# Q2 · Pure ALOHA throughput

Subject: Computer Networks

Exam: M2

Reader: https://deep-learning-study-room.vercel.app/cn/#m2/mocks/mock-a/q2

Canonical: https://deep-learning-study-room.vercel.app/read/cn/m2/mocks/mock-a--q2/

Q1(a)+Q1(b) compulsory; choose any two of Q2/Q3/Q4 · 10 marks attempted

### Question. A Pure ALOHA channel sends 500-bit frames at 500 kbps. The aggregate attempt rate is 500 frames/s. Find G, normalized throughput and successful frames/s. [3 practice marks]

#### Answer

## Given

| Symbol | Given |
| --- | --- |
| L | 500 bits |
| R | 500 kbps = 500,000 bits/s |
| λ | 500 aggregate attempts/s |

## To find

T, G, S, successful frames/s and bit/s.

## Formula and why



$$
T=L/R
$$





$$
G=\lambda T
$$





$$
S=G e^{-2G}
$$





$$
F_{\mathrm{success}}=S/T,\qquad B_{\mathrm{success}}=SR
$$



First 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 = 500/500,000 = 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 × 500,000 ≈ 91969.860293 bits/s ≈ 91.969860 kbps.

**Final answer:** G=0.5, S≈0.183940, successful rate≈183.940 frames/s = 91.970 kbps.

#### What this result means

The input is 500 attempts/s; collisions reduce the delivered frame rate to about 183.940/s. If the paper calls generated traffic the offered traffic, use that stated value as λ.