CSE 120: Homework #2

Spring 2019

Out: Tuesday April 16
Due: Saturday April 27 at 11:59pm

For the homework questions below, if you believe that you cannot answer a question without making some assumptions, state those assumptions in your answer.

The Intel x86 instruction set architecture provides an atomic instruction called XCHG for implementing synchronization primitives. (If you are curious, this reference page shows the full syntax and semantics of the instruction.) Semantically, XCHG works as follows (although keep in mind it is executed atomically):
```
void XCHG (bool *X, bool *Y) {
  bool tmp = *X;
  *X = *Y;
  *Y = tmp;
}
```
Show how XCHG can be used instead of test-and-set to implement the acquire() and release() functions of the spinlock data structure described in the "Synchronization" lecture.
```
struct lock {
  ...
}

void acquire (struct lock *) {
  ...
}

void release (struct lock *) {
  ...
}
```
One of the goals of this question is to give you practice with context switching and thread queue manipulation in Nachos. Consider the following test program for an implementation of KThread.join in Nachos. It begins when the main Nachos thread calls KThread.selfTest. You do not need to know the details of how join is implemented. All you need to know is that when a parent thread calls join on a child thread, the parent does one of two things: (1) if the child is still running, the parent blocks until the child finishes (at which point the parent is placed on the ready queue); (2) if the child has finished, the parent continues to execute without blocking. Assume join uses a wait queue of some kind in its implementation.
```
private static class A implements Runnable {
    A () {}
    public void run () {
        KThread t2 = new KThread (new B()).setName ("B");
	System.out.println ("foo");
	t2.fork ();
	System.out.println ("far");
	t2.join ();
	System.out.println ("fum");
    }
}
    
private static class B implements Runnable {
    B () {}
    public void run () {
        System.out.println ("fie");
    }
}

public static void selfTest() {
    KThread t1 = new KThread (new A()).setName ("A");
    System.out.println ("fee");
    t1.fork ();
    System.out.println ("foe");
    t1.join ();
    System.out.println ("fun");
}
```
Assume that the scheduler runs threads in FIFO order with non-preemptive scheduling (no preemptive time-slicing), and threads are placed on wait queues in FIFO order. Trace the execution of this program until it returns from selfTest and (a) write the sequence of context switches that occurred up this point, (b) write the output of the program, and (c) list the queues that the threads are on, and their relative order if more than one thread is on a queue.
a. Context switches: main →
b. Output:
c. Thread queues when selfTest returns:
currentThread:
readyQueue:
join wait queue:

[Hint: First try the problem by following the code manually, keeping track of which queues threads are on using paper. Then, once you have implemented join, try adding the code as a test in KThread.java and running it to check your answer.]
A common pattern in parallel scientific programs is to have a set of threads do a computation in a sequence of phases. In each phase i, all threads must finish phase i before any thread starts computing phase i+1. One way to accomplish this is with barrier synchronization. At the end of each phase, each thread executes Barrier::Done(n), where n is the number of threads in the computation. A call to Barrier::Done blocks until all of the n threads have called Barrier::Done. Then, all threads proceed. You may assume that the process allocates a new Barrier for each iteration, and that all threads of the program will call Done with the same value.
a. Write a monitor that implements Barrier using Mesa semantics.
```
monitor Barrier {
  ...
}
```
b. Implement Barrier using an explicit lock and condition variable. The lock and condition variable have the semantics described at the end of the "Semaphore and Monitor" lecture in the ping_pong example, and as implemented by you in Project 1.
```
class Barrier {
  ...private variables...
  void Done (int n) {
    ...
  }
  ...
}
```
Microsoft .NET provides a synchronization primitive called a CountdownEvent. Programs use CountdownEvent to synchronize on the completion of many threads (similar to CountDownLatch in Java). A CountdownEvent is initialized with a count, and a CountdownEvent can be in two states, nonsignalled and signalled. Threads use a CountdownEvent in the nonsignalled state to Wait (block) until the internal count reaches zero. When the internal count of a CountdownEvent reaches zero, the CountdownEvent transitions to the signalled state and wakes up (unblocks) all waiting threads. Once a CountdownEvent has transitioned from nonsignalled to signalled, the CountdownEvent remains in the signalled state. In the nonsignalled state, at any time a thread may call the Decrement operation to decrease the count and Increment to increase the count. In the signalled state, Wait, Decrement, and Increment have no effect and return immediately.
Use pseudo-code to implement a thread-safe CountdownEvent using locks and condition variables by implementing the following methods:
```
class CountdownEvent {
  ...private variables...
  CountdownEvent (int count) { ... }
  void Increment () { ... }
  void Decrement () { ... }
  void Wait () { ... }
}
```
Notes:
- The CountdownEvent constructor takes an integer count as input and initializes the CountdownEvent counter with count. Positive values of count cause the CountdownEvent to be constructed in the nonsignalled state. Other values of count will construct it in the signalled state.
- Increment increments the internal counter.
- Decrement decrements the internal counter. If the counter reaches zero, the CountdownEvent transitions to the signalled state and unblocks any waiting threads.
- Wait blocks the calling thread if the CountdownEvent is in the nonsignalled state, and otherwise returns.
- Each of these methods is relatively short.
[Silberschatz] Consider a system running ten I/O-bound tasks and one CPU-bound task. Assume that the I/O-bound tasks issue an I/O operation once for every millisecond of CPU computing and that each I/O operation takes 10 milliseconds to complete. Also assume that the context-switching overhead is 0.1 millisecond and that all processes are long-running tasks. What is the CPU utilization for a round-robin scheduler when:
a. The time quantum is 1 millisecond
b. The time quantum is 10 milliseconds
Annabelle, Bertrand, Chloe and Dag are working on their term papers in CSE 120, which is a 10,000 word essay on My All-Time Favorite Race Conditions. To help them work on their papers, they have one dictionary, two copies of Roget's Thesaurus, and two coffee cups.
- Annabelle needs to use the dictionary and a thesaurus to write her paper;
- Bertrand needs a thesaurus and a coffee cup to write his paper;
- Chloe needs a dictionary and a thesaurus to write her paper;
- Dag needs two coffee cups to write his paper (he likes to have a cup of regular and a cup of decaf at the same time to keep himself in balance).
Consider the following state:
- Annabelle has a thesaurus and needs the dictionary.
- Bertrand has a thesaurus and a coffee cup.
- Chloe has the dictionary and needs a thesaurus.
- Dag has a coffee cup and needs another coffee cup.
Is the system deadlocked in this state? Explain using a resource allocation graph as a reference.