I/O Mapping MCQ Quiz - Objective Question with Answer for I/O Mapping - Download Free PDF
Last updated on Apr 8, 2025
Latest I/O Mapping MCQ Objective Questions
I/O Mapping Question 1:
Process P1, P2, and P3 with execution time of 6 ms, 4 ms, and 2 ms respectively enter in ready state together in order P1, P2, P3. Calculate the waiting and turnaround time of Process P1. Assuming no wait time due to I/O and round-robin scheduling with time slot of 2 ms.
Answer (Detailed Solution Below)
I/O Mapping Question 1 Detailed Solution
Concept:
- Scheduling of a process in operating systems is done to finish the work in time.
- Different timing description with respect to a process is as shown:
Arrival time: Time at which the process arrives in the ready Queue.
Completion/Execution Time: Time at which the process completes its execution.
Burst Time: Time Required by a process for CPU execution
Turn-Around Time: Time difference between completion time and arrival time.
Turn around time = Completion Time – Arrival Time.
Calculation:
Given Wait time due to I/O = 0s
Completion / Execution Time of P1 = 6ms
Completion / Execution Time of P2 = 4ms
Completion / Execution Time of P3 = 2ms
Waiting Time of Process P1 = Execution time of P1 + Execution time of P2 + Execution time fo P3 + Wait time due to I/O.
= 6ms + 4ms + 2ms + 0
= 12ms
Turn around time of process P1 = Execution Time - Arrival time.
= 6 ms - 0
= 6ms
So, Option (1) is Correct.
I/O Mapping Question 2:
In a microprocessor-based system, a DMA facility is required to increase the speed of the data transfer between the
Answer (Detailed Solution Below)
I/O Mapping Question 2 Detailed Solution
- Direct memory access (DMA) is a feature of computer systems that allows certain hardware subsystems to access main system memory (random-access memory), independent of the central processing unit (CPU).
- DMA avoids using the CPU thus allowing the CPU to attend another job
- It increases the speed of the data transfer between the memory and the I/O devices
- DMA is used when a large amount of data is to be printed out from the memory of a computer
- The problem of slow data transfer between input-output port and memory or between two memory is avoided by implementing the Direct Memory Access (DMA) technique
- It is implemented by using 8257 DMA controller
Top I/O Mapping MCQ Objective Questions
In a microprocessor-based system, a DMA facility is required to increase the speed of the data transfer between the
Answer (Detailed Solution Below)
I/O Mapping Question 3 Detailed Solution
Download Solution PDF- Direct memory access (DMA) is a feature of computer systems that allows certain hardware subsystems to access main system memory (random-access memory), independent of the central processing unit (CPU).
- DMA avoids using the CPU thus allowing the CPU to attend another job
- It increases the speed of the data transfer between the memory and the I/O devices
- DMA is used when a large amount of data is to be printed out from the memory of a computer
- The problem of slow data transfer between input-output port and memory or between two memory is avoided by implementing the Direct Memory Access (DMA) technique
- It is implemented by using 8257 DMA controller
Process P1, P2, and P3 with execution time of 6 ms, 4 ms, and 2 ms respectively enter in ready state together in order P1, P2, P3. Calculate the waiting and turnaround time of Process P1. Assuming no wait time due to I/O and round-robin scheduling with time slot of 2 ms.
Answer (Detailed Solution Below)
I/O Mapping Question 4 Detailed Solution
Download Solution PDFConcept:
- Scheduling of a process in operating systems is done to finish the work in time.
- Different timing description with respect to a process is as shown:
Arrival time: Time at which the process arrives in the ready Queue.
Completion/Execution Time: Time at which the process completes its execution.
Burst Time: Time Required by a process for CPU execution
Turn-Around Time: Time difference between completion time and arrival time.
Turn around time = Completion Time – Arrival Time.
Calculation:
Given Wait time due to I/O = 0s
Completion / Execution Time of P1 = 6ms
Completion / Execution Time of P2 = 4ms
Completion / Execution Time of P3 = 2ms
Waiting Time of Process P1 = Execution time of P1 + Execution time of P2 + Execution time fo P3 + Wait time due to I/O.
= 6ms + 4ms + 2ms + 0
= 12ms
Turn around time of process P1 = Execution Time - Arrival time.
= 6 ms - 0
= 6ms
So, Option (1) is Correct.