4360 Final Exam Cards

EstimatedRTT Formula

((1 - a) x EstimatedRTT) + (SampleRTT x a)

A user requests a web page that consists of some text and three images. For this page, the client will send on request message and receive four response messages

False

Application layer is not accessible by the users of a computer

False

Encapsulation occurs at the receiver of a message

False

Generalized forwarding and destination based forwarding are synonymous

False

HOL blocking occurs in a router's output ports

False

HTTP is a push protocol while SMTP is a pull protocol

False

If an IPv6 datagram received by a router is too larger to be forwarded over the outgoing link, the router fragments the datagram

False

If two packets arrive at two different input ports of a router at the same time and these two packets are to e forwarded to the same output port of the router, then these two packets can be switched through the fabric at the same time only when a fabric is a crossbar

False

In IPv6 datagram structure, the fragmentation offset field is set a value in 8 byte blocks

False

In TCP 3-way handshake, when the segment with the ACK for the SYNACK is sent, no application data can be sent with this segment

False

In network security, DDoS stand for Distributed Duplication of Service

False

In store and forward packet switch -- formula is (P + N -1) (R/L)(P+N-1)RL

False

One of the golden rules in network traffic engineering is to design the system so that the traffic intensity is always greater than one

False

Only the client can intitate the closure of a TCP connection

False

Propagation delay is the time taken to put all bits of a packet on the link and transmission delay is the time taken by a bit to travel from source to destination

False

Queuing delays occur at the network edges

False

Ranking of home access technologies by speed from lowest t highest is: dialup, HFC, DSL, and FTTH

False

Reno is older than Tahoe

False

Root servers of the DNS are running on the premises of your ISP

False

Selective Repeat uses a single timer for timeouts

False

Suppose two packets arrive to two different input ports of a router at exactly the same time. Also suppose there are no other packets anywhere in the router. A shared bus switch fabric router can forward these two packets to different output ports at the same time.

False

TCAM returns the data at an address in the memory in variable time - that is, once it may return in 50 ns and the next time it may return the data in 100 ns

False

Tahoe supports fast recovery

False

The formula for traffic intensity is R/La

False

The transport layer is present in all devices in the core of the internet

False

Throughput and bandwidth are synonymous

False

To apply the link-state algorithm a node in the network needs to know the cost of connections to only its own neighbors and does not need the information of the cost of all links in the network

False

When Dijkstra's algorithm is applied at a node in a network, first obtain the networking table and then the routing table

False

When a large datagram is fragmented into multiple smaller datagrams, all the fragments follow the same route from source to destination.

False

When you read emails you use the SMTP protocol

False

Akamai is an example of third-party CDN

True

All computers on a subnet have the same subnet mask

True

Dijkstra's algorithm is an example of link-state routing algorithm

True

During the TCP slow start phase, the number of segments transmitted doubles every transmission round

True

Host A send two segments to host B, the first containing sequence number 92 and the second containing sequence number 100; this means that the first segment contained 8 bytes of data.

True

In GBN, if a timeout occurs, the sender resends all packets that have been previously sent but that have not yet been acknowledged

True

In IPv4 datagram, the header length field has a value of 5 for the standard header (without options)

True

In a network , we need to apply Dijkstra's algorithm at every node to get the routing and forwarding tables for each node

True

In a packet switched network, when a router in the path of a packet drops the packet then that packet is lost

True

In the first lab we saw that uttyler.edu used TLS; this stands for Transport Layer Security

True

Maximum payload in an IPv6 datagram is 65535 bytes

True

The DASH video streaming protocol encodes a video into several different versions with each version having a different bit rate

True

The first step in the DHP protocol is the DHCP server discovery step

True

The problem of oscillations can occur in the link state routing algorithm

True

The product of bandwidth and transmission delay for a link gives the number of bits currently on a link

True

There are 13 root server in the DNS system

True

Transport layer multiplexes several messages leaving the system and demultiplexes messages received from outside

True

Tunnel is used to connect IPv6 through IPv4

True

UDP has a 8-byte header

True

UDP stands for User Datagram Protocol

True

When a network drops packets due to congestion they are actually dropped at the input and output buffers of router interfaces due to the finite size of these buffers

True

Wireshark is a packet sniffer

True

A router with 48 intefaces needs only one P address for itself

False

How big is the IPv6 address space?

128 bit

IP overhead?

20 bytes

TCP overhead?

20 bytes

How big is the IPv4 address space?

32 bit

How big is the MAC address space?

48 bit

Propagation Delay Formula

D(length of physical cable)/s(propagation speed)

Packet Transmission Delay Formula

L(bits)/R(bits/sec)

Suppose datagrams are limited to 1500 bytes (including header) between source Host A and destination Host B Assuming a 20-byte IP header, how many datagrams would be required to send an MP3 consisting of 5 million bytes using UDP?

To solve: 20 byte IP header + 8 byte UDP header gives 1472 available application data bytes in each packet. 5,000,000 / 1472 = 3396.73, round up to 3397.

A web browser requests a 5000 byte HTML file over a 10Mbps link, RTT is 150ms and DNS lookup is 200ms. Ignore processing, queuing, assume browser doesn't know IP address.

To solve: ((File x byte-to-bit) / Link Rate) + (RTT x 2) + DNS Lookup ((5000 x 8)/10,000,000)+(.150 x 2)+.200

Consider transferring an enormous file of L bytes from Host A to Host B. Assume a MSS of 800 bytes. Given that L is the maximum number that can be supported by the TCP sequence field (4 bytes wide) how many TCP segments will be needed for transferring this file?

To solve: 4 bytes wide means 32 bits. Maximum number would then be 2^32 for file size, divide by 800 and round up. 2^32 / 800 = 5368709.12 = 5368710

1's Complement of sum of two bytes 11011010 and 01100101?

To solve: Add bits 11011010 + 01100101 = (1) 00111111, carry the 1 by adding from the right = 01000000, flip all digits = 10111111

A UDP segment is carrying 946 bytes of application data. What is the value in the length field of this segment in binary? If a TCP segment is carrying this same application data, what will be the length of the segment assuming no options?

To solve: Application data + UDP header (946 + 8 = 954). 0000 0011 1011 1010 in binary (must be 16 bit) Application data + TCP header (946 + 20 = 966)

Two hosts A and B are connected by link 56 kbps Separated by 1000km and propagation speed along link 2.5 x 10*8. A sends packet size 120 bits to B. Dprop in whole number ms?

To solve: D/S. ( 1,000,000m / 2.5 x 10*8 )

When a packet arrives at a router, one packet is halfway done and three are waiting to be transmitted. Packets are transmitted in order Suppose all packets are 1000 byes and ink rate is 10Mbps. What is queuing delay for the packet in milliseconds?

To solve: Keep in mind it is asking for queuing delay, or when the packet is question will arrive at the the front of the queue to be transmitted. The entire previous queue must first clear, so we must calculate how long it will take for the half packet + 3 packets to transfer. ((1/2P + P x 3) byte-to-bit) / Link Rate ((500 + 1000 x 3) 8) / 10,000,000

An IPv4 router that uses longest prefix matching has an entry in its forwarding table that reads "if the prefix is 1110 then forward to interface 3". If this is the only entry that forwards datagrams to interface 3, how many IP addresses will be forwarded by this router to interface 3?

To solve: Know that longest prefix matching means that in a 32 bit IP address, if the first 4 bits are 1110, all IP addresses from 1110 0000 00000000 00000000 00000000 to 1110 1111 11111111 11111111 11111111 will forward to interface 3. The easiest way to calculate this is to take all the free bits and plug them into 2^x (2^28), giving 268,435,456.

The IP address for a host is 192.169.225.87/26. What are the network address, broadcast address, and subnet mask? What is the maximum number of hosts that can be on this subnet?

To solve: Network Address: All bytes are calculated in powers of 2, from 0 to 256. .87 is between two of the powers, 64 (00100000) and 128 (01000000), and is part of the lesser power. Therefore, a .87 address will belong to the 12.169.255.64 Network Address. Broadcast Address: the greatest IP address that doesn't reach the next power. Since .128 would be the next power, the broadcast address will end in .127, so 192.169.255.127. Supported Hosts: 2^(32-26)-2 = 62. Follows formula 2^(32 - (number after slash in IP address)) - 2 Subnet Mask: 255.255.255.193. It's simply 255 - supported hosts in the last slot of 255.255.255.???

A DDoS attack uses a botnet to attack a target

True