Scenario 2 Part 1.2.1&your Digital Footprint

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ELG 3120 Signals and Systems Chapter 2 3/2 Yao Example: Consider the LTI system with impulse responsehn and input xn, as illustrated in Fig. N hn 1 0 1 2 1 1 n.

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8.2.1.4 Packet Tracer – Designing and Implementing a VLSM Addressing Scheme

Packet Tracer – Designing and Implementing a VLSM Addressing Scheme (Answer Version)

Answer Note: Red font color or gray highlights indicate text that appears in the Answer copy only.

Page 4 of 5 Packet Tracer - Subnetting Scenario 1 Question Possible Earned Activity Section Location Points Points Part 1: Design an IP Step 1a 1 Addressing Scheme Step 1b 1 Step 1c 1 Step 1d 1 Step 1e 4 Step 1f 2 Complete Subnet Table Step 1g 10 Assign Subnets Step 2 10 Document Addressing Step 3 40 Part 1 Total 70 Packet Tracer Score 30 Total. The case study is described in two parts: 1) decreasing cycle time and 2) decreasing effort. Step 1: Define the Problem Following the recommend practice of TQM, the project team began by defining the beginning and end of the process – in this case, the start and finish of the recruitment process. View Scenarios Part 2.docx from EDUCATION 500 at University of West Alabama. Scenario 1: Student 1 is a 5th grader and uses inappropriate language at school. This is the third offense for this.

Topology

You will receive one of three possible topologies.

Addressing Table

DeviceInterfaceIP AddressSubnet MaskDefault Gateway
[[R1Name]]G0/0[[R1G0Add]][[R1G0Sub]]N/A
G0/1[[R1G1Add]][[R1G1Sub]]N/A
S0/0/0[[R1S0Add]][[R1S0Sub]]N/A
[[R2Name]]G0/0[[R2G0Add]][[R2G0Sub]]N/A
G0/1[[R2G1Add]][[R2G1Sub]]N/A
S0/0/0[[R2S0Add]][[R2S0Sub]]N/A
[[S1Name]]VLAN 1[[S1Add]][[S1Sub]][[R1G0Add]]
[[S2Name]]VLAN 1[[S2Add]][[S2Sub]][[R1G1Add]]
[[S3Name]]VLAN 1[[S3Add]][[S3Sub]][[R2G0Add]]
[[S4Name]]VLAN 1[[S4Add]][[S4Sub]][[R2G1Add]]
[[PC1Name]]NIC[[PC1Add]][[PC1Sub]][[R1G0Add]]
[[PC2Name]]NIC[[PC2Add]][[PC2Sub]][[R1G1Add]]
[[PC3Name]]NIC[[PC3Add]][[PC3Sub]][[R2G0Add]]
[[PC4Name]]NIC[[PC4Add]][[PC4Sub]][[R2G1Add]]

Objectives

Part 1: Examine the Network Requirements

Part 2: Design the VLSM Addressing Scheme

Part 3: Assign IP Addresses to Devices and Verify Connectivity

Background

In this activity, you are given a /24 network address to use to design a VLSM addressing scheme. Based on a set of requirements, you will assign subnets and addressing, configure devices and verify connectivity.

Part 1: Examine the Network Requirements

Step 1: Determine the number of subnets needed.

You will subnet the network address [[DisplayNet]]. The network has the following requirements:

  • [[S1Name]] LAN will require [[HostReg1]] host IP addresses
  • [[S2Name]] LAN will require [[HostReg2]] host IP addresses
  • [[S3Name]] LAN will require [[HostReg3]] host IP addresses
  • [[S4Name]] LAN will require [[HostReg4]] host IP addresses

How many subnets are needed in the network topology? 5

Step 2: Determine the subnet mask information for each subnet.

  1. Which subnet mask will accommodate the number of IP addresses required for [[S1Name]]?
    • How many usable host addresses will this subnet support?
  2. Which subnet mask will accommodate the number of IP addresses required for [[S2Name]]?
    • How many usable host addresses will this subnet support?
  3. Which subnet mask will accommodate the number of IP addresses required for [[S3Name]]?
    • How many usable host addresses will this subnet support?
  4. Which subnet mask will accommodate the number of IP addresses required for [[S4Name]]?
    • How many usable host addresses will this subnet support?
  5. Which subnet mask will accommodate the number of IP addresses required for the connection between [[R1Name]] and [[R2Name]]?

Part 2: Design the VLSM Addressing Scheme

Step 1: Divide the [[DisplayNet]] network based on the number of hosts per subnet.

  1. Use the first subnet to accommodate the largest LAN.
  2. Use the second subnet to accommodate the second largest LAN.
  3. Use the third subnet to accommodate the third largest LAN.
  4. Use the fourth subnet to accommodate the fourth largest LAN.
  5. Use the fifth subnet to accommodate the connection between [[R1Name]] and [[R2Name]].

Step 2: Document the VLSM subnets.

Complete the Subnet Table, listing the subnet descriptions (e.g. [[S1Name]] LAN), number of hosts needed, then network address for the subnet, the first usable host address, and the broadcast address. Repeat until all addresses are listed.

Subnet Table

Note: The correct answers for this table are variable depending on the scenario received. Refer to the Answer Notes at the end of these instructions for further information. The format here follows what the student used in Designing and Implementing a VLSM Addressing Scheme.

Subnet DescriptionNumber of Hosts NeededNetwork Address/CIDRFirst Usable Host AddressBroadcast Address

Step 3: Document the addressing scheme.

  1. Assign the first usable IP addresses to [[R1Name]] for the two LAN links and the WAN link.
  2. Assign the first usable IP addresses to [[R2Name]] for the two LANs links. Assign the last usable IP address for the WAN link.
  3. Assign the second usable IP addresses to the switches.
  4. Assign the last usable IP addresses to the hosts.

Part 3: Assign IP Addresses to Devices and Verify Connectivity

Most of the IP addressing is already configured on this network. Implement the following steps to complete the addressing configuration.

Step 1: Configure IP addressing on [[R1Name]] LAN interfaces.

Step 2: Configure IP addressing on [[S3Name]], including the default gateway.

Step 3: Configure IP addressing on [[PC4Name]], including the default gateway.

Step 4: Verify connectivity.

You can only verify connectivity from [[R1Name]], [[S3Name]], and [[PC4Name]]. However, you should be able to ping every IP address listed in the Addressing Table.

Suggested Scoring Rubric

Note: The majority of points are allocated to designing and documenting the addressing scheme. Implementation of the addresses in Packet Tracer is of minimal consideration.

Activity SectionQuestion LocationPossible PointsEarned Points
Part 1: Examine the Network RequirementsStep 11
Step 24
Part 1 Total5
Part 2: Design the VLSM Addressing Scheme
Complete Subnet Table25
Document Addressing40
Part 2 Total65
Packet Tracer Score30
Total Score100

ID:[[indexAdds]][[indexNames]][[indexTopos]]

Answer Notes:

The following addressing tables represent the three possible addressing scenarios the student may get. Note that the Device column is independent of the addressing scheme. For example, a student could receive the device names from Scenario 1 and the addressing scheme from Scenario 3. In addition, the three possible topologies are also independent of the device names and the addressing scheme (click reset in the activity to see the different topologies). Therefore, this activity uses three independent variables with three possible values each for a total of 27 possible combinations (3 device names x 3 addressing schemes x 3 topologies = 27 isomorphs).

Scenario 1 – Network Address: 10.11.48.0/24

Subnet Table

Subnet DescriptionNumber of Hosts NeededNetwork Address/CIDRFirst Usable Host AddressLast Usable Host AddressBroadcast Address
Host-D LAN6010.11.48.0/2610.11.48.110.11.48.6210.11.48.63
Host-B LAN3010.11.48.64/2710.11.48.6510.11.48.9410.11.48.95
Host-A LAN1410.11.48.96/2810.11.48.9710.11.48.11010.11.48.111
Host-C LAN610.11.48.112/2910.11.48.11310.11.48.11810.11.48.119
WAN Link210.11.48.120/3010.11.48.12110.11.48.12210.11.48.123
DeviceInterfaceAddressSubnet MaskDefault Gateway
Building1G0/010.11.48.97255.255.255.240N/A
G0/110.11.48.65255.255.255.224N/A
S0/0/010.11.48.121255.255.255.252N/A
Building2G0/010.11.48.113255.255.255.248N/A
G0/110.11.48.1255.255.255.192N/A
S0/0/010.11.48.122255.255.255.252N/A
ASW1VLAN 110.11.48.98255.255.255.24010.11.48.97
ASW2VLAN 110.11.48.66255.255.255.22410.11.48.65
ASW3VLAN 110.11.48.114255.255.255.24810.11.48.113
ASW4VLAN 110.11.48.2255.255.255.19210.11.48.1
Host-ANIC10.11.48.110255.255.255.24010.11.48.97
Host-BNIC10.11.48.94255.255.255.22410.11.48.65
Host-CNIC10.11.48.118255.255.255.24810.11.48.113
Host-DNIC10.11.48.62255.255.255.19210.11.48.1

Building 1

ASW3

Scenario 2 – Network Address: 172.31.103.0/24

Subnet Table

Subnet DescriptionNumber of Hosts NeededNetwork Address/CIDRFirst Usable Host AddressLast Usable Host AddressBroadcast Address
PC-A LAN27172.31.103.0/27172.31.103.1172.31.103.30172.31.103.31
PC-B LAN25172.31.103.32/27172.31.103.33172.31.103.62172.31.103.63
PC-C LAN14172.31.103.64/28172.31.103.65172.31.103.78172.31.103.79
PC-D LAN8172.31.103.80/28172.31.103.81172.31.103.94172.31.103.95
WAN Link2172.31.103.96/30172.31.103.97172.31.103.98172.31.103.99
DeviceInterfaceAddressSubnet MaskDefault Gateway
Branch1G0/0172.31.103.1255.255.255.224N/A
G0/1172.31.103.33255.255.255.224N/A
S0/0/0172.31.103.97255.255.255.252N/A
Branch2G0/0172.31.103.65255.255.255.240N/A
G0/1172.31.103.81255.255.255.240N/A
S0/0/0172.31.103.98255.255.255.252N/A
Room-114VLAN 1172.31.103.2255.255.255.224172.31.103.1
Room-279VLAN 1172.31.103.34255.255.255.224172.31.103.33
Room-312VLAN 1172.31.103.66255.255.255.240172.31.103.65
Room-407VLAN 1172.31.103.82255.255.255.240172.31.103.81
PC-ANIC172.31.103.30255.255.255.224172.31.103.1
PC-BNIC172.31.103.62255.255.255.224172.31.103.33
PC-CNIC172.31.103.78255.255.255.240172.31.103.65
PC-DNIC172.31.103.94255.255.255.240172.31.103.81

Branch 1

Room-312

Scenario 3 – Network Address: 192.168.72.0/24

Subnet Table

Subnet DescriptionNumber of Hosts NeededNetwork Address/CIDRFirst Usable Host AddressLast Usable Host AddressBroadcast Address
User-4 LAN58192.168.72.0/26192.168.72.1192.168.72.62192.168.72.63
User-3 LAN29192.168.72.64/27192.168.72.65192.168.72.94192.168.72.95
User-2 LAN15192.168.72.96/27192.168.72.97192.168.72.126192.168.72.127
User-1 LAN7192.168.72.128/28192.168.72.129192.168.72.142192.168.72.143
WAN Link2192.168.72.144/30192.168.72.145192.168.72.146192.168.72.147
DeviceInterfaceAddressSubnet MaskDefault Gateway
Remote-Site1G0/0192.168.72.129255.255.255.240N/A
G0/1192.168.72.97255.255.255.224N/A
S0/0/0192.168.72.145255.255.255.252N/A
Remote-Site2G0/0192.168.72.65255.255.255.224N/A
G0/1192.168.72.1255.255.255.192N/A
S0/0/0192.168.72.146255.255.255.252N/A
Sw1VLAN 1192.168.72.130255.255.255.240192.168.72.129
Sw2VLAN 1192.168.72.98255.255.255.224192.168.72.97
Sw3VLAN 1192.168.72.66255.255.255.224192.168.72.65
Sw4VLAN 1192.168.72.2255.255.255.192192.168.72.1
User-1NIC192.168.72.142255.255.255.240192.168.72.129
User-2NIC192.168.72.126255.255.255.224192.168.72.97
User-3NIC192.168.72.94255.255.255.224192.168.72.65
User-4NIC192.168.72.62255.255.255.192192.168.72.1

Remote-Site1

Sw-3

Below are multiple fraction calculators capable of addition, subtraction, multiplication, division, simplification, and conversion between fractions and decimals. Fields above the solid black line represent the numerator, while fields below represent the denominator.


Mixed Numbers Calculator


Simplify Fractions Calculator

Scenario 2 Part 1.2.1&your Digital Footprint Software


1.2.1&your

Decimal to Fraction Calculator


Fraction to Decimal Calculator


Big Number Fraction Calculator

Use this calculator if the numerators or denominators are very big integers.


In mathematics, a fraction is a number that represents a part of a whole. It consists of a numerator and a denominator. The numerator represents the number of equal parts of a whole, while the denominator is the total number of parts that make up said whole. For example, in the fraction
3
8
, the numerator is 3, and the denominator is 8. A more illustrative example could involve a pie with 8 slices. 1 of those 8 slices would constitute the numerator of a fraction, while the total of 8 slices that comprises the whole pie would be the denominator. If a person were to eat 3 slices, the remaining fraction of the pie would therefore be
5
8
as shown in the image to the right. Note that the denominator of a fraction cannot be 0, as it would make the fraction undefined. Fractions can undergo many different operations, some of which are mentioned below.

Addition:

Unlike adding and subtracting integers such as 2 and 8, fractions require a common denominator to undergo these operations. One method for finding a common denominator involves multiplying the numerators and denominators of all of the fractions involved by the product of the denominators of each fraction. Multiplying all of the denominators ensures that the new denominator is certain to be a multiple of each individual denominator. The numerators also need to be multiplied by the appropriate factors to preserve the value of the fraction as a whole. This is arguably the simplest way to ensure that the fractions have a common denominator. However, in most cases, the solutions to these equations will not appear in simplified form (the provided calculator computes the simplification automatically). Below is an example using this method.

Worksheet
a
b
 +
c
d
 =
a×d
b×d
 +
c×b
d×b
 =
ad + bc
bd
EX:
3
4
 +
1
6
 =
3×6
4×6
 +
1×4
6×4
 =
22
24
 =
11
12

This process can be used for any number of fractions. Just multiply the numerators and denominators of each fraction in the problem by the product of the denominators of all the other fractions (not including its own respective denominator) in the problem.

EX:
1
4
 +
1
6
 +
1
2
 =
1×6×2
4×6×2
 +
1×4×2
6×4×2
 +
1×4×6
2×4×6
=
12
48
 +
8
48
 +
24
48
 =
44
48
 =
11
12

An alternative method for finding a common denominator is to determine the least common multiple (LCM) for the denominators, then add or subtract the numerators as one would an integer. Using the least common multiple can be more efficient and is more likely result in a fraction in simplified form. In the example above, the denominators were 4, 6, and 2. The least common multiple is the first shared multiple of these three numbers.

Multiples of 2: 2, 4, 6, 8 10, 12
Multiples of 4: 4, 8, 12
Multiples of 6: 6, 12

The first multiple they all share is 12, so this is the least common multiple. To complete an addition (or subtraction) problem, multiply the numerators and denominators of each fraction in the problem by whatever value will make the denominators 12, then add the numerators.

EX:
1
4
 +
1
6
 +
1
2
 =
1×3
4×3
 +
1×2
6×2
 +
1×6
2×6
=
3
12
 +
2
12
 +
6
12
 =
11
12

Subtraction:

Fraction subtraction is essentially the same as fraction addition. A common denominator is required for the operation to occur. Refer to the addition section as well as the equations below for clarification.

a
b
 –
c
d
 =
a×d
b×d
 –
c×b
d×b
 =
ad – bc
bd
EX:
3
4
 –
1
6
 =
3×6
4×6
 –
1×4
6×4
 =
14
24
 =
7
12

Multiplication:

Multiplying fractions is fairly straightforward. Unlike adding and subtracting, it is not necessary to compute a common denominator in order to multiply fractions. Simply, the numerators and denominators of each fraction are multiplied, and the result forms a new numerator and denominator. If possible, the solution should be simplified. Refer to the equations below for clarification.

a
b
 ×
c
d
 =
ac
bd
EX:
3
4
 ×
1
6
 =
3
24
 =
1
8

Division:

The process for dividing fractions is similar to that for multiplying fractions. In order to divide fractions, the fraction in the numerator is multiplied by the reciprocal of the fraction in the denominator. The reciprocal of a number Scenario 2 Part 1.2.1&your Digital Footprinta is simply
1
a
. When a is a fraction, this essentially involves exchanging the position of the numerator and the denominator. The reciprocal of the fraction
3
4
would therefore be
4
3
. Refer to the equations below for clarification.
a
b
 /
c
d
 =
a
b
 ×
d
c
 =
ad
bc
EX:
3
4
 /
1
6
 =
3
4
 ×
6
1
 =
18
4
 =
9
2

Simplification:

It is often easier to work with simplified fractions. As such, fraction solutions are commonly expressed in their simplified forms.
220
440
for example, is more cumbersome than
1
2
. The calculator provided returns fraction inputs in both improper fraction form, as well as mixed number form. In both cases, fractions are presented in their lowest forms by dividing both numerator and denominator by their greatest common factor.

Converting between fractions and decimals:

Converting from decimals to fractions is straightforward. It does however require the understanding that each decimal place to the right of the decimal point represents a power of 10; the first decimal place being 101, the second 102, the third 103, and so on. Simply determine what power of 10 the decimal extends to, use that power of 10 as the denominator, enter each number to the right of the decimal point as the numerator, and simplify. For example, looking at the number 0.1234, the number 4 is in the fourth decimal place which constitutes 10

Scenario 2 Part 1.2.1&your Digital Footprints

4, or 10,000. This would make the fraction
1234
10000
, which simplifies to
617
5000
, since the greatest common factor between the numerator and denominator is 2. Similarly, fractions with denominators that are powers of 10 (or can be converted to powers of 10) can be translated to decimal form using the same principles. Take the fraction
1
2
for example. To convert this fraction into a decimal, first convert it into the fraction
5
10
. Knowing that the first decimal place represents 10-1,
5
10
can be converted to 0.5. If the fraction were instead
5
100
, the decimal would then be 0.05, and so on. Beyond this, converting fractions into decimals requires the operation of long division.

Common Engineering Fraction to Decimal Conversions

In engineering, fractions are widely used to describe the size of components such as pipes and bolts. The most common fractional and decimal equivalents are listed below.

64th32nd16th8th4th2ndDecimalDecimal
(inch to mm)
1/640.0156250.396875
2/641/320.031250.79375
3/640.0468751.190625
4/642/321/160.06251.5875
5/640.0781251.984375
6/643/320.093752.38125
7/640.1093752.778125
8/644/322/161/80.1253.175
9/640.1406253.571875
10/645/320.156253.96875
11/640.1718754.365625
12/646/323/160.18754.7625
13/640.2031255.159375
14/647/320.218755.55625
15/640.2343755.953125
16/648/324/162/81/40.256.35
17/640.2656256.746875
18/649/320.281257.14375
19/640.2968757.540625
20/6410/325/160.31257.9375
21/640.3281258.334375
22/6411/320.343758.73125
23/640.3593759.128125
24/6412/326/163/80.3759.525
25/640.3906259.921875
26/6413/320.4062510.31875
27/640.42187510.715625
28/6414/327/160.437511.1125
29/640.45312511.509375
30/6415/320.4687511.90625
31/640.48437512.303125
32/6416/328/164/82/41/20.512.7
33/640.51562513.096875
34/6417/320.5312513.49375
35/640.54687513.890625
36/6418/329/160.562514.2875
37/640.57812514.684375
38/6419/320.5937515.08125
39/640.60937515.478125
40/6420/3210/165/80.62515.875
41/640.64062516.271875
42/6421/320.6562516.66875
43/640.67187517.065625
44/6422/3211/160.687517.4625
45/640.70312517.859375
46/6423/320.7187518.25625
47/640.73437518.653125
48/6424/3212/166/83/40.7519.05
49/640.76562519.446875
50/6425/320.7812519.84375
51/640.79687520.240625
52/6426/3213/160.812520.6375
53/640.82812521.034375
54/6427/320.8437521.43125
55/640.85937521.828125
56/6428/3214/167/80.87522.225
57/640.89062522.621875
58/6429/320.9062523.01875
59/640.92187523.415625
60/6430/3215/160.937523.8125
61/640.95312524.209375
62/6431/320.9687524.60625
63/640.98437525.003125
64/6432/3216/168/84/42/2125.4




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