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Q. Explain about Deadlocks?
Deadlocks
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];
At present there are two resources available. It is an system is in an unsafe state as process P1 could complete thereby freeing a total of four resources. But we can't guarantee that processes P0 and P2 can complete. Nevertheless it is possible that a process may release resources before requesting any further. For instance process P2 could release a resource thereby increasing the total number of resources to five. This permit process P0 to complete which would free a total of nine resources thereby allowing process P2 to complete as well.
Explain Wait for Graph a. Use a resource allocation graph to derive a wait-for graph. b. Wait-for graph acquired by making an edge from p 1 to p 2 iff p 1 is waiting for
Ok so i think i got this right by not 100% on this. what I want to figure out is if there are x visitors that can line up for a ride but only 1 passenger can get on the ride at a t
Deadlocks can be detected while the program is running, by running cycle detection algorithms on the graph that de?nes the current use of resources. De?ne this graph as follows:
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