Networking Chapter 4 There Are Many Fields Can Matched

Chapter 4 Review Questions

1. A network-layer packet is a datagram. A router forwards a packet based on the

-layer switch forwards a packet based on the

2. The main function of the data plane is packet forwarding, which is to forward

datagrams from their input links to their output links.

input ports perform physical layer function of terminating an incoming physical link

3. The key differences between routing and forwarding is that f

local action of transferring packets from its input interfaces to its output interfaces,

4. The role of the forwarding table within a router is to hold entries to determine the

5. -effort service. With this

6. Input port, switching fabric, and output ports are implemented in hardware, because

their datagram-processing functionality is far too fast for software implementation. A

routing processor inside a traditional router uses software for executing routing

7. With the shadow copy, the forwarding lookup is made locally, at each input port,

8. Destination-based forwarding means that a datagram arriving at a router will be

forwarded to an output interface based on only the final destination of the datagram.

9. A router uses longest prefix matching to determine which link interface a packet will

10. Switching via memory; switching via a bus; switching via an interconnection

11. If the rate at which packets arrive to the fabric exceeds switching fabric rate, then

packets will need to queue at the input ports. If this rate mismatch persists, the queues

12. Assuming input and output line speeds are the same, packet loss can still occur if the

rate at which packets arrive to a single output port exceeds the line speed. If this rate

13. HOL blocking: Sometimes a packet that is first in line at an input port queue must

wait because there is no available buffer space at the output port to which it wants to

14. (A typo in this question: the first question mark should be replaced by a period).

Only FIFO can ensure that all packets depart in the order in which they arrived.

15. For example, a packet carrying network management information should receive

16. (A typo in the question: different difference)

With RR, all service classes are treated equally, i.e., no service class has priority over

RR.

17. The 8-bit protocol field in the IP datagram contains information about which transport

layer protocol the destination host should pass the segment to.

18. Time-to-live.

19.

20. The reassembly of the fragments of an

destination host.

25. 50% overhead.

26. Typically the wireless router includes a DHCP server. DHCP is used to assign IP

27. Route aggregation means that an ISP uses a single prefix to advertise multiple

28. A plug-and-play or zeroconf protocol means that the protocol is able to automatically

-related aspects in order to connect the host into a network.

29. A private network address of a device in a network refers to a network address that is

only meaningful to those devices within that network. A datagram with a private

30. IPv6 has a fixed length header, which does not include most of the options an IPv4

header can include. Even though the IPv6 header contains two 128 bit addresses

32. Forwarding has two main operations: match and action. With destination-based

forwarding, the match operation of a router looks up only the destination IP address

of the to-be-forwarded datagram, and the action operation of the router involves

33. Each entry in the forwarding table of a destination-based forwarding contains only an

IP header field value and the outgoing link interface to which a packet (that matches

34. a switch tries to find a match between

some of the header values of a packet with some entry in a flow table, and then based

on that match, the router decides to which interface(s) the packet will be forwarded

35. Three example header fields in an IP datagram that can be matched in OpenFlow 1.0

generalized forwarding are IP source address, TCP source port, and source MAC

Chapter 4 Problems

Problem 1

a) Data destined to host H3 is forwarded through interface 3

Problem 2

a) No, you can only transmit one packet at a time over a shared bus.

Problem 3

Problem 4

The minimal number of time slots needed is 3. The scheduling is as follows.

Slot 1: send X in top input queue, send Y in middle input queue.

Problem 5

a)

Prefix Match Link Interface

Problem 6

Destination Address Range Link Interface

00000000

through 0

00111111

Problem 7

Destination Address Range Link Interface

11000000

through (32 addresses) 0

Problem 8

223.1.17.0/26

Problem 9

Destination Address Link Interface

200.23.16/21 0

Problem 10

Destination Address Link Interface

Problem 11

Any IP address in range 128.119.40.128 to 128.119.40.191

Problem 12

From 214.97.254/23, possible assignments are

a) Subnet A: 214.97.255/24 (256 addresses)

b) To simplify the solution, assume that no datagrams have router interfaces as

ultimate destinations. Also, label D, E, F for the upper-right, bottom, and upper-

left interior subnets, respectively.

Router 1

Longest Prefix Match Outgoing Interface

Router 2

Longest Prefix Match Outgoing Interface

Router 3

Longest Prefix Match Outgoing Interface

Problem 13

The IP address blocks of Polytechnic Institute of New York University are:

The IP address blocks University of Washington are:

NetRange: 140.142.0.0 – 140.142.255.255

CIDR: 140.142.0.0/16

No, the whois services cannot be used to determine with certainty the geographical

location of a specific IP address.

Problem 14

The maximum size of data field in each fragment = 680 (because there are 20 bytes IP

Problem 15

MP3 file size = 5 million bytes. Assume the data is carried in TCP segments, with each

TCP segment also having 20 bytes of header. Then each datagram can carry 1500–

Problem 16

a) Home addresses: 192.168.1.1, 192.168.1.2, 192.168.1.3 with the router interface

being 192.168.1.4

b)

NAT Translation Table

WAN Side LAN Side

24.34.112.235, 4000 192.168.1.1, 3345

Problem 17

a) Since all IP packets are sent outside, so we can use a packet sniffer to record all IP

-8, 2002, Marseille, France.

–

Weifeng Chen, Yong Huang, Bruno F. Ribeiro, Kyoungwon Suh, Honggang Zhang,

Problem 18

It is not possible to devise such a technique. In order to establish a direct TCP connection

Problem 19

S2 Flow Table

Match Action

Ingress Port = 1; IP Src = 10.3.*.*; IP Dst = 10.1.*.*

Forward (2)

Ingress Port = 2; IP Src = 10.1.*.*; IP Dst = 10.3.*.*

Forward (1)

Ingress Port = 2; IP Dst = 10.2.0.4

Forward (4)

Ingress Port = 3

Forward (4)

Problem 20

S2 Flow Table

Match Action

Ingress Port = 3; IP Dst = 10.1.*.*

Ingress Port = 3; IP Dst = 10.3.*.*

Forward (2)

Problem 21

S1 Flow Table

Match Action

IP Src = 10.2.*.*; IP Dst = 10.1.0.1 Forward (2)

Problem 22

S2 Flow Table

Match Action

IP Src = 10.1.0.1; IP Dst = 10.2.0.3 Forward (3)

IP Src

= 10.3.0.6; IP Dst = 10.2.0.4

Forward (4)

S2 Flow Table

Match Action

IP Src =.*.*.*.*; IP Dst = 10.2.0.3; port = TCP

Forward (3)

IP Src =.*.*.*.*; IP Dst = 10.2.0.4; port = TCP

Forward (4)

S2 Flow Table

Match Action

IP Src = 10.1.0.1; IP Dst = 10.2.0.3; port = UDP

Forward (3)