# XYZ Company -- TCP_IP Design with VLSM_ Supernetting_ and CIDR

Document Sample

```					     VLSM CIDR                                                               1 of 3                                                                  wyousif
A          B            C            D        E         F            G        H          I         J     K   L   M   N   O       P            Q
1

2                                          XYZ Company -- TCP/IP Design with VLSM, Supernetting, and CIDR
3
4    Answers for the Tampa Branch Office
5    Assigned addresses from the upstream provider:
6          200         100         48          0 11001000 01100100 00110000 00000000
7          200         100         49          0 11001000 01100100 00110001 00000000
8          200         100         50          0 11001000 01100100 00110010 00000000
9
10
11   Tampa:             200          100          48        0
12
13   Most populated LAN:      60 nodes
14
15   Map the number of nodes to the closest group (1,2,4,8,16,32,64,128,256,512,1024,etc….): 64 is the closest
16   Subnet network 200.100.48.0 creating groups of 64 hosts per subnet
17   Use the following process:
18   Number of hosts per subnet = 2^(number of host bits)
19   64 = 2^(number of host bits)
20   Number of host bits = 6. Why? Because 2^6 = 64
21   From here you can calculate the bit mask used to create subnets with 64 host per subnet:
22   bit mask = 32 - host bits
25   From here you can calculate how many groups of 64 hosts were created (This is normally referred to as number of subnets created)
26   Number of subnets created = 2^ (number of host bits borrowed)
27   Number of host bits borrowed = current bit mask - default bit mask
28   Number of host bits borrowed = 26-24. why 24? Because the default bit mask of a class C network is 24
29   Number of host bits borrowed = 2
30   Now you can go back to the formula in row 26 and calculate the number of subnets created which should equal 2^2 or 4
31   Now, Look at this and make sure you understand it:               200       100         48          0 /26 Was subnetted as follows:
32
34       0            200            100          48         0       200        100       48           63   /26                   address the
35       1            200            100          48        64       200        100       48          127   /26                   needs of the 60-
36       2            200            100          48       128       200        100       48          191   /26                   nodes LAN
37       3            200            100          48       192       200        100       48          255   /26
38
39   Let's say that subnet # 0 or subnet 200.100.48.0/26 will be used to address the need of the 60-nodes LAN
40
VLSM CIDR                                                               2 of 3                                                                  wyousif
A           B          C           D          E          F          G          H           I       J    K    L   M   N    O        P          Q

2                                      XYZ Company -- TCP/IP Design with VLSM, Supernetting, and CIDR
41   Now we will have to repeat the same steps again for the second most populated LAN which is the 28-nodes LAN
42   Applying the same rules as before (for the 60-nodes LAN) and going back to row 15, we should be able to arrive at the following:
43   Closest group is 32
44   Now we are going to take one of the 64-hosts subnets that was created with a 26 bit mask and further subnet it into sub-subnets of 32 hosts
45   Since subnet # 0 has already been used let's use subnet # 1 or subnet 200.100.48.64 "And yes, you can use subnet # 2 or subnet # 3 instead"
46   Repeating the same steps in row 17-20 you should be able to conclude that we need 5 host bits to create sub-subnets of 32 hosts
47   Repeating the same steps in row 21-24, the bit mask now should be 27
48   From here you can calculate how many sub groups of 32 hosts were created (This is normally referred to as number of sub-subnets created)
49   Number of sub-subnets created = 2^ (number of host bits borrowed)
51   Number of host bits borrowed = 27-26.
52   Number of host bits borrowed = 1
53   Now you can go back to the formula in row 49 and calculate the number of sub-subnets created which should equal 2^1 or 2
54   Now, Look at this and make sure you understand it:               200         100          48       64 /26 Was sub-subnetted as follows:
55
Sub-
Will be used to
57     1.0            200        100           48         64        200         100          48       95 /27
of the 28-nodes
58     1.1            200        100           48         96        200         100          48      127 /27
LAN
59
60 Let's say that sub-subnet # 1.0 or subnet 200.100.48.64/27 will be used to address the need of the 28-nodes LAN
61 Now we will have to repeat the same steps again for the third most populated LAN which is the 12-nodes LAN
62 Applying the same rules as before (for the 28-nodes LAN) and going back to row 43, we should be able to arrive at the following:     Closest group is 16
63 Now we are going to take the only 32-host sub-subnet left which was created with a 27 bit mask and further subnet it into sub-sub-subnets of 16 hosts
64 Sub-subnet # 1.1 or subnet 200.100.48.96/27 will now be sub-sub-subnetted
65 Repeating the same steps in row 17-20 you should be able to conclude that we need 4 host bits to create sub-subnets of 16 hosts
66 Repeating the same steps in row 21-24, the bit mask now should be 28
67 From here you can calculate how many sub groups of 16 hosts were created (This is normally referred to as number of sub-sub-subnets created)
68 Number of sub-subnets created = 2^ (number of host bits borrowed)
70 Number of host bits borrowed = 28-27. why 27?             Number of host bits borrowed = 1
71 Now you can go back to the formula in row 49 and calculate the number of sub-subnets created which should equal 2^1 or 2
72 Now, Look at this and make sure you understand it:               200         100          48       96 /27 Was sub-sub-subnetted as follows:
73 Subnet # Subnet ID                                        t Address
74    1.1.0           200        100           48         96        200         100          48      111 /28                      the needs of the 12-
75    1.1.1           200        100           48        112        200         100          48      127 /28                      nodes LAN
76
77 Let's say that sub-sub-subnet # 1.1.0 or subnet 200.100.48.96/28 will be used to address the need of the 12-nodes LAN
VLSM CIDR                                                                3 of 3                                                                   wyousif
A           B           C           D          E           F          G          H           I       J    K    L   M    N    O       P          Q

2                                      XYZ Company -- TCP/IP Design with VLSM, Supernetting, and CIDR
78   Now we will have to repeat the same steps again for the fourth most populated LAN which is the 5-nodes LAN
79   Applying the same rules as before (for the 12-nodes LAN) and going back to row 43, we should be able to arrive at the following:     Closest group is 8
80   Take the only 16-host sub-subnet left which was created with a 28 bit mask and further subnet it into sub-sub-sub-subnets of 8 hosts
81   Sub-subnet # 1.1.1 or subnet 200.100.48.96/28 will now be sub-sub-subnetted
82   Repeating the same steps in row 17-20 you should be able to conclude that we need 3 host bits to create sub-subnets of 8 hosts
83   Repeating the same steps in row 21-24, the bit mask now should be 29
84   From here you can calculate how many sub groups of 8 hosts were created (This is normally referred to as number of sub-sub-sub-subnets created)
85   Number of sub-subnets created = 2^ (number of host bits borrowed)
87   Number of host bits borrowed = 29-28.
88   Now you can go back to the formula in row 49 and calculate the number of sub-subnets created which should equal 2^1 or 2
89   Now, Look at this and make sure you understand it:

90  Subnet # Subnet ID                                        t Address                                                           the needs of the 5-
91 1.1.1.0              200         100         48        112        200         100          48       119 /29                    nodes LAN
92 1.1.1.1              200         100         48        120        200         100          48       127 /29
93
94 Let's say that sub-sub-subnet # 1.1.1.0 or subnet 200.100.48.112/28 will be used to address the need of the 5-nodes LAN
95 Now we will have to repeat the same steps again for the least populated network which is the serial link (2 nodes only)
96 Applying the same rules as before (for the 5-nodes LAN) and going back to row 43, we should be able to arrive at the following:      Closest group is 4
97 Take the only 8-host sub-subnet left which was created with a 29 bit mask and further subnet it into sub-sub-sub-subnets of 4 hosts
98 Sub-subnet # 1.1.1.1 or subnet 200.100.48.120/29 will now be sub-sub-subnetted
99 Repeating the same steps in row 17-20 you should be able to conclude that we need 2 host bits to create sub-subnets of 4 hosts
100 Repeating the same steps in row 21-24, the bit mask now should be 30
101 From here you can calculate how many sub groups of 4 hosts were created (This is normally referred to as number of sub-sub-sub-subnets created)
102 Number of sub-subnets created = 2^ (number of host bits borrowed)
104 Number of host bits borrowed = 30-29.
105 Now you can go back to the formula in row 49 and calculate the number of sub-subnets created which should equal 2^1 or 2
106 Now, Look at this and make sure you understand it:
107 Subnet # Subnet ID                                        t Address                                                           Will be used to address
108 1.1.1.1.0           200         100         48        120        200         100          48       123 /30                    the needs of the 5-
109 1.1.1.1.1           200         100         48        124        200         100          48       127 /30                    nodes LAN
110 Both sub-subnets # 1.1.1.1.0 and 1.1.1.1.1 will be used to address the two WAN links
111 Look at the detailed break of the subnets for the Tampa branch office then apply what you learned to Orlando and Miami offices

```
DOCUMENT INFO
Shared By:
Categories:
Stats:
 views: 24 posted: 9/21/2011 language: English pages: 3
Description: VLSM (Variable Length Subnet Mask) to the effective use of Classless Inter-Domain Routing (CIDR) and route summarization to control the size of routing tables, network administrators use advanced IP addressing techniques, VLSM is one of the commonly used methods, can be hierarchical subnet addressing, so the most effective use of existing address space.
How are you planning on using Docstoc?