# Q4(b) · HTTP/2 multiplexing versus HTTP/1.1

Subject: Computer Networks

Exam: TEE

Reader: https://deep-learning-study-room.vercel.app/cn/#tee/papers/paper-2023-jan-ai-reexam/q4b

Canonical: https://deep-learning-study-room.vercel.app/read/cn/tee/papers/paper-2023-jan-ai-reexam--q4b/

Exact historical variant. Current-syllabus questions solved; excluded topics and source limitations stated.

### Q4(b). RTT=4 time units; all four objects are requested at t=0 on an established connection. O1 has 10 frames, O2 has 1, O3 has 2, O4 has 1. One frame takes 1 time unit. HTTP/2 cycles O1,O2,O3,O4, skipping finished objects; HTTP/1.1 sends complete objects FCFS O1,O2,O3,O4. Find each completion time and average object download delay for both methods, including request and response propagation. [10 marks]

### Answer

### Given

RTT4; frame time1; O1/O2/O3/O4 have10/1/2/1 frames; established connection.

### To find

Every object completion time and average delay.

### Formula and why



$$
T_j=RTT+\text{position of the last frame of object }j
$$



Requests need2 units to reach the server and final frame bits need2 to return. Thus add4 to each last-frame transmission count; do not add a new RTT per frame.

### Solution

### Step 1. Write transmission orders and last-frame positions

| Method | Frame order | O1 completion | O2 completion | O3 completion | O4 completion | Average |
| --- | --- | --- | --- | --- | --- | --- |
| HTTP/2 | O1, O2, O3, O4, O1, O3, O1, O1, O1, O1, O1, O1, O1, O1 | 18 | 6 | 10 | 8 | 10.5 |
| HTTP/1.1 | O1, O1, O1, O1, O1, O1, O1, O1, O1, O1, O2, O3, O3, O4 | 14 | 15 | 17 | 18 | 16.0 |

### Step 2. Calculate averages explicitly

HTTP/2: (18+6+10+8)/4=42/4=**10.5**. HTTP/1.1: (14+15+17+18)/4=64/4=**16**. All objects finish by18 in both cases; multiplexing makes short objects finish earlier.