1 Lecture #24 Shared Objects and Concurrent Programming This material is not available in the textbook. The online powerpoint presentations contain the

1

Lecture #24Shared Objects and Concurrent

Programming

This material is not available in the textbook. The online powerpoint presentations contain the text explanations given in class.

2

A Multiprocessor Machine

P1

P2

P10

...

P1 memory

Shared memory

Uniprocessor

Multiprocessor

3

Concurrent Programming

object

object

Shared Memory

Challenge: coordinating access

4

Persistent vs Transient Communication

•Persistent Communication medium: the sending of information changes the state of the medium forever.Example: Blackboard.

•Transient communication medium: the change of state is only for some limited time period.Example: Talking.

5

Parallel Primality Testing

Task: Print all primes from 1 to 1010 in some order Avaliable: A machine with 10 processors

Solution: Speed work up 10 times, that is, new time to print all primes will be 1/10 of time for single processor

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P1 P2 P10

1 109 2x109 1010

Split the work among processors!

Each processor Pi gets 109 numbers to test.

…

…

7


(define (P i) (let ((counter (+ 1 (* (- i 1) (power 10 9)))) (upto (* i (power 10 9)))) (define (iter) (if (< counter upto) (begin (if (prime? counter) (display counter) #f) (increment-counter) (iter)) 'done)) (iter)))

(parallel-execute (P 1) (P 2) ... (P 10))

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Problem: work is split unevenly

Some processors have less primes to test… Some composite numbers are easier to test…

P1 P2 P10

1 109 2x109 1010

Need to split the work range dynamically!

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A Shared Counter Object

(define (make-shared-counter value) (define (fetch) value) (define (increment) (set! value (+ 1 value)) (define (dispatch m) (cond (((eq? m 'fetch) (fetch)) (eq? m 'increment) (increment)) (else (error “unknown request”)))) dispatch)

(define shared-counter (make-shared-counter 1))

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Using the Shared Counter

(define (P i) (define (iter) (let ((index (shared-counter 'fetch))) (if (< index (power 10 10)) (begin (if (prime? index) (display index) #f) (shared-counter 'increment) (iter)) 'done)) (iter)))

(parallel-execute (P 1) (P 2) ... (P 10))

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This Solution Doesn’t Work

time

Increment: (set! value (+ 1 value))

P1 read value77

77

P2 increment 10 times

87 set! value78 Error!

(let ((index (shared-counter 'fetch)))

77P1 fetch

P2 fetch

77

77Error!

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It Will Never Work!?

Real World Example: Walking in the Street

Look / Move Fetch / Increment

Computers: Accessing Memory

Fischer Lynch & Patterson: Impossible to solve!!!

Need to “glue” the Fetch / Increment pair into one indivisible operation: Fetch-and-Increment

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The Fetch-and-Increment Operation

(define (make-shared-counter value) (define (fetch-and-increment) (let ((old value)) (set! value (+ old 1)) old)) (define (dispatch m) (cond (((eq? m 'fetch-and-increment) (fetch-and-increment)) (else (error ``unknown request -- counter'' m)))) dispatch)

Instanteneous

Shared CounterFetch-and-inc

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A Correct Shared Counter

(define shared-counter (make-shared-counter 1))(define (P i) (define (iter) (let ((index (shared-counter 'fetch-and-increment))) (if (< index (power 10 10)) (begin (if (prime? index) (display index) #f) (iter)) 'done)) (iter))) (parallel-execute (P 1) (P 2) ... (P 10))

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Implementing Fetch-and-Inc

To make the program work we need an “intantaneous” implementation of fetch-and-increment. How can we do this:

• Special Hardware. Built-in synchronization instructions. • Special Software. Use regular instructions -- the solution

will involve waiting.

Software: Mutual Exclusion

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Mutual Exclusion

(mutex 'start) (let ((old value)) (set! value (+ old 1)) old)(mutex 'end))

Only one process at a time can execute these instructions

P1

P2

P10

...11 P2

returns 1Mutex count

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The Story of Alice and Bob

Bob Alice

Yard

* As told by Leslie Lamport

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The Mutual Exclusion Problem

Requirements: • Mutual Exclusion: there will never be two dogs

simultaneously in the yard.• No Deadlock: if only one dog wants to be in the yard it will

succeed, and if both dogs want to go out, at least one of them will succeed.

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Cell Phone Solution

Bob Alice

Yard

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Coke Can Solution

Bob Alice

Yard

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Flag Solution -- Alice

(define (Alice) (loop ;; ``repeat forever'' (set! Alice-flag 'up) ;; Alice wants to enter (do ((= Bob-flag 'up)) (skip)) ;; loop until Bob lowers flag (Alice-dog-in-yard) ;; Dog can enter the yard (set! Alice-flag 'down) ;; Alice is leaving ))


Bob Alice

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Flag Solution -- Bob

(define (Bob) (loop ;; ``repeat forever'' (set! Bob-flag 'up) ;; Bob wants to enter (do ((= Alice-flag 'up)) ;; If Alice wants to enter (set! Bob-flag 'down) ;; Bob is a gentleman (do ((= Alice-flag 'up)) (skip)) ;; loop (skip) till Alice leaves (set! Bob-flag 'up) ;; raise flag ) ;; and go through the do again (Bob-dog-in-yard) ;; Dog can enter yard (set! Bob-flag 'down) ;; Bob is leaving ))


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Flag Solution -- Both





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Intuition: Why Mutual Exclusion is Preserved

Each perform: • First raise the flag, to signal interest. Then• look to see if the other one has raised the flag.

One can claim that the following flag principle holds:

since Alice and Bob each raise their own flag and then look at the others flag, the last one to start looking must notice that both flags are up.

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Why is there no Deadlock?

Since Alice has priority over Bob…if neither is entering the critical section, both are repeatedly trying, and Bob will give Alice priority.

Unfortunately, the algorithm is not a fair one, and Bob's dogs might eventually grow very anxious :-)

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The Morals of our Story

• The Mutual Exclusion problem cannot be solved using transient communication. (I.e. Cell-phones.)

• The Mutual Exclusion problem cannot be solved using interrupts or interrupt bits (I.e. Cans)

• The Mutual Exclusion problem can be solved with one bit registers (I.e. Flags), memory locations that can be read and written (set!-ed).

We cheated a little: the arbiter problem…

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The Solution and Conclusion

(define (Alice) (loop (mutex 'begin) (Alice-dog-in-yard) ;; critical section (mutex 'end) ))

Question: then why not execute all the code of the parallel prime-printing algorithm in a critical section?

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Answer: Amdahl’s Law

Speedup Overall = Execution_time_Old / Execution_time_New =

1 Fraction_Enhanced Speedup_Enhanced

1- Fraction_Enhanced +

If 40\% of the execution time and can be sped-up by a factor of 10 by using 10 processors. Then

Speedup Overall = 1/(0.6 + (0.4/10)) = 1/0.64 = 1.56

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Length of Critical Sections

We want Linear Speedup: 100 processors = 100 times faster By Amdahl's law that means that Fraction_Enhanced = 1 No Sequential Parts!!!!!

For a speedup of only 99 times: 1-Fraction_Enhacned= 0.0001

In other words, we must try and make the parts that are sequential (the critical sections) as short as possible!!!!!

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Summarizing it all

To summarize it all, our world is asynchronous, and yet with a bit of luck we have ways of overcoming this asynchrony to create powerful concurrent algorithms. This was best summarized by the quote which some attribute to Tommy Lasorda:

“Life is the synchronicity of chance”

Good Luck on Your Final Exam!

Documents

1 Lecture #24 Shared Objects and Concurrent Programming This material is not available in the textbook. The online powerpoint presentations contain the