Computer Networks · TEE · mocks
Q6B · Random access
TEE mock B · full paper
TEE · 100 attempted marks · solutions for every choice · current syllabus 702AI0C023
TEE practice mock · Q6B. A pure ALOHA network transmits 200-bit frames on a shared channel of 200 kbps. What is the throughput if the system (all stations together) produces (a) 1000 frames per second, (b) 500 frames per second, (c) 250 frames per second? [5 mock marks]
Answer
Interpret the stated aggregate production rates as offered attempt rates in the standard textbook model. The following working includes normalized throughput, successful frames/s and successful frame bits/s for each case.
Case a · 1000 frames/s
Given
| Symbol | Given |
|---|---|
| L | 200 bits |
| R | 200 kbps = 200,000 bits/s |
| λ | 1000 aggregate attempts/s |
To find
T, G, S, successful frames/s and bit/s.
Formula and why
LaTeX: T=L/R
LaTeX: G=\lambda T
LaTeX: S=G e^{-2G}
LaTeX: F_{\mathrm{success}}=S/T,\qquad B_{\mathrm{success}}=SRFirst normalize attempts by frame time. Then apply the correct ALOHA model and convert S back to requested rate units.
Step-by-step solution
Step 1. Frame transmission time
T = 200/200,000 = 0.001 s = 1 ms.
Step 2. Offered load
G = 1000 × 0.001 = 1 attempts/frame time.
Step 3. Exponent and exponential
−2G = -2; e to that power ≈ 0.135335283.
Step 4. Normalized throughput
S = 1 × 0.135335283 ≈ 0.135335283.
Step 5. Successful frames each second
S/T = 0.135335283/0.001 ≈ 135.335283 frames/s.
Step 6. Successful frame bit rate
SR = 0.135335283 × 200,000 ≈ 27067.056647 bits/s ≈ 27.067057 kbps.
Final answer: G=1, S≈0.135335, successful rate≈135.335 frames/s = 27.067 kbps.
Case b · 500 frames/s
Given
| Symbol | Given |
|---|---|
| L | 200 bits |
| R | 200 kbps = 200,000 bits/s |
| λ | 500 aggregate attempts/s |
To find
T, G, S, successful frames/s and bit/s.
Formula and why
LaTeX: T=L/RLaTeX: G=\lambda TLaTeX: S=G e^{-2G}LaTeX: F_{\mathrm{success}}=S/T,\qquad B_{\mathrm{success}}=SRFirst normalize attempts by frame time. Then apply the correct ALOHA model and convert S back to requested rate units.
Step-by-step solution
Step 1. Frame transmission time
T = 200/200,000 = 0.001 s = 1 ms.
Step 2. Offered load
G = 500 × 0.001 = 0.5 attempts/frame time.
Step 3. Exponent and exponential
−2G = -1; e to that power ≈ 0.367879441.
Step 4. Normalized throughput
S = 0.5 × 0.367879441 ≈ 0.183939721.
Step 5. Successful frames each second
S/T = 0.183939721/0.001 ≈ 183.939721 frames/s.
Step 6. Successful frame bit rate
SR = 0.183939721 × 200,000 ≈ 36787.944117 bits/s ≈ 36.787944 kbps.
Final answer: G=0.5, S≈0.183940, successful rate≈183.940 frames/s = 36.788 kbps.
Case c · 250 frames/s
Given
| Symbol | Given |
|---|---|
| L | 200 bits |
| R | 200 kbps = 200,000 bits/s |
| λ | 250 aggregate attempts/s |
To find
T, G, S, successful frames/s and bit/s.
Formula and why
LaTeX: T=L/RLaTeX: G=\lambda TLaTeX: S=G e^{-2G}LaTeX: F_{\mathrm{success}}=S/T,\qquad B_{\mathrm{success}}=SRFirst normalize attempts by frame time. Then apply the correct ALOHA model and convert S back to requested rate units.
Step-by-step solution
Step 1. Frame transmission time
T = 200/200,000 = 0.001 s = 1 ms.
Step 2. Offered load
G = 250 × 0.001 = 0.25 attempts/frame time.
Step 3. Exponent and exponential
−2G = -0.5; e to that power ≈ 0.606530660.
Step 4. Normalized throughput
S = 0.25 × 0.606530660 ≈ 0.151632665.
Step 5. Successful frames each second
S/T = 0.151632665/0.001 ≈ 151.632665 frames/s.
Step 6. Successful frame bit rate
SR = 0.151632665 × 200,000 ≈ 30326.532986 bits/s ≈ 30.326533 kbps.
Final answer: G=0.25, S≈0.151633, successful rate≈151.633 frames/s = 30.327 kbps.