Computer Networks · TEE · mocks

Q1C · ALOHA throughput

TEE mock A · full paper

TEE · 100 attempted marks · solutions for every choice · current syllabus 702AI0C023

TEE practice mock · Q1C. In a Pure ALOHA system, the offered load is G=0.25. Find normalized throughput, percentage throughput, and the theoretical maximum. [5 mock marks]

Answer

Given

Protocol = Pure ALOHA; G = 0.25 attempts/frame time.

To find

S, 100S and the maximum possible S.

Formula and why

S=G e^{-2G}LaTeX: S=G e^{-2G}

Select the formula from the protocol name. The exponent is −2G because Pure ALOHA has a 2T vulnerable period.

Step-by-step solution

Step 1. Calculate the exponent

−2 × 0.25 = -0.5.

Step 2. Evaluate the exponential

e^{-0.5} \approx 0.606530660LaTeX: e^{-0.5} \approx 0.606530660

Step 3. Multiply by G

S = 0.25 × 0.606530660 ≈ 0.151632665 successful frames/frame time.

Step 4. Convert to percentage

0.151632665 × 100 ≈ 15.1633%.

Step 5. State the maximum separately

The maximum is 1/(2e) ≈ 0.183940 = 18.3940%, at G=0.5.

Final answer: S ≈ 0.151633; percentage ≈ 15.1633%; Smax ≈ 0.183940 at G=0.5.