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Concurrent Programming, JAVA Programming

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Problem 1
A savings account object holds a non-negative balance, and provides deposit(k )
and withdraw(k ) methods, where deposit(k ) adds k to the balance, and
withdraw(k ) subtracts k, if the balance is at least k, and otherwise blocks until
the balance becomes k or greater. getbalance() gives the current balance.
1. Implement this savings account using locks and conditions (use
java.util.concurrent.locks.ReentrantLock). Test by using the 3
functions.
2. Now suppose there are two kinds of withdrawals: ordinary and preferred.
Devise an implementation that ensures that no ordinary withdrawal occurs
if there is a preferred withdrawal waiting to occur.
Problem 2
Consider the following conditions: An enqueuer waiting on a full-queue or
a dequeuer waiting on an empty queue sleep indefinitely, unless woken up by
another thread. A thread must send a signal ONLY when it adds an element
to an empty queue or removes an element from a full-queue.
1. Implement a bounded partial queue using a signaling mechanism that
signals to all waiting dequeuers.
2. Implement the bounded partial queue by using a signaling mechanism
(your own scheme) that signals to only one waiting dequeuer or enqueuer,
and ensure that the lost-wake-up problem does not happen.
1
Problem 3
We have n threads, each of which executes method foo() followed by bar().
We want to add synchronization to ensure that no thread starts executing bar()
until all threads have finished executing foo(). To achieve this, we will insert
some barrier code between the two methods. Implement the following two
schemes for barrier code:
• Use a shared counter protected by test-and-test-and-set lock. Each thread
locks the counter, increments it, releases the lock, and repeatedly reads
the counter until it reaches n.
• Use an n-element Boolean array A. Initially all entries are 0. When thread
0 executes its barrier, it sets b[0] to 1, and repeatedly reads b[n - 1]
until it becomes 1. Every other thread i, repeatedly reads b[i - 1] until
it becomes 1, then it sets b[i] to 1, and repeatedly reads b[n - 1] until
it becomes 1.
Implement both these schemes in Java. Each of the methods foo() and
bar() just sleeps for 20 milliseconds. Test the two schemes for n = 16. Run
each scheme at least ten times, measure the total runtime in each test run,
discard the highest and lowest values, take the average, and use it to compare
the two schemes. Which performs better? Can you explain the reason? You
can run the experiments on ecen5033.colorado.edu. Submit the code as well as
experimental results.

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