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Q. What is Stack and register manipulation?
Stack and register manipulation: If we create stacks, stack instructions prove to be useful. LOAD IMMEDIATE is a good illustration of register manipulation (the value being loaded is part of the instruction). Every CPU has multiple registers when instruction set is designed; one has to specify that which register instruction is referring to.
No operation (or idle) is needed when there is nothing to run on a computer.
The EFI community has formed an open source shell environment. Rather than booting directly into a full OS, on some executions, the user can boot to the EFI shell. The shell is an
Non-Uniform Memory Access Model (NUMA) In shared memory multiprocessor systems, local memories are able to be connected with every processor. The collection of all local
Q. Features of read-only memory? ROMs are memories on which it's not possible to write data when they are on-line to computer. They can only be read. This is reason why it is k
Advantages of MPI: Every process has its own local variables It can be used on a broader range of problems than OpenMP It runs on either distributed or shared memor
The demand placed on a system is explained by a lognormally distributed random variable with mean 50 and standard deviation of 10. The capacity of the system is modeled by a Weibul
explain network operating system and design issues?
Telnet It is a terminal emulation program for TCP/IP networks like as the Internet. The Telnet program runs on your computer and attaches your PC to a server on the network. On
Q Consider the following expression. Assume that complement inputs are available. F(A,B,C,D) = ∑m (1,2,6,9,10,14) + ∑d (4,7,8,11,12) a. Find minimal expression for SOP. Draw
Micro-instructions are stored in control memory. Address register for control memory comprises the address of subsequent instruction which is to be read. Control memory Buffer Regi
Q. Find simplified function F and implement that function using only NAND gates. 1. F(A,B,C) = (A+B) (A'+B+C') (A'+B'+C') 2. F (A,B,C,D) = A'B'C'+B'CD'+A'BCD'+AB'C' 3.
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