows in the table below.)
Subnet
# Subnet ID First Host IP Last Host IP Broadcast IP
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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Intermediate Subnetting
pg. 11
Worksheet 4: Continued
Problem 6: You have been given the IP address space 10.0.0.0 /8. Subnet the space so that you create equal sized
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? 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
# Subnet ID First Host IP Last Host IP Broadcast IP
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
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16
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Worksheet 4: Continued
Problem 7: You have been given the IP address space 10.0.0.0 /8. Subnet the space so that you create equal sized
subnets that can accommodate the needs to 16 LANS. (Note, you may not need all of the rows in the table below.)
Subnet
# Subnet ID First Host IP Last Host IP Broadcast IP
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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Intermediate Subnetting
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
Worksheet 5: Continued
Problem 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 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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CSCO 120: CCNA Networ Fun amenta s Wor oo 6
Intermediate Subnetting
pg. 15
Worksheet 5: Continued
Problem 2: You have been given the 192.168.0.0 /20 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 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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CSCO 120: CCNA Networ Fun amenta s Wor oo 6
Intermediate Subnetting
pg. 17
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CSCO 120: CCNA Networ Fun amenta s Wor oo 6
Intermediate Subnetting