Practice Mid-Semester Test 1

Total Marks: 22 | Format: Short Answer & Math

The Questions

Question 1: Layers and Addressing (4 marks)

In Lab 1, you inspected your network configuration and found both an IP address and a MAC address. a) Explain the primary difference between how an IP address and a MAC address are assigned to a device. (2 marks) b) State exactly which OSI layer each of these two addresses operates on. (2 marks)

Question 2: Network Delay (4 marks)

Host A sends a 128-byte packet to Host B over a single link.

The link bandwidth is 2 Mbps ( $2, 000, 000$ bps).
The propagation speed of the link is $2 \times 1 0^{8}$ m/s.
The distance between the hosts is 50 km ( $50, 000$ meters).

Calculate the total end-to-end delay in milliseconds (ms). (Show your working for $T_{t r an s}$ and $T_{p ro p}$ ).

Question 3: Error Detection / CRC (4 marks)

A sender wants to transmit the data 1011 using CRC for error detection. The agreed-upon divisor is 110. Calculate the exact final bit pattern that the sender will transmit over the link. (Show your modulo-2 division steps).

Question 4: Stop-and-Wait Efficiency (3 marks)

A network uses the Stop-and-Wait protocol.

The frame transmission time ( $T_{t r an s}$ ) is 2 ms.
The one-way propagation delay ( $T_{p ro p}$ ) is 24 ms.

Calculate the channel utilization (efficiency) of this connection. (Express your answer as a percentage or a decimal).

Question 5: Medium Access Control (3 marks)

Two stations on a classic Ethernet network collide for the second time. They both apply the Binary Exponential Backoff (BEB) algorithm. What is the exact probability that they will pick the same random wait time and collide again on their third attempt?

Question 6: Distance Vector Routing (4 marks)

Router X is using Distance Vector Routing. It knows the cost to reach its two immediate neighbors:

Cost to reach Router Y = 4
Cost to reach Router Z = 2

Router X receives routing table updates from both neighbors regarding a faraway destination, Router D:

Router Y says: “My cost to reach D is 3.”
Router Z says: “My cost to reach D is 8.”

a) Calculate the total cost for Router X to reach Router D via Router Y, and the total cost via Router Z. (2 marks) b) Based on the algorithm, which router (Y or Z) will Router X choose as its next hop, and what will be the final cost stored in its routing table? (2 marks)

(Scroll down for solutions)

Solutions & Explanations

Question 1: Layers and Addressing

a) A MAC address is a physical address permanently assigned to the device’s Network Interface Card (NIC) by the manufacturer during assembly. An IP address is a logical address that is assigned dynamically when a device connects to a network, meaning it changes depending on the device’s location in the network topology. b)

MAC Address: Data Link Layer (Layer 2)
IP Address: Network Layer (Layer 3)

Question 2: Network Delay

Total Delay = 0.762 ms.

Step 1: Convert Bytes to Bits! $L = 128 Bytes \times 8 = 1024 bits$ .
Step 2: Transmission Delay ( $T_{t r an s}$ ): $T_{t r an s} = \frac{L}{R} = \frac{1024}{2 , 000 , 000} = 0.000512 seconds$ .
Step 3: Propagation Delay ( $T_{p ro p}$ ): $T_{p ro p} = \frac{d}{s} = \frac{50 , 000}{2 \times 1 0 ^{8}} = 0.00025 seconds$ .
Step 4: Total Latency in seconds: $0.000512 + 0.00025 = 0.000762 seconds$ .
Step 5: Convert to milliseconds: $0.000762 \times 1000 = 0.762 ms$ .

Question 3: Error Detection / CRC

Final Transmitted Pattern: 101110

Step 1: Padding. Divisor 110 has 3 bits. Append 2 zeros to the data. Padded Data = 101100.
Step 2: Modulo-2 Division (XOR).

   101100
 ^ 110      (XOR first 3 bits)
   ---
   0111     (Bring down the 1)
 ^  110     (XOR)
    ---
    0010    (Bring down the 0)
 ^   000    (Leading bit is 0, so XOR with 000)
     ---
     0100   (Bring down final 0)
 ^    110   (XOR)
      ---
       10   <-- REMAINDER

Step 3: Replace padding. Original data 1011 + remainder 10 = 101110.

Question 4: Stop-and-Wait Efficiency

Utilization ( $U$ ) = 0.04 (or 4%)

Formula: $U = \frac{T _{t r an s}}{T _{t r an s} + 2 \times T _{p ro p}}$
Calculation: $U = \frac{2}{2 + ( 2 \times 24 )} = \frac{2}{2 + 48} = \frac{2}{50} = 0.04$

Question 5: Medium Access Control (BEB)

Probability = 0.25 (or 25% or 1/4).

Explanation: After the second collision ( $i = 2$ ), the contention window size is $2^{i} - 1$ , which means the stations choose a random wait time slot from the range $[0, 3]$ .
There are 4 possible choices (0, 1, 2, or 3) for Station A, and 4 for Station B, meaning $4 \times 4 = 16$ possible combinations.
They will only collide if they pick the exact same number: (0,0), (1,1), (2,2), or (3,3).
4 collision scenarios out of 16 total scenarios = $4/16 = 1/4 = 0.25$ . (Shortcut formula: $\frac{1}{2 ^{i}}$ ).

Question 6: Distance Vector Routing

a) Total costs:

Path via Y = (Cost to Y) + (Y’s cost to D) = $4 + 3 = 7$ .
Path via Z = (Cost to Z) + (Z’s cost to D) = $2 + 8 = 10$ .

b) Final Routing Table Entry:

Next Hop: Router Y
Final Cost: 7
(Explanation: The router selects the path with the minimum total cost).

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