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You administer your company's network shown in the exhibit. You have just added a new host to the networ (Host A). You need to manually configure the TCP/IP properties for Host A. At SRV1, you run ipconfig and discover that the subnet mask is 255.255.255.224. Assuming there are no other hosts on the network other than those shown in the exhibit, which of the follow valid IP addresses for Host A? (Choose all that apply.)

Solve Ip Address Problem By Yapa Wijeratne

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Page 1: Solve Ip Address Problem By Yapa Wijeratne

You administer your company's network shown in the exhibit. You have just added a new host to the networ (Host A). You need to manually configure the TCP/IP properties for Host A. At SRV1, you run ipconfig and discover that the subnet mask is 255.255.255.224.Assuming there are no other hosts on the network other than those shown in the exhibit, which of the follow valid IP addresses for Host A? (Choose all that apply.)

Page 2: Solve Ip Address Problem By Yapa Wijeratne

1) 192.168.1.602) 192.168.1.113) 192.168.1.1244) 192.168.1.155) 192.168.1.56) 192.168.1.31

Page 3: Solve Ip Address Problem By Yapa Wijeratne

Answers are:

192.168.1.15

192.168.1.5

Page 4: Solve Ip Address Problem By Yapa Wijeratne

Now let’s see how we can reach the answer.

• The only possible addresses in this list are 192.168.1.15 and 192.168.1.5.

• 192.168.1.11 is already in use by another host.

• 192.168.1.31 is the broadcast address for the subnet and should not be assigned to a specific host.

This is not enough, let’s find more

Page 5: Solve Ip Address Problem By Yapa Wijeratne

To answer this question, you must first identify the subnet address and the range of IP addresses on the subnet. To find the subnet address:

• 1. Convert the subnet mask to binary. Focus on the octet that is different than 255 or 0. In this example, 224 converts to 11100000 binary.

224 11100000

Page 6: Solve Ip Address Problem By Yapa Wijeratne

2. Convert a known IP address on the subnet to binary. Focus on the last octet masked by the subnet mask. In this example,

you know the IP address for SRV1 is 192.168.1.11. The last octet converts to 1011 binary.

11 1011

Add 0's to the front to complete the octet (00001011).

1011 00001011

Page 7: Solve Ip Address Problem By Yapa Wijeratne

3. Compare the binary forms of the IP address and masks. In this example you are comparing 11100000 with 00001011

From subnet mask 11100000

From server 00001011

Page 8: Solve Ip Address Problem By Yapa Wijeratne

4. For every bit in the mask that is a 0, convert the corresponding bit in the IP address to 0. For every bit in the mask that is a 1, leave the value in the IP address alone. In this example, you are left with 00000000.

From subnet mask 11100000

From server 00001011

00000000

Page 9: Solve Ip Address Problem By Yapa Wijeratne

5. Convert the octet you are working with to decimal. Change this value in the IP address. If there are any octets to the right of the one you are working with, set them to 0 as well. In this example, you have the subnet address of 192.168.1.0.

• 00000000 0

• 192.168.1.0 subnet address

Now that you have the subnet address, you have to find the range of addresses on the subnet.

Page 10: Solve Ip Address Problem By Yapa Wijeratne

1. Convert the subnet mask to binary. Focus on the octet that is different than 255 or 0. In this example, 224 converts to 11100000 binary.

224 11100000

Page 11: Solve Ip Address Problem By Yapa Wijeratne

2. Identify the right-most 1 bit. Convert all other 1's to 0's. In this example, this leaves you with 100000.

• 11100000 100000

Page 12: Solve Ip Address Problem By Yapa Wijeratne

3. Convert this value to decimal. This is the increment value. In this example, the increment value is 32.

• 100000 32

to decimal

increment value

Means that you can add 32 clients.

Page 13: Solve Ip Address Problem By Yapa Wijeratne

4. Add the increment value to the subnet address to find the next subnet address. In this example, the nextsubnet address is 192.168.1.32. All addresses between 192.168.1.0 and 192.168.1.31 belong to the firstsubnet.

• So the IP range will be

192.168.1.0 192.168.1.31

but….

Page 14: Solve Ip Address Problem By Yapa Wijeratne

5. Remove the subnet address (192.168.1.0) and the broadcast address (192.168.1.31) from the range. You cannot assign these addresses to hosts.

This leaves you with possible host addresses between 192.168.1.1 and 192.168.1.30.

Page 15: Solve Ip Address Problem By Yapa Wijeratne

6. Remove any addresses already assigned to other hosts. You should not assign the same address to two

hosts.

Then you will find the answer.

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