For the memory of Von-Neumann, if it store 8 bits/cell and the address width is 7 bits. Maximum memory size is ----
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A: Here word size=64bit = 64/8=8 Bytes.
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A:
Q: 34. Von Neumann architecture is a. SISD b. SIMD c. MIMD d. MISD
A: Von Neumann architecture is There are four options given.
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A: Answer
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Q: Do not copy from other websites Correct and detailed answer will be Upvoted else downvoted. Thank…
A: Here I am trying to provide you the correct solution It is given address space is 32768.
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Q: List The Components Of The Ven Neumann Architecture?
A: Given that: List The Components Of The Ven Neumann Architecture?
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A: This is The flow of instruction execution MAR<-- PC MBR<--Memory[MAR] PC=PC+1 IR<-- MBR
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- Processor R is a 64-bit RISC processor with a 2 GHz clock rate. The average instruction requires one cycle to complete, assuming zero wait state memory accesses. Processor C is a CISC processor with a 1.8 GHz clock rate. The average simple instruction requires one cycle to complete, assuming zero wait state memory accesses. The average complex instruction requires two cycles to complete, assuming zero wait state memory accesses. Processor R can’t directly implement the complex processing instructions of Processor C. Executing an equivalent set of simple instructions requires an average of three cycles to complete, assuming zero wait state memory accesses. Program S contains nothing but simple instructions. Program C executes 70% simple instructions and 30% complex instructions. Which processor will execute program S more quickly? Which processor will execute program C more quickly? At what percentage of complex instructions will the performance of the two processors be equal?A computer system has a memory access time of 120 ns. The hit rate is 96% and memory and cache accesses don' t lap and affect each other. In order AMAT to be under 12 ns, what should the maximum cache access time be?Consider a 512 KB cache system used in our laptop. The access time for the cache is 25 ns, and the memory access time is 200 ns. Assuming a miss rate of 25%, what would be the average memory access time?
- Q2) Given a physical memory of 8 k and a cache memory of 512 bytes with block size 64 bytes. The system uses associative mapping with set size 2 lines per setA- How the memory address will be split to indicate tag, and offset B- What is the size of tag directory.Consider a memory system with a cache access time of 100ns and a memory access timeof 1200ns. If the effective access time is 10% greater than the cache access time, what is thehit ratio H?Physical address will be:::--.
- A computer employs RAM chips of 512 x8 and ROM chips of 1024x8. The computer system needs 2K bytes of RAM, 2K bytes of ROM. Draw the complete diagram for such system. Show how the address is organizedConsider a computer with the following characteristics: total of 1Mbyte of main memory; word size of 1 byte; block size of 16 bytes; and cache size of 64 Kbytes. For the main memory addresses of F0010 and CABBE, give the corresponding tag, cache set, and offset values for a two-way set-associative cache.Q.In a computer instruction format, the instruction length is 11 bits and the size of an address field is 4 bits. Is it possible to have 32 zero-address instructions using the specified format? Justify your answer.
- Q.2. The content of PC in the basic computer architecture (as shown in the figure below) is 3AF. The content of AC is 7EC3. The content of memory at address 3AF is A32E. The content of memory at address 32E is 09AC. The content of memory at 9AC is 8B9F. The next two contents of PC are 3B0 and 3B1. The content of memory at address 3B0 is 1340 and the content of memory at 340 is 10B0. The content of memory at address 3B1 is 3345. Identify each register values mentioned in the question and update accordingly according to the sequence of PC value.A computer employs RAM chips of 512 x 8 and ROM chips of 256 x 8. The computer system needs1K bytes of RAM, 2K bytes of ROM, and eight interface units, each with 2 registers. A memory-mapped1/0 configuration is used. The two highest-order bits of the 16 bit address bus are assigned 10 for RAM,11 for ROM, and 00 for interface registers.a. How many RAM and ROM chips are needed?b. Draw a memory-address map for the system.c. Give the address range in hexadecimal for RAM, ROM, and interface.. A computer employs RAM chips of 512 x 8 and ROM chips of 256 x 8. The computer system needs1K bytes of RAM, 2K bytes of ROM, and eight interface units, each with 2 registers. A memory-mapped1/0 configuration is used. The two highest-order bits of the 16 bit address bus are assigned 10 for RAM,11 for ROM, and 00 for interface registers.a. How many RAM and ROM chips are needed?b. Draw a memory-address map for the system.c. Give the address range in hexadecimal for RAM, ROM, and interface please sir i need full ansewr with full details