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Explain Load Balancing Client Server Components
When migration functionality from the client - only model to the client - server model, care must be taken not over-or underutilize either the client or the server node. If the client node remains over- loaded, there is the potential for the server to be underutilized. In this case, no additional benefits would be gained by the use of client - server technology.
At the point that the client node is fully utilized (but not over utilized), the client - server model has provided additional benefits. Overloading the server computer also has hazardous ramifications because many clients will be slowed by the resource- constrained machine.
Convert the following from hex to binary and draw it on the memory map. RAM = 0000 -> 00FF EPROM = FF00 -> FFFF Answer: 0000 0000 0000 0000 (0) RAM sta
Explain in detail about ipc in linux
What are the five major activities of an operating system in regard to file management? The creation and deletion of files The creation and deletion of directories The s
Consider the following snapshot of a system, answer the following questions using the banker's algorithm: 1. What is the content of the matrix Need? Is the system in a safe stat
Write a brief note on demand paging. A demand paging is alike to a paging system with swapping. The Processes reside on the secondary memory while we want to implement a proces
What is a Real-time system? A Real-time system is used when inflexible time requirements have placed on the operation of processor or the flow of data so it is often used as a
Q. What are the three main activities of an operating system in regard to secondary-storage management? Answer: 1) Free-space management 2) Storage allocation 3) Disk
Worst Fit Algorithm Here we obtain the largest space available for the smallest. Therefore after that process entered there will be much more space remaining. In this space we
EXPLAIN MULTITASKING IN DETAIL
Q. Explain about Deadlocks? Deadlocks for (int i = 0; i // first find a thread that can finish for (int j = 0; j if (!finish[j]) { boolean temp = true; for
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