A firm is assigned the network part 128.171. It selects a 10-bit subnet part. a) Draw the bits for the four octets of the IP address of the first host on the first subnet. (Hint: Use Windows Calculator.)
b) Convert this answer into dotted decimal notation.
c) Draw the bits for the second host on the third subnet. (In binary, 2 is 10, while 3 is 11.)
d) Convert this into dotted decimal notation.
e) Draw the bits for the last host on the third subnet.
f) Convert this answer into dotted decimal notation.
a) To find the bits for the four octets of the IP address of the first host on the first subnet, we start with the network part 128.171 and add the subnet part. Since the subnet part is 10 bits, the first host on the first subnet will have all zeros in the subnet part. Thus, the bits for the four octets are:
Network: 10000000.10101011
Subnet: 00000000.00000000
Host: 00000000.00000000
b) Converting this into dotted decimal notation, we get the IP address of the first host on the first subnet as:
128.171.0.0
c) To find the bits for the second host on the third subnet, we keep the same network and subnet parts but increment the host part by 2. Since the subnet is 10 bits and we need the second host, it would have a binary value of 10 in the last two bits. Thus, the bits for the four octets are:
Network: 10000000.10101011
Subnet: 00000000.00000011
Host: 00000000.00000010
d) Converting this into dotted decimal notation, we get the IP address of the second host on the third subnet as:
128.171.0.2
e) To find the bits for the last host on the third subnet, we keep the same network and subnet parts but increment the host part by 2 multiplied by the number of hosts per subnet (which is 2^10 = 1024). Since we need the last host, the binary value for the host part would be 1023 (which is 1111111111 in binary). Thus, the bits for the four octets are:
Network: 10000000.10101011
Subnet: 00000000.00000011
Host: 00111111.11111111
f) Converting this into dotted decimal notation, we get the IP address of the last host on the third subnet as:
128.171.0.1023
a) The first subnet will have a subnet ID of 0000000000, which gives us the subnet mask of 255.255.255.192.
To find the first host address on the subnet, we need to add 1 to the subnet ID. So the bits for the four octets of the IP address of the first host on the first subnet would be:
128.171.0.1
b) Converting this answer into dotted decimal notation, the first host address on the first subnet is 128.171.0.1.
c) To find the second host on the third subnet, we need to add 2 (which in binary is 10) to the subnet ID. So the bits for the four octets of the IP address of the second host on the third subnet would be:
128.171.0.10
d) Converting this answer into dotted decimal notation, the second host address on the third subnet is 128.171.0.10.
e) To find the last host on the third subnet, we need to subtract 1 from the next subnet's ID. The next subnet's ID would be 0000001100 (12 in decimal), so the bits for the four octets of the IP address of the last host on the third subnet would be:
128.171.0.11
f) Converting this answer into dotted decimal notation, the last host address on the third subnet is 128.171.0.11.
To solve this problem, we need to understand how IP addressing works and how to calculate subnet addresses.
In this case, the network part provided is 128.171. The subnet part requires 10 bits, which means we have 2^10 = 1024 subnets available (since each bit can be either 0 or 1). Let's now break down the problem into smaller steps.
a) To find the bits for the four octets of the IP address of the first host on the first subnet, we need to go through these steps:
1. Convert 128.171 to binary: 10000000.10101011
2. Add 10 bits for the subnet: 10000000.10101011.0000000000
3. The first subnet will always have all zeros in the subnet part, so the complete IP address would be: 10000000.10101011.0000000000.00000001
b) To convert the above answer into dotted decimal notation, we divide the binary into four octets:
128.171.0.1
c) To find the bits for the second host on the third subnet, we follow the same steps as before:
1. Convert 128.171 to binary: 10000000.10101011
2. Add 10 bits for the subnet: 10000000.10101011.0000000010
3. The third subnet has a subnet ID of binary 10 (decimal 2), so the second host would have an ID of binary 11 (decimal 3), resulting in the address: 10000000.10101011.0000000010.00000011
d) To convert the above answer into dotted decimal notation, we divide the binary into four octets:
128.171.0.3
e) To find the bits for the last host on the third subnet, we follow the same steps as before:
1. Convert 128.171 to binary: 10000000.10101011
2. Add 10 bits for the subnet: 10000000.10101011.0000000011
3. The third subnet has a subnet ID of binary 11 (decimal 3), so the last host would be the highest possible value in this subnet, which is all 1's: 10000000.10101011.0000000011.11111111
f) To convert the above answer into dotted decimal notation, we divide the binary into four octets:
128.171.0.255
To perform these calculations, one way to do so is by using the Windows Calculator application:
1. Open Windows Calculator.
2. Switch to the "Programmer" mode by clicking on "View" and selecting "Programmer".
3. Click on the "Bin" radio button to switch to binary mode.
4. Enter the decimal part of the IP address in decimal mode.
5. Convert it to binary by clicking on the "Bin" radio button.
6. Add the necessary subnet bits.
7. Repeat steps 4-6 for each part of the IP address and the desired subnet address.
8. Convert the final binary answer back to decimal for dotted decimal notation.