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21 votes

Consider three concurrent processes $P_1, P_2$ and $P_3$ as shown below, which access a shared variable $D$ that has been initialized to $100.$ $$\begin{array}{|c|c|c|} \hline P_1 & P_2 & P_3 \\ \hline : & : & : \\ : & : & : \\ D=D+20 & D=D-50 & D=D+10 \\ : & : & : \\ : & : & : \\ \hline \end{array}$$ The processes are executed on a uniprocessor system running a time-shared operating system. If the minimum and maximum possible values of $D$ after the three processes have completed execution are $X$ and $Y$ respectively, then the value of $Y-X$ is ______

1

36 votes

Best answer

$D=100$

Arithmetic operations are not ATOMIC.

These are three step process:

- Read
- Calculate
- Update

Maximum value:

- Run $P_{2}$ for read and Calculate. $D = 100$
- Run $P_{1}$ for read and calculate. $D = 100$
- Run $P_{2}$ update. $D = 50$
- Run $P_{1}$ update. $D = 110$
- Run $P_{2}$ read, calculate and update. $D = 130$

Minimum Value:

- Run $P_{1}, P_{2}, P_{3}$ for Read and Calculate. $D = 100$
- Run $P_{1}$ update. $D = 110$
- Run $P_{3}$ update. $D = 120$
- Run $P_{2}$ update. $D = 50$

Difference between Maximum and Minimum $= 130-50 = 80.$

5

edited
Sep 12, 2019
by Kaurbaljit

In multiprocessor environment, let's assume we have 3 processors for each of the processes. So each process will execute as a whole(if there exists no other processes) in a processor it was assigned. So output certainly depends on the order in which those processes were being assigned to them.

Again we gotta assume the assignments which can result in maximum and minimum value of "D".

1

79 votes

For minimum value-

P2 should write back the result after P1 and P3 only,but P2 should read the value of "D" before its being modified by P1 and P3(i.e minimum).

So,

P1 | P2 | P3 |

Read(D):100 | ||

Read(D): 100 | ||

D= D+20: 120 | ||

Write(D): 120 | ||

Read(D):120 | ||

D=D+10 : 130 | ||

Write(D): 130 | ||

D= D-50 : 50 Write(D) :50 |

X = 50

Similarly for maximum value-

P1 | P2 | P3 |

Read(D): 100 | ||

D= D+20 :120 | ||

Write(D): 120 | ||

Read(D): 120 | ||

Read(D): 120 | ||

D= D-50: 70 | ||

Write(D): 70 | ||

D=D +10 : 130 Write(D): 130 |

So Y = 130

Hence Y-X = 130-50 = 80

0

1

8 votes

$P1:$

- $Load\ R_{p1},M[D]$
- $Add\ 20$
- $Store M[D],R_{p1}$

$P2:$

- $Load\ R_{p2},M[D]$
- $Sub\ 50$
- $Store M[D],R_{p2}$

$P3:$

- $Load\ R_{p3},M[D]$
- $Add\ 10$
- $Store M[D],R_{p3}$

**Minimum Value: **

$P2\rightarrow 1|2|preempt$

$P1\rightarrow 1|2|3(D=120)$

$P3\rightarrow 1|2|3(D=130)$

$P2\rightarrow 3$ $(D=50)$

**Maximum Value: **

$P1\rightarrow 1|2|3(D=120)$

$P3\rightarrow 1|2|preempt$

$P2\rightarrow 1|2|3(D=70)$

$P3\rightarrow 3$ $(D=130)$

$ANS: 130-50=80$

1 vote

$\text{(a) - For Max Value}$

$P_1$ | $P_2$ | $P_3$ |

$W(S)$ | $W(S)$ | $W(S)$ |

$R(D)$ | $R(D)$ | $R(D)$ |

$\color{red}{C.S\space to \space P_2}\color{blue}{\space D = 100 \space here}$ | ||

$D = D+ 20$ | $D = D- 50$ | $D = D+ 10$ |

$W(D)$ | $W(D)$ | $W(D)$ |

$P(S)$ | $P(S)$ | $P(S)$ |

Start from either $P_1$ or $P_3$ In my case starting from $P_1$ so $P_1$ have $D = 100$ after read it context switched and given charge to $P_2$ ${\color{green}{\text{Why $P_2$ because it can decrease D's value }}}$ so $D's$ value became $50$ now charge returned to $P_1$ but here in $P_1$ D's value is aleady set so here $\text{LOST UPDATE}$ problem arises, so $D = 100+20 = 120$ , Now $P_3$ will start execution. Eventually $D = 120+10 = 130$ so $Y = 130$

$\text{(b) - For Min Value}$

$P_1$ | $P_2$ | $P_3$ |

$W(S)$ | $W(S)$ | $W(S)$ |

$R(D)$ | $R(D)$ | $R(D)$ |

$\color{red}{C.S\space to \space P_1}\color{blue}{\space D = 100 \space here}$ | ||

$D = D+ 20$ | $D = D- 50$ | $D = D+ 10$ |

$W(D)$ | $W(D)$ | $W(D)$ |

$P(S)$ | $P(S)$ | $P(S)$ |

We'll start from $P_2$ because it contains negative operation and similiar execution done as $(a)$ so $D = 100 - 50$, $X = 50$

Then the value of $Y - X = 130 - 50 = 80$