A company Bit-by-Bit Inc. has been assigned a network IP address of 200.108.18.0

Question

A company Bit-by-Bit Inc. has been assigned a network IP address of 200.108.18.0. It has three divisions that need a total of 27 subnets. You are the network administrator for the company and are responsible for creating the subnets. Please answer the following questions.

1.What is the subnet mask in binary representation?

2. What is the subnet mask in decimal representation?

3. Please indicate the following four details for the first three subnets and last possible subnet in the network

*Subnet Number Subnet Address (Decimal Format Only)

*Range of host addresses that can be assigned to a host device. (Decimal Format Only)

*Direct broadcast address (Decimal Format Only) 1 2 3 Last 9.

4.Can you assign the IP address 200.108.18.16 to a host device in the network?

5.What subnet is 200.108.18.26 on?

Explanation / Answer

The address provided 200.108.18.0 is of Class C and default subnet mask is 255.255.255.0. In order to create 27 subnets, we need take few bits from host part. If we take 4 bits, no of subnets will be 2^4 - 2=14 which is less than what we want. Let us increase bits to 5, no of subnets will 2^5-2 = 30. Here, if we take 4 bits then we get fewer subnets and if we take 5 we get more. So we don't have any other option in between, so we will borrow 5 bits from host bits.

Hence subnet mask will become- 11111111.11111111.11111111.11111000

1. Subnet mask will be 11111111.11111111.11111111.11111000

2.Subnet mask in decimal, 255.255.255.248

3. We already know that 5 bits are needed for subnets. Therefore, in the last octet(8 bits) of the IP address, first 5 bits are for subnet and 3 are for hosts.

To calculate the first subnet id we put 00001000 in the last octet of network IP address. We cannot put 00000 for the first subnet and 11111 for the last as both are reserved. That's why we subtracted 2 while calculating the number of subnets above.

=>200.108.18.00001000(replaced first 5 bits with 00001)

=>200.108.18.8(First subnet in Decimal)

The range of host addresses-->No of host bits is 3. Therefore 001 will be first assignable host address and 110 will be last. 000 and 111 can be used as they will be used for subnet id and direct broadcast id.

For the first subnet, no of hosts will be-

=>200.108.18.00001001 To 200.108.00001110(Last 3 host bits 001 to 110)

=>200.108.18.9 To 200.108.14(Range of host addresses can be assigned in the first subnet.)

To calculate DBA of the first subnet, make all host bit 1.

First subnet- 200.108.18.00001111

=>200.108.18.15(DBA of first subnet)

Like wise 2nd, 3rd and last subnet address and details can be calculated.

4. Lets convert last octet of address 200.108.18.16 to binary 200.108.18.00010000. Here we can see all the host bits(3bits in this case) is set to zero which is not a valid host address(all 0's and all 1's not allowed) as this will be a subnet address. So this address cannot be assigned to host.

5. To calculate subnet of 200.108.18.26.Let us convert the last octet to decimal- 200.108.18.00011010. Making host bits zero will give us subnet id. Therefore, 200.108.18.00011000. In decimal, 200.108.18.24 is subnet id.

Related Questions

Navigate

• Browse (All)
• Browse A
• Subjects

Integrity-first tutoring: explanations and feedback only — we do not complete graded work. Learn more.