Java puzzle, low latency queue

There a number of ways to use queues. A simple and very fast queue is to use AtomicReference. (It stores 0 or 1 element)

1. Write a producer and consumer thread which communicates via an AtomicReference. [Click for sample solution]

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Took an average of 48 ns per object

public class Main {
    public static void main(String... args) throws InterruptedException {
        AtomicReference<Long> queue = new AtomicReference<Long>();
        final Consumer<Long> consumer = new Consumer<Long>(queue);
        Thread t = new Thread(new Runnable() {
            @Override
            public void run() {
                while(consumer.take() >= 0);
            }
        });
        t.start();

        Producer<Long> producer = new Producer<Long>(queue);

        long start = System.nanoTime();
        int runs = 10*1000*1000;
        for(long i=0;i<runs;i++)
            producer.push(i);
        producer.push(-1L); // poison pill
        t.join();
        long time = System.nanoTime() - start;
        System.out.printf("Took an average of %,d ns per object%n", time/runs);
    }
}

class Producer<T> {
    final AtomicReference<T> queue;

    public Producer(AtomicReference<T> queue) {
        this.queue = queue;
    }

    public void push(T l) {
        while (!queue.compareAndSet(null, l)) ;
    }
}

class Consumer<T> {
    final AtomicReference<T> queue;

    public Consumer(AtomicReference<T> queue) {
        this.queue = queue;
    }

    public T take() {
        T t;
        do {
            t = queue.getAndSet(null);
        } while(t == null);
        return t;
    }
}

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2. Have the consumer send messages back to the producer using another AtomicReference. [left as an exercise]
The average round trip time taken was 156 ns if you have one message at a time, or 98 ns if you have two messages at a time.

3. Time the throughput for one second repeatedly. What variation do you see?

4. Pad the AtomicReference to make sure they are at least 64 bytes apart in memory and measure again.