OS Synchronization

Question

Consider the following program:

Const int n= 10 
int Count= 0 
Void A( ) 
{ 
    int i; 
    for(i= 1 to n) 
    Count= Count + 1; 
} 

Main ( ) 
{ 
    Par begin 
A( ); 
A( ); 
A( ); 
A( ); 
    Par end 
    
}

5 What is the minimum and maximum possible value of count after the completion of the program?

(a) 1, 40                               (b) 2, 40

(c) 3, 40                               (d) 4, 40
 

Comments

is it A?

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Ab

Answer 1

MIN = 2

MAX = 40

Maximum is obvious.

For Min -

Lets consider increment of count is in 3 micro instruction 

step 1  Load Ri M[count]

step 2 INCR Ri

step 3 M[count] Ri

Where Ri is the local register allocated by each function called and used for temporary storage.

Lets called   first A() is A1()

Second A() is A2()

Third A() is A3()

Fourth A() is A4().

Now every function from A1() to A4() will execute concurrently and each function will execute all 3 micro instruction; 

1> Now consider A1() call first, it increment COUNT from 0 to 1 in 3 micro instruction but after executing 2 micro instruction and gets preempted so at this time the value of R1(temprrary register ) = 1 but the count value will be still 0.

 2> Now let A2() and A3() complete the execution so count value will be 20.

3> A4() will start it execution and let it iterate for "9" times so the count value will be 29 and preempt the A4().

4> Let A1() will start it's execution and it will execute its 3 micro instruction and overwrite count value from 29 to 1 in it's first iteration and again preempt.

5> lets A4() start it's execution now the count value is 1 so it will iterate its 10 'th iteration and execute 2 micro instruction and then preempt A4(). at this time R4(temporary register = 2) and count = 1.

6> Now execute remaining iteration of A1 form i=2 to 10 so it will make count value as 10 and complete it's execution.

7>In final step A4() will start it's execution and will execute 3 micro instruction for 9'th iteration and overwrite count = 1 and in the 10 iteration it will increment count from 1 to 2 and finish the execution.

So Min = 2

Max = 40. 

Comments

hello sir,

actually i have a doubt regarding ur explaination!

As u have said in 3rd step, A4() will iterate till 9th iteration then only its value becomes 29.

In 5th step u have said that the 10th iteration of A4() started and executed 2 micro instructuions.

Now in step 7th u said it will execute its 3rd micro instruction of 9th iteration and started 10th iteration?

how it is possible because 9th iteration has already completed. ??

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step 7 is given wrong .

STEP 7:In final step A4() will start it's execution and will execute it's 10th iteration  3rd micro instruction and replace count =10 to count =2

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finding min is like Doctor Strange playing with the TIME stone lol!!

Answer 2

Option B --2,40 

Devid count=count+1 into lower lvl instruction..

​plz follow the fig...and correct me if I am wrong..Thanks..

Comments

Really awesome explaination thanks..

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What is the meaning of par begin and par end?
Please tell.

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@Rajesh Pradhan If at Step 3 , P4 would have completed execution completely, then only, P1 would be left with the last step, hence count value would have been 1. Is my statement wrong?

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All that between par begin and par end execute at the same time.