Exams

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Image /page/0/Picture/1 description: The image shows the logo of TU Delft. The logo consists of a stylized flame above the letters 'TU' in black and blue, respectively, followed by the word 'Delft' in black.

Examiner responsible: Fernando Kuipers Examination reviewer: Otto Visser

2025-07-14

Please read the following information carefully! (The same information is on the exam and on the answer sheet)

• Opening the exam or answer sheet before you are instructed to start is strictly prohibited.
• This exam consists of 20 questions, worth 20 points in total. Partial points may be given. You need at least 58% of the points to pass.
• You have 120 minutes to complete this exam.
• Before you hand in your answers, check that your answer form contains your name and student number on every page, also filled in using the boxes.
• . Please check carefully which question should be answered where and stick to the indicated space.
• The use of books, notes, smart watches, and other aids is strictly prohibited. You may use a non-graphical calculator.
• With respect to the multiple-choice questions: You can select ONLY one answer. If you consider multiple or no options to be correct, then select the best (i.e., most specific) answer.
• . Any answer to an open question should be backed up by an explanation and, if applicable, the accompanying maths! Try to be short and concise. The provided answer space should be too big for most questions.
• Fill in the answer sheet with blue/black pen. If you make a mistake on the answer form, there is extra space on the back. If you use this: indicate this clearly in the original question box that your answer is or continues in the extra space.
• You can solve the exercises in any order, and the order of the exercises is not determined by their difficulty.

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These answers are not meant as model answers, but rather to give an idea of what is the correct solution. To get full points for the questions, we expect work shown and detailed explanations.

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1. (1 point) Which of the 7 layers of the OSI stack operate only between the end-nodes?

Solution: The transport layer, session layer, presentation layer, and application layer. All lower layers are needed on the other nodes along the path as well.

• 1. (1 point) Why is Wi-Fi using collision avoidance instead of collision detection?
  • A. It is better to avoid collisions than to detect them
  • B. Wi-Fi nodes cannot send and receive data at the same time
  • C. Because collision detection already takes places through ARQ
  • D. Because of the hidden node problem where certain hidden Wi-Fi nodes cannot be detected
• 1. (1 point) The IPv6 address space is x times larger than the IPv4 address space, where x equals:
  • A. 2 $B.4$
  • $C.96$
  • $D. 2^{96}$

Solution: IPv4 is 32 bits, IPv6 is 128 bits.

1. (1 point) Most SOHO (Small Office, Home Office) routers use NAT for IPv4. Using the terms internal/external and address/port, explain what NAT is doing.

Solution: NAT replaces the IP address and port number in packets and keeps a mapping of external port numbers to a combination of internal IP address and port number.

• 1. (1 point) When Gwen sends an email to Owen, she uses the following protocol:
  • A. POP3
  • B. IMAP
  • C. XMPP
  • D. SMTP

Solution: POP and IMAP are for receiving/reading email, XMPP is for instant messaging, SMTP is for sending.

1. (1 point) Assume byte stuffing is used with $FLAG = 011110$ and $ESC = 111000$ . The following bit sequence is received: 011110 110011 111000 111000 111000 011110 011110. Write down the original bit sequence (before byte stuffing was applied).

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Solution: 110011 111000 011110

1. (1 point) Assume a 4-port Ethernet switch, where the ports are labelled with identifiers 1, 2, 3, 4, respectively. Initially the switch table is empty. Assume now that the following packets arrive (in the displayed order) with source $s_x$ and destination $d_y$ on port |z: [ $s_2$ , | 2], [ , | 3], Give the switch table after the packets have been processed.

Solution: $s_2$ , port 2 $s_1$ , port 3 $s_4$ , port 3 , port 1 This replaces an earlier entry.

1. (1 point) Alice sends an IPv4 packet of size 12000 bytes. The packet passes through three networks: first network A, then B, and last C. Network A has a maximum transmission unit (MTU) of 6000 bytes, B has an MTU of 3000 bytes, and C has an MTU of 1500 bytes. Assume that there is no packet loss and that all fragments take the same route. How many fragments arrive at B, at C, and at the destination with and without MTU discovery (i.e., 6 answers)?

Solution: The simplified answer: 2 from A to B, 4 from B to C, 8 from C to destination without MTU discovery and 12000/1500=8 at B, 8 at C, and 8 at the destination with MTU discovery. A more accurate answer would take into account that IP headers occupy space and that the size of all but the last fragment needs to be divisable by 8.

1. (1 point) Alice wants to send the data 1001 1010 1011 1100 to Bob. What is the Internet checksum (1's complement) of this data, given that the block length is 4?

Solution: $1001+1010+1011+1100 = 101010 \rightarrow 10 + 1010 = 1100 \rightarrow 0011$

1. (1 point) Alice wants to send the data 010 100 111 to Bob. For error detection, they decided to use a CRC with the generator 1111. Compute the message that Alice sends.

Solution: 010 100 111 000 / 1111 gives remainder 001, so the message is 010 100 111 001

1. (1 point) Compute an (11,7) Hamming Code for 1111111 using odd parity.

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• Solution: $a b 1 c 1 1 1 d 1 1 1$ $a+1+1+1+1+1$ is odd -> is odd -> is odd c=0 is odd d=0 > 00101110111

Image /page/4/Figure/5 description: The image shows a graph with nine nodes labeled 1 through 9. The nodes are connected by edges with associated weights. Node 1 is connected to node 2 with a weight of 5 and to node 3 with a weight of 2. Node 2 is connected to node 3 with a weight of 1, to node 4 with a weight of 9, and to node 5 with a weight of 3. Node 4 is connected to node 7 with a weight of 1 and to node 6 with a weight of 2. Node 5 is connected to node 6 with a weight of 1. Node 6 is connected to node 8 with a weight of 5 and to node 9 with a weight of 3. Node 7 is connected to node 8 with a weight of 10.

1. (1 point) Raya wants to send data at a rate of 15 bytes per second to other members in her party. They are using a protocol which allows packets to remain in the network for 10 seconds, and a clock that advances by $2^{12}$ ticks a second. The initial sequence number Raya picks is 2, and she knows that the sequence numbers of her packets would enter the forbidden region in approximately 22.08 seconds. How many bits does the protocol use for the sequence numbers?

Solution: 15 \times 22.08 = 2^{12} \times (22.08 + 10) - 2^b331.2 = 131399.68 - 2<sup>b</sup>b=17

1. (1 point) A few questions ago, Alice sent an IPv4 packet that got fragmented. She is now sending a new one with a TCP segment in it. This one gets fragmented in 3 packets (X, Y, and Z). They arrive in this order: Z, X, Y. What are the earliest moments that the receiver can check each of the checksums in the TCP and IP headers?

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Solution: After receiving: Z: check IP checksum of packet Z X: check IP checksum of packet X Y: check IP checksum of packet Y and check TCP checksum of reassembled segment.

• 1. (1 point) Damla is communicating with Elvan over the Internet. The connection seems to be quite slow, so she wants to know whether the packets are routed via satellite. Damla sends one ICMP Echo (ping) to Elvan. Elvan also sends one ICMP Echo to Damla. Damla sees a time of 866 ms; Elvan sees a time of 42 ms. For each of the following, state why they collectively (so with this combined information) can or cannot draw any decisive conclusion on:
  • (a) whether there is a satellite used in the route
  • (b) processing time at Elvan
  • (c) Elvan has a faster Internet subscription than Damla
  • (d) the route Damla -> Elvan is not using the same hops as the route Elvan -> Damla

Solution:

• 1. one echo can not reveal the route
• 1. echo gives total RTT, this cannot be split in processing time and delay
• 1. one RTT measurement can not say anything about speed
• 1. we do not know the route from an echo

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1. (1 point) Enna and Snuffles are still using a sliding window protocol, this time with k=4, for their communication. They send 300-byte frames. The header and trailer together make up $x$ fraction of these 300 bytes. They furthermore have a duplex channel that can send up to 1500 bytes per second in each direction. The latency from Enna to Snuffles is 0.5 seconds, whereas the latency from Snuffles to Enna is 0.55 seconds. Assume that the channel is error- and jitter-free, the acknowledgements are very short, and processing time can be ignored. Furthermore, assume Enna always sends data when the protocol allows it.

If the goodput from Enna to Snuffles is 816 bytes per second, what is the value of $x$ ? Include all steps of the computation.

Solution: Solution: 0.15.

We compute the throughput and then compare to the goodput. First, compute the time to transmit one frame, which is $300/1500 = 0.2$ seconds. Second, compute the time until the first ACK is received, which is $300/1500 + 0.5 + 0.55 = 1.25$ seconds. Last, check whether that is less than the time to transmit 4 frames, i.e., $4 \cdot 300/1500 = 0.8$ seconds. The first ACK only arrives after the 4 frames have been sent, so we get the throughput by dividing the number of bytes in the 4 frames (namely ) by the time needed to receive the first ACK: . However, x fraction of the sent data is overhead, so we need to multiply the above value with 1-x to obtain the goodput of 816. This means 1-x is , and x is 0.15

• 1. (1 point) Leo recently became the head of a rather big organisation. He decided that the internal data should be protected better, so he ordered to have a stateful firewall installed to monitor all traffic coming in from the outside. It uses signature-based detection.
  • (a) What is an advantage of signature-based detection?
  • (b) Klaus claims the biggest threat comes from the inside of the organisation and this firewall is pointless. He wants to add an anomaly-based detection firewall for connections from inside the organisation to the outside world. Instead of using TCP reset, he wants this firewall to just discard. What is a disadvantage of discarding?

Solution:

• (a) you would be able to detect any threat that the firewall knows
• (b) if the packets get discarded, TCP will retransmit, so you'll have to discard multiple times.
• 1. (1 point) Zirconia wants to send a message to Raven through TOR. Explain the encryption process Zirconia has to complete in order to send the message.

Solution: The message is encrypted with the key of every relay on the way, starting with the last relay (reverse order). This ensures that every relay only learns the information of what needs to be sent as the next step in the process.

1. (1 point) Orsik and Ath did not like using an S-box, so these days they use a P-box to encrypt their communication. The P-box maps the bit in position 0 to the bit in position 1, the bit in position 1 to the bit in position 2, the bit in position 2 to the bit in position 3, and the bit in position 3 to the bit in position 0. Bits are counted from the left starting at 0. As encryption mode, they use CBC.

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Orsik receives the cyphertext 0110 1110, and the IV 1011. Derive the plaintext. Include intermediary steps.

Solution: Solution: 0111 1011. Make sure to decrypt and not encrypt. Note that when decrypting the first bit becomes the fourth, the second becomes the first, the third becomes the second, and the fourth the third, i.e., the inverse mapping to the one described in the exercise.

Take the first 0110 and apply the inverse P-box to get 1100. As the initial vector has been XOR-ed to the first plaintext block, we retrieve the plaintext block by XOR-ing 1100 with initial vector 1011 and get 0111. For the second block 1110, we get 1101 after applying the P-box. We now XOR the first cyphertext block, 0110, to get 1011.

• 1. (1 point) In secure communications, one often speaks of CIA: Confidentiality, Integrity, and Availability. What does "Integrity" refer to?
  • A. Ensure that only authorised parties are able to use the network.
  • B. Ensure that only authorised parties can access information.
  • C. Ensure that information and services are available when needed.
  • D. Ensure that users can verify that they have the correct information.

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Question 1

• 3 points: Application layer
• 2 points: Session and presentation layer
• 3 points: Transport layer
• 2 points: Explanation
• -2 points: One wrong layer

Question 2

$\bullet$ 10 points: B

Question 3

$\bullet$ 10 points: D

Ouestion 4

• 8 points: Map internal IP + port to external port
• 5 points: Map internal IP to external port (+ address) (forgot internal or external port)
• 2 points: Multiple internal addresses are "mapped" to one external address
• 2 points: Replace internal IP (and port) in outgoing packet with external IP (and port) (and vice versa)

Question 5

$\bullet$ 10 points: D

Ouestion 6

• 3 points: Remove start and end flag
• 4 points: Remove esc flags, keep escaped bytes
• 3 points: Mentions which specific bytes to remove and why

Question 7

• 3 points: Table mapping address to port
• 2 points: s1 and s4 $\rightarrow$ 3
• 3 points: $s2 \rightarrow 1$

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• 2 points: Explain at least why s2 not on port 2

Question 8

Without MTU discovery:

• 2 points: $A \rightarrow B$ : 2
• 2 points: $B \rightarrow C$ : 4
• 2 points: $C \rightarrow$ destination: 8
• -3 points: No explanation/calculation
• -2 points: Incorrect calculation method (e.g., instead of '4 $\times$ 3000 / 1500', '12000 / 1500' is calculated)

With MTU discovery:

• 2 points: $12000 / 1500 = 8$
• 2 points: All: 8

Question 9

• 4 points: Add message in blocks
• 4 points: Add overflow
• 2 points: Take 1's complement
• -2 points: Minor (math) mistake

Ouestion 10

• 10 points: Correct answer with provided working out
• 5 points: Correct approach, small mistake, consistent conclusion
• 3 points: Correct answer, incomplete explanation/working out

Question 11

• 10 points: Correct + good explanation
  • Indication of what is a parity bit
  • How they are filled (table/adding them up, etc.)
• 7 points: Correct + incomplete explanation (e.g., dots & lines as explanation, or some indication of what was done but not in full)
• 5 points: Just the answer, but no justification $\bullet$
• 7 points: One wrong/small mistake + good explanation

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• 2 points: One wrong/mistake $+$ incomplete explanation
• 5 points: Flipped parity + good explanation

Question 12

• 7 points: Correct costs
• 3 points: Correct lines
• -2 points: Provided path instead of line
• -2 points: One row wrong
• -4 points: Two rows wrong

Question 13

• 10 points: Correct answer with working out
• 8 points: Minor mistake in provided answer

Ouestion 14

• 5 points: IP checksum at arrival of each fragment
• 5 points: TCP checksum: after reassembly of all fragments

Question 15

• 2 points: a: NO: one ping is not a traceroute
• 3 points: b: NO: round trip time cannot be split in delays and processing time
• 3 points: c: NO: delay and speed are not the same
• 2 points: d: NO: one ping does not reveal the route

Question 16

• 2 points: Time to transmit one frame
• 2 points: Time for first ACK
• 2 points: Time for transmit 4 frames
• 2 points: Correct throughput
• 2 points: Correct calculation of x
• -2 points: Minor mistake or terminology error

Ouestion 17

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A:

• 5 points: Detect any threat of which the signature is known $\bullet$
• 5 points: Faster than anomaly-based detection $\bullet$
• O points: Confused threat signature with encryption

B:

• 5 points: TCP will retransmit, so discard multiple times $\bullet$
• 5 points: TCP connection stays open
• 3 points: Explained risk of discarding false positives, not discard vs reset

Question 18

• 3 points: Encrypt for each relay at the start
• 2 points: With (not public) key of each relay
• 5 points: Starting with last relay/each relay removes one layer $\bullet$
• -2 points: Additional incorrect information

Question 19

• 10 points: Correct with explanation
• 8 points: Minor mistake with explanation
• 5 points: Encrypting P-box instead of decrypting, calculations good $\bullet$
• 5 points: Wrong order (XOR then P-box), calculations good
• 5 points: XOR with wrong thing, calculations good $\bullet$

Question 20

10 points: D $\bullet$

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Image /page/12/Picture/1 description: The image shows the logo for TU Delft. The logo consists of the letters "TU" in bold, with the "U" in a lighter blue color. Above the letters is a stylized flame. To the right of the letters is the word "Delft" in black.

Examiner responsible: Fernando Kuipers Examination reviewer: Otto Visser

2025-06-23

Please read the following information carefully! (The same information is on the exam and on the answer sheet)

• Opening the exam or answer sheet before you are instructed to start is strictly prohibited.
• This exam consists of 20 questions, worth 20 points in total. Partial points may be given. You need at least 58% of the points to pass.
• You have 120 minutes to complete this exam.
• Before you hand in your answers, check that your answer form contains your name and student number on every page, also filled in using the boxes.
• . Please check carefully which question should be answered where and stick to the indicated space.
• The use of books, notes, smart watches, and other aids is strictly prohibited. You may use a non-graphical calculator.
• With respect to the multiple-choice questions: You can select ONLY one answer. If you consider multiple or no options to be correct, then select the best (i.e., most specific) answer.
• . Any answer to an open question should be backed up by an explanation and, if applicable, the accompanying maths! Try to be short and concise. The provided answer space should be too big for most questions.
• Fill in the answer sheet with blue/black pen. If you make a mistake on the answer form, there is extra space on the back. If you use this: indicate this clearly in the original question box that your answer is or continues in the extra space.
• You can solve the exercises in any order, and the order of the exercises is not determined by their difficulty.

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These answers are not meant as model answers, but rather to give an idea of what is the correct solution. To get full points for the questions, we expect work shown and detailed explanations.

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• 1. (1 point) Give the correct name of what each OSI layer is sending. Choose from the following options:
  • $bit$
  • container
  • $-$ frame
  • message
  • · packet
  • segment
  • $-$ unit
  • $\bullet$ Or: N/A if the layer does not have a specific name for what it is sending.

Solution: physical: bit data link: frame network laver: packet transport layer: segment session, presentation, and application layers do not have a specific name for what is being transmitted.

• 1. (1 point) Different protocols can be used for channel allocation. The ALOHA protocol:
  • A. waits until the channel is idle and then sends the data.
  • B. checks if the channel is busy, if it is then it waits a random amount of time otherwise it transmits the data.
  • C. tries to transmit the data, if a collision occurred it retries after a random delay.
  • D. tries to transmit the data at the beginning of a time slot, if a collision occurred it retries after a random delay.
• 1. (1 point) Assume communication between two machines (A and B). The time between the moment a very short message leaves machine A until it arrives at machine B is called:

A. The propagation delay

• B. The queuing delay
• C. The transmission delay
• D. The round-trip time
• 1. (1 point) Assume a wireless setting with CSMA/CA and 3 stations A, B, C, that all want to send a message to a 4th station D. All 4 stations are in range of each other. At t=0 station A sends its message which lasts 8s, followed immediately by an ACK from D of 2s. B has computed a backoff of 20s and C has computed a backoff of 15s. The messages from both B and C also last 8s and the ACK also immediately follows and lasts 2s. At what time can B begin to send its message?

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• 1. (1 point) Why does Ethernet prescribe a minimum frame size?
  • A. To make sure users only send when they have accumulated data
  • B. To make sure the header does not exceed the payload, leading to inefficiency
  • C. Else potential collisions might go undetected
  • D. Else IP packets would have to be fragmented in too many parts
• 1. (1 point) When Explicit Congestion Notification (ECN) is used and a router along the path experiences congestion, it:
  • A. sends an explicit message to the source to reduce the sending rate
  • B. randomly drops packets, causing the source to lower its TCP window and hence sending rate
  • C. marks the packets and then the destination receiving the packet(s) signals the source to reduce its sending rate
  • D. notifies the previous router to detour packets to a different path
• 1. (1 point) (a) What information can be gathered by using DHCP (mention at least 2 things) (b) What information is provided to the DHCP server to get this?

Solution: gathered: IP address, gateway, DNS server, clock server, and more To be provided : MAC address

1. (1 point) A message 1 0010 0001 has been received that contains a Cyclic Redundancy Check (CRC) at the end that was made with generator $G = 1101$ . Has the message been received correctly?

Solution: CRC = 1 0010 0001 / 1101 = 11 1101 with remainder 000, which means the message was received correctly.

• 1. (1 point) During a recent meeting in Rome, some people were worried that a voting process would be influenced by external sources. To prevent that, they installed wireless (WiFi, GSM, Bluetooth, etc.) jammers and made sure the only possible communication was with smoke signals.
  • Participant A thinks the jammers work on OSI layer 7 and can be circumvented by using VPN.
  • Participant B thinks the jammers work on OSI layer 4 and can be circumvented by SSL (like using HTTPS instead of HTTP).
  • Participant C thinks the jammers work on OSI layer 1 and all hope of communication with the outside is lost.

Purely from a computer networks perspective: which candidate is more correct and as such should get the votes?

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Solution: Layer 7 is the application layer. If they are jamming the application, they would have access to either the senders device, or the recipient or some server in between. In either case, VPN would encode the data to that place, but would not prevent a jamming operation on the application layer. Layer 4 (transport layer) jammer would block based on e.g. IP address and/or port number. It is possible they would allow port 443 (https) and block port 80 (http), but that would be extremely unlikely. Jamming WiFi and GSM at the physical layer (1) sounds like a likely scenario and then indeed all hope of communication is lost.

1. (1 point) Isaac has decided to sneak off without waiting for his friends. After quite a while of walking, he finally decides to send them a message containing the directions he's taken, in hopes that they will come find him. To ensure the directions are more likely to be correct when his friends get them, he decides to convert his message into a (19, 14)-Hamming code. What is the Hamming code (message $+$ parity bits) he should send when encoding the following message using odd parity: 0100 1011 1000 11.

Enumerate the bits from the left, starting at 1.

Solution:
           M: 0100 1011 1000 11 -> ab0c 100d 1011 100e 011
a+0+1+0+1+1+1+0+0+1 is odd -> a=0
b+0+0+0+0+1+0+0+1+1 is odd -> b=0c+1+0+0+1+1+0+0 is odd \rightarrow c=0d+1+0+1+1+1+0+0 is odd \rightarrow d=1
e+0+1+1 is odd \rightarrow e=1Solution: 0000 1001 1011 1001 011

1. (1 point) Router D receives the following distance vectors from its neighbours: A, C and F.

Costs are symmetric. You can assume that the transmission of distance vectors is error-free. If there are multiple options for an outgoing line in a routing table, routers choose the alphabetically first neighbour. Give D's routing table; no further explanation is required.

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• 1. (1 point) The UDP header is:
  • A. 4 bytes long
  • B. 8 bytes long
  • C. 20 bytes long
  • D. of variable length
• 1. (1 point) Alice starts a TCP handshake with Bob using the initial sequence number 101. After receiving Bob's SYN-ACK and his choice of initial sequence number, 404, Alice sends three segments of size 12, 34, and 56 bytes. Which sequence number will the fourth segment have?

Solution: $101 + 1 + 12 + 34 + 56 = 204$

1. (1 point) The wireless encryption protocol uses a 64-bit key with 24-bit initialisation vector. Many routers allowed users to provide this key by typing 5 characters. Explain why this is problematic.

Solution: The 64-bit key with 24-bit IV means we need to provide bits for a key of 40 bits or 5 bytes. But 5 ASCII chars do not use the full range of 256 options per byte, most people only use 26 (or double that if they use capitals) $+$ 10 numbers $+$ maybe 10-ish special characters. In any case, not more than half of the possible range.

1. (1 point) QUIC was introduced by Google to solve the problem of head-of-line-blocking. Which protocol was suffering from this, and how does QUIC fix this?

Solution: The problem of head-of-line blocking is present in TCP and QUIC solves this by having multiple independent streams over UDP. This way e.g. a video stream is not blocked by a slower part of the webpage like comments or advertisements.

1. (1 point) Enna and Snuffles use a sliding window protocol with $k=7$ for their communication. They always send 200-byte frames. The header and trailer together make up 40 of these 200 bytes. They furthermore have a duplex channel that can send up to 1300 bytes per second in each direction. The latency from Enna to Snuffles is 0.5 seconds, whereas the latency from Snuffles to Enna is 0.6 seconds. Assume that the channel is error- and jitter-free, the acknowledgements are very short, and processing time can be ignored. Furthermore, assume Enna always sends data when the protocol allows it.

What is the goodput from Enna to Snuffles in bytes per second? Include all steps of the computation. (You may round to two decimal digits)

Solution: Solution: 896 bytes/second.

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We compute the throughput and then the goodput. First, compute the time to transmit one frame, which is $200/1300 \approx 0.15$ seconds. Second, compute the time until the first ACK is received, which is 200/1300 + 0.5 + 0.6 $\approx$ 1.25 seconds. Last, check whether that is less than the time to transmit 7 frames, i.e., $7 \cdot 200/1300 \approx 1.08$ seconds. The first ACK only arrives after the 7 frames have been sent, so we get the throughput by dividing the number of bytes in the 7 frames (namely $$ ) by the time needed to receive the first ACK: . However, 20% of the sent data is overhead, so we need to multiply the above value with 0.8 to obtain the goodput, which gives 896 bytes/second.

1. (1 point) Zac chooses $p = 3$ and $q = 11$ for RSA. Furthermore, his public key part e is equal to $7$ and his secret key part $d$ is 3. Otto sends the ciphertext 14 to Zac. What is the plaintext?

Solution: Plaintext: 5. $n = p \cdot q = 3 \cdot 11 = 33.$ $c^d \mod n = 14^3 \mod 33 = 5.$

1. (1 point) Assuming the Chord Distributed Hash Table is used with m=5, i.e.: the Chord ring has $2^m$ locations. Compute for the following ring (with users marked in blue) the finger table of node 24.

Image /page/18/Figure/6 description: The image shows a circular diagram with several gray circles arranged in a ring. There are five blue circles with numbers next to them: 30, 3, 7, 17, and 24. An arrow points to the top of the circle, labeled as 'Location 0'.

Figure 1: Chord ring

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1. $(1 \text{ point})$ Table $\overline{1}$ displays the Base64 conversion table.

Table 1: Base64 conversion table

Kilwic desperately needs to send a binary message to his friend, Taremin, who is across the river. Sadly, he doesn't have time to cross the river himself. Tethis the boatman offers to deliver the message for him, however, Tethis does not have the best memory and will not remember the full binary message. Therefore, Kilwic decides to tell him the base64 encoding of the message, in hopes that this is short enough for Tethis to remember. What is the encoding Kilwic should tell Tethis, if the message that should be delivered is:

01010100 01101111 01110111 01100101 01110010

Solution: VG93ZXI=

1. (1 point) Orsik and Ath use an S-box to encrypt their communication. The S-box is secret and only known by Orsik and Ath. No additional key is used. The S-box they use takes an integer between 0 and 3 as input and maps 0 to 0, 1 to 3, 2 to 1, and 3 to 2. Orsik wants to send the message 01 11 and uses CBC mode with initial vector 10 to encrypt it. Derive the ciphertext. Include intermediary steps. The integers 0, 1, 2, and 3 correspond to the binary 00, 01, 10, and 11, respectively.

Solution: ciphertext is 10 11. XOR 01 with 10: 11. S-box on 11 (3 in decimal): 10 (2 in decimal). XOR 11 with 10: 01. S-box on 01 (1 in decimal): 11 (3 in decimal).

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Question 1

• 10 points: 4 correct answers
• 7 points: 3 correct answers
• -3 points: 1 extra answer
• -7 points: 2 extra answers or 1 invalid OSI layer
• O points: Incorrect too many incorrect/extra answers/invalid OSI layers
• 3 points: Two options were swapped

Question 2

$\bullet$ 10 points: $\subset$

Question 3

$\bullet$ 10 points: A

Question 4

• 10 points: Correct
• 5 points: Only a small mistake (e.g., "C starts at 15")
• O points: Correct number but wrong reasoning (e.g., "B backs off again")

Question 5

$\bullet$ 10 points: $\circ$

Question 6

$\bullet$ 10 points: $\subset$

Question 7

• 5 points: a) correct (2 things)
• 5 points: b) correct

Question 8

• 10 points: Correct with good reasoning/mathematics

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• 5 points: Correct approach, small mistake, consistent conclusion

Ouestion 9

• 8 points: C correctly explained
• 2 points: C incorrectly explained
• 1 point: B correctly explained
• 1 point: A correctly explained

Ouestion 10

• 10 points: Correct + good explanation
  • Indication of what is a parity bit
  • How they are filled (table/adding them up, etc.)
• 7 points: Correct + incomplete explanation (e.g., dots & lines as explanation, or some indication of what was done but not in full)
• 5 points: Just the answer, but no justification
• 7 points: One wrong/small mistake + good explanation
• 2 points: One wrong/mistake + incomplete explanation
• 5 points: Flipped parity + good explanation

Ouestion 11

• 2 points: A correct
• 2 points: B correct
• 2 points: C correct
• 2 points: E correct
• 2 points: F correct
• -5 points: Provided paths

Ouestion 12

$\bullet$ 10 points: B

Ouestion 13

• 10 points: Correct, with explanation
• 5 points: Correct reasoning, small mistake

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Question 14

• 3 points: 40 bit key
• 7 points: 5 characters do not give the full range
• 0 points: Brute forcing mentioned without further explanation

Question 15

• 5 points: Correct protocol (TCP)
• 5 points: Correct reasoning

Question 16

• 2 points: Time to send a frame
• 2 points: Time to send 7 frames
• 2 points: Time until first ACK
• 2 points: Correct throughput
• 2 points: Correct goodput

Ouestion 17

• 10 points: Correct with explanation
• 5 points: Correct approach, small mistake

Question 18

• 5 points: Correct, but no explanation
• O points: Correct, with explanation:
  • 2 points: $i=0$
  • 2 points: $i=1$
  • 2 points: $i=2$
  • 2 points: $i=3$

Question 19

• 10 points: Correct with explanation
• 5 points: Correct approach, small mistake

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• 5 points: Correct, but no explanation

Question 20

• 10 points: Correct with explanation
• 5 points: Correct CBC order, small mistake; often 1000 (XOR second part with 11, not with 10)
• 5 points: Wrong order (S-box and then XOR), calculations good. 0111 (always S-box M) or 0011 (always S-box IV/C1)

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Image /page/24/Picture/1 description: The image shows the logo for TU Delft. The logo consists of the letters "TU" in bold, with the "U" in a light blue color. To the left of the "TU" is a black flame. To the right of the "TU" is the word "Delft" in black.

Examiner responsible: Fernando Kuipers Examination reviewer: Otto Visser

2024-07-16

Please read the following information carefully! (The same information is on the exam and on the answer sheet)

• Opening the exam or answer sheet before you are instructed to start is strictly prohibited.
• This exam consists of 20 questions, worth 23 points. Partial points may be given. You need at least 58% of the points to pass.
• You have 120 minutes to complete this exam.
• Before you hand in your answers, check that your answer form contains your name and student number on every page, also filled in using the boxes.
• . Please check carefully which question should be answered where and stick to the indicated space.
• The use of books, notes, smart watches, and other aids is strictly prohibited. You may use a non-graphical calculator.
• With respect to the multiple-choice questions: You can select ONLY one answer. If you consider multiple or no options to be correct, then select the best (i.e., most specific) answer.
• Any answer that you provide to an open question should be backed up by an explanation and the accompanying maths (if applicable)!
• Fill in the answer sheet with blue/black pen. If you make a mistake on the answer form, there is extra space on the back. If you use this: indicate this clearly in the original question box that your answer is or continues in the extra space.
• You can solve the exercises in any order, and the order of the exercises is not determined by their difficulty.

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These answers are not meant as model answers, but rather to give an idea of what is the correct solution. To get full points for the questions, we expect work shown and detailed explanations.

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1. (1 point) The OSI model is split in multiple layers. Explain in just a few sentences the advantage of such a layering.

Solution: Important points are:

• Details of the working of lower layers are abstracted away from higher layers. This makes it possible to write e.g. applications without having to know whether the connection is WiFi or Ethernet, tunneled or not, etc. Similarly: the lower layers don't need to know what is happening on higher layers.

• Protocols and even hardware can be changed/swapped, only affecting the layers above and helow

• 1. (1 point) TCP over IP is:
  • A. A connection-oriented protocol over a connection-oriented protocol
  • B. A connection-oriented protocol over a connection-less protocol
  • C. A connection-less protocol over a connection-oriented protocol
  • D. A connection-less protocol over a connection-less protocol
• 1. (1 point) Assume communication between two machines (A and B). Transmission delay is then defined as:
  • A. The time between the moment a very short message leaves machine A until it arrives at machine B.
  • B. The time between machine A sending a very short message to machine B and getting a very short response back.
  • C. The time that it takes machine A to transmit a message, i.e., the time between when the first bit leaves machine A to when the last bit leaves machine A.
  • D. The time that a packet waits in the queue before it can be transmitted to machine B.
• 1. (1 point) In which case would it be better to use the Go-Back-N protocol than the Selective Repeat protocol?
  • A. Abundant bandwidth but limited buffer at the receiver.
  • B. Limited bandwidth but large buffer at the receiver.
  • C. A very lossy (e.g., radio) channel.
  • D. Frames have very different sizes.

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Solution: Flags at start and end; escaping the flag and escape byte already in the message. 01111110 11111111 11100000 11100000 11100000 0111111 00000111 11101010 01111110

• 1. (1 point) Why is slotted ALOHA twice as efficient as pure ALOHA?
  • A. In slotted ALOHA each sender is assigned a dedicated slot in which it can send, which avoids the collisions of pure ALOHA.
  • B. The RTS and CTS messages used in slotted ALOHA (and not in pure ALOHA) prevent that others start transmitting when a transmission is ongoing.
  • C. In slotted ALOHA the messages are longer than in pure ALOHA, which leads to a higher efficiency.
  • D. In slotted ALOHA messages are send at the start of fixed time slots, which reduces the chance of collisions compared to pure ALOHA.
• 1. (1 point) With binary exponential backoff, after the 4th collision, a station chooses to back-off for a time that is uniformly and randomly selected from the range $[0, x]$ timeslots. What is the value of x here?

Solution: $2^4 - 1 = 15$

• 1. (1 point) Why does Ethernet require that frames are not shorter than a prescribed minimum frame size (72 bytes)?
  • A. To avoid that a sender finishes sending before receiving a potential collision signal.
  • B. To have high transmission efficiency.
  • C. To accommodate the 48-bit MAC addresses.
  • D. To, besides the header, also have room for the (CRC) trailer.
• 1. (1 point) Assume A knows its cost to its neighbours and costs are symmetric, i.e., the same in both directions. Initially the routing table of A is empty. Then A receives the distance vector [A:1, B:-, C:2, D:5, E:6] from its neighbour B and the distance vector [A:5, B:6, C:7, D:4, E:-] from its neighbour E. (Notation X:Y indicates a cost of Y to reach X). Based on this information, construct the routing table of A.