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The first version creates a mutex that is initially unowned In the second version, if initiallyOwned is true, the initial state of the mutex is owned by the calling thread Otherwise, it is unowned To acquire the mutex, your code will call WaitOne( ) on the mutex This method is inherited by Mutex from the ThreadWaitHandle class Here is its simplest form: public bool WaitOne( ); It waits until the mutex on which it is called can be acquired Thus, it blocks execution of the calling thread until the specified mutex is available It always returns true When your code no longer needs ownership of the mutex, it releases it by calling ReleaseMutex( ), shown here: public void ReleaseMutex( ) This releases the mutex on which it is called, enabling the mutex to be acquired by another thread To use a mutex to synchronize access to a shared resource, you will use WaitOne( ) and ReleaseMutex( ), as shown in the following sequence:

Mutex myMtx = new Mutex(); // myMtxWaitOne(); // wait to acquire the mutex // Access the shared resource myMtxReleaseMutex(); // release the mutex

When the call to WaitOne( ) takes place, execution of the thread will suspend until the mutex can be acquired When the call to ReleaseMutex( ) takes place, the mutex is released and another thread can acquire it Using this approach, access to a shared resource can be limited to one thread at a time The following program puts this framework into action It creates two threads, IncThread and DecThread, which both access a shared resource called SharedResCount IncThread increments SharedResCount and DecThread decrements it To prevent both threads from accessing SharedResCount at the same time, access is synchronized by the Mtx mutex, which is also part of the SharedRes class

// Use a Mutex using System; using SystemThreading; // This class contains a shared resource (Count), // and a mutex (Mtx) to control access to it class SharedRes { public static int Count = 0; public static Mutex Mtx = new Mutex(); } // This thread increments SharedResCount

We divide by 5000 to obtain 2 = eK 4 Taking the natural logarithm of both sides yields ln 2 = ln( eK 4 ) = 4K We conclude that K = [ln 2]/4 As a result, B( t) = 5000 e( [ln 2]/4) t We simplify this equation by noting that e( [ln 2]/4) t = ( eln 2 ) t/4 = 2t/4 In conclusion, B( t) = 5000 2t/4 The number of bacteria at noon (time t = 3) is then given by B( 3) = 5000 23/4 8409

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Part II:

class IncThread { int num; public Thread Thrd; public IncThread(string name, int n) { Thrd = new Thread(thisRun); num = n; ThrdName = name; ThrdStart(); } // Entry point of thread void Run() { ConsoleWriteLine(ThrdName + " is waiting for the mutex"); // Acquire the Mutex SharedResMtxWaitOne(); ConsoleWriteLine(ThrdName + " acquires the mutex"); do { ThreadSleep(500); SharedResCount++; ConsoleWriteLine("In " + ThrdName + ", SharedResCount is " + SharedResCount); num--; } while(num > 0); ConsoleWriteLine(ThrdName + " releases the mutex"); // Release the Mutex SharedResMtxReleaseMutex(); } } // This thread decrements SharedResCount class DecThread { int num; public Thread Thrd; public DecThread(string name, int n) { Thrd = new Thread(new ThreadStart(thisRun)); num = n; ThrdName = name; ThrdStart(); } // Entry point of thread void Run() { ConsoleWriteLine(ThrdName + " is waiting for the mutex");

23:

It is important to realize that population growth problems cannot be described using just arithmetic Exponential growth is nonlinear, and advanced analytical ideas (such as calculus) must be used to understand it

// Acquire the Mutex SharedResMtxWaitOne(); ConsoleWriteLine(ThrdName + " acquires the mutex"); do { ThreadSleep(500); SharedResCount--; ConsoleWriteLine("In " + ThrdName + ", SharedResCount is " + SharedResCount); num--; } while(num > 0);

ConsoleWriteLine(ThrdName + " releases the mutex"); // Release the Mutex SharedResMtxReleaseMutex(); } } class MutexDemo { static void Main() { // Construct three threads IncThread mt1 = new IncThread("Increment Thread", 5); ThreadSleep(1); // let the Increment thread start DecThread mt2 = new DecThread("Decrement Thread", 5); mt1ThrdJoin(); mt2ThrdJoin(); } }

The output is shown here:

Suppose that a certain petri dish contains 6000 bacteria at 9:00 pm and 10,000 bacteria at 11:00 pm How many of the bacteria were there at 7:00 pm