phần 1 bài tập chương 6

Semester 1

I. Subnet Excercises

Note: You need to thoroughly understand the structure of Class A's, B's, and C's network address in order to handle the following questions. Good lucks. I do not have answer keys to these questions. You are welcome to ask me to check your answers.

1. Given the subnet mask, 255.255.255.0, and an IP address 66.68.100.1,

a. How many bits are used for subnet ID? Why?

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b. How many IP addresses are lost due to the subnetting?

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2. In dealing with a network configuration problem. An IT manager found that the following information on a station: IP address was 225.101.10.97 and the subnet mask was 255.255.255.224. What might be the problem?

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3. In dealing with a network configuration problem. An IT manager found that the following information on a station: IP address was 199.101.10.32 and the subnet mask was 255.255.255.192. Assume the network ID is correct. What might be the problem?

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4. For a particular Class C subnet, if the subnet ID is 0100, answer the following questions:

a. Define the subnet mask.

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b. Does 212.34.23.129 belong to this subnet?

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c. How many subnets can the organization have?

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d. How many IP addresses are available in each segment?

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e. How many IP addresses are lost due to subnetting (in terms of the whole

organizations)?

5. For a particular Class B subnet, if the subnet ID is 001000:

a. How many bits are used for subnet ID?

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b. Define the subnet mask.

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c. Does 144.33.48.254 belong to this subnet?

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d. How many subnets can the organization have?

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e. How many IP addresses are available in each segment?

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f. How many IP addresses are lost due to subnetting (in terms of the whole organizations)?

7. Divide the IP network address, 195.25.2.0, so that it is capable of supporting at least 9 network segments.

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8. Divide the IP network address, 140.25.0.0, so that it is capable of supporting 254 network segments.

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9. Divide the IP network address, 140.25.0.0, so that it is capable of supporting 128 network segments.

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10. Divide IP network address, 25.25.0.0, so that it is capable of supporting 65,354 network segments.

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11. Create subnets for IP network address, 222.22.22.0, so that one segment can support at least 56 attachments and 5 segments at least 12 attachments each. You also need to define the subnet mask for each segment.

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12. Create subnets for IP network address, 200.10.10.0, so that one segment can support at least 48 attachments and 3 segments at least 30 attachments each. You also need to define the subnet mask for each segment.

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II. Calculating VLSM

Scenario: Your company has been given the network address 172.16.32.0/19.

10101100 . 00010000 . 001 00000 . 00000000

After careful planning, looking at current needs and expansion, you realize you need a maximum of three

subnets of 1,000 hosts, three subnets of 250 hosts, and several subnets for serial point-to-point links.

There are several ways to do this, but you have decided that you will variably subnet your network as follows: (We did it this way just for the exercise and to keep it somewhat simple.)

Step 1: The maximum number of hosts any of your subnets will need is 1,000, so you decide to make the initial subnets 172.16.32.0/22. Write out the eight /22 subnets in binary and decimal form:

Step 2: You only need three /22 subnets, so you reserved the first three for those subnets needing 1,000 hosts. Which ones are they?

Step 3: You decide to sub-subnet the fourth subnet (172.16.44.0/22) for the three subnets each needing 250 hosts. A /24 will work well for this. What are the three /24 sub-subnets you will use?

Step 4: The last (fourth) of the /24 sub-subnets (172.16.47.0/24), you decide to use for you serial point-to-point links. List the first five /30 sub-sub-subnets (if there is such a term):

phần 2

IV. Open Lab2

Major network: 172.16.0.0 /17

Task 1: IP Subnetting (15 points, for all team)

Given the network as in the diagram above, you are to perform the tasks:

(Note: write down subnet address to "......" zone):

- IP planning:

o Lan 1 : 128 hosts

o Lan 2 : 10 hosts - loopback

o Lan 3 : 100 hosts - looback

o Lan 4 : 10 host

o Lan 5 : 30 hosts

o Lan 6 : 534 hosts - loopback

o Lan 7 : 64 hosts - loopback

o Lan 8 : 3 hosts

o WanLink : 2 hosts

Network Subnet Address Host Address Range Broadcast Address Subnet Mask

LAN1

LAN2

LAN3

LAN4

LAN5

LAN6

LAN7

LAN8

WAN link