Q1 | App Development Resources

Q1

Suppose a client machine A is communicating with a data center B located 8000 km away from A.
The speed of light in the cable is assumed to be 2 × 10^8 m/s (instead of the default 3 × 10^8 m/s).

How long will it take for a request to reach the server and for the response to return back to the client (round-trip time)?

Solution:

             request ————>>———————
   Client A                          Server B
             <<—————— response ————
   Distance between A and B = 8000 km

Latency is the delay between sending a request and receiving a response in a system.

Formula:
Speed = Distance / Time
Time = Distance / Speed
One-way distance:
8000 × 1000 = 8 × 10^6 m
One-way time:
(8 × 10^6) / (2 × 10^8) = 0.04 s
Round-trip time (RTT):
2 × 0.04 = 0.08 s
Converting to milliseconds:
0.08 × 1000 = 80 ms

Final Answer:
Latency (Round-trip latency) = 80 ms

Q2

Suppose a client machine C is communicating with a data center D located 15000 km away from C.
Assume that the TCP connection has already been established and is kept alive.
If the client can send a new request only after receiving an acknowledgement (response) for the previous request, what is the maximum number of successful requests that can be completed from C to D in one second?

(Assume the speed of light in cable is 2 × 10^8 m/s.)

Solution:

             request ————>>———————
   Client C                          Server D
             <<—————— response ————
   Distance between C and D = 15000 km

Formula:
Speed = Distance / Time
Time = Distance / Speed
One-way distance:
15000 × 1000 = 1.5 × 10^7 m
One-way time:
(1.5 × 10^7) / (2 × 10^8) = 0.075 s = 75 ms
Round-trip time (RTT):
2 × 75 ms = 150 ms
Number of requests possible in 1 second (1000 ms):
1000 / 150 ≈ 6.67
Only 6 full request–response cycles can complete in 1 second.
(The 7th would not complete within the 1 second window.)

Final Answer:
Maximum number of successful requests = 6 per second

Q3

Consider a client located 6000 km from a server.
The client makes a request through a cable, but after the request reaches the server, the cable breaks and the response is sent to the client via air.

If this change of medium causes an additional delay of 75 ms at the server, how long will the client have to wait before receiving the response?

(Assume speed of light in cable = 2 × 10^8 m/s, in air = 3 × 10^8 m/s)

Assume RTT = latency.

Solution:

             request ————>>——————— (cable)
   Client A                          Server B
             <<—————— response ———— (air)
   Distance between A and B = 6000 km

Formula:
Latency = request time + response time + server delay
Request time (cable):
6000 × 1000 / 2 × 10^8 = 0.03 s = 30 ms
Response time (air):
6000 × 1000 / 3 × 10^8 = 0.02 s = 20 ms
Server delay:
75 ms = 0.075 s ( you can choose either second or milisecond , but we will prefer ms in this course)
Total client wait (latency):
0.03 + 0.02 + 0.075 = 0.125 s = 125 ms

Final Answer:
Client wait time (latency) = 125 ms

Q4

A client C and a server S, located 12,000 km apart, are connected via a cable.
A request is sent by the client to the server, but while sending back the response, the cable breaks and the response is now sent via air.

This change in medium causes an additional delay of 20 ms.

As compared to the Round Trip Time (RTT) of the healthy network, the RTT of the faulty system would ________.

Options:
A. Increase
B. Decrease
C. Remain Same
D. Insufficient Information

(Speed of light in cable = 2 × 10^8 m/s, in air = 3 × 10^8 m/s)

Solution:

             request ————>>——————— (cable)
   Client C                          Server S
             <<—————— response ———— (air)
   Distance between A and B = 12000 km

Case 1: Healthy network
RTT1 = 2 × (12000 × 1000 / 2 × 10^8) = 0.12 s = 120 ms
Case 2: Faulty network
RTT2 = (12000 × 1000 / 2 × 10^8) + (12000 × 1000 / 3 × 10^8) + 0.02
= 0.06 + 0.04 + 0.02
= 0.12 s = 120 ms

Conclusion:
RTT remains the same despite the medium change and server delay.

Answer: C. Remain Same

Q5

A network has a bandwidth of 8 Gbps. A client wants to send 5000 requests per second over this network.
What should be the size of each request in MB?

Use the following unit conversions:

1 Byte = 8 bits
1 KB = 1000 Bytes
1 MB = 1000 KB
1 GB = 1000 MB

Hint: Bandwidth can be expressed in two equivalent ways:

Bandwidth = Number of requests per second × Size of each request
Bandwidth = Total data per second

Solution:

Step 1: Using requests/sec × size per request

Bandwidth (bits/sec) = Number of requests/sec × Size per request (bits)

8,000,000,000 = 5000 × S

S = 8,000,000,000 / 5000 = 1,600,000 bits

Convert to bytes:

S = 1,600,000 / 8 = 200,000 Bytes = 0.2 MB

Step 2: Using total data per second analogy

Total data sent per second = 5000 requests × 0.2 MB/request = 1000 MB = 1 GB/s

Convert to bits/sec:

1 GB = 8 × 10^9 bits → Bandwidth = 8 Gbps

This confirms the calculation matches the given network bandwidth.

Answer:
Size of each request = 0.2 MB

Key Takeaway:

Bandwidth = requests/sec × size per request → emphasizes request granularity
Bandwidth = total data per second → emphasizes total data transferred
Both approaches give the same result and help understand network data flow.

Q8

Step 1: Extract bandwidth segments from the graph

From the graph you shared:

0–4 h: BW = 8 Mb/s
4–8 h: BW = 6 Mb/s
8–12 h: BW = 8 Mb/s
12–16 h: BW = 2 Mb/s
16–20 h: BW = 2 Mb/s
20–24 h: BW = 4 Mb/s

(These are the orange steps in the graph representing total network usage.)

Step 2: Formula

 Data= Bandwidth × Time

Time must be in seconds (1 hour = 3600 s)
Convert bits to bytes by dividing by 8

Step 3: Calculate total data

Segment 0–4 h: 8 Mb/s × 4 h = 8 × 4 × 3600 = 115,200 Mb
Segment 4–8 h: 6 Mb/s × 4 h = 6 × 4 × 3600 = 86,400 Mb
Segment 8–10 h: 8 Mb/s × 2 h = 8 × 2 × 3600 = 57,600 Mb
Segment 10–14 h: 2 Mb/s × 4 h = 2 × 4 × 3600 = 28,800 Mb
Segment 14–18 h: 4 Mb/s × 4 h = 4 × 4 × 3600 = 57,600 Mb

Sum = 115,200 + 86,400 + 57,600 + 28,800 + 57,600 = 345,600 Mb

Convert to MB: 345,600 / 8 = 43,200 MB = 43.2 GB

Answer: 43.2 GB → Option C is correct.