Chapter 4 Review Questions
1. A network-layer packet is a datagram. A router forwards a packet based on the
2. Datagram-based network layer: forwarding; routing. Additional function of VC-based
network layer: call setup.
3. Forwarding is about moving a packet from a router’s input port to the appropriate
5. Single packet: guaranteed delivery; guaranteed delivery with guaranteed maximum
delay. Flow of packets: in-order packet delivery; guaranteed minimal bandwidth;
6. Interactive live multimedia applications, such as IP telephony and video conference,
could benefit from ATM CBR’s service, which maintains timing.
7. With the shadow copy, the forwarding lookup is made locally, at each input port,
8. Switching via memory; switching via a bus; switching via an interconnection
9. 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
10. 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
11. HOL blocking: Sometimes the 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
13. 11011111 00000001 00000011 00011100.
15. 8 interfaces; 3 forwarding tables.
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. Typically the wireless router includes a DHCP server. DHCP is used to assign IP
19. 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
21. Link state algorithms: Computes the least-cost path between source and destination
using complete, global knowledge about the network. Distance-vector routing: The
22. Routers are organized into autonomous systems (ASs). Within an AS, all routers run
the same intra-AS routing protocol. The problem of scale is solved since an router in
23. No. Each AS has administrative autonomy for routing within an AS.
24. No. The advertisement tells D that it can get to z in 11 hops by way of A. However, D
25. With OSPF, a router periodically broadcasts routing information to all other routers in
the AS, not just to its neighboring routers. This routing information sent by a router
27. Policy: Among ASs, policy issues dominate. It may well be important that traffic
originating in a given AS not be able to pass through another specific AS. Similarly, a
given AS may want to control what transit traffic it carries between other ASs. Within
an AS, everything is nominally under the same administrative control and thus policy
28. ISP C can use the BGP Multi-Exit Descriptor to suggest to ISP B that the preferred
route to ISP D is through the east coast peering point. For example, the east coast
29. A subnet is a portion of a larger network; a subnet does not contain a router; its
boundaries are defined by the router and host interfaces. A prefix is the network
30. Routers use the AS-PATH attribute to detect and prevent looping advertisements;
they also use it in choosing among multiple paths to the same prefix. The NEXT–
31. A tier-1 ISP B may not to carry transit traffic between two other tier-1 ISPs, say A
32. N-way unicast has a number of drawbacks, including:
33. a) uncontrolled flooding: T; controlled flooding: T; spanning-tree: F
b) uncontrolled flooding: T; controlled flooding: F; spanning-tree: F
35. IGMP is a protocol run only between the host and its first-hop multicast router.
IGMP allows a host to specify (to the first-hop multicast router) the multicast group it
36. In a group-shared tree, all senders send their multicast traffic using the same routing
tree. With source-based tree, the multicast datagrams from a given source are routed
Chapter 4 Problems
Problem 1
a) With a connection-oriented network, every router failure will involve the routing of
that connection. At a minimum, this will require the router that is “upstream” from
the failed router to establish a new downstream part of the path to the destination
node, with all of the requisite signaling involved in setting up a path. Moreover, all
of the routers on the initial path that are downstream from the failed node must take
down the failed connection, with all of the requisite signaling involved to do this.
Problem 2
a) Maximum number of VCs over a link = 28 = 256.
b) The centralized node could pick any VC number which is free from the set
Problem 3
Problem 4
a) Data destined to host H3 is forwarded through interface 3
Destination Address Link Interface
H3 3
b) No, because forwarding rule is only based on destination address.
Problem 5
a) No VC number can be assigned to the new VC; thus the new VC cannot be
Problem 6
In a virtual circuit network, there is an end-to-end connection in the sense that each router
Problem 7
a) No, you can only transmit one packet at a time over a shared bus.
Problem 8
Problem 9
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 10
a)
Prefix Match Link Interface
Problem 11
Destination Address Range Link Interface
00000000
through 0
00111111
Problem 12
Destination Address Range Link Interface
11000000
through (32 addresses) 0
Problem 13
223.1.17.128/25
Problem 14
Destination Address Link Interface
200.23.24/24 1
Problem 15
Destination Address Link Interface
Problem 16
128.119.40.112/28
Problem 17
a) Subnet A: 214.97.255/24 (256 addresses)
Subnet B: 214.97.254.0/25 – 214.97.254.0/29 (128-8 = 120 addresses)
Subnet C: 214.97.254.128/25 (128 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 18
The IP address blocks of Polytechnic Institute of New York University are:
The IP address blocks Stanford University are:
The IP address blocks University of Washington are:
No, the whois services cannot be used to determine with certainty the geographical
location of a specific IP address.
Locations of the Web server at Polytechnic Institute of New York University is
Locations of the Web server Stanford University is
Locations of the Web server at University of Massachusetts is
Problem 19
The maximum size of data field in each fragment = 680 (because there are 20 bytes IP
Problem 20
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 21
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, 4001 192.168.1.1, 3346
24.34.112.235, 4003 192.168.1.2, 3446
24.34.112.235, 4005 192.168.1.3, 3546
Problem 22
a) Since all IP packets are sent outside, so we can use a packet sniffer to record all IP
packets generated by the hosts behind a NAT. As each host generates a sequence of
IP packets with sequential numbers and a distinct (very likely, as they are randomly
For more practical algorithms, see the following papers.
“A Technique for Counting NATted Hosts”, by Steven M. Bellovin, appeared in
b) However, if those identification numbers are not sequentially assigned but randomly
Problem 23
It is not possible to devise such a technique. In order to establish a direct TCP connection
Problem 24
Problem 25
x to z:
x-y-z, x-y-w-z,
x-w-z, x-w-y-z,