Computer Networks · TEE · papers
Q6(a) · Pure ALOHA rates
2024-12-06 · Final 2024-25 / Re-exam 2023-24
Actual paper; 100 marks attempted. Current-syllabus alignment is stated per question.
Q6(a). A pure ALOHA network transmits 400-bit frames on a shared channel of 100 kbps. Calculate the throughput, and how many frames will survive if the system (all stations together) produces the following number of frames per second: i. 1200 frames per second ii. 600 frames per second iii. 300 frames per second. [10 marks]
Answer
Given
Frame length L = 400 bits; link rate R = 100,000 bit/s; attempted rates λ = 1200, 600, 300 frames/s.
To find
Normalized throughput S and successful frames per second for each offered load.
Formula and why
LaTeX: T_f=L/R,\quad G=\lambda T_f,\quad S=Ge^{-2G},\quad \lambda_{success}=S/T_f=\lambda e^{-2G}G counts attempted frames per frame time, including retransmissions if present. S is not measured in frames/s. Multiply by the frame rate capacity 1/Tf to obtain surviving frames/s.
Step 1. Convert the link rate and compute frame time
Tf = 400/100000 = 0.004 s. Capacity = 1/Tf = 250 frames/s.
Step 2. Offered rate 1200 frames/s
G = 1200 × 0.004 = 4.8.
e−2G = e−9.6 = 6.77287365e-05.
S = 4.8 × 6.77287365e-05 = 0.000325097935.
Successful frames/s = 1200 × 6.77287365e-05 = 0.0812744838. Successful bit/s = 0.0812744838 × 400 = 32.5097935.
Step 3. Offered rate 600 frames/s
G = 600 × 0.004 = 2.4.
e−2G = e−4.8 = 0.00822974705.
S = 2.4 × 0.00822974705 = 0.0197513929.
Successful frames/s = 600 × 0.00822974705 = 4.93784823. Successful bit/s = 4.93784823 × 400 = 1975.13929.
Step 4. Offered rate 300 frames/s
G = 300 × 0.004 = 1.2.
e−2G = e−2.4 = 0.0907179533.
S = 1.2 × 0.0907179533 = 0.108861544.
Successful frames/s = 300 × 0.0907179533 = 27.215386. Successful bit/s = 27.215386 × 400 = 10886.1544.