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Increment Thread is waiting for the mutex Increment Thread acquires the mutex Decrement Thread is waiting for the mutex In Increment Thread, SharedResCount is 1 In Increment Thread, SharedResCount is 2 In Increment Thread, SharedResCount is 3 In Increment Thread, SharedResCount is 4 In Increment Thread, SharedResCount is 5 Increment Thread releases the mutex Decrement Thread acquires the mutex In Decrement Thread, SharedResCount is 4 In Decrement Thread, SharedResCount is 3 In Decrement Thread, SharedResCount is 2 In Decrement Thread, SharedResCount is 1 In Decrement Thread, SharedResCount is 0 Decrement Thread releases the mutex

Part II:

Instead of using parentheses, you might choose a different approach You might use a glossary, sidebar, footnotes, or endnotes, for example, to explain or de ne all terminology Kathleen, the administrative assistant to the chief executive of cer (CEO) of a physical therapy provider, decided to use a sidebar in her newsletter The newsletter goes to patients ROM means range of motion and is a much-used term in the industry But if you ve just started working with us to get back full use of your arm, you may not be familiar with the term ROM To de ne it in every article, or even in every issue of the newsletter, was a pain and seemed like overkill But I knew that some people receiving each issue wouldn t know the jargon What I did was create a sidebar on the cover I called it Jargon and de ned all the terms used in that issue That way I didn t have to worry about it at all Titling the glossary Jargon is a nice touch; it lets readers know that ignorance is expected Successful business writing demands clarity Make it easy, not hard, for your readers to get your message

As the output shows, access to SharedResCount is synchronized, with only one thread at a time being able to change its value To prove that the Mtx mutex was needed to produce the preceding output, try commenting out the calls to WaitOne( ) and ReleaseMutex( ) in the preceding program When you run the program, you will see the following sequence (the actual output you see may vary):

In In In In In In In In In Increment Decrement Increment Decrement Increment Decrement Increment Decrement Increment Thread, Thread, Thread, Thread, Thread, Thread, Thread, Thread, Thread, SharedResCount SharedResCount SharedResCount SharedResCount SharedResCount SharedResCount SharedResCount SharedResCount SharedResCount is is is is is is is is is 1 0 1 0 1 0 1 0 1

The algebra is always simpler if we take one of the times in the initial data to correspond to t = 0 So let us say that 9:00 pm is t = 0 Then 11:00 pm is t = 2 and 7:00 pm is t = 2 The initial data then tell us that 6000 = P eK 0 10000 = P eK 2 ( ) ( )

As this output shows, without the mutex, increments and decrements to SharedResCount are interspersed rather than sequenced The mutex created by the previous example is known only to the process that creates it However, it is possible to create a mutex that is known systemwide To do so, you must create a named mutex, using one of these constructors: public Mutex(bool initiallyOwned, string name) public Mutex(bool initiallyOwned, string name, out bool createdNew) In both forms, the name of the mutex is passed in name In the first form, if initiallyOwned is true, then ownership of the mutex is requested However, because a systemwide mutex might already be owned by another process, it is better to specify false for this parameter In the second form, on return createdNew will be true if ownership was requested and acquired It will be false if ownership was denied (There is also a third form of the Mutex constructor that allows you to specify a MutexSecurity object, which controls access) Using a named mutex enables you to manage interprocess synchronization One other point: It is legal for a thread that has acquired a mutex to make one or more additional calls to WaitOne( ) prior to calling ReleaseMutex( ), and these additional calls will succeed That is, redundant calls to WaitOne( ) will not block a thread that already owns the mutex However, the number of calls to WaitOne( ) must be balanced by the same number of calls to ReleaseMutex( ) before the mutex is released