Deferred exam

CITS3002: Deferred Mid-Semester Test (2026) - Explanations & Study Guide

This note breaks down the official 2026 Deferred Mid-Semester Test. This version of the test features some notoriously tricky edge-cases (especially Question 3). Under each question, you will find the official solution followed by a “How to understand this” section that explains the exact steps and logic.

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Question 1: CRC Error Checking at Receiver (10%)

Question: Consider a CRC error detection scheme with a divisor of 1010. If the received data is 10110011001, determine whether any error has occurred during transmission.

Official Solution:

      1010101
 1010|10110011001
      1010
      ----
      0001001
         1010
         ----
         001110
           1010
           ----
           01000
            1010
            ----
            00101

Since the remainder is not 000, the receiver detects an error.

How to understand this: Unlike the sender, the receiver does NOT add padding. It simply takes the exact string it received from the wire and divides it by the divisor using XOR.

• If the remainder at the very end is zero, the data is flawless.
• If the remainder is anything other than zero (in this case, it’s 101), bits were flipped during transmission and the receiver flags an error.

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Question 2: Hamming Code Construction (20%)

Question: Given the data bits 11011011 (8 bits): a) How many redundant (parity) bits are required to detect and correct a single-bit error? b) Using even parity, construct the Hamming codeword.

Official Solution: a) $(8+r+1)\le 2^r$ implies that $r$ must be greater or equal to 4. b) Total bits = $8+4=12$. Parity positions: 1, 2, 4, 8. Data positions: 3, 5, 6, 7, 9, 10, 11, 12.

• P1 = B3 ⊕ B5 ⊕ B7 ⊕ B9 ⊕ B11 = 1
• P2 = B3 ⊕ B6 ⊕ B7 ⊕ B10 ⊕ B11 = 1
• P4 = B5 ⊕ B6 ⊕ B7 ⊕ B12 = 1
• P8 = B9 ⊕ B10 ⊕ B11 ⊕ B12 = 1 Final result: 110111011111

How to understand this:

• Part a: Just plug $m=8$ into the inequality $(m+r+1)\le 2^r$ and test numbers. If $r=3$, $12\le 8$ (False). If $r=4$, $13\le 16$ (True).
• Part b: Map out a 12-bit table. Put your data bits 1 1 0 1 1 0 1 1 into the non-power-of-2 slots (12, 11, 10, 9, 7, 6, 5, 3). Then, calculate the parity bits by XOR-ing the specific data bits they are responsible for. (Remember: XOR just counts the 1s. If there is an odd number of 1s, the parity bit is 1. If even, it’s 0).

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Question 3: Network Delay (The “First Bit” Trap) (10%)

Question: Hosts A and B are connected via two links (L1 and L2) with a Router in between. Node A sends a 16-byte packet to Node B. How long does it take for node B to receive the first bit of the packet?

• L1 = 1000 bps, 40 m.
• L2 = 2000 bps, 10 m.
• Propagation speed = $5\times 10^8$ m/s.

Official Solution:

• Trans delay A $\to$ R1 (whole packet) = $(16\times 8)/1000=0.128\text{ s}$
• Prop delay A $\to$ R1 = $40/(5\times 10^8)=0.00000008\text{ s}$
• Trans delay R1 $\to$ B (for the first bit) = $1/2000=0.0005\text{ s}$
• Prop delay R1 $\to$ B = $10/(5\times 10^8)=0.00000002\text{ s}$
• Total Delay = $0.128+0.00000008+0.0005+0.00000002=$ 0.1285001 s

How to understand this: MASSIVE EXAM TRAP! Read the question carefully: “receive the first bit”. Because the router uses Store-and-Forward, it must wait to receive the entire 128-bit packet from A before it can do anything. So Link 1 has a full transmission delay. However, once the router starts pushing the packet onto Link 2, the question stops the clock the microsecond the first bit hits B. Therefore, the transmission delay on Link 2 is calculated for only 1 bit ($1/2000$), not the whole packet!

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Question 4: Go-Back-N vs Selective Repeat (10%)

Question: Explain the key difference between Go-Back-N and Selective Repeat protocols that results in improved performance for one of them.

Official Solution: In Go-Back-N, when a packet is lost or corrupted, the sender retransmits that packet and all subsequent packets, even if some were received correctly (because the receiver window size is 1). In contrast, Selective Repeat retransmits only the specific lost or corrupted packets while buffering correctly received ones at the receiver (receiver window size > 1). This drastically improves performance on noisy channels.

How to understand this: This is the textbook definition. Focus on two keywords for full marks: Window size (1 vs >1) and Buffering (discarding vs saving out-of-order frames).

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Question 5: Ethernet Efficiency Variables (15%)

Question: CSMA/CD Ethernet. 40 stations. Probability of success in a slot ($A$) is $0.4$. a) Find the mean number of contention slots per contention interval. b) Does this mean number affect efficiency? Explain. c) How does the duration of a contention slot affect efficiency?

Official Solution: a) Mean number of slots = $1/A=1/0.4=2.5$. b) Channel efficiency is given by $\frac{P}{P+T/A}$. As the mean number of slots ($1/A$) increases, the total wasted time increases, which reduces channel efficiency. c) Increasing $T$ (the duration of each slot) increases the time wasted per collision, which also reduces channel efficiency.

How to understand this: This question tests your understanding of the formula $\eta =\frac{P}{P+\text{Wasted Time}}$. Wasted time is (Number of tries) × (Length of a try).

• If it takes more tries to get a success ($1/A$ goes up), efficiency drops.
• If the awkward pause of a collision lasts longer ($T$ goes up), efficiency drops.

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Question 6: Distance Vector Routing (15%)

Question: Node C connects to A (cost 5), B (cost 7), D (cost 13). Node C receives vectors from A, B, and D. Construct the routing table for Node C. (Vectors provided in prompt).

Official Solution:

• Destination A: Via A = 5. Via B = 7+12=19. Via D = 13+11=24. $\to$ Min: 5 via A.
• Destination B: Via A = 5+12=17. Via B = 7. Via D = 13+6=19. $\to$ Min: 7 via B.
• Destination D: Via A = 5+11=16. Via B = 7+6=13. Via D = 13. $\to$ Min: 13 via D (or B).
• Destination E: Via A = 5+3=8. Via B = 7+14=21. Via D = 13+8=21. $\to$ Min: 8 via A.

How to understand this: For every destination, C asks itself: “What is my cost to my neighbor + my neighbor’s advertised cost to the destination?” It calculates this for A, B, and D, and simply chooses the minimum resulting number.

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Question 7: Dijkstra’s Algorithm (20%)

Question: Graph provided: (A,B)=3, (A,C)=2, (B,C)=8, (B,E)=4, (B,D)=9, (D,E)=6. Find shortest paths from source node C.

Official Solution:

• Step 1: Start at C. Update neighbors: A(2), B(8). Make C permanent.
• Step 2: A is the minimum (2). Make A permanent. A connects to B. Path to B via A is 2+3=5. (5 is better than 8, so update B to 5).
• Step 3: B is the minimum (5). Make B permanent. B connects to E and D. Path to E via B is 5+4=9. Path to D via B is 5+9=14.
• Step 4: E is the minimum (9). Make E permanent. E connects to D. Path to D via E is 9+6=15. (15 is worse than 14, so keep 14).
• Step 5: D is the minimum (14). Make D permanent.

How to understand this: This tests the exact “Relaxation” rule of Dijkstra. In Step 1, Node C thinks the best way to get to B is the direct road (cost 8). But in Step 2, when Node A becomes permanent, we discover a “shortcut”: going through A to get to B only costs 5! Because $5<8$, we cross out the 8 and update B’s tentative distance to 5.