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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 start
0000 0000 1111 1111 (255 byte) RAM stop
1111 1111 0000 0000 (63.75K) Eprom start
1111 1111 1111 1111 (64K) Eprom stop The memory map is shown below:-
Due to the tri-state nature of the I.C's (Ram, Rom etc.) they may be placed in parallel and 'turned on ' when they are required. This process is termed decoding. There are two types of decoding:- 1. Partial decoding 2. Absolute decoding
While it is hard to resolve a deadlock which has been detected, fortunately it is fairly easy to prevent deadlocks from ever happening. The key is that the conditions above for dea
Q. Why is rotational latency typically not considered in disk scheduling? How would you alter SSTF, SCAN and C-SCAN to comprise latency optimization? Answer: Most disks don't
What are the services provided by the operating systems? Program Development It provides a variety of services and facilities, such as debuggers and editor, to help the
Q. Dynamically loadable kernel modules give elasticity when drivers are added to a system however do they have disadvantages too? Under what situations would a kernel be compiled
Resolution of externally defined symbols is carried out by Resolution of externally defined symbols is carried out by Linker
Compare 2 different operating systems.
a) The barber shop has one barber, one barber chair, and n chairs for waiting customers, to sit on. If there are no customers present, the barber sits down in the barber chair and
please help us to write a code in c for tree directory structures.
what is application software?
Question 1 Explain the following with respect to Resource Management in Distributed Systems- Task assignment Approach Load - Balancing Approach Load - Sharing Approach
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