UGC NET CSE | June 2014 | Part 2 | Question: 33

Question

Consider a system with five processes $P_{0}$ through $P_{4}$ and three resource types $R_{1}$, $R_{2}$ and $R_{3}$ . Resource type $R_{1}$ has $10$ instances, $R_{2}$ has $5$ instances and $R_{3}$ has $7$ instances. Suppose that at time $T_{0}$, the following snapshot of the system has been taken :

          Allocation

         $R_{1}$         $R_{2}$          $R_{3}$

$P_{0}$     0            1             0

$P_{1}$     2            0             0

$P_{2}$     3            0             2

$P_{3}$     2            1             1

$P_{4}$     0            2             2

            Max

$R_{1}$         $R_{2}$           $R_{3}$ 

7            5              3

3            2              2

9            0              2

2            2              2

4            3              3

           Available

$R_{1}$         $R_{2}$            $R_{3}$

3             3               2

Assume that now the process $P_{1}$ requests one additional instance of type $R_{1}$ and two instances of resource type $R_{3}$. The state resulting after this allocation will be 

• A. Ready state 
• B. Safe state 
• C. Blocked state
• D. Unsafe state 

Comments

Kindly tag the source of this question. Also, here is a source that is worth the re-tagging.

UGC-NET 2014 - June - Paper II - Qn 33

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done

Answer 1

Answer is B).

By taking one instance of R1 , P1 will fulfill its need of maximum R1

and By taking two instances of R3 , P1 will fulfill its need of maximum R3

Now , it needs two instances of R2 after which it can complete execution  and release all of its resources,

hence it is in safe state .

Comments

wrong question

Answer 2

option (b)

process no	alocation (R1 R2 R3)	max  need(R1 R2 R3)	reaminig need (R1 R2 R3)	available(R1 R2 R3)
p0	010	753	743	332
p1	200	424	224
p2	302	902	600
P3	211	222	011
P4	022	433	411

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initially need for process is 322 but process P1 requests one additional instance of type R1 and two instances of resource type R3. with 332 we can satisfy p3 need therefore giving all its allocated resource back resulting 5 4 3 now available resources. with 5 4 3 we can satisfy either p1 need or  p4 need here we cant satisfy p0 or p2 need becauz remaining need resouce are more than available resource  so on ..... therefore there is safe state with safe sequence as P3, P4, P1 , P2. P0

Comments

okk thx..

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This answer is wrong. This question is taken from Galvin book theory section. Please validate.

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with 5 4 3 we can satisfy either p1 need or  p4 need here

 

This answer is completely wrong, You cannot satisfy p1(i.e. 7 4 3) with 5 4 3.  Since 7 4 3 > 5 4 3

Answer 3

Need=Max-Allocation

Need

           R₁        R₂           R₃

 P₀      7            4              3

 P₁       1            2              2

 P₂       6            0              0

 P₃       0            1              1

 P₄       4            1              1

If Available<=Need

then Available[P_i]=Available[P_i]+Need[P_i]

Safe state is P₁->P₃->P₄->P₀->P₂

Answer 4

Ans (B) Safe State

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