CS 540 Operating Systems Test 2 - Name: Date: Thursday, April 25 ...

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2. Suppose that the following processes arrive for execution at time 0 in the order A, B, C:Process Run Time PriorityA 6 2B 2 3 = lowC 4 1 = high(a) Draw four Gantt charts illustrating the execution of these processes using FCFS, SJF, a nonpreemptivepriority (a smaller priority number implies a higher priority), and RR (quantum = 2)scheduling.(b) What is the waiting time of each process for each of the scheduling algorithms?(c) What is the turnaround time of each process for each of the scheduling algorithms?(12 points)Ans:a. The four Gantt charts are0 6 8 12 0 2 6 12+---+-+--+ +-+--+---+| A |B| C| FCFS |B| C| A | SJF+---+-+--+ +-+--+---+0 4 10 12 0 2 4 6 8 10 12+--+---+-+ +-+-+-+-+-+-+|C | A |B| Priority |A|B|C|A|C|A| RR+--+---+-+ +-+-+-+-+-+-+b. Waiting time:| FCFS | SJF | Priority | RR---+------+-----+----------+----A | 0 | 6 | 4 | 6B | 6 | 0 | 10 | 2C | 8 | 2 | 0 | 6c. Turnaround time:| FCFS | SJF | Priority | RR---+------+-----+----------+----A | 6 | 12 | 10 | 12B | 8 | 2 | 12 | 4C | 12 | 6 | 4 | 103. Suppose we have 4 processes: A, B, C, and D. A must finish executing before B and C start. B and Cmust finish executing before D starts. Show your solution. Remember to indicate the initial value ofeach semaphore. (10 points)Ans:S1 S2A ----> B ----> D| ^| S3 S4 |+-----> C ------+Then, the processes may be synchronized using semaphores:
semaphore S1, S2, S3, S4 = 0,0,0,0;Process A:----------- do work of AUP(S1); /* Let B start */UP(S3); /* Let C start */Process B:----------DOWN(S1); /* Block until A is finished */- do work of BUP(S2);Process C:----------DOWN(S3);- do work of CUP(S4);Process D:----------DOWN(S2);DOWN(S4);- do work of D4. Consider a swapping system in which memory consists of the following hole sizes in memory order:10K, 4K, 20K, 15K, and 9K. Which hole is taken for successive segment requests of:(a) 8K (b) 12K (c) 10Kfor first fit, best fit, and worst fit. (8 points)Ans: First fit : (a) 10K (b) 20K (c) 15KBest fit : (a) 9K (b) 15K (c) 10KWorst fit : (a) 20K (b) 15K (c) leftover of 20K5. P is a set of processes. R is a set of resources. E is a set of request or assignment edges. The sets P,R, and E are as follows: (10 points)P = {P 1 , P 2 , P 3 } R = {R 1 , R 2 , R 3 } E = {P 1 → R 1 , P 2 → R 2 , P 3 → R 3 , R 1 → P 3 , R 2 → P 1 , R 3 → P 2 }R1 has one instance. R2 has one instances. R3 has one instance.(a) Draw the resource-allocation graph.(b) Is there any deadlock in this situation? Briefly Explain.Ans:b) Consider the resource-allocation graph. One cycle exist in the system.P1 -> R1 -> P3 -> R3 -> P2 -> R2 -> P1P1, P2, and P3 are deadlocked.
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CS 540 Operating Systems Test 2 - Name: Date: Thursday, April 25 ...

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