Network Subnet 34 Subnet 35 Subnet 36 Subnet 37 Subnet 38 in Objective-C

Creation PDF 417 in Objective-C Network Subnet 34 Subnet 35 Subnet 36 Subnet 37 Subnet 38

Network Subnet 34 Subnet 35 Subnet 36 Subnet 37 Subnet 38
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Table 6-2: Subnet Ranges in the VLSM Example Address/Mask Range Hosts 172161760/20 172161920/20 172162080/20 172162240/20 172162400/20 172161761 17216191254 172161921 17216207254 172162081 17216223254 172162241 17216239254 172162401 17216255254 4,094 4,094 4,094 4,094 4,094
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Now let's take a slightly more complicated example: four groups of buildings connected together with WAN links, each with multiple buildings and different numbers of hosts per building, as shown in Figure 6-46
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Figure 6-46: Initial requirements for the complicated VLSM example In this situation, we need to do a little more complicated address grouping The first group of buildings, Campus Area Network (CAN) 1, has four buildings Building 1 needs 16,000 hosts, Building 2 needs 8,000 hosts, Building 3 needs 6,000 hosts, and Building 4 needs 2,000 hosts, for a grand total of 32,000 hosts CAN 2 has two buildings Building 5 needs 4,000 hosts, and Building 6 needs 12,000 hosts, for a grand total of 16,000 hosts CAN 3 has three buildings Building 7 needs 4,000 hosts, Building 8 needs 3,000 hosts, and Building 9 needs 1,000 hosts, for a grand total of 8,000 hosts Finally, CAN 4 has five buildings Building 10 needs 1,000
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hosts, Building 11 needs 500 hosts, Building 12 needs 250 hosts, and Buildings 13 and 14 need 100 hosts each, for a grand total of 2,000 hosts The first step in dividing the IP address space here is to take the default class-based network address (a class B address of 1721600, in this example) and split it into a small number of pieces (two, if possible) In this case, CAN 1 needs 32,000 addresses, while CANs 2 through 4 need 26,000 addresses The easiest solution here is to initially split the networks off at this point with a 17-bit mask, giving the first 32,000 addresses to CAN 1, and splitting the other 32,000 addresses among CANs 2, 3, and 4 This is shown in Figure 6-47
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Figure 6-47: Initial subdivision of the address space Next, we will subdivide the 1721600/17 network for CAN 1 into another group of two, with Building 1 in one group (it needs exactly half of the addresses) and the other three buildings in the second group We do this by adding another bit to the mask, making the address for the first group (Building 1) 1721600/18, and the base address for the second group 17216640/18 This is shown in Figure 6-48
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Figure 6-48: Subdividing CAN 1 Now we will subdivide the second address group in CAN 1 (17216640/18) into two more groups, using a 19-bit mask The first group will contain only Building 2 because it needs 8,000 hosts, and the second group will contain Buildings 3 and 4 Building 2's IP address space will therefore be 17216640/19, and the group containing Buildings 3 and 4 will have a base address of 17216960/19 This is shown in Figure 6-49
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Figure 6-49: Subdividing the second group in CAN 1 Now we need to subdivide the group that Buildings 3 and 4 are included in, but we can't do it quite as simply as we have been Building 3 requires 6,000 hosts, while Building 4 requires 2,000 Obviously, just dividing the address space in half will not work However, dividing the address space by 4 will To do this, we add two bits to the 19-bit mask to make a 21-bit mask, giving us four networks with 2,000 hosts each We then assign three of them to Building 3 (17216960/21, 172161040/21, and 172161120/21) and one of them to Building 4 (172161200/21) This is shown in Figure 6-50
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Figure 6-50: Subdividing the second /19 grouping in CAN 1 Now for the second /17 group (172161280/17), which included CANs 2 through 4 Once again, the easiest way to proceed here is to divide the address space right down the middle by adding a single bit to the mask (See Figure 6-51)
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Figure 6-51: Subdivision of the second /17 group This creates 172161280/18 and 172161920/18 We will assign 172161280/18 to CAN 2 because it requires 16,000 total hosts We will then subdivide this network into four smaller networks of 4,000 hosts, each with a /20 mask, and assign three of these (172161280/20, 172161440/20, and 172161600/20) to Building 6 and assign one (172161760/20) to Building 5 This is shown in Figure 6-52
Figure 6-52: Subdivision of CAN 2 We will split the 172161920/18 group down the middle with a 19-bit mask and assign 172161920/19 to CAN 3; and we will assign 172162240/19 to another group, including CAN 4 and three other CANs (that do not currently exist) for future expansion (as shown in Figure 6-53)
Figure 6-53: Subdivision of the 172161920/18 group For CAN 3, we will subdivide the 172161920/19 group with a 20-bit mask We will use the 172161920/20 space for Building 7 and the 172162080/20 space for the group of buildings including Buildings 8 and 9 We will then subdivide the 172162080/ 20 space with a 22-bit mask, creating four subnets with 1,000 hosts each We will assign three of these (172162080/22, 172162120/22, and 172162160/22) to Building 8, and one of these (172162200/22) to Building 9 This is shown in Figure 6-54
Figure 6-54: Subdivision of CAN 3 Finally, we will take the final address space, 172162240/19, and subdivide it into four groups of 2,000 hosts by using a /21 mask One of these groups (172162240/21) will be used for CAN 4, and the rest (172162320/21, 172162400/21, and 172162480/21) will be used for future needs (perhaps another small CAN at a later date) As for CAN 4, we will take the 172162240/21 space and subdivide it with a 22-bit mask, giving us two networks with 1,000 hosts per network The first network (17216 2240/22) we will assign to Building 10, and the other network (172162280/22) we will assign to the group of buildings including Buildings 11 through 14 This is shown in Figure 6-55
Figure 6-55: Subdivision of CAN 4 We will then take the 172162280/22 network and subdivide it into two networks with 510 hosts each using a 23-bit mask This creates subnet 172162280/23, which we will assign to
Building 11, and 172162300/23, which we will assign to the group of buildings containing Buildings 12 through 14 This is shown in Figure 6-56
Figure 6-56: Subdivision of the 172162280/22 network Finally, we will take the 172162300/23 space and divide it with a 24-bit mask, creating two networks with 254 hosts each We will then assign 172162300/24 to Building 12, and we will assign 172162310/24 to the group of buildings containing Buildings 13 and 14 We will then split the 172162310/24 network into two separate networks using a 25-bit mask, creating two networks with 126 hosts per network We will assign the 172162310/25 network to Building 13 and the 17216231128/25 network to Building 14 This last subdivision is shown in Figure 6-57
Figure 6-57: The final subdivision That's all there is to VLSM It's not very complicated; it just requires a good understanding of the binary behind TCP/IP
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