A disk head is initially positioned at track 32. The disk I/O request queue, in arrival order, is: 98, 37, 14, 124, 65, 67. Using the FCFS disk-scheduling algorithm, find the total number of head movements (in tracks) required to service all requests.
Pick an option to check yourself, or open the answer below.
Show answer & explanation
Correct answer: C. 321
Explanation
FCFS services requests strictly in their arrival order, without any reordering by proximity. Starting at track 32: |98−32|=66, |37−98|=61, |14−37|=23, |124−14|=110, |65−124|=59, |67−65|=2. Summing these: 66+61+23+110+59+2 = 321 total track movements, option (iii).More System Software and Operating System questions
- Arrange the following levels of interrupt protection within the Linux Kernel in order of increasing priority: (A) user mode programs (B) bo…
- Given below are an Assertion (A) and a Reason (R): Assertion (A): I/O protection is ensured by a hardware trap. Reason (R): I/O interrupts…
- In a multi-user operating system, requests for a particular resource arrive at an average rate of 20 per hour (a Poisson arrival process).…
- User level threads are threads that are visible to the programmer and are unknown to the kernel. The operating system kernel supports and m…
- One of the disadvantages of user-level threads compared to kernel-level threads is:
- Match List I with List II regarding operating system I/O management. List I: (a) Device controller (b) Device driver (c) Interrupt handler…