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CLASS FEATURE JANUARY 5, 2004 ISSUE

By Paul Krugman

DECEMBER 18, 2003

m

The Death of Horatio Alger

Our political leaders are doing everything they can to fortify class

inequality.

The other day I found myself reading a leftist rag that made

outrageous claims about America. It said that we are becoming a

society in which the poor tend to stay poor, no matter how hard

they work; in which sons are much more likely to inherit the

socioeconomic status of their father than they were a generation

ago.

The name of the leftist rag? Business Week, which published an

article titled “Waking Up From the American Dream.” The article

summarizes recent research showing that social mobility in the

United States (which was never as high as legend had it) has

declined considerably over the past few decades. If you put that

research together with other research that shows a drastic increase

in income and wealth inequality, you reach an uncomfortable

conclusion: America looks more and more like a class-ridden

society.

And guess what? Our political leaders are doing everything they can

to fortify class inequality, while denouncing anyone who

complains–or even points out what is happening–as a practitioner

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of “class warfare.”

Let’s talk first about the facts on income distribution. Thirty years

ago we were a relatively middle-class nation. It had not always been

thus: Gilded Age America was a highly unequal society, and it stayed

that way through the 1920s. During the 1930s and ’40s, however,

America experienced what the economic historians Claudia Goldin

and Robert Margo have dubbed the Great Compression: a drastic

narrowing of income gaps, probably as a result of New Deal

policies. And the new economic order persisted for more than a

generation: Strong unions; taxes on inherited wealth, corporate

profits and high incomes; close public scrutiny of corporate

management–all helped to keep income gaps relatively small. The

economy was hardly egalitarian, but a generation ago the gross

inequalities of the 1920s seemed very distant.

Now they’re back. According to estimates by the economists

Thomas Piketty and Emmanuel Saez–confirmed by data from the

Congressional Budget Office–between 1973 and 2000 the average

real income of the bottom 90 percent of American taxpayers

actually fell by 7 percent. Meanwhile, the income of the top 1

percent rose by 148 percent, the income of the top 0.1 percent rose

by 343 percent and the income of the top 0.01 percent rose 599

percent. (Those numbers exclude capital gains, so they’re not an

artifact of the stock-market bubble.) The distribution of income in

the United States has gone right back to Gilded Age levels of

inequality.

Never mind, say the apologists, who churn out papers with titles

like that of a 2001 Heritage Foundation piece, “Income Mobility

and the Fallacy of Class-Warfare Arguments.” America, they say,

isn’t a caste society–people with high incomes this year may have

low incomes next year and vice versa, and the route to wealth is

open to all. That’s where those commies at Business Week come in:

As they point out (and as economists and sociologists have been

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pointing out for some time), America actually is more of a caste

society than we like to think. And the caste lines have lately become

a lot more rigid.

The myth of income mobility has always exceeded the reality: As a

general rule, once they’ve reached their 30s, people don’t move up

and down the income ladder very much. Conservatives often cite

studies like a 1992 report by Glenn Hubbard, a Treasury official

under the elder Bush who later became chief economic adviser to

the younger Bush, that purport to show large numbers of

Americans moving from low-wage to high-wage jobs during their

working lives. But what these studies measure, as the economist

Kevin Murphy put it, is mainly “the guy who works in the college

bookstore and has a real job by his early 30s.” Serious studies that

exclude this sort of pseudo-mobility show that inequality in average

incomes over long periods isn’t much smaller than inequality in

annual incomes.

It is true, however, that America was once a place of substantial

intergenerational mobility: Sons often did much better than their

fathers. A classic 1978 survey found that among adult men whose

fathers were in the bottom 25 percent of the population as ranked

by social and economic status, 23 percent had made it into the top

25 percent. In other words, during the first thirty years or so after

World War II, the American dream of upward mobility was a real

experience for many people.

Now for the shocker: The Business Week piece cites a new survey of

today’s adult men, which finds that this number has dropped to

only 10 percent. That is, over the past generation upward mobility

has fallen drastically. Very few children of the lower class are

making their way to even moderate affluence. This goes along with

other studies indicating that rags-to-riches stories have become

vanishingly rare, and that the correlation between fathers’ and sons’

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incomes has risen in recent decades. In modern America, it seems,

you’re quite likely to stay in the social and economic class into

which you were born.

Business Week attributes this to the “Wal-Martization” of the

economy, the proliferation of dead-end, low-wage jobs and the

disappearance of jobs that provide entry to the middle class. That’s

surely part of the explanation. But public policy plays a role–and

will, if present trends continue, play an even bigger role in the

future.

Put it this way: Suppose that you actually liked a caste society, and

you were seeking ways to use your control of the government to

further entrench the advantages of the haves against the have-nots.

What would you do?

One thing you would definitely do is get rid of the estate tax, so that

large fortunes can be passed on to the next generation. More

broadly, you would seek to reduce tax rates both on corporate

profits and on unearned income such as dividends and capital gains,

so that those with large accumulated or inherited wealth could

more easily accumulate even more. You’d also try to create tax

shelters mainly useful for the rich. And more broadly still, you’d try

to reduce tax rates on people with high incomes, shifting the

burden to the payroll tax and other revenue sources that bear most

heavily on people with lower incomes.

Meanwhile, on the spending side, you’d cut back on healthcare for

the poor, on the quality of public education and on state aid for

higher education. This would make it more difficult for people with

low incomes to climb out of their difficulties and acquire the

education essential to upward mobility in the modern economy.

And just to close off as many routes to upward mobility as possible,

you’d do everything possible to break the power of unions, and

you’d privatize government functions so that well-paid civil

servants could be replaced with poorly paid private employees.

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Paul Krugman Paul Krugman, an economics professor at Princeton and a columnist at the New York Times, is the author, most recently, of The Great Unraveling: Losing Our

Way in the New Century (Norton).

To submit a correction for our consideration, click here. For Reprints and Permissions, click here.

It all sounds sort of familiar, doesn’t it?

Where is this taking us? Thomas Piketty, whose work with Saez has

transformed our understanding of income distribution, warns that

current policies will eventually create “a class of rentiers in the U.S.,

whereby a small group of wealthy but untalented children controls

vast segments of the US economy and penniless, talented children

simply can’t compete.” If he’s right–and I fear that he is–we will end

up suffering not only from injustice, but from a vast waste of human

potential.

Goodbye, Horatio Alger. And goodbye, American Dream.

©The McGraw-Hill Companies, Inc., 2000 © Adapted for use at JMU by Mohamed Aboutabl, 200311

Chapter 5

Subnetting/Supernetting and

Classless Addressing

• SUBNETTING • SUPERNETTING • CLASSLESS ADDRSSING

©The McGraw-Hill Companies, Inc., 2000 © Adapted for use at JMU by Mohamed Aboutabl, 200322

SUBNETTING5.15.1

©The McGraw-Hill Companies, Inc., 2000 © Adapted for use at JMU by Mohamed Aboutabl, 200333

A network with two levels of hierarchy (not subnetted)

• All hosts in such a large network must be laid out as ONE physical network

• May not always be feasible (due to geographic constraints)

©The McGraw-Hill Companies, Inc., 2000 © Adapted for use at JMU by Mohamed Aboutabl, 200344

A network with three levels of hierarchy (subnetted)

• 3-step delivery: site, subnet, host.

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Default mask and subnet mask

Subnetwork

©The McGraw-Hill Companies, Inc., 2000 © Adapted for use at JMU by Mohamed Aboutabl, 200366

Finding the Subnet Address

Given an IP address, we can find the subnet address the same way we found the network address in the previous chapter. We apply the mask to the address. We can do this in two ways: straight or short-cut.

©The McGraw-Hill Companies, Inc., 2000 © Adapted for use at JMU by Mohamed Aboutabl, 200377

Straight Method

In the straight method, we use binary notation for both the address and the mask and then apply the AND operation to find the subnet address.

Example 1Example 1

What is the subnetwork address if the destination address is 200.45.34.56 and the subnet mask is 255.255.240.0?

11001000 00101101 00100010 00111000

11111111 11111111 11110000 00000000

11001000 00101101 001000000000 0000000000000000

The subnetwork address is 200.45.32.0.

©The McGraw-Hill Companies, Inc., 2000 © Adapted for use at JMU by Mohamed Aboutabl, 200388

Short-Cut Method

** If the byte in the mask is 255, copy the byte in the address.

** If the byte in the mask is 0, replace the byte in the address with 0.

** If the byte in the mask is neither 255 nor 0, we write the mask and the address in binary and apply the AND operation.

©The McGraw-Hill Companies, Inc., 2000 © Adapted for use at JMU by Mohamed Aboutabl, 200399

Comparison of a default mask and a subnet mask

• #of Subnetworks = 23 (always a power of 2) • #of Addresses/subnet = 213 (always a power of 2) • Special Addresses:

– Hostid 0 : Subnetwork Address – HostId all 1’s: Limited Broadcast inside subnetwork

©The McGraw-Hill Companies, Inc., 2000 © Adapted for use at JMU by Mohamed Aboutabl, 20031010

Designing Subnets

• The number of subnets N must be a power of 2.

1. Find the Subnet Mask

• 1-Bits: 1’s from default mask + Log2 N

• 0-Bits: The remaining bits (of the 32-bit IP address)

2. Number of Addresses in Each Subnet = 2 #of 0-Bits.

3. Range of Addresses in ach Subnet.

©The McGraw-Hill Companies, Inc., 2000 © Adapted for use at JMU by Mohamed Aboutabl, 20031111

Example 3Example 3

A company is granted the site address 201.70.64.0 (class C). The company needs six subnets. Design the subnets.

SolutionSolution

• The number of 1s in the default mask is 24 (class C).

• The company needs six subnets. This number 6 is not a power of 2. The next number that is a power of 2 is 8 (23). We need 3 more 1s in the subnet mask. The total number of 1s in the subnet mask is 27 (24 + 3).

• The total number of 0s is 5 (32 - 27). The mask is 11111111 11111111 11111111 11100000 or 255.255.255.224

• The number of subnets is 8. • The number of addresses in each subnet is 25

(5 is the number of 0s) or 32.

©The McGraw-Hill Companies, Inc., 2000 © Adapted for use at JMU by Mohamed Aboutabl, 20031212

Example 4Example 4

A company is granted the site address 181.56.0.0 (class B). The company needs 1000 subnets. Design the subnets.

SolutionSolution

• The company needs 1000 subnets. This number is not a power of 2. The next number that is a power of 2 is 1024 (210). We need 10 more 1s in the subnet mask.

• The total number of 1s in the subnet mask is 26 (16 + 10).

• The total number of 0s is 6 (32 - 26). The mask is 11111111 11111111 11111111 11000000 or 255.255.255.192

• The number of subnets is 1024. • The number of addresses in each subnet is 26

(6 is the number of 0s) or 64.

©The McGraw-Hill Companies, Inc., 2000 © Adapted for use at JMU by Mohamed Aboutabl, 20031313

Variable-length subnetting • Granted a Class C address • Needs 5 subnets of 60, 60, 60,

30, 30 hosts, respectively • 2-bit Subnet mask is too small • 3-bit Subnet mask → #hosts too

small

©The McGraw-Hill Companies, Inc., 2000 © Adapted for use at JMU by Mohamed Aboutabl, 20031414

SUPERNETTING

5.25.2

©The McGraw-Hill Companies, Inc., 2000 © Adapted for use at JMU by Mohamed Aboutabl, 20031515

A supernetwork • Combine several Class C blocks to create a larger superblock 1. #of blocks N is a power of 2 2. Blocks are contiguous 3. Byte 3 of starting address is divisible by N.

©The McGraw-Hill Companies, Inc., 2000 © Adapted for use at JMU by Mohamed Aboutabl, 20031616

Example 5 Example 5

A company needs 600 addresses. Which of the following set of class C blocks can be used to form a supernet for this company? 198.47.32.0 198.47.33.0 198.47.34.0

198.47.32.0 198.47.42.0 198.47.52.0 198.47.62.0

198.47.31.0 198.47.32.0 198.47.33.0 198.47.52.0

198.47.32.0 198.47.33.0 198.47.34.0 198.47.35.0

©The McGraw-Hill Companies, Inc., 2000 © Adapted for use at JMU by Mohamed Aboutabl, 20031717

Comparison of subnet, default, and supernet masks

In supernetting, we need the first address of the supernet and the supernet mask to define the range of addresses.

©The McGraw-Hill Companies, Inc., 2000 © Adapted for use at JMU by Mohamed Aboutabl, 20031818

Example 6 Example 6

We need to make a supernetwork out of 16 class C blocks. What is the supernet mask?

SolutionSolution

We need 16 blocks. For 16 blocks we need to change four 1s to 0s in the default mask. So the mask is

11111111 11111111 11110000 00000000 or

255.255.240.0

©The McGraw-Hill Companies, Inc., 2000 © Adapted for use at JMU by Mohamed Aboutabl, 20031919

Example 7 Example 7 A supernet has a first address of 205.16.32.0 and a supernet mask of 255.255.248.0. A router receives three packets with the following destination addresses:

205.16.37.44 205.16.42.56 205.17.33.76

Which packet belongs to the supernet?

We apply the supernet mask to see if we can find the beginning address.

205.16.37.44 AND 255.255.248.0 205.16.32.0

205.16.42.56 AND 255.255.248.0 205.16.40.0

205.17.33.76 AND 255.255.248.0 205.17.32.0

Only the first address belongs to this supernet.

©The McGraw-Hill Companies, Inc., 2000 © Adapted for use at JMU by Mohamed Aboutabl, 20032020

Example 8 Example 8 A supernet has a first address of 205.16.32.0 and a supernet mask of 255.255.248.0. How many blocks are in this supernet and what is the range of addresses?

SolutionSolution

The supernet has 21 1s. The default mask has 24 1s. Since the difference is 3, there are 23 or 8 blocks in this supernet. The blocks are 205.16.32.0 to 205.16.39.0. The first address is 205.16.32.0. The last address is 205.16.39.255.

©The McGraw-Hill Companies, Inc., 2000 © Adapted for use at JMU by Mohamed Aboutabl, 20032121

CLASSLESS ADDRESSING

5.35.3

©The McGraw-Hill Companies, Inc., 2000 © Adapted for use at JMU by Mohamed Aboutabl, 20032222

Variable-length blocks

Number of Addresses in a Block There is only one condition on the number of addresses in a block; it must be a power of 2 (2, 4, 8, . . .). A household may be given a block of 2 addresses. A small business may be given 16 addresses. A large organization may be given 1024 addresses.

Beginning Address

The beginning address must be evenly divisible by the number of addresses. For example, if a block contains 4 addresses, the beginning address must be divisible by 4. If the block has less than 256 addresses, we need to check only the rightmost byte. If it has less than 65,536 addresses, we need to check only the two rightmost bytes, and so on.

©The McGraw-Hill Companies, Inc., 2000 © Adapted for use at JMU by Mohamed Aboutabl, 20032323

Example 9 Example 9

Which of the following can be the beginning address of a block that contains 16 addresses?

205.16.37.32 190.16.42.44 17.17.33.80 123.45.24.52

SolutionSolution

The address 205.16.37.32 is eligible because 32 is divisible by 16. The address 17.17.33.80 is eligible because 80 is divisible by 16.

©The McGraw-Hill Companies, Inc., 2000 © Adapted for use at JMU by Mohamed Aboutabl, 20032424

Example 10 Example 10

Which of the following can be the beginning address of a block that contains 1024 addresses?

205.16.37.32 190.16.42.0 17.17.32.0 123.45.24.52

SolutionSolution

To be divisible by 1024, the rightmost byte of an address should be 0 and the second rightmost byte must be divisible by 4. Only the address 17.17.32.0 meets this condition.

©The McGraw-Hill Companies, Inc., 2000 © Adapted for use at JMU by Mohamed Aboutabl, 20032525

Slash notation

Attach the #of 1s in the mask (a.k.a. Attach the #of 1s in the mask (a.k.a. prefix length) to a classless address.prefix length) to a classless address.

Slash notation is also called Slash notation is also called CIDRCIDR

notation. notation.

©The McGraw-Hill Companies, Inc., 2000 © Adapted for use at JMU by Mohamed Aboutabl, 20032626

Example 11 Example 11

A small organization is given a block with the beginning address and the prefix length 205.16.37.24/29 (in slash notation). What is the range of the block?

SolutionSolution

The beginning address is 205.16.37.24. To find the last address we keep the first 29 bits and change the last 3 bits to 1s.

Beginning:11001111 00010000 00100101 00011000 Ending :11001111 00010000 00100101 00011111

There are only 8 addresses in this block.

©The McGraw-Hill Companies, Inc., 2000 © Adapted for use at JMU by Mohamed Aboutabl, 20032727

Example 13 Example 13

What is the network address if one of the addresses is 167.199.170.82/27?

SolutionSolution

The prefix length is 27, which means that we must keep the first 27 bits as is and change the remaining bits (5) to 0s. The 5 bits affect only the last byte. The last byte is 01010010. Changing the last 5 bits to 0s, we get 01000000 or 64. The network address is 167.199.170.64/27.

©The McGraw-Hill Companies, Inc., 2000 © Adapted for use at JMU by Mohamed Aboutabl, 20032828

Subnetting a Classless Address

Example 14 Example 14

An organization is granted the block 130.34.12.64/26. The

SolutionSolution

The suffix length is 6. This means the total number of addresses in the block is 64 (26). If we create four subnets, each subnet will have 16 addresses.

©The McGraw-Hill Companies, Inc., 2000 © Adapted for use at JMU by Mohamed Aboutabl, 20032929

Solution (Continued)Solution (Continued)

Let us first find the subnet prefix (subnet mask). We need four subnets, which means we need to add two more 1s to the site prefix. The subnet prefix is then /28.

Subnet 1:

130.34.12.64/28 → 130.34.12.79/28.

Subnet 2 :

130.34.12.80/28 → 130.34.12.95/28. Subnet 3:

130.34.12.96/28 → 130.34.12.111/28. Subnet 4:

130.34.12.112/28 → 130.34.12.127/28.

©The McGraw-Hill Companies, Inc., 2000 © Adapted for use at JMU by Mohamed Aboutabl, 20033030

Example 15 Example 15

An ISP is granted a block of addresses starting with 190.100.0.0/16. The ISP needs to distribute these addresses to three groups of customers as follows:

1. The first group has 64 customers; each needs 256 addresses.

2. The second group has 128 customers; each needs 128 addresses.

3. The third group has 128 customers; each needs 64 addresses.

Design the subblocks and give the slash notation for each subblock. Find out how many addresses are still available after these allocations.

©The McGraw-Hill Companies, Inc., 2000 © Adapted for use at JMU by Mohamed Aboutabl, 20033131

Solution Solution

Group 1

For this group, each customer needs 256 addresses. This means the suffix length is 8 (28 = 256). The prefix length is then 32 − 8 = 24.

01: 190.100.0.0/24 190.100.0.255/24

02: 190.100.1.0/24 190.100.1.255/24

…………………………………..

64: 190.100.63.0/24 190.100.63.255/24

Total = 64 × 256 = 16,384

©The McGraw-Hill Companies, Inc., 2000 © Adapted for use at JMU by Mohamed Aboutabl, 20033232

Solution (Continued) Solution (Continued)

Group 2

For this group, each customer needs 128 addresses. This means the suffix length is 7 (27 = 128). The prefix length is then 32 − 7 = 25. The addresses are:

001: 190.100.64.0/25 190.100.64.127/25

002: 190.100.64.128/25 190.100.64.255/25

003: 190.100.127.128/25 190.100.127.255/25

Total = 128 × 128 = 16,384

©The McGraw-Hill Companies, Inc., 2000 © Adapted for use at JMU by Mohamed Aboutabl, 20033333

Solution (Continued)Solution (Continued) Group 3

For this group, each customer needs 64 addresses. This means the suffix length is 6 (26 = 64). The prefix length is then 32 − 6 = 26.

001:190.100.128.0/26 190.100.128.63/26

002:190.100.128.64/26 190.100.128.127/26

…………………………

128:190.100.159.192/26 190.100.159.255/26

Total = 128 × 64 = 8,192

Number of granted addresses : 65,536

Number of allocated addresses: 40,960

Number of available addresses: 24,576

  • Designing Subnets
  • Variable-length subnetting

Expand your network by supernetting IP

addresses

By Rick Vanover

May 19, 2003, 7:00am PDT

Put simply, supernetting a TCP/IP network address is the opposite of subnetting it. Supernetting

is also known as CIDR (classless interdomain routing) as defined by RFCs 1517, 1518, 1519,

and 1520. In IPv4, CIDR is one way of attempting to manage the shortage of TCP/IP addresses

until IPv6 takes over.

Supernetting in itself does not give you more TCP/IP addresses; however, it provides larger

single networks for use. Here's how to implement supernetting on your network or support a

supernetted network that you may have inherited.

How supernetting works

Supernetting acts to bridge the gap between a Class C network that is limited to 254 addresses

and a Class B network that is too large, with over 65,000 addresses. In this way, it's possible to

have a "logical" network that offers the number of hosts that best suits your situation.

Supernetting achieves this by making a single network that has your specified number of hosts

and corresponding supernet (like a subnet mask). A supernetted address will look like any other

TCP/IP address in dotted decimal format (XXX.XXX.XXX.XXX), but it will have a supernetted

subnet mask. This looks like a normal subnet mask, but the last octet is not 0 (however, the

leading octets of the supernet mask are still 255). Supernetted addresses will require a default

gateway that needs to be supernetted as well.

Address ranges, or blocks, are important in supernetting. They allow you to identify the valid

addresses in a tabular format that helps identify boundaries on networks. There are many tables

you can create or find on the Internet to plan your networks when using supernetting. Figure A

shows a supernetting chart using an example configuration that we'll examine in this article.

Figure A

Supernetting Class C addresses

This represents part of the CIDR/supernetting chart to help determine which supernet option to choose.

CIDR Block Supernet Mask # of Networks* # of Hosts**

/17 255.255.128.0 128 32766

/18 255.255.192.0 64 16382

/19 255.255.224.0 32 8190

/20 255.255.240.0 16 4094

/21 255.255.248.0 8 2046

/22 255.255.252.0 4 1022

/23 255.255.254.0 2 510

/24 255.255.255.0 1 254

/25 255.255.255.128 Less than 1* 126

/26 255.255.255.192 Less than 1* 62

/27 255.255.255.224 Less than 1* 30

/28 255.255.255.240 Less than 1* 14

/29 255.255.255.248 Less than 1* 6

/30 255.255.255.252 Less than 1* 2

*Number of full Class C networks—256 or more available addresses **Available addresses—network and broadcast addresses excluded

This is a chart of the /17 through the /30 block of Class C supernets. These ranges are scalable,

helping you select how many networks and hosts you would like to use. You may notice that /24

CIDR block looks familiar, as that is really not a supernetted network but a subnetted single

Class C network with a standard 24-bit subnet.

Calculating supernet addresses

Calculating a supernet address is easy if the approach is organized. Using the chart in Figure A,

determine how many hosts you want to have available on your network and reference that

against the # of Hosts column to select the best match. Then, once you select the appropriate

number of hosts, you can look across the chart and see the corresponding supernet mask. With

that, you will need to determine a valid starting network.

This starting network must meet certain criteria:

 All networks are consecutive from your starting network.

 The third octet of the first network must be an even number (zero is valid for certain

situations).

 When combining eight networks (like the example below), the third octet of the network

number must be evenly divisible by eight.

 Create a table listing the available networks(s), addresses, supernet mask(s), default

gateway(s), and other networking objects to outline the network.

Usage scenario

In this example, we'll need approximately 1,220 IP addresses for a training lab scenario that

involves 150 people, each of whom requires two servers, five network-attached, multiport serial

devices, and their own laptop. We'll also need extra addresses for a few routers (including one

for Internet access) and addresses for the instructors. This example would be a good candidate

for using CIDR. I'll use the 192.168.16.0 network for our starting address.

To satisfy the 1,220 TCP/IP addresses for this scenario, we can use many of the different CIDR

blocks. We will use eight Class C networks, or CIDR /21, to give us 2,048 possible addresses.

The 2,048 possible addresses are calculated by taking eight networks that will have 256

addresses each (8 x 256 = 2048). We have to subtract two for the network and broadcast

addresses (as in a subnetted network), giving us 2048 – 2 = 2046 possible addresses. Starting

with 192.168.16.0, all "connected" networks must be consecutive in the numbering of the third

octet. Table A outlines the networks and available addresses.

Table A

Network Available Addresses Usage Circumstances

192.168.16.0 1-255 First address not available

192.168.17.0 0-255 All addresses in range available

192.168.18.0 0-255 All addresses in range available

192.168.19.0 0-255 All addresses in range available

192.168.20.0 0-255 All addresses in range available

192.168.21.0 0-255 All addresses in range available

192.168.22.0 0-255 All addresses in range available

192.168.23.0 0-254 Last address not available

Note that certain IP addresses are valid with atypical numbers in the last octet of the address. For

example, both 192.168.19.0 and 192.168.22.255 are valid addresses for a client, but they may not

be available for use by all clients that connect to this network. This is because certain operating

systems may not allow these types of addresses to be assigned as an IP address, since they may

view the address as a network or broadcast address and as invalid for use as a client address

(based on standard TCP/IP usage).

Specifically, Windows NT and 2000 do not allow the use of the X.X.X.255 or X.X.X.0 IP

addresses. (For more information on this, see Microsoft Knowledge Base Article 281579.)

Because the available hosts for this range of addresses will exceed our requirements, the loss of

these few addresses will not be an issue.

The resulting networks will start at 192.168.16.0 and increase in single increments up to

192.168.23.0. The supernet mask (functions as a subnet mask for all involved network

devices/systems) for these networks will be 255.255.248.0. This same supernet and default

gateway will be used for all of the networks on this supernet. Rendering the 255.255.248.0

supernet mask is easy from the chart in Figure A, but we will now prove how this is achieved.

We obtain our example supernet mask by taking the number of Class C networks we would like

(eight in our case) and subtracting that from 256. This result is 248. We take this value and place

it into the third octet of the mask, making our result 255.255.248.0. If we want to have 256 or

more Class C networks, this quick rule will not work. The addresses listed in Table A will all be

on the same network. For example, there is no route necessary for host 192.168.17.49 to access

192.168.19.244, or any other hosts in the range.

Implementing a supernetted network

I set up a supernetted network in a lab that I have access to. While I do not have over 1,000

computers, I did allocate all of my computers and virtual machines to reside on each network of

this supernetted network. The supernetted network required no settings beyond the IP address,

supernet mask, and default gateway options of the operating systems in question. There were no

special routing requirements or hardware necessary to quickly implement this network. The

supernetted network was implemented easily, and I was able to perform all network activities as

if it were a more typical 24-bit subnet mask (255.255.255.0). Addressing, name resolution, and

network-based applications all performed without incident.

You can also use supernetting in a reverse fashion by decreasing the number of hosts per

network. This is common in ISP situations where you need only a limited number of addresses

on the Internet, and the carrier provides you with a subnet mask of 255.255.255.248, for

example. This particular supernet means that you will have six available hosts on the network. In

this scenario, the first and last addresses are removed for the network and broadcast addresses, so

dividing 256 hosts by 32 gives us eight hosts in 32 networks. Removing the first and last

addresses for each network gives us six available hosts per network.

Why would I want to use supernetting?

ISPs frequently use supernetting to allocate IP addresses most effectively. There may be

scenarios where you have many LANS, WLANs, or VLANs that might be optimally suited for

supernetting to best administer your network needs. Keep in mind that supernetting introduces

complexity to network administration that needs thorough planning, testing, documentation, and

administrator competence.

Most new routing equipment and current operating systems support CIDR in their

implementation of the TCP/IP protocol. However, before a supernetting implementation, it is

critical to ensure that all components of your network are supernetting-aware. This includes

operating systems, network services, routers, routing protocols (RIP2, for example does not

support CIDR), and any network-based services used on your network.

For more information

I've shown how you can take advantage of the flexibility offered by CIDR, commonly know as

supernetting. These links can provide you with more examples, usage situations, and details on

using supernetting:

Subnetting

A Class A, B, or C TCP/IP network can be further divided, or subnetted, by a system administrator. This

becomes necessary as you reconcile the logical address scheme of the Internet (the abstract world of IP

addresses and subnets) with the physical networks in use by the real world.

A system administrator who is allocated a block of IP addresses may be administering networks that are

not organized in a way that easily fits these addresses. For example, you have a wide area network with

150 hosts on three networks (in different cities) that are connected by a TCP/IP router. Each of these three

networks has 50 hosts. You are allocated the class C network 192.168.123.0. (For illustration, this address is

actually from a range that is not allocated on the Internet.) This means that you can use the addresses

192.168.123.1 to 192.168.123.254 for your 150 hosts.

Two addresses that cannot be used in your example are 192.168.123.0 and 192.168.123.255 because

binary addresses with a host portion of all ones and all zeros are invalid. The zero address is invalid

because it is used to specify a network without specifying a host. The 255 address (in binary notation, a

host address of all ones) is used to broadcast a message to every host on a network. Just remember that

the first and last address in any network or subnet cannot be assigned to any individual host.

You should now be able to give IP addresses to 254 hosts. This works fine if all 150 computers are on a

single network. However, your 150 computers are on three separate physical networks. Instead of

requesting more address blocks for each network, you divide your network into subnets that enable you

to use one block of addresses on multiple physical networks.

In this case, you divide your network into four subnets by using a subnet mask that makes the network

address larger and the possible range of host addresses smaller. In other words, you are 'borrowing' some

of the bits usually used for the host address, and using them for the network portion of the address. The

subnet mask 255.255.255.192 gives you four networks of 62 hosts each. This works because in binary

notation, 255.255.255.192 is the same as 1111111.11111111.1111111.11000000. The first two digits of the

last octet become network addresses, so you get the additional networks 00000000 (0), 01000000 (64),

10000000 (128) and 11000000 (192). (Some administrators will only use two of the subnetworks using

255.255.255.192 as a subnet mask. For more information on this topic, see RFC 1878.) In these four

networks, the last 6 binary digits can be used for host addresses.

Using a subnet mask of 255.255.255.192, your 192.168.123.0 network then becomes the four networks

192.168.123.0, 192.168.123.64, 192.168.123.128 and 192.168.123.192. These four networks would have as

valid host addresses: 192.168.123.1-62 192.168.123.65-126 192.168.123.129-190 192.168.123.193-254

Remember, again, that binary host addresses with all ones or all zeros are invalid, so you cannot use

addresses with the last octet of 0, 63, 64, 127, 128, 191, 192, or 255.

You can see how this works by looking at two host addresses, 192.168.123.71 and 192.168.123.133. If you

used the default Class C subnet mask of 255.255.255.0, both addresses are on the 192.168.123.0 network.

However, if you use the subnet mask of 255.255.255.192, they are on different networks; 192.168.123.71 is

on the 192.168.123.64 network, 192.168.123.133 is on the 192.168.123.128 network.

Classes

The following are the classes of IP addresses.

 Class A—The first octet denotes the network address, and the last three octets are the host portion. Any IP address whose first octet is between 1 and 126 is a Class A address. Note that 0 is reserved as a part of the default address, and 127 is reserved for internal loopback testing.

1-126

 Class B—The first two octets denote the network address, and the last two octets are the host portion. Any address whose first octet is in the range 128 to 191 is a Class B address.

128-191

 Class C—The first three octets denote the network address, and the last octet is the host portion. The first octet range of 192 to 223 is a Class C address.

192-223

 Class D—Used for multicast. Multicast IP addresses have their first octets in the range 224 to 239.

224-239 (Multicast)

 Class E—Reserved for future use and includes the range of addresses with a first octet from 240 to 255.

240-255 (Reserved)

Subnetting

Subnetting is the concept of dividing the network into smaller portions called subnets. This is done by borrowing bits from the host portion of the IP address, enabling more efficient use of the network address. A subnet mask

defines which portion of the address is used to identify the network and which denotes the hosts.

The following tables show all possible ways a major network can be subnetted, and, in each case, how many effective subnets and hosts are possible.

There are three tables, one for each class of addresses.

 The first column shows how many bits are borrowed from the host portion of the address for subnetting.

 The second column shows the resulting subnet mask in dotted decimal format.

 The third column shows how many subnets are possible.

 The fourth column shows how many valid hosts are possible on each of these subnets.

 The fifth column shows the number of subnet mask bits.

Class A Host/Subnet Table

Class A

Number of

Bits Borrowed Subnet Effective Number of Number of Subnet

from Host Portion Mask Subnets Hosts/Subnet Mask Bits

------- --------------- --------- ------------- -------------

1 255.128.0.0 2 8388606 /9

2 255.192.0.0 4 4194302 /10

3 255.224.0.0 8 2097150 /11

4 255.240.0.0 16 1048574 /12

5 255.248.0.0 32 524286 /13

6 255.252.0.0 64 262142 /14

7 255.254.0.0 128 131070 /15

8 255.255.0.0 256 65534 /16

9 255.255.128.0 512 32766 /17

10 255.255.192.0 1024 16382 /18

11 255.255.224.0 2048 8190 /19

12 255.255.240.0 4096 4094 /20

13 255.255.248.0 8192 2046 /21

14 255.255.252.0 16384 1022 /22

15 255.255.254.0 32768 510 /23

16 255.255.255.0 65536 254 /24

17 255.255.255.128 131072 126 /25

18 255.255.255.192 262144 62 /26

19 255.255.255.224 524288 30 /27

20 255.255.255.240 1048576 14 /28

21 255.255.255.248 2097152 6 /29

22 255.255.255.252 4194304 2 /30

23 255.255.255.254 8388608 2* /31

Class B Host/Subnet Table

Class B Subnet Effective Effective Number of Subnet

Bits Mask Subnets Hosts Mask Bits

------- --------------- --------- --------- -------------

1 255.255.128.0 2 32766 /17

2 255.255.192.0 4 16382 /18

3 255.255.224.0 8 8190 /19

4 255.255.240.0 16 4094 /20

5 255.255.248.0 32 2046 /21

6 255.255.252.0 64 1022 /22

7 255.255.254.0 128 510 /23

8 255.255.255.0 256 254 /24

9 255.255.255.128 512 126 /25

10 255.255.255.192 1024 62 /26

11 255.255.255.224 2048 30 /27

12 255.255.255.240 4096 14 /28

13 255.255.255.248 8192 6 /29

14 255.255.255.252 16384 2 /30

Class C Host/Subnet Table

Class C Subnet Effective Effective Number of Subnet

Bits Mask Subnets Hosts Mask Bits

------- --------------- --------- --------- --------------

1 255.255.255.128 2 126 /25

2 255.255.255.192 4 62 /26

3 255.255.255.224 8 30 /27

4 255.255.255.240 16 14 /28

5 255.255.255.248 32 6 /29

6 255.255.255.252 64 2 /30

Subnetting Example

The first entry in the Class A table (/10 subnet mask) borrows two bits (the leftmost bits) from the host portion of the network for subnetting, then with two bits you have four (2

2 ) combinations, 00, 01, 10, and 11. Each of these will

represent a subnet.

Binary Notation Decimal Notation

-------------------------------------------------- -----------------

xxxx xxxx. 0000 0000.0000 0000.0000 0000/10 ------> X.0.0.0/10

xxxx xxxx. 0100 0000.0000 0000.0000 0000/10 ------> X.64.0.0/10

xxxx xxxx. 1000 0000.0000 0000.0000 0000/10 ------> X.128.0.0/10

xxxx xxxx. 1100 0000.0000 0000.0000 0000/10 ------> X.192.0.0/10

Out of these four subnets, 00 and 11 are called subnet zero and the all-ones subnet, respectively. Prior to Cisco IOS

® Software Release 12.0, the ip subnet-zero global configuration command was required to be able to configure

subnet zero on an interface. In Cisco IOS 12.0, ip subnet-zero is enabled by default.

Note: The subnet zero and all-ones subnet are included in the effective number of subnets as shown in the third column.

Since the host portion has now lost two bits, the host portion will have only 22 bits (out of the last three octets). This means the complete Class A network is now divided (or subnetted) into four subnets, and each subnet can have 2

22 hosts (4194304). A host portion with all zeros is network number itself, and a host portion with all ones is reserved

for broadcast on that subnet, leaving the effective number of hosts to 4194302 (2 22

– 2), as shown in the fourth column.

IPv6 neighbor discovery

Neighbor Discovery Protocol (NDP) itself does not describe a wire-level protocol or packet structure,

but rather it establishes directions for accomplishing routine tasks using certain algorithms and five

ICMPv6 message types.

Many of the capabilities provided by NDP are very similar to those found in IPv4's ARP and ICMPv4,

while others are new implementations available only under IPv6. RFC 4861 describes the nine

functions of NDP in detail.

Router Discovery

Whereas IPv4 hosts must rely on manual configuration or DHCP to provide the address of a default

gateway, IPv6 hosts can automatically locate default routers on the link. This is accomplished

through the use of two ICMPv6 messages: Router Solicitation (type 133) and Router

Advertisement (type 134). When first joining a link, an IPv6 host multicasts a router solicitation to

the all routers multicast group, and each router active on the link responds by sending a router

advertisement with its address to the all nodes group.

Router advertisements indicate paths out of the local link, but they also specify additional information

necessary to assist other NDP operations.

Prefix Discovery

One of the options typically carried by a router advertisement is the Prefix Information option (type

3). Each prefix information option lists an IPv6 prefix (subnet) reachable on the local link. Remember

that it is not uncommon for multiple IPv6 prefixes to reside on the same link, and routers may include

more than one prefix in each advertisement. A host which knows what prefixes are reachable on the

link can communicate directly with destinations in those prefixes without passing its traffic through a

router.

Parameter Discovery

Another option included in router advertisements is the MTU option (type 5), which informs hosts of

the IP MTU to use. For example, this value is typically set to 1500 for Ethernet networks. However,

not all link types have a standardized MTU size. Including this option ensures all hosts know the

correct MTU to use.

Router advertisements also specify the default value hosts should use for the IPv6 hop count. This

isn't an option, but a field built into the router advertisement message header.

Address Autoconfiguration

NDP provides mechanisms for a host to automatically configure itself with an address from a prefix

learned from a local router through prefix discovery. This is done by concatenating a candidate

learned prefix with the EUI-64 address of the host's interface (embedding the MAC address inserting

FF:FE). In this manner, a host can achieve stateless autoconfiguration.

Address Resolution

The function of address resolution was handled by ARP for IPv4, but is handled by ICMPv6 for IPv6.

In a process very similar to router discovery, two ICMPv6 messages are used: Neighbor

Solicitation (type 135) and Neighbor Advertisement (type 136). A host seeking the link layer

address of a neighbor multicasts a neighbor solicitation and the neighbor (if online) responds with its

link layer address in a neighbor advertisement.

Next-Hop Determination

As in IPv4, next-hop determination is simply a procedure for performing longest-match lookups on

the host routing table and, for off-link destinations, the selection of a default router.

Neighbor Unreachability Detection

NDP is able to determine the reachability of a neighbor by examining clues from upper-layer

protocols (for example, received TCP acknowledgments), or by actively re-performing address

resolution (via ICMPv6) when certain thresholds are reached.

Duplicate Address Detection

When a host first joins a link, it multicasts neighbor solicitations for its own IPv6 address for a short

period before attempting to use that address to communicate. If it receives a neighbor advertisement

in response, the host realizes that another neighbor on the link is already using that address. The

host will mark the address as a duplicate and will not use it on the link.

Note that this process is similar to IPv4 gratuitous ARP requests, but NDP elegantly allows for

detection of two hosts with the same address before both hosts are actively sending traffic from the

address.

Redirection

A fifth type of ICMPv6 message, the Redirect (type 137), is used by routers to either point hosts

toward a more preferable router, or to indicate that the destination actually resides on link. ICMPv4

provides the same capability with its own redirect message.

IPv6 Address structure Hexadecimal Number System Before introducing IPv6 Address format, we shall look into Hexadecimal Number System. Hexadecimal is positional number system which uses radix (base) of 16. To represent the values in readable format, this system uses 0-9 symbols to represent values from zero to nine and A-F symbol to represent values from ten to fifteen. Every digit in Hexadecimal can represent values from 0 to 15.

Address Structure An IPv6 address is made of 128 bits divided into eight 16-bits blocks. Each block is then converted into 4-digit Hexadecimal numbers separated by colon symbol.

For example, the below is 128 bit IPv6 address represented in binary format and divided into eight 16-bits blocks:

0010000000000001 0000000000000000 0011001000110100 1101111111100001 0000000001100011 0000000000000000 0000000000000000 1111111011111011

2001:0000:3238:DFE1:0063:0000:0000:FEFB

Rules to shorten the IPv6 address:

Even after converting into Hexadecimal format, IPv6 address remains long. IPv6 provides some rules to shorten the address. These rules are:

2001:0000:3238:DFE1:0063:0000:0000:FEFB

Rule:1 Discard leading Zero(es):

In Block 5, 0063, the leading two 0s can be omitted, such as (5th block):

2001:0000:3238:DFE1:63:0000:0000:FEFB

Rule:2 If two of more blocks contains consecutive zeroes, omit them all and replace with double colon sign ::, such as (6th and 7th block): 2001:0000:3238:DFE1:63::FEFB

Consecutive blocks of zeroes can be replaced only once by :: so if there are still blocks of zeroes in the address they can be shrink down to single zero, such as (2nd block):

2001:0:3238:DFE1:63::FEFB

Interface ID IPv6 has three different types of Unicast Address schemes.

The second half of the address (last 64 bits) is always used for Interface ID.

MAC address of a system is composed of 48-bits and represented in Hexadecimal. MAC address is considered to be uniquely assigned worldwide. Interface ID takes advantage of this uniqueness of MAC addresses. A host can auto-

-64) format.

First, a Host divides its own MAC address into two 24-bits halves. Then 16-bit Hex value 0xFFFE is sandwiched into those two halves of MAC address, resulting in 64-bit Interface ID.

Global Unicast Address

and uniquely addressable.

Global Routing Prefix: The most significant 48-bits are designated as Global Routing Prefix which is assigned to specific Autonomous System. Three most significant bits of Global Routing Prefix is always set to 001.

Link-Local Address Auto-configured IPv6 address is known as Link-Local address. This address always starts with FE80. First 16 bits of Link-Local address is always set to 1111 1110 1000 0000 (FE80). Next 48-bits are set to 0, thus:

Link-Local addresses are used for communication among IPv6 hosts on a link (broadcast segment) only. These addresses are not routable so a Router never forwards these addresses outside the link.

Unique-Local Address This type of IPv6 address which is though globally unique, but it should be used in local communication. This address has second half of Interface ID and first half is divided among Prefix, Local Bit, Global ID and Subnet ID.

Prefix is always set to 1111 110. L bit, which is set to 1 if the address is locally assigned. So far the meaning of L bit

SCOPE OF IPV6 UNICAST ADDRESSES:

The scope of Link-local address is limited to the segment.

Unique Local Address are boundary.

Global Unicast addresses are globally unique and recognizable.

 

The Network Layer

While the OSI Reference Model refers to this layer as the Network Layer, TCP/IP refers to it as the Internet Layer. Commonly defined by the Internet Protocol (RFC 791)

• Provides the basic packet delivery service on which TCP/IP networks are built. • All TCP/IP data flows thru IP regardless of it’s final destination.

Internet Protocol

• Defines the Internet Addressing scheme • Moves data between the data link layer and the transport layer • Defines the datagram. • Performs fragmentation and re-assembly of datagrams • A connectionless protocol (unreliable), utilizing best effort delivery with some

control messages (ICMP). 1. Internet Addressing

• Allocation managed by the Internet Assigned numbers Authority (IANA) by Jon Postel until 1998 and the Internet Corporation for Assigned Names and Numbers (ICANN) since 1998. In the U.S, numbers are allocated through the American Registry of Internet Numbers (ARIN).

• Logical addresses

• 32 bit IP address expressed in dotted-decimal notation.

1000 0011 1011 0111 1101 1001 1111 1110 131 . 183 . 217 . 254

• Identifies the network and the host on the network.

• Local addresses are delivered directly.

• If the address is not local it is passed to gateway. A gateway is a device that switches packets between different physical networks. Deciding which gateway to use is called routing.

The IP address consists of a network portion and a host portion. How do we identify the network or the host? Where is the network boundary in the IP address? Looking at the decimal dot notation, one logical place to draw the boundary is after the first octet: Result: Not enough networks….. Looking at the decimal dot notation once again, a second logical place to draw the boundary is after the second octet: Result: Not enough networks….. Looking at the decimal dot notation yet again, a third logical place to draw the boundary is after the third octet: Result: Not enough hosts…… Possible solution: Use a combination of the three logical boundaries. This provides for a large number of networks for those with few hosts and a large number of hosts for big networks. Problem: Using an address of 131.183.217.1, which portion is the network portion and which portion is the host portion? Ugh, this is one ugly table to maintain! In order to simplify the tables, networks were divided into classes based on the three byte boundaries. In order to determine the class of a network, the First Octet Rule is applied:

First Octet Class Decimal Values Network bits Host bits 0xxx xxxx A 0 – 127 8 24 10xx xxxx B 128 – 191 16 16 110x xxxx C 192 – 223 24 8 1110 xxxx D 224 – 239 1111 xxxx E 240 – 255

Another method of specifying the boundary between the network and the host is to use a netmask. The netmask is 32 bits and a 1 bit marks the network portion and a 0 bit marks the host portion of an address. Class A addresses would have an 8 bit netmask (255.0.0.0).

2

Class B addresses would have a 16 bit netmask (255.255.0.0). Class C addresses would have a 24 bit netmask (255.255.255.0). Since the netmask coincides with the byte boundaries of each of the address classes, they are referred to as the “natural” mask.

131.183.0.0 16 bit network, 16 bit host 131.183.0.0 255.255.0.0 natural mask 205.133.127.0 24 bit network, 8 bit host 205.133.127.0 255.255.255.0 natural mask

The host portion on Class A networks is 24 bits. This means that there could be 2^24 hosts on this network. Class B networks will support 2^16 hosts, and Class C networks allow 2^8 hosts. In many instances, dividing a large network into smaller networks with fewer hosts is desirable. This process is called sub-netting and requires an increase in the netmask. U.T. Address Space: 131.183.0.0 Netmask: 255.255.0.0 Address Range: 131.183.0.0 - 131.183.255.255 Sub-dividing the space: Network: 131.183.217.0 Sub-netmask: 255.255.255.0 Range: 131.183.217.0 - 131.183.217.255 The use of a mask to determine the destination network is called Classless Inter-Domain Routing (CIDR). CIDR requires modification to routes and routing protocols

- Expressed as: o Address/prefix length (# of bits in network address) instead of network

and subnet via longhand. o 131.183.217.0/24 vs. 131.183.217.0 with a subnet mask of

255.255.255.0 - RFC 1878

Reserved Network Addresses

0.0.0.0/8 = default route 127.0.0.0/8 = loopback address

Private Networks (RFC1918)

3

10/8 172.16/12 = 172.16.0.0 thru 172.31.255.255 192.168/16 = 192.168.0.0 thru 192.255.255.255 169.254/16 = 169.254.0.0 thru 169.255.255.255

Reserved Host Addresses

0 the network, the wire 255 broadcast address, flooded to all hosts Thus, usable addresses = 2n-2

Gateways

If the destination host is not on the same network as the source host, the packet is delivered to a gateway (generally a “router”), which forwards the packet to the proper destination.

2. The Datagram

The compete information package from the transport layer is encapsulated in the data portion of the Ethernet frame. The Ethernet header identifies the target machine.

IP Header IP Data (payload)

Ethernet Data (payload) Ethernet Header

Once the frame reaches the target host and is extracted, the ethernet header is stripped and the packet is passed up to the transport layer.

4

A summary of the contents of the Internet header follows:

0 1 2 3 0 1 2 3 4 5 6 7 8 9 0 1 2 3 4 5 6 7 8 9 0 1 2 3 4 5 6 7 8 9 0 1

+-+-+-+-+-+-+-+-+-+-+-+-+-+-+-+-+-+-+-+-+-+-+-+-+-+-+-+-+-+-+-+-+ |Version| IHL |Type of Service| Total Length | +-+-+-+-+-+-+-+-+-+-+-+-+-+-+-+-+-+-+-+-+-+-+-+-+-+-+-+-+-+-+-+-+ | Identification |Flags| Fragment Offset | +-+-+-+-+-+-+-+-+-+-+-+-+-+-+-+-+-+-+-+-+-+-+-+-+-+-+-+-+-+-+-+-+ | Time to Live | Protocol | Header Checksum | +-+-+-+-+-+-+-+-+-+-+-+-+-+-+-+-+-+-+-+-+-+-+-+-+-+-+-+-+-+-+-+-+ | Source Address | +-+-+-+-+-+-+-+-+-+-+-+-+-+-+-+-+-+-+-+-+-+-+-+-+-+-+-+-+-+-+-+-+ | Destination Address | +-+-+-+-+-+-+-+-+-+-+-+-+-+-+-+-+-+-+-+-+-+-+-+-+-+-+-+-+-+-+-+-+ | Options | Padding | +-+-+-+-+-+-+-+-+-+-+-+-+-+-+-+-+-+-+-+-+-+-+-+-+-+-+-+-+-+-+-+-+

Example Internet Datagram Header

Version: 0100h (version 4) Internet Header Length: Type of Service: specifies the “priority” of the packet Total Length: of the IP datagram (up to 65,535 octets) Identification, Flags, Fragment Offset: Time to Live: Protocol: Header Checksum: Source Address: Destination Address:

5

6

3. Moving data between the data link layer and the transport layer.

Moving the data between the data link layer and the transport layer requires coordination of the addressing schemes in each of the layers. Recall:

• Transport layer: 32 bit IP addresses • Data-link layer: 48 bit MAC addresses

Two machines can communicate only if they know each other’s physical network address. Data from applications are transported between logical hosts. In other words, the “application” uses the IP address to identify the target machine, while the “network” uses the MAC address to identify the target machine.

The address resolution problem: How do you map an IP address to physical address.

• Keep tables. What if hardware fails and is replaced? How do you re-map? • Encode h/w address in high level address. However, you can’t do direct

mapping – IP address is 32 bits, the Ethernet address is 48 bits so it can’t be encapsulated in the IP address.

• Dynamic binding: address resolution protocol RFC 826: Address Resolution Protocol (ARP) ARP binds addresses dynamically (no direct or static mapping)

Checks cache table Broadcasts to all hosts Host answers MAC address is inserted into table

Why not broadcast all packets instead of asking? Efficiency Cache timeout –20 minutes. Why? No other way to determine downtime since no guarantee of delivery Delay in updates (20 minutes…) Improvements sender inserts it’s IP in it’s broadcast request both intended and other hosts can extract

arp broadcasts can be sent at boot to update arp tables.

IPv6 Address Types

::/128

::1/128

::ffff/96

Example: ::ffff:192.0.2.47

fc00::/7

Example: fdf8:f53b:82e4::53

fe80::/10

Example: fe80::200:5aee:feaa:20a2

Prefix Designation and Explanation IPv4 Equivalent

Unspecified This address may only be used as a source address by an initialising host before it has learned its own address.

Loopback This address is used when a host talks to itself over IPv6. This often happens when one program sends data to another.

IPv4-Mapped These addresses are used to embed IPv4 addresses in an IPv6 address. One use for this is in a dual stack transition scenario where IPv4 addresses can be mapped into an IPv6 address. See RFC 4038 for more details.

Unique Local Addresses (ULAs) These addresses are reserved for local use in home and enterprise environments and are not public address space.

These addresses might not be unique, and there is no formal address registration. Packets with these addresses in the source or destination fields are not intended to be routed on the public Internet but are intended to be routed within the enterprise or organisation.

See RFC 4193 for more details.

Link-Local Addresses These addresses are used on a single link or a non-routed common access network, such as an Ethernet LAN. They do not need to be unique outside of that link.

Link-local addresses may appear as the source or destination of an IPv6 packet. Routers must not forward IPv6 packets if the source or destination contains a link- local address.

Link-local addresses may appear as the source or destination of an IPv6 packet. Routers must not forward IPv6 packets if the source or destination contains a link- local address.

0.0.0.0

127.0.0.1

There is no equivalent. However, the mapped IPv4 address can be looked up in the relevant RIR’s Whois database.

Private, or RFC 1918 address space:

10.0.0.0/8 172.16.0.0/12 192.168.0.0/16

169.254.0.0/16

This sheet is available at www.ripe.net/ipv6-address-types • Produced by the RIPE NCC in cooperation with ICANN • www.ripe.net • www.icann.org

IPv6 Address Types

2001:0000::/32

Example: 2001:0000:4136:e378: 8000:63bf:3fff:fdd2

2001:0002::/48

Example: 2001:0002:6c::430

2001:0010::/28

Example: 2001:10:240:ab::a

2002::/16

Example: 2002:cb0a:3cdd:1::1

2001:db8::/32

Example: 2001:db8:8:4::2

2000::/3

ff00::/8

Example: ff01:0:0:0:0:0:0:2

Prefix Designation and Explanation IPv4 Equivalent

Teredo This is a mapped address allowing IPv6 tunneling through IPv4 NATs. The address is formed using the Teredo prefix, the server’s unique IPv4 address, flags describing the type of NAT, the obfuscated client port and the client IPv4 address, which is probably a private address. It is possible to reverse the process and identify the IPv4 address of the relay server, which can then be looked up in the relevant RIR’s Whois database.

You can do this on the following webpage: http://www.potaroo.net/cgi-bin/ipv6addr

Benchmarking These addresses are reserved for use in documentation. They should not be used as source or destination addresses.

Orchid These addresses are used for a fixed-term experiment. They should only be visible on an end-to-end basis and routers should not see packets using them as source or destination addresses.

6to4 A 6to4 gateway adds its IPv4 address to this 2002::/16, creating a unique /48 prefix. As the IPv4 address of the gateway router is used to compose the IPv6 prefix, it is possible to reverse the process and identify the IPv4 address, which can then be looked up in the relevant RIR’s Whois database.

You can do this on the following webpage: http://www.potaroo.net/cgi-bin/ipv6addr

Documentation These addresses are used in examples and documentation. They should never be source or destination addresses.

Global Unicast Other than the exceptions documented in this table, the operators of networks using these addresses can be found using the Whois servers of the RIRs listed in the registry at: http://www.iana.org/assignments/ipv6- unicast-address-assignments

Multicast These addresses are used to identify multicast groups. They should only be used as destination addresses, never as source addresses.

No equivalent

198.18.0.0/15

No equivalent

There is no equivalent but 192.88.99.0/24 has been reserved as the 6to4 relay anycast address prefix by the IETF.

192.0.2.0/24 198.51.100.0/24 203.0.113.0/24

No equivalent single block

224.0.0.0/4

This sheet is available at www.ripe.net/ipv6-address-types • Produced by the RIPE NCC in cooperation with ICANN • www.ripe.net • www.icann.org

Before a device on a TCP/IP network can effectively communicate, it needs to know its IP address. While a conventional network host can read this information from its internal disk, some devices have no storage, and so do not have this luxury. They need help from another device on the network to provide them with an IP address and other information and/or software they need to become active IP hosts. This problem of getting a new machine up and running is commonly called bootstrapping, and to provide this capability to IP hosts, the TCP/IP Bootstrap Protocol (BOOTP) was created.

Without a form of internal storage, a device must rely on someone or something to tell it “who it is” (its address) and how to function each time it is powered up. When a device like this is turned on, it is in a difficult position: it needs to use IP to communicate with another device that will tell it how to communicate using IP! This process, called bootstrapping or booting, comes from an analogy to a person “pulling himself up using his own bootstraps”.

The Reverse Address Resolution Protocol (RARP) was the first attempt to resolve this “bootstrap problem”. Created in 1984, RARP is a direct adaptation of the low-level Address Resolution Protocol (ARP) that binds IP addresses to link-layer hardware addresses. RARP is capable of providing a diskless device with its IP address, using a simple client/server exchange of a request and reply between a host and an RARP server.

The difficulty with RARP is that it has so many limitations. It operates at a fairly low level using hardware broadcasts, so it requires adjustments for different hardware types. An RARP server is also required on every physical network to respond to layer-two broadcasts. Each RARP server must have address assignments manually provided by an administrator. And perhaps worst of all, RARP only provides an IP address to a host and none of the other information a host may need.

RARP clearly wasn't sufficient for the host configuration needs of TCP/IP. To support both the needs of diskless hosts and other situations where the benefits of autoconfiguration were required, the Bootstrap Protocol (BOOTP) was created. BOOTP was standardized in RFC 951, published September 1985. This relatively straight- forward protocol was designed specifically to address the shortcomings of RARP:

BOOTP Deals With the First Phase of Bootstrapping

It should be noted that even though the name of BOOTP implies that it defines everything needed for a storageless device to “boot”, this isn't really the case. As the BOOTP standard itself describes, “bootstrapping” generally requires two phases. In the first, the client is provided with an address and other parameters. In the second, the client downloads software, such as an operating system and drivers, that let it function on the network and perform whatever tasks it is charged with. BOOTP really only deals with the first of these phases: address assignment and configuration. The second is assumed to take place using a simple file transfer protocol like the Trivial File Transfer Protocol (TFTP).

Changes to BOOTP and the Development of DHCP

BOOTP was the TCP/IP host configuration of choice from the mid-1980s through the end of the 1990s. The vendor extensions introduced in RFC 1048 were popular, and over the years, additional vendor extensions were defined; RFC 1048 was replaced by RFCs 1084, 1395 and 1497 in succession. Some confusion also resulted over the years in how some sections of RFC 951 should be interpreted, and how certain features of BOOTP work.

RFC 1542, Clarifications and Extensions for the Bootstrap Protocol, was published in October 1993 to address this, and also made some slight changes to the protocol's operation. (RFC 1542 is actually a correction of the nearly-identical RFC 1532 that had some small errors in it.)

While BOOTP was obviously quite successful, it also had certain weaknesses of its own. One of the most important of these is lack of support for dynamic address assignment. The need for dynamic assignment became much more pronounced when the Internet really started to take off in the late 90s. This led directly to the development of the Dynamic Host Configuration Protocol (DHCP).

While DHCP replaced BOOTP as the TCP/IP host configuration protocol of choice, it would be inaccurate to say that BOOTP is “gone”. It is still used to this day in various networks. Furthermore, DHCP was based directly on BOOTP, and they share many attributes, including a common message format. BOOTP vendor extensions were used as the basis for DHCP options, which work in the same way but include extra capabilities. In fact, the successor to RFC 1497 is RFC 1533, which officially merges BOOTP vendor extensions and BOOTP options into the same standard.

PC & Industrial Networks

70 Points

Answer each of the following questions in complete detail:

1. (10 points) Assume a host computer has the following configuration:

IP Address: 200.110.84.176

Subnet Mask: 255.255.248.0

Default Gateway: 200.110.84.1

a. What is the Class of this ‘network’?

b. Is this a Network, Subnetwork, Supernetwork or something else? How do you know?

c. How many possible hosts would there be on the above network if all usable addresses were assigned?

d. How would this IP address be expressed using CIDR notation?

e. Using CIDR notation, what is the range of the block of addresses this Host belongs to?

2. (6 points) Explain everything that can be determined from the following:

a. fe80::9890:96ff:fea1:53ed%12

b. 2002:77fe:8921::77fe:8921

c. ::01

d. ::

3. (4 points) What is the significance of the ninth octet in the header of the IP datagram?

4. (5 points) SSK Corp has offices in Toledo, Detroit and Columbus. Each office has 127

Computers. The IT plan calls for connecting all offices using data lines. The Toledo site will also connect to the Internet. SSK Corp. has elected to use PUBLIC IP address space on all computers at each of its sites.

Their ISP has restricted the IP ranges to the ones below. The ISP’s Network

Administrator is on vacation – you have been asked to fill-in and select a range of addresses that will satisfy SSK Corp.’s needs with the least amount of wasted IP addresses. Propose a range of addresses for SSK Corp. and explain your answer.

225.113.8.0/24

225.113.9.0/24

192.168.0.0/16

221.127.136.0/24

221.128.135.0/24

221.128.136.0/24

221.128.137.0/24

221.128.138.0/24

206.122.148.0/24

10.0.0.0/8

221.125.138.0/24

221.126.137.0/25

221.128.139.0/24

5. (5 points) Using the IP ranges below:

a) What IP range would an ISP provide to a customer, if the customer wanted a range of Public IP’s for use on the Internet? Explain your choice and why you feel the other choices are not adequate?

b) Using the Range you selected in ‘5a’ above - subnet the range into as many /28 networks as possible – show your work and each /28 range of addresses. (Show the network address and the broadcast address for each /28 subnet)

225.113.8.0/24

225.113.9.0/25

192.168.0.0/16

201.127.136.0/24

172.16.0.0/24

10.0.0.0/8

169.254.137.0/25

245.125.1378.0/24

10.0.0.0/24

245.0.0.0/8

127.0.0.0/8

6. (15 points) The following information was extracted from an Ethernet Frame:

IP Datagram Header: 45 00 00 44 5b d2 00 00 80 11 ef 4d 83 b7 75 39 83 b7 72 e1

Based on the above information, describe everything that can be determined about this packet (give the actual data value for each field). Convert each field to its normally displayed value (i.e. Hexadecimal/Decimal/Binary).

Example: Version is 4 which is IPv4

7. (10 points) Bob obtained the following information from a workstation: C:\>ipconfig

Ethernet adapter Local Area Connection:

IP Address. . . . . . . ……... . : 169.254.10.105 Subnet Mask . . . . . . ……... : 255.255.0.0 Default Gateway . . . . ……..:

Link-Local IPv6 Address…..fe80::9890:96ff:fea1:53ed%12

a. Explain everything that can be determined about this host.

After waiting 5 minutes and making no changes to his workstation. Bob obtained the following information from his workstation.

C:\>ipconfig

Ethernet adapter Local Area Connection:

IP Address. . . . . . . . ………………: 138.110.10.50 Subnet Mask . . . . . . …………….. : 255.255.0.0

Default Gateway . . . …………….. : 138.110.10.1

Link-Local IPv6 Address………….:fe80::9890:96ff:fea1:53ed%12

b. How would you explain the changes to the IP stack? Explain in detail what occurred.

8. (5 points) Explain each of the good ‘Network Design goals’ as discussed in class.

9. (10 points) You have been given the task of changing the IP Address and enabling telnet remote access on a CISCO 2950 enterprise switch.

The current IP address is 172.25.2/16 the new IP address is 10.0.0.2/24

The enterprise switch has no password configured.

a. Explain all of the commands needed for you to successfully accomplish the IP Address change and telnet remote access.

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