Suppose in a web browser, you click on the (formula) . — GATE CS 2023 Q42

Suppose in a web browser, you click on the $\textsf{www. gate-2023.in}\; \text{URL}$ . The browser cache is empty. The $\text{IP address}$ for this $\text{URL}$ is not cached in your local host, so a $\text{DNS}$ lookup is triggered (by the local $\text{DNS}$ server deployed on your local host) over the $3$ -tier $\text{DNS}$ hierarchy in an iterative mode. No resource records are cached anywhere across all $\text{DNS}$ servers.

Let $\text{RTT}$ denote the round trip time between your local host and $\text{DNS}$ servers in the $\text{DNS}$ hierarchy. The round trip time between the local host and the web server hosting $\textsf{www.gate-2023.in}$ is also equal to $\text{RTT}$ . The $\text{HTML}$ file associated with the $\text{URL}$ is small enough to have negligible transmission time and negligible rendering time by your web browser, which references $10$ equally small objects on the same web server.

Which of the following statements is/are $\text{CORRECT}$ about the minimum elapsed time between clicking on the $\text{URL}$ and your browser fully rendering it?

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Answer: C;D

To find the minimum elapsed time, we calculate the time taken for DNS resolution and the subsequent HTTP object retrieval.

1. DNS Resolution Time:
Since the local DNS server is deployed on the local host, the communication between them takes $0\text{ RTT}$ . The local DNS server queries the $3$ -tier DNS hierarchy (Root, TLD, and Authoritative servers) in iterative mode. With no cached records anywhere: $\text{DNS Time} = 1\text{ RTT (Root)} + 1\text{ RTT (TLD)} + 1\text{ RTT (Authoritative)} = 3\text{ RTT}$

2. Non-persistent HTTP with $5$ parallel connections (Options A & C):

Base HTML: $1\text{ RTT}$ (TCP handshake) + $1\text{ RTT}$ (HTTP request/response) = $2\text{ RTT}$ .
10 Objects: Since we have $5$ $5$ parallel connections, we can fetch the $10$ $10$ objects in $2$ $2$ consecutive batches of $5$ $5$ objects each:
- Batch 1 ( $5$ objects): $1\text{ RTT}$ (handshake) + $1\text{ RTT}$ (transfer) = $2\text{ RTT}$
- Batch 2 ( $5$ objects): $1\text{ RTT}$ (handshake) + $1\text{ RTT}$ (transfer) = $2\text{ RTT}$

\text{Total HTTP Time} = 2\text{ RTT} + 2\text{ RTT} + 2\text{ RTT} = 6\text{ RTT}

\text{Total Elapsed Time} = 3\text{ RTT (DNS)} + 6\text{ RTT (HTTP)} = 9\text{ RTT}

Thus, Option C is correct.

3. Persistent HTTP with pipelining (Options B & D):

Base HTML: $1\text{ RTT}$ (TCP handshake) + $1\text{ RTT}$ (HTTP request/response) = $2\text{ RTT}$ .
10 Objects: The connection remains open (persistent), and all $10$ $10$ requests are sent back-to-back without waiting for replies (pipelined):
- Request and receive all $10$ objects: $1\text{ RTT}$

\text{Total HTTP Time} = 2\text{ RTT} + 1\text{ RTT} = 3\text{ RTT}

\text{Total Elapsed Time} = 3\text{ RTT (DNS)} + 3\text{ RTT (HTTP)} = 6\text{ RTT}

Thus, Option D is correct.

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GATE CS 2023 Q42 — Fragmentation and IP Addressing, IPv4, CIDR

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