CS4410- hw2-soln

CS4410- hw2-soln - CS4410 Fall 2008 Homework 2 Solution...

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CS4410 - Fall 2008 Homework 2 Solution Due September 23, 11:59PM Q1. Explain what goes wrong in the following version of Dekker’s Algorithm: CSEnter(int i) { inside[i] = true; while(inside[j]) { inside[i] = false; while(turn == j) continue; inside[i] = true; } } CSExit(int i) { turn = j; inside[i] = false; } Answer: There is a possibility of starvation here. Suppose process j is in critical section and process i is busy waiting at the inner while loop. Now when process j exits the critical section it sets turn = i, but it immediately tried to access the critical section and so sets inside[j] = true. Right before process j executes “while (inside[i])”, scheduler schedules process i. Now process i exits from the inner while loop as turn = = i now, but it finds inside[j] to be true so continue the outer while loop. Then after it executes “inside[i]=false”, the scheduler schedules process j and it finds the condition at outer while loop to be false and enters the critical section. As a result, process i busy waits at outer while loop. These steps can repeat arbitrary number of times and that starves process i and it may never enter the critical section. In the original algorithm, there is an extra checking “if (turn==j)” which ensures that this starvation never happens. Grading guide: -5 : Failing to mention starvation / violation of bounded waiting; mentioning it violates mutual exclusion (two threads inside the critical section simultaneously) which actually does not happen here. -1 : Mentioning starvation, but no explanation how it happens -5 : Just mentioning it does not have "if (turn==j)", but no explanation on why it is wrong (i.e., starvation/violation of bounded waiting)
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Q2. Round-robin schedulers normally maintain a list of all runnable processes, with each
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CS4410- hw2-soln - CS4410 Fall 2008 Homework 2 Solution...

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