Consider a file of size 4 million bytes being transferred between two hosts connected via a path consisting of three consecutive links of bandwidth 2 Mbps, 500 kbps, and 1 Mbps, respectively. All processing delays and propagation delays are negligible. Assume that there is no other background traffic over the path and no other additional overhead to transfer the file.
Which one of the following is the total time (in seconds) to transfer the file?
Note: $1M=10^6$, $1k=10^3$

Question

A database has 25,000 fixed-length records of 100 bytes (primary key 15 bytes). The data file is block-aligned so each record is fully contained within a block. The file is indexed by a primary index file, also block-aligned and ordered. Block size is 1024 bytes and a block pointer is 5 bytes. Each index entry stores the block’s anchor key (15 bytes) and a pointer (5 bytes). Binary search on an index file of $b$ blocks needs $\lceil \log_2 b \rceil$ block accesses in the worst case. Given a key, the number of block accesses required to identify the data file block that may contain the record, in the worst case, is ____________.

Answer 1

The trick in this question is recognizing that a chain of links with different capacities behaves like a single pipe whose overall carrying capacity is set by its narrowest section, the link with the least bandwidth.

Work out the data size in bits: the file is $4 \times 10^6$ bytes. Multiplying by 8 bits per byte gives $32 \times 10^6$ bits that must cross the path.
List the three link capacities in a common unit (bps): $2 \times 10^6$ bps, $0.5 \times 10^6$ bps, and $1 \times 10^6$ bps for the first, second and third links respectively.
Pick the bottleneck: among $2 \times 10^6$, $0.5 \times 10^6$, and $1 \times 10^6$, the smallest is $0.5 \times 10^6$ bps, the 500 kbps link. Because propagation and processing delays are zero and there is no competing traffic, the sustained end-to-end throughput of the whole three-link path equals this smallest capacity; a faster link before or after it cannot push data through the slow middle link any quicker.
Divide total bits by the bottleneck rate: $\dfrac{32 \times 10^6 \text{ bits}}{0.5 \times 10^6 \text{ bps}} = 64$ seconds.

This matches option (B). The other listed values correspond to using the wrong link's rate on its own, 16 s from the 2 Mbps link or roughly 32 s from the 1 Mbps link, rather than the true bottleneck, so they do not represent the actual achievable transfer rate of the path.

Let's summarize:

A file's transfer time across several links in series, with no meaningful store-and-forward delay, is set by the link with the smallest bandwidth, the bottleneck.
Total time equals total bits divided by that bottleneck bandwidth, here $32 \times 10^6 / 0.5 \times 10^6 = 64$ s.

So the total time to transfer the file is 64 seconds.

Answer 2

The trick in this question is recognizing that a chain of links with different capacities behaves like a single pipe whose overall carrying capacity is set by its narrowest section, the link with the least bandwidth.

Work out the data size in bits: the file is $4 \times 10^6$ bytes. Multiplying by 8 bits per byte gives $32 \times 10^6$ bits that must cross the path.
List the three link capacities in a common unit (bps): $2 \times 10^6$ bps, $0.5 \times 10^6$ bps, and $1 \times 10^6$ bps for the first, second and third links respectively.
Pick the bottleneck: among $2 \times 10^6$, $0.5 \times 10^6$, and $1 \times 10^6$, the smallest is $0.5 \times 10^6$ bps, the 500 kbps link. Because propagation and processing delays are zero and there is no competing traffic, the sustained end-to-end throughput of the whole three-link path equals this smallest capacity; a faster link before or after it cannot push data through the slow middle link any quicker.
Divide total bits by the bottleneck rate: $\dfrac{32 \times 10^6 \text{ bits}}{0.5 \times 10^6 \text{ bps}} = 64$ seconds.

This matches option (B). The other listed values correspond to using the wrong link's rate on its own, 16 s from the 2 Mbps link or roughly 32 s from the 1 Mbps link, rather than the true bottleneck, so they do not represent the actual achievable transfer rate of the path.

Let's summarize:

A file's transfer time across several links in series, with no meaningful store-and-forward delay, is set by the link with the smallest bandwidth, the bottleneck.
Total time equals total bits divided by that bottleneck bandwidth, here $32 \times 10^6 / 0.5 \times 10^6 = 64$ s.

So the total time to transfer the file is 64 seconds.

Show Hint

The Correct Option is B

Solution and Explanation

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