Note: This case study utilizes Packet Tracer. Please see the Chapter 4 Packet Tracer file located in Supplemental Materials.

Transcription

1 Part 1 Variable Length Subnet Mask (VLSM) Note: This case study utilizes Packet Tracer Please see the Chapter 4 Packet Tracer file located in Supplemental Materials An organization has been assigned the network number /16 and it plans to deploy VLSM The diagram shows how the addressing range is to be applied /

2 To arrive at this design, the first step of the subnetting process divides the base network address into eight equally sized address blocks Subnet #2 is divided into 32 equally sized address blocks and Subnet #6 divided into 16 equally sized address blocks Finally, Subnet #6 14 is divided into eight equally sized address blocks (see diagram above) 1 Specify the eight subnets of /16 2 List the network address and the range of host addresses that can be assigned to Subnet #3 3 Identify the broadcast address for Subnet #3 4 Specify the 16 subnets addresses of Subnet #6 5 List the host addresses that can be assigned to Subnet #6 5 6 Identify the broadcast address for Subnet #6 5 7 Specify the eight subnets of Subnet # List the subnet address and host addresses that can be assigned to Subnet # Identify the broadcast address for Subnet # Part 2 CIDR Exercises 1 List the individual network numbers defined by the CIDR block /21 2 List the range of individual network numbers defined by the CIDR block 19324/13 3 Aggregate the following set of four /24 IP network addresses to the highest degree possible / / / /24 4 Aggregate the following set of four /24 IP network addresses to the highest degree possible / / / /24

3 5 Aggregate the following set of 64 /24 IP network addresses to the highest degree possible a / / /24 through / /24 b / /24 through / /24 6 How could you express the entire Class A address space as a single CIDR advertisement? 7 The entire Class B address space as a single CIDR advertisement? 8 The entire Class C address space as a single CIDR advertisement? Answers: VLSM Q1 If the /16 address is to be sub netted into 8 further subnets, then 3 extra bits are required in the subnet field These are taken from the leftmost bits of the /16 host field bits This can be shown as *************

4 Where the two last digits are shown in binary form (we don t need to consider the first two as they are untouched!) The boxed bits are the new address bits and will count in the normal binary pattern from 000 to 111 for each subnet The * represent the host field bits Remember, the first host is when the rightmost bit is a 1 and the broadcast address is when all the host field bits are 1 So first subnet (subnet #0) is / 19 (slash 19 as three extra bits have been used for subnet field) Second subnet (subnet #1) is ************** (last two digits again shown in binary) Which is / 19 We can see the pattern here! Subnet # / 19 Subnet # / 19 Subnet # / 19 Subnet # / 19 Subnet # / 19 Subnet # / 19 Q2 From the table, subnet #3 is Host addresses are from through or to Q3 The broadcast address is when all host field bits are set to (which, of course, is last host address + 1 Q4 Subnet #6 is to be subnetted again into 16 smaller subnets so we need four more network field bits Subnet #6 is /19 so the first sub subnet will be / 23 Using the same mixed binary format as before ********* The second four boxed bits count through 0000 to 1111 (0 through 15)

5 The corresponding addresses are: # / 23 # / 23 # / 23 # / 23 # / 23 # / 23 # / 23 # / 23 # / 23 # / 23 # / 23 # / 23 # / 23 # / 23 # / 23 # / 23 Q5 Subnet 6 5 has the network address / 23 (see above) Hosts are through Q6 The broadcast address is last host plus 1 or Q7 Since subnet 6 14 is to be subnetted into a further 8 subnets, we need another 3 bits for the subnet field ****** Note these extra bits span the decimal dot between the third and fourth octet

6 Again, these bits will count from 000 to 111 (0 through 7) The subnet addresses are: # / 26 # / 26 # / 26 # / 26 # / 26 # / 26 # / 26 # / 26 Q8 From the table (above), subnet is / 26 The host addresses are through Q9 From the table (above), subnet is /26 Subnet is / 26 This means the last or broadcast address in the required subnet 6 is one less than the network address of subnet 7 or To check, 191(the last octet) is in binary and since the last six bits are all ones, this IS indeed the broadcast address of the 6 subnet CIDR Q1 The block / 21 is 3 bits short of the natural address Write the address block in binary (we only need to consider the third and fourth octets): *** ******** The last three bits of the third octet are those that would be included in the natural addresses Since there are three bits, there are 8 network addresses as these three bits count through 000,001,010,011,100,101,110,and 111 (0 through 7)

7 Hence the eight addresses are: / / / / / / / / 24 Q2 The block / 13 expands to / 13 Since the natural addresses in this range would have a /24 mask, there are = 11 bits being used for the CIDR block Hence, there are 2 11 = 2048 addresses in the block Write in binary format (only need consider the second and third octets) ***********0 The starred bits are those define the individual addresses The range is: = / 24 through = / 24

8 Q3 The four addresses have the last two bits varying in their third octet / 24 = / 24 = / 24 = / 24 = Since these are contiguous, the block address is / 22 Q4 The four addresses have the last three bits in the third octet varying However, since three bits vary, there would have to be eight addresses (2 3 = 8) for a contiguous block Here, there are only four This means there cannot be a /21 block Express the addresses in binary (only consider the third octet) / 24 = / 24 = / 24 = / 24 = The first two addresses vary in their last bit of the third octet and would aggregate to: / 23 The last two addresses vary in their last bit and would aggregate to: / 23 (To aggregate to / 21 would have required all eight addresses / 24 through / 24 to be present in the block)

9 Q5 The addresses are /24 so we need to consider the degree of commonality in the third octet The addresses in binary (only the third octet) are = = At this point the first bit changes = = a In the first block 996 through 127), all the last FIVE bits change between and The aggregated address is / 19 b In the second block, again all five last bits vary from to The aggregated address block is / 19 Q6 The Class A block is from 0000 / 8 through / 8 The defining factor here is that the first bit of the first octet is always 0 Since this is the only bit that is static, the CIDR aggregation is 0000 / 1 (it can also be written 0 / 1) Q7 As with Q6, Class B addresses have the first two bits of the first octet as 10 CIDR block is / 2 Q8 Again, the first three bits of Class C addresses are 110, so CIDR block is / 3