Intro to Operating Systems

Exercise 8: Compare and contrast multiprocessing and concurrent processing.

Describe the role of process synchronization for both systems. Exercise 10: SEE ANSWERS Exercise 13: SEE ANSWERS Advanced Exercise 16: SEE ANSWERS ANSWERS: Exercise 2: a. Example of Deadlock: When two people are about to buy the same product at the same time. b. Example of Starvation: When one person borrowed a pen from his classmate and his classmate gets his pen back.

c. Example of Race: When two guys have the same girlfriend.Deadlocks can be resolved by (in this example) one of the two people substituting the one product that they are about to buy for a generic version of the same product. Exercise 9: 1. Under the above policy, 2 processes can be run simultaneously or active at once. 2.

The minimum number of tape drives that may be idle as a result of this policy is 3 (see diagram). The maximum number of tape drives that may be idle as a result of this policy is 5 (see diagram). [pic] B1. The maximum number of jobs that can be in progress at once is 3.

B2.The minimum number of tape drives that may be idle as a result of this policy is zero (0) (see diagram). The maximum number of tape drives that may be idle as a result of this policy is one (1) (see diagram). [pic] Advanced Exercise 14: An unsafe state may not necessarily lead to deadlock, it just means that we cannot guarantee that deadlock will not occur. Thus, it is possible that a system in an unsafe state may still allow all processes to complete without deadlock occurring. Consider the situation where a system has 12 resources allocated among processes P0, P1, and P2.There sources are allocated according to the following policy: Max Current Need P0 10 5 5 P1 4 2 2 P2 9 3 6 for (int i = 0; i < n; i++) { // first find a thread that can finish for (int j = 0; j < n; j++) { if (! finish[j]) { boolean temp = true; for (int k = 0; k < m; k++) { if (need[j][k] > work[k]) temp = false; } if (temp) { // if this thread can finish finish[j] = true; for (int x = 0; x < m; x++) work[x] += work[j][x]; } } } }