ISRO2011-60

Question

A total of 9 units of a resource type available, and given the safe state shown below, which of the following sequence will be a safe state?

Process	Used	Max
$P_1$	2	7
$P_2$	1	6
$P_3$	2	5
$P_4$	1	4

• A. $\langle P_4, P_1, P_3, P_2\rangle$
• B. $\langle P_4, P_2, P_1, P_3\rangle $
• C. $\langle P_4, P_2, P_3, P_1\rangle $
• D. $\langle P_3, P_1, P_2, P_4 \rangle$

Answer 1

Process	Used	Max	Need
P1	2	7	5
P2	1	6	5
P3	2	5	3
P4	1	4	3

We have total 9 resources and out of 9, 6 are already allocated to the processes. Now we are left with 3 resource which can fulfill the need of process P3 and P4. Now if we check options A, B and C where first P4 is being served then after completion of P4 we will have total 3 + 1 = 4 resources but now we can't fulfill the need of either P1 or P2, because each needs 5 resource so option A, B and C can not be correct ans. Similarly we can check for option D where after P3 we can fulfill any combination. So Ans- D. <P3,P1,P2,P4><P3,P1,P2,P4><P3,P1,P2,P4>

Comments

You forget to mention how we decided among Process p3 and p4 when they have equal NEED of resource. Note: When Arrival time is not specified expliclity then consider process naming sequence. P1, P2, P3 and P4

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^ please read the solution again .. I think, I have mentioned that point ...actually sometimes reverse engineering helps in getting ans quickly. ..so I used options filtering ...

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@vijaycs  other than given options is Sequence <P4,P3,P2,P1> could be the also??

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That is a really good solution.

Thanks.

Answer 2

According to me answer should be

⟨P3,P1,P2,P4⟩

Available resourses A=3

By this we can only fulfill the future need of the P3

Now new availbility Anew=3+2=5,

By Anew we can fill the need of P1 or P2,Suppose we fill need of P1,then Anew=5+2=7

Anew=7,Now for P2;

Anew=7+1=8

Finally with P4;

Anew=8+1=9

So our system is safe and safe sequence is P3---->P1------->P2------>P4

Comments

Perfect