A new host (Host A) needs an IP address configured on the network. The document determines the valid IP address range based on the subnet mask of 255.255.255.224 discovered from an existing host (SRV1). This subnet mask allows for a maximum of 30 IP addresses, with a range of 192.168.1.1 to 192.168.1.30 after removing the subnet address of 192.168.1.0 and broadcast address of 192.168.1.31. Of the options provided, 192.168.1.15 and 192.168.1.5 are valid since the other addresses are either already in use or are the subnet/broadcast address.

1. 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.)

2. 1) 192.168.1.60
2) 192.168.1.11
3) 192.168.1.124
4) 192.168.1.15
5) 192.168.1.5
6) 192.168.1.31

3. Answers are:
192.168.1.15
192.168.1.5

4. 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

5. 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

6. 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

7. 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

8. 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

9. 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.

10. 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

11. 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

12. 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.

13. 4. Add the increment value to the subnet address to
find the next subnet address. In this example, the next
subnet address is 192.168.1.32. All addresses between
192.168.1.0 and 192.168.1.31 belong to the first
subnet.
• So the IP range will be
192.168.1.0 192.168.1.31
but….

14. 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.

15. 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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