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The Cigarette-Smokers difficulty Consider a system with three smoker processes and one agent process. Every smoker continuously rolls a cigarette and then smokes it. But to roll as well as smoke a cigarette the smoker needs three ingredients: paper, tobacco and matches. One of the smoker progression has paper another has tobacco and the third has matches. The agent has an endless supply of all three materials. The agent put two of the ingredients on the table. The smoker who has the residual ingredient then makes and smokes a cigarette signalling the agent on completion. The agent then places out another two of the three ingredients and the cycle repeats. Write a program to synchronize the agent as well as the smokers using Java synchronization.
Question 1: (a) Discuss about the evolution of operating systems from mainframe batch systems to mainframe time sharing systems. (b) Explain your understanding of the "sus
Q. Why page sizes always powers of 2? Answer: Recall that paging is executed by breaking up an address into a page and offset number. It is most competent to break the address
Q. What is the major difficulty that a programmer should overcome in writing an operating system for a real-time environment? Answer: The major difficulty is keeping the oper
Define properties of Time sharing operating system Time sharing:- T his systems uses CPU scheduling and multiprogramming to provide economical interactive use of a system. The
Explain what semaphores are, their usage, implementation given to avoid busy waiting and binary semaphores. Semaphore definition Usage for mutual exclusion and process synch
Semispace works by maintaining two disjoint areas from which memory can be allocated. These areas are called the from-space and the to-space. At ?rst, the algorithm allocates memor
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What is the difference between multiprocessing and multitasking? Multi programming means implementing more than one program but here whenever the processor is idle during execu
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What is internal fragmentation? Consider holes of 20k assume the process requests 18 bites. If we allocate accurately the request block, we are left with a hole of 2k. The over
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