Computers and Technology, 03.10.2019 02:30, jumeljean123oythxy
Adigital computer has a memory unit with 16 bits per word. the instruction set consists of 122 different operations. all instructions have an operation code part (opcode) and an address part (allowing for only one address). each instruction is stored in one word of memory.
1. how many bits are needed for the opcode?
2. how many bits are left for the address part of the instruction?
3. what is the maximum allowable size for memory?
4. what is the largest unsigned binary number that can be accommodated in one word of memory?
Answers: 3
Computers and Technology, 22.06.2019 22:20, gingerham1
Avariable of the data type arrays is storing 10 quantities. what is true about these quantities? a. the quantities all have different characteristics. b. the quantities all have the same characteristics. c. five quantities have the same and five have different characteristics. d. it is necessary for all quantities to be integers. e. it is necessary for all quantities to be characters.
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Computers and Technology, 23.06.2019 16:00, natasniebow
Kenny works with an it company. his company is about to launch new software in the market. he has to ensure that this new software is functional and meets all of the quality standards set up at the planning stage. which job profile is kenny likely to have? kenny is likely to have the job profile of a blank .
Answers: 2
Computers and Technology, 23.06.2019 17:30, Annlee23
When making changes to optimize part of a processor, it is often the case that speeding up one type of instruction comes at the cost of slowing down something else. for example, if we put in a complicated fast floating-point unit, that takes space, and something might have to be moved farther away from the middle to accommodate it, adding an extra cycle in delay to reach that unit. the basic amdahl's law equation does not take into account this trade-off. a. if the new fast floating-point unit speeds up floating-point operations by, on average, 2ฤโ, and floating-point operations take 20% of the original program's execution time, what is the overall speedup (ignoring the penalty to any other instructions)? b. now assume that speeding up the floating-point unit slowed down data cache accesses, resulting in a 1.5ฤโ slowdown (or 2/3 speedup). data cache accesses consume 10% of the execution time. what is the overall speedup now? c. after implementing the new floating-point operations, what percentage of execution time is spent on floating-point operations? what percentage is spent on data cache accesses?
Answers: 2
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