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9618 · 13.2

File organisation and access — practice questions

Practice and worked examples for 9618 File organisation and access. Short previews only — attempt the full question in MarkScheme against the official scheme.

Worked example 1

A sequential file on a magnetic tape stores 20,000 employee records. The system takes 0.8 ms to read a single record. Calculate the time taken, in seconds, to access the 15,000th record. State one assumption you have made.

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To access the 15,000th record sequentially, the system must read all records from the beginning up to that point.

Working:

  1. Number of records to read = 15,000
  2. Time per record = 0.8 ms
  3. Total time in milliseconds = 15,000 records * 0.8 ms/record = 12,000 ms
  4. Convert to seconds: 12,000 ms / 1000 = 12 seconds

Answer: 12 seconds.

Assumption: The time to read each record is constant and there are no other delays (e.g., from tape acceleration/deceleration or system interrupts).

Worked example 2

A random access file is used to store 1000 patient records, indexed from 0 to 999. The hashing algorithm used is Address=PatientIDMOD1000Address = PatientID MOD 1000.

(a) Calculate the storage address for a patient with ID 45821. (b) A new patient with ID 77821 needs to be added. Calculate their address and identify the problem that arises.

Show solution outline

(a) Address for Patient ID 45821:

  1. Apply the hashing algorithm: Address=45821MOD1000Address = 45821 MOD 1000
  2. The result of the MOD operator is the remainder after division. The remainder of 45821 divided by 1000 is 821.
  3. Address = 821

(b) Address for Patient ID 77821 and the problem:

  1. Apply the hashing algorithm: Address=77821MOD1000Address = 77821 MOD 1000
  2. The remainder of 77821 divided by 1000 is also 821.
  3. Address = 821

Problem: Both Patient IDs (45821 and 77821) hash to the same storage address (821). This is known as a collision or a synonym. The system cannot store two different records in the same physical location. A collision handling technique, such as placing the second record in an overflow area and creating a pointer to it, would be required.

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