☰ 🔍
🖊️Ask a Question
Electrical Engineering

Consider the fractional knapsack instance n = 4, (p1, p2, p3, p4) = (10, 10, 12, 18), (w1, w2, w3, w4) = (2, 4, 6, 9) and M = 15. The maximum profit is given by (Assume p and w denotes profit and weight of objects respectively) (A) 40 (B) 38 (C) 32 (D) 30

by 8 views
Consider the fractional knapsack instance n = 4, (p1, p2, p3, p4) = (10, 10, 12, 18), (w1, w2, w3, w4) = (2, 4, 6, 9) and M = 15. The maximum profit is given by (Assume p and w denotes profit and weight of objects respectively) (A) 40 (B) 38 (C) 32 (D) 30
0 like 0 dislike

1 Answer

by
(B) 38
0 like 0 dislike

Related questions

by anonymous asked Feb 7, 2021 18 views
Suppose there are four processes in execution with 12 instances of a Resource R in a system. The maximum need of each process and current allocation are given below: Image With reference to current allocation, is system safe? If so, what is the safe sequence? (A) No (B) Yes, P1 P2 P3 P4 (C) Yes, P4 P3 P1 P2 (D) Yes, P2 P1 P3 P4
image

Suppose there are four processes in execution with 12 instances of a Resource R in a system. The maximum need of each process and current allocation are given below: With reference to current allocation, is system safe? If so, ... B) Yes, P1 P2 P3 P4 (C) Yes, P4 P3 P1 P2 (D) Yes, P2 P1 P3 P4
Answer : (C) Yes, P4 P3 P1 P2...

View solution
0 like 0 dislike
1 answer
by anonymous asked Feb 7, 2021 29 views
Consider three CPU intensive processes P1, P2, P3 which require 20, 10 and 30 units of time, arrive at times 1, 3 and 7 respectively. Suppose operating system is implementing Shortest Remaining Time first (preemptive scheduling) algorithm, then .............. context switches are required (suppose context switch at the beginning of Ready queue and at the end of Ready queue are not counted). (A) 3 (B) 2 (C) 4 (D) 5

Consider three CPU intensive processes P1, P2, P3 which require 20, 10 and 30 units of time, arrive at times 1, 3 and 7 respectively. Suppose operating system is implementing Shortest Remaining Time first (preemptive scheduling) ... end of Ready queue are not counted). (A) 3 (B) 2 (C) 4 (D) 5
Answer : (A) 3...

View solution
0 like 0 dislike
1 answer
by anonymous asked Feb 7, 2021 22 views
Consider a system having ‘m’ resources of the same type. These resources are shared by three processes P1, P2 and P3 which have peak demands of 2, 5 and 7 resources respectively. For what value of ‘m’ deadlock will not occur? (A) 70 (B) 14 (C) 13 (D) 7

Consider a system having ‘m’ resources of the same type. These resources are shared by three processes P1, P2 and P3 which have peak demands of 2, 5 and 7 resources respectively. For what value of ‘m’ deadlock will not occur? (A) 70 (B) 14 (C) 13 (D) 7
Answer : (B) 14...

View solution
0 like 0 dislike
1 answer
by anonymous asked Feb 7, 2021 16 views
Consider a system with twelve magnetic tape drives and three processes P1, P2 and P3. Process P1 requires maximum ten tape drives, process P2 may need as many as four tape drives and P3 may need upto nine tape drives. Suppose that at time t1, process P1 is holding five tape drives, process P2 is holding two tape drives and process P3 is holding three tape drives. At time t1, system is in: (A) safe state (B) unsafe state (C) deadlocked state (D) starvation state

Consider a system with twelve magnetic tape drives and three processes P1, P2 and P3. Process P1 requires maximum ten tape drives, process P2 may need as many as four tape drives and P3 may need upto nine ... , system is in: (A) safe state (B) unsafe state (C) deadlocked state (D) starvation state
Answer : Answer: B...

View solution
0 like 0 dislike
1 answer
by anonymous asked Feb 6, 2021 24 views
There are three processes P1, P2 and P3 sharing a semaphore for synchronising a variable. Initial value of semaphore is one. Assume that negative value of semaphore tells us how many processes are waiting in queue. Processes access the semaphore in following order: (a) P2 needs to access (b) P1 needs to access (c) P3 needs to access (d) P2 exits critical section (e) Pi exits critical section The final value of semaphore will be: (1) 0 (2) 1 (3) -1 (4) -2

There are three processes P1, P2 and P3 sharing a semaphore for synchronising a variable. Initial value of semaphore is one. Assume that negative value of semaphore tells us how many processes are waiting in queue. Processes access ... The final value of semaphore will be: (1) 0 (2) 1 (3) -1 (4) -2 
Answer : Answer: 1...

View solution
0 like 0 dislike
1 answer

50.5k questions

47.1k answers

240 comments

7.0k users

© 2025 Question Answer