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Computer Science Engineering (CSE) Exam  >  Computer Science Engineering (CSE) Questions  >  Six files F1, F2, F3, F4, F5 and F6 have 100,... Start Learning for Free
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Six files F1, F2, F3, F4, F5 and F6 have 100, 200, 50, 80, 120, 150 records respectively. In what order should they be stored so as to optimize act. Assume each file is accessed with the same frequency
  • a)
    F3, F4, F1, F5, F6, F2
  • b)
    F2, F6, F5, F1, F4, F3
  • c)
    F1, F2, F3, F4, F5, F6
  • d)
    Ordering is immaterial as all files are accessed with the same frequency.
Correct answer is option 'A'. Can you explain this answer?
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Verified Answer
Six files F1, F2, F3, F4, F5 and F6 have 100, 200, 50, 80, 120, 150 re...
This question is based on the Optimal Storage on Tape problem which uses greedy approach to find the optimal time to retrieve them.
There are n programs of length L that are to be stored on a computer tape. Associated with each program i is a length Li. So in order to
retrieve these programs most optimally, we need to store them in the non-decreasing order of length Li.
So, the correct order is F3, F4, F1, F5, F6, F2
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Six files F1, F2, F3, F4, F5 and F6 have 100, 200, 50, 80, 120, 150 re...
Understanding the Problem

The problem is to determine the optimal order in which to store six files (F1, F2, F3, F4, F5, and F6) in order to optimize access. Each file has a different number of records, and all files are accessed with the same frequency.

Approach

To optimize access, we need to arrange the files in a way that minimizes the total number of accesses required. The idea is to place the files with the highest frequency of access closer to the beginning of the storage, reducing the overall access time.

Analyzing the Options

Let's analyze each option to determine the most optimal order:

Option A: F3, F4, F1, F5, F6, F2
- F3 (50 records) and F4 (80 records) are placed at the beginning, which have the least number of records. This arrangement helps in reducing the initial access time.
- F1 (100 records) is placed next, followed by F5 (120 records) and F6 (150 records).
- Finally, F2 (200 records) is placed at the end, as it has the highest number of records.

Option B: F2, F6, F5, F1, F4, F3
- F2, with the highest number of records, is placed at the beginning, which increases the initial access time.
- F6, F5, and F1 are placed next, which is reasonable considering their record counts.
- F4 and F3, with the least number of records, are placed at the end, which increases the overall access time.

Option C: F1, F2, F3, F4, F5, F6
- This option arranges the files in ascending order of their record counts, which does not consider the frequency of access.
- Placing F1 and F2, which have the highest number of records, at the beginning increases the initial access time significantly.

Option D: Ordering is immaterial as all files are accessed with the same frequency.
- This option assumes that the order of storage does not affect the access time, which is not correct given the different number of records in each file.

Conclusion

Based on the analysis, Option A: F3, F4, F1, F5, F6, F2 is the most optimal arrangement as it considers the frequency of access and places files with fewer records at the beginning. This arrangement minimizes the overall access time and optimizes the storage order.
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Community Answer
Six files F1, F2, F3, F4, F5 and F6 have 100, 200, 50, 80, 120, 150 re...
This question is based on the Optimal Storage on Tape problem which uses greedy approach to find the optimal time to retrieve them.
There are n programs of length L that are to be stored on a computer tape. Associated with each program i is a length Li. So in order to
retrieve these programs most optimally, we need to store them in the non-decreasing order of length Li.
So, the correct order is F3, F4, F1, F5, F6, F2
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An instruction pipeline has five stages, namely, instruction fetch (IF), instruction decode and register fetch (ID/RF), instruction execution (EX), memory access (MEM), and register writeback (WB) with stage latencies 1 ns, 2.2 ns, 2 ns, 1 ns, and 0.75 ns, respectively (ns stands for nanoseconds). To gain in terms of frequency, the designers have decided to split theID/RF stage into three stages (ID, RF1, RF2) each of latency 2.2/3 ns. Also, the EX stage is split into two stages (EX1, EX2) each of latency 1 ns. The new design has a total of eight pipeline stages. A program has 20% branch instructions which execute in the EX stage and produce the next instruction pointer at the end of the EX stage in the old design and at the end of the EX2 stage in the new design. The IF stage stalls after fetching a branch instruction until the next instruction pointer is computed. All instructions other than the branch instruction have an average CPI of one in both the designs. The execution times of this program on the old and the new design are P and Q nanoseconds, respectively. The value of P/Q is __________. Correct answer is '1.54'. Can you explain this answer?

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Six files F1, F2, F3, F4, F5 and F6 have 100, 200, 50, 80, 120, 150 records respectively. In what order should they be stored so as to optimize act. Assume each file is accessed with the same frequencya)F3, F4, F1, F5, F6, F2b)F2, F6, F5, F1, F4, F3c)F1, F2, F3, F4, F5, F6d)Ordering is immaterial as all files are accessed with the same frequency.Correct answer is option 'A'. Can you explain this answer?
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