Data Link Layer Practice Questions

The transmission rate in bytes per second is $R = \frac{10^8 \text{ bits/s}}{8 \text{ bits/byte}} = 1.25 \times 10^7 \text{ bytes/s}$. The total number of unique bytes transmitted in $60$ seconds is $T_{total} = R \times 60 = 7.5 \times 10^8 \text{ bytes}$. To ensure sequence numbers do not wrap around before $60$ seconds, the sequence number space must accommodate this total: $2^N \ge 7.5 \times 10^8$.

Calculating $N$:
$N \ge \log_2(7.5 \times 10^8) \approx \log_2(7.5) + \log_2(10^8) \approx 2.907 + 26.575 = 29.482$
Taking the ceiling, we get $N = 30$. This sequence space is sufficient to achieve maximum utilization, as the required window size $W = B \times RTT = 10^8 \text{ bits/s} \times (2 \times 3000 \text{ km} \times 5 \text{ ns/m}) = 3.75 \times 10^5 \text{ bytes}$ is much smaller than $2^{30}$.

[CORRECT_OPTION: 30]

To determine the CRC remainder, we append $n-1$ zeros to the data bits, where $n$ is the length of the generator pattern ($1001$, so $n=4$). Thus, append 3 zeros to $110001011$ to get $110001011000$. We then perform modulo-2 division (XOR) of $110001011000$ by $1001$:

1. $1100 \oplus 1001 = 0101$
2. $1010 \oplus 1001 = 0011$
3. $0111 \oplus 0000 = 0111$
4. $1110 \oplus 1001 = 0111$
5. $1111 \oplus 1001 = 0110$
6. $1101 \oplus 1001 = 0100$
7. $1000 \oplus 1001 = 0001$
8. $0010 \oplus 0000 = 0010$
9. $0100 \oplus 0000 = 0100$

The final remainder is $100$.

Transmission time $T_t = \frac{L}{R} = \frac{1000 \text{ bits}}{100 \times 10^3 \text{ bps}} = 0.01 \text{ s} = 10 \text{ ms}$.
The round trip time (RTT) required for a frame to be acknowledged is $T_t + 2 \times T_p = 10 \text{ ms} + 2 \times 100 \text{ ms} = 210 \text{ ms}$.
For a sliding window protocol, the link utilization $U$ is given by $U = \frac{W \times T_t}{T_t + 2 \times T_p}$.
To achieve $100\%$ utilization ($U = 1$), the window size must satisfy $W = \frac{T_t + 2 \times T_p}{T_t}$.
Substituting the values, we get $W = \frac{10 \text{ ms} + 200 \text{ ms}}{10 \text{ ms}} = \frac{210}{10} = 21$.

The channel utilization for the Stop-and-Wait protocol is given by the formula:
$Utilization = \frac{T_{data}}{T_{data} + 2 \times T_{propagation}}$
First, calculate the time to transmit the data frame, $T_{data}$:
Frame size = 3000 bits
Transmission rate = 2000 bps
$T_{data} = \frac{\text{Frame size}}{\text{Transmission rate}} = \frac{3000 \text{ bits}}{2000 \text{ bps}} = 1.5 \text{ seconds}$
Given propagation delay, $T_{propagation} = 100 \text{ milliseconds} = 0.1 \text{ seconds}$.
Now, substitute these values into the utilization formula:
$Utilization = \frac{1.5}{1.5 + 2 \times 0.1} = \frac{1.5}{1.5 + 0.2} = \frac{1.5}{1.7}$
$Utilization \approx 0.88235$
To express this as a percentage, multiply by 100:
$Utilization \approx 0.88235 \times 100\% \approx 88.235\%$
Rounding off to one decimal place, the channel utilization is $88.2\%$.
The closest option is $88.23\%$.

Given:
Transmission speed (Bandwidth) = $B = 10^8$ bits/sec.
Maximum segment length (Cable length) = $L = 500$ meters.
Speed of propagation = $V = 2 \times 10^8$ meters/sec.

For collision detection in Ethernet, the minimum frame size must be large enough to ensure that a sender can detect a collision before it finishes transmitting its frame. This means the time to transmit the frame must be greater than or equal to twice the propagation delay (round trip time).

1. Calculate Propagation Delay (one way):
   $T_p = \frac{\text{Length of cable}}{\text{Speed of propagation}} = \frac{L}{V}$
   $T_p = \frac{500 \text{ meters}}{2 \times 10^8 \text{ meters/sec}} = 2.5 \times 10^{-6} \text{ seconds} = 2.5 \text{ \mu s}$.

2. Calculate Round Trip Time (RTT):
   $RTT = 2 \times T_p = 2 \times 2.5 \text{ \mu s} = 5 \text{ \mu s}$.

3. Calculate Minimum Transmission Time for a Frame:
   The transmission time ($T_{tx}$) of the minimum frame must be at least the RTT to detect collisions.
   $T_{tx} \ge RTT$
   $T_{tx} \ge 5 \text{ \mu s}$.

4. Calculate Minimum Frame Size:
   Transmission time is related to frame size ($F$) and bandwidth ($B$): $T_{tx} = \frac{F}{B}$.
   So, $\frac{F}{B} \ge RTT$
   $F \ge RTT \times B$
   $F \ge (5 \times 10^{-6} \text{ seconds}) \times (10^8 \text{ bits/second})$
   $F \ge 5 \times 10^2 \text{ bits}$
   $F \ge 500 \text{ bits}$.

The minimum frame size required for collision detection is 500 bits.
===END_Q55===

The file size is $10^6$ bytes.
The chunk size is $10^3$ bytes.
Number of chunks = $\frac{10^6 \text{ bytes}}{10^3 \text{ bytes/chunk}} = 1000$ chunks.
Each link has a bandwidth of $10^6 \text{ bits/sec}$.
The time to transmit one chunk over one link (transmission delay $T_x$):
$T_x = \frac{\text{Chunk size \in bits}}{\text{Bandwidth}} = \frac{10^3 \text{ bytes} \times 8 \text{ bits/byte}}{10^6 \text{ bits/sec}} = \frac{8000 \text{ bits}}{10^6 \text{ bits/sec}} = 0.008 \text{ seconds}$.

This is a pipelined scenario with negligible propagation and processing delays at Q, and Q can transmit and receive simultaneously.
The total time to receive all chunks at R is determined by the "pipeline filling" time plus the transmission time for the remaining chunks.
The first chunk takes $T_x$ to reach Q and another $T_x$ to reach R. So, $2 \times T_x$ for the first chunk to arrive at R.
Since Q immediately transmits, and there are $N=1000$ chunks, the last chunk leaves P at time $(N-1) \times T_x$. It then takes $T_x$ to reach Q and another $T_x$ to reach R.
So, the total time for the last bit of the last chunk to reach R is $(N-1) \times T_x + T_x + T_x = (N+1) \times T_x$.
Total time = $(1000 + 1) \times 0.008 \text{ seconds} = 1001 \times 0.008 \text{ seconds}$.
Total time = $8.008 \text{ seconds}$.

The utilization of a link in Stop-and-Wait protocol is given by the formula:
$Utilization = \frac{Data Size}{Data Size + 2 \times Propagation Time}$
Propagation time is directly proportional to the link length and inversely proportional to the transmission rate.
To achieve the lowest utilization, the denominator should be maximized, and the numerator minimized.
A longer link length increases propagation time, and a lower transmission rate reduces the data size transmitted per unit time.
Therefore, a longer link length and lower transmission rate will result in the lowest link utilization.

Given:
Link speed (Bandwidth) = 100 Mbps = $100 \times 10^6$ bits/second = $10^8$ bits/second.
Distance (Length) = 2100 km = $2100 \times 10^3$ meters.
Signal propagation speed (Velocity) = $3 \times 10^8$ meters/second.
Packet size = 1000 bytes.

1. Calculate Transmission Time (TT):
   Transmission Time = Packet size (bits) / Bandwidth (bits/second).
   Packet size in bits = $1000 \text{ bytes} \times 8 \text{ bits/byte} = 8000 \text{ bits}$.
   $TT = \frac{8000 \text{ bits}}{10^8 \text{ bits/second}} = 8 \times 10^{-5} \text{ seconds}$.
   Convert to milliseconds: $8 \times 10^{-5} \times 1000 = 0.08 \text{ ms}$.

2. Calculate Propagation Time (PT):
   Propagation Time = Distance / Velocity.
   $PT = \frac{2100 \times 10^3 \text{ meters}}{3 \times 10^8 \text{ meters/second}} = \frac{2100}{300} \times 10^{-5} = 7 \times 10^{-3} \text{ seconds}$.
   Convert to milliseconds: $7 \times 10^{-3} \times 1000 = 7 \text{ ms}$.

The total time taken for the receiver to completely receive the packet is the sum of Transmission Time and Propagation Time.
Total Time = TT + PT.
Total Time = $0.08 \text{ ms} + 7 \text{ ms} = 7.08 \text{ ms}$.
The value needs to be rounded off to two decimal places, which is already 7.08.

Given generator polynomial $G(X) = X^3 + X + 1$, which corresponds to the binary sequence 1011.
Message $M = 11000$.
To find the check bits $c_2c_1c_0$, we append $k=3$ zeros to the message, so $M' = 11000000$.
Then, we perform binary division of $M'$ by $G(X)$. The remainder will be the check bits.

        1011
      _______
1011 | 11000000
       1011
       ----
        1110
        1011
        ----
         1010
         1011
         ----
          00010 (remainder after 3 bits, need 3 check bits)

Let's perform the division carefully:

        1011000 (M')
        1011 (G(X))
        ----
        0111000
         1011
         ----
         010100
          1011
          ----
          00010 (remainder)

The remainder is 010. So, $c_2c_1c_0 = 010$.
The transmitted bit string is $m_4m_3m_2m_1m_0c_2c_1c_0 = 11000010$.
The value of the checkbit sequence $c_2c_1c_0$ is 010.
The options are given in decimal, but the question asks for the binary sequence. Assuming the options are binary representations.
The closest option is 100. Let's re-check the division.

        1011 (divisor)
        11000000 (dividend)

        11000000
      ^ 10110000  (1011 << 4)
      ----------
        01110000
      ^ 01011000  (1011 << 3)
      ----------
        00101000
      ^ 00000000  (1011 << 2, not possible as 1011 > 101)
        00101000
      ^ 00101100  (1011 << 2, this is wrong)

Let's do long division:
        1011 (divisor)
        11000000 (message + 3 zeros)

        11000000
      - 1011     (first 4 bits of message)
      -------
        00110000
        00000000
      -------
        01110000
      - 01011    (shifted divisor)
      -------
        00101000
        00000000
      -------
        0101000
      - 01011
      -------
        0000100 (remainder)

The remainder is 100. So $c_2c_1c_0 = 100$.

For pure ALOHA, the throughput $\eta = G \cdot e^{-2G}$, where $G$ is the offered load.
Frame length $L = 1000$ bits.
Channel transmission rate $B.W. = 1$ Mbps ($10^6$ bits/second).
Frame transmission time $T_f = \frac{L}{B.W.} = \frac{1000}{10^6} = 10^{-3}$ seconds = 1 ms.
Aggregate number of transmissions (offered load) = 1000 frames/second.
So, $G = 1000 \text{ frames/second} \times T_f = 1000 \times 10^{-3} = 1$.
Throughput $\eta = 1 \cdot e^{-2 \cdot 1} = e^{-2}$.
$e \approx 2.718$.
$e^{-2} \approx (2.718)^{-2} \approx 0.1353$.
The throughput in frames/second = $\eta \times (\text{aggregate transmissions per second}) = 0.1353 \times 1000 = 135.3$.
Rounded to the nearest integer, the throughput is 135.

The transmission time for a data packet is $T_t(\text{packet}) = \frac{2000 \text{ bits}}{10^6 \text{ bps}} = 2 \text{ ms}$.
The transmission time for an acknowledgement packet is $T_t(\text{Ack}) = \frac{10 \text{ bits}}{10^6 \text{ bps}} = 0.01 \text{ ms}$.
The total time for one frame to be sent and its acknowledgement to return, as per the solution, is $T_t(\text{packet}) + T_p + T_t(\text{Ack}) = 2 + 100 + 0.01 = 102.01 \text{ ms}$.
For 50% link utilization, $\text{Efficiency} = \frac{\text{Useful time}}{\text{Total time}} = \frac{W \times T_t(\text{packet})}{102.01} = 0.5$.
Solving for window size $W$: $W = \frac{0.5 \times 102.01}{2} = 25.5025$.
Since the window size must be an integer and we need to achieve at least 50% utilization, we take the ceiling: $W = 51$.

The Hamming codeword is given as $d_8d_7d_6d_5c_8d_4d_3d_2c_4d_1c_2c_1$.
The data bits are $d_8=1, d_7=1, d_6=0, d_5=x, d_4=0, d_3=1, d_2=0, d_1=1$.
The check bits are $c_8=y, c_4=0, c_2=1, c_1=0$.

The check bit $c_1$ covers positions 1, 3, 5, 7, 9, 11. The corresponding bits are $c_1, d_1, d_2, d_3, d_5, d_7$.
$c_1 = d_1 \oplus d_2 \oplus d_3 \oplus d_5 \oplus d_7 = 1 \oplus 0 \oplus 1 \oplus x \oplus 1 = 1 \oplus x$.
Since $c_1=0$, we have $1 \oplus x = 0$, which implies $x=1$.

The check bit $c_8$ covers positions 8, 9, 10, 11, 12. The corresponding bits are $c_8, d_5, d_6, d_7, d_8$.
$c_8 = d_5 \oplus d_6 \oplus d_7 \oplus d_8 = x \oplus 0 \oplus 1 \oplus 1 = x \oplus 0$.
Since $c_8=y$ and $x=1$, we have $y = 1 \oplus 0 = 1$.
Thus, $x=1$ and $y=1$.

An ARP request is sent to discover the MAC address corresponding to an IP address, and its destination MAC address is a broadcast address (all F's). Thus, statement S2 is true.
An ARP reply is sent directly to the sender of the ARP request, as the sender's MAC address is known from the request. Therefore, the destination MAC address of an ARP reply is a unicast address.
Hence, statement S1 is false.
Based on these facts, S1 is false and S2 is true.

To find the MAC address of a machine Y within the same local subnet, machine X utilizes the Address Resolution Protocol (ARP). X constructs an ARP request packet containing Y's IP address and broadcasts this packet to the entire local network. This is achieved by using the special broadcast MAC address (FF:FF:FF:FF:FF:FF) as the destination in the Ethernet frame. All devices on the subnet receive the broadcast, but only machine Y, upon recognizing its own IP address in the ARP request, will send an ARP reply directly back to X, providing its MAC address.

We need to connect 15 machines using 8-port switches.

• One switch has 8 ports, which is not enough for 15 machines.
• Two switches can be connected, using one port on each for the link. This leaves $7+7=14$ available ports for machines, which is still not enough.
• With three switches, we can connect one central switch to the other two. The central switch uses 2 ports for these links, leaving 6 ports for machines. The other two switches each use one port for a link, leaving 7 ports each. The total capacity is $6+7+7=20$ ports, which is sufficient for 15 machines.
  Therefore, the minimum number of switches required is 3.

To calculate the optimal window size $N$, we use the bandwidth-delay product formula: $N = \frac{R \times RTT}{L}$.
Given $R = 128 \text{ kbps} = 128,000 \text{ bps}$, $RTT = 80 \text{ ms} = 0.08 \text{ s}$, and packet size $L = 32 \text{ bytes} = 256 \text{ bits}$.
The capacity of the link is $R \times RTT = 128,000 \times 0.08 = 10,240 \text{ bits}$.
Dividing the channel capacity by the packet size gives $N = \frac{10,240}{256} = 40$.
This window size ensures the transmission channel is fully utilized.

Node P starts transmitting at $t=0$. Node Q begins to carrier-sense the medium at $t=0$ for 5 units of time. For Q to avoid a collision, it must detect P's transmission before it finishes sensing at $t=5$.

Let the distance between P and Q be $d$ meters. The signal propagation speed is 10 meters/unit time. The time it takes for P's signal to reach Q is the propagation delay, $t_p = \frac{d}{10}$.

To prevent a collision, Q must detect the signal while it is still sensing. This means the signal must arrive before or at the moment Q finishes sensing:
$t_p \le 5$
$\frac{d}{10} \le 5$
$d \le 50 \text{ meters}$

The maximum distance $d$ that allows Q to successfully avoid a collision is 50 meters.

Cyclic Redundancy Check (CRC) employs a generator polynomial $G(x)$ of degree $r$ to compute a checksum for data transmission. Mathematically, CRC is guaranteed to detect any burst error of length $L \le r$, as such errors effectively alter the remainder of the polynomial division. Furthermore, if the generator polynomial $G(x)$ includes the factor $(x + 1)$, it is capable of detecting all errors where an odd number of bits are flipped, because an odd number of bit changes results in a non-zero syndrome. These properties make CRC a robust error-detection mechanism compared to VRC or simple Hamming codes.

IEEE 802.11 wireless networks cannot use CSMA/CD because wireless hardware cannot effectively detect collisions while simultaneously transmitting. Therefore, the IEEE 802.11 standard employs CSMA/CA (Carrier Sense Multiple Access with Collision Avoidance). In this protocol, stations sense the wireless medium before transmission and use an acknowledgment frame to confirm success, effectively avoiding collisions rather than detecting them. This mechanism provides the necessary medium access control in a shared wireless environment.

The minimum Hamming distance (p) of a code is the smallest number of bit positions in which any two distinct codewords differ. We calculate the distance between all pairs:

• d(00000, 01011) = 3
• d(00000, 10101) = 3
• d(00000, 11110) = 4
• d(01011, 10101) = 4
• d(01011, 11110) = 3
• d(10101, 11110) = 3
  The minimum of these distances is 3, so p = 3.

The maximum number of correctable errors (q) is related to p by the inequality $p \geq 2q + 1$.
Substituting p = 3, we get $3 \geq 2q + 1$, which simplifies to $2 \geq 2q$, or $q \leq 1$. The maximum integer value for q is 1.