• Table of Contents

<aside> <img src="/icons/map-pin_gray.svg" alt="/icons/map-pin_gray.svg" width="40px" /> The midterm will be a mixture of true or false question, short answers, calculations, and other variations.

• No coding involved.
• Less focus on mathematical/calculation question.
• When solving a problem, it’s best to list out the formulas used and steps, for part marks.

The following formulas provided are:

• Transmission latency:

  $T_f = M/R$ ($M$ data size in bits and $R$ bit rate)

• Propagation latency:

  $T_c$ the ratio between link length and the propagation speed

• Link efficiency of wait and stop protocol:

  $1/(1+2a)$ where $a = T_c/T_f$

• Link efficiency of CSMA/CD:

  $1/(1+5a)$ where $a = T_c/T_f$

The following metric units provided are:

• $1$ K(ilo) bit = $10^3$ bits
• $1$ M(ega) bit = $10^6$ bits
• $1$ G(iga) bit = $10^9$ bits
• Kbps: Kilobits per second
• Mbps: Megabits per second
• Second ($\text{s}$), Millisecond ($\text{ms}$) and Microsecond ($\mu\text{s}$)
  • $1\;\text{s} = 10^3\;\text{ms}$
  • $1\;\text{s} = 10^6\;\mu\text{s}$ </aside>

<aside> <img src="/icons/map-pin_gray.svg" alt="/icons/map-pin_gray.svg" width="40px" /> Things that are not in the formula sheet, that I think are at least worth memorizing are:

1. Byte to bit conversion is $1\;\text{byte} = 8\;\text{bit}$
2. For bytes:
   1. $1$ K(ilo) Byte = $1024$ Byte
   2. $1$ M(ega) Byte = $1024$ K(ilo) Byte
   3. $1$ G(iga) Byte = $1024$ M(ega) Byte
3. Propagation speed (if not given) is two-thirds the speed of light, $2 \times 10^8\;\text{m/s}$
4. Round-trip Time: $\text{RTT} = 2 \times T_c$
5. Alternatively, propagation delay can be calculated given $\text{RTT}$
6. Bandwidth-delay Product: $\text{BDP} = \text{BW }\times \text{RTT}$ ($\text{BW}$ bandwidth in bits per second) </aside>

Link Efficiency

Given the channel data $R = 10\;\text{Mbps}$ and frame size $D = 1000\;\text{byte}$. Determine link efficiencies pf stop and wait protocol for a $1-\text{km}$ link.

1. Convert any bytes ($\text{B}$) to bits ($\text{b}$) and standardize other units.

   $$ 1000\;\text{byte} \times \frac{8\;\text{bit}}{1\;\text{byte}} = 8000\;\text{bit} $$

   $$ 1\;\text{km} \times \frac{1000\;\text{m}}{1\;\text{km}} = 1000\;\text{m} $$

2. Calculate the transmission delay.

   $$ T_f = \frac{8000\;\text{bit}}{1 \times 10^7\;\text{bit/s}} = 8 \times 10^{-4}\;\text{s} $$

3. Calculate the propagation delay. Since the propagation speed is not given, assume it’s two-thirds the speed of light.

   $$ T_c = \frac{1000\;\text{m}}{2 \times 10^8\;\text{m/s}} = 5 \times 10^{-6}\;\text{s} $$

4. Calculate the ratio $a$ and link efficiency.

   $$ a = \frac{T_c}{T_f} = \frac{5 \times 10^{-6}\;\text{s}}{8 \times 10^{-4}\;\text{s}} = 6.25 \times 10^{-3} $$

   $$ \text{Link Efficiency} = \frac{1}{1 + 2a} \approx 0.9877 = 98.77\% $$

If the bandwidth of the line is $1.5\;\text{Mbps}$, RTT (Round Trip Time) is $45\;\text{msec}$ and packet size is $1\;\text{KB}$, then find the link efficiency in stop and wait.

1. Convert any bytes ($\text{B}$) to bits ($\text{b}$) and standardize other units.

   $$ 1.5\;\text{Mbps} = 1.5 \times 10^6\;\text{bps} $$

   $$ 45\;\text{ms} = 0.045\;\text{s} $$

   $$ 1\;\text{KB} \times \frac{1024\;\text{B}}{1\;\text{KB}} = 1024\;\text{B} \qquad 1.5\;\text{KB} \times \frac{1024\;\text{B}}{1\;\text{KB}} = 1536\;\text{B} $$

   $$ 1024\;\text{B} \times \frac{8\;\text{b}}{1\;\text{B}} = 8192\;\text{b} \qquad 1536\;\text{B} \times \frac{8\;\text{b}}{1\;\text{B}} = 12288\;\text{b} $$

2. Calculate the transmission delay and propagation delay.

   $$ T_f = \frac{8192\;\text{b}}{1.5 \times 10^6\;\text{b/s}} \approx 5.461 \times 10^{-3}\;\text{s} $$

   $$ T_c = \frac{\text{RTT}}{2} = \frac{0.045\;\text{s}}{2} = 0.0225\;\text{s} $$

3. Calculate the ratio $a$ and link efficiency.

   $$ a = \frac{T_c}{T_f} = \frac{0.0225\;\text{s}}{5.461 \times 10^{-3}\;\text{s}} \approx 4.12 $$

   $$ \text{Link Efficiency} = \frac{1}{1 + 2a} \approx 0.1082 = 10.82\% $$

Message Latency