As the MIPS stack grows in size when more items are push onto it, the $sp register wi O contain increaseing memory address values O contain decreasing memory address values
Q: t is shown a fully associative cache with four lines. The block size is 8 bytes. What is the memory…
A: Given, Block offset bits = 100 Tag bits = 0001000111011
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A: Below is the answer to above question. I hope this will helpful for you...
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A: Here page size = 2K words Thus page offset bits is 11 bits Total number of pages=8 Thus page number…
Q: A cache is set up with a block size of 32 words. There are 16 blocks in cache and set up to be…
A: Block size = 32 words Block Offset = 5 bits Total # of blocks inside cache = 16 So index bits = 4…
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A: Because the writing buffer is writing back to memory, the cache will be able to satisfy the request;…
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A: Since no of page frames is not given, I am considering 4 page frames LRU algorithm works on the…
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A: Answer : I attached an answer please have a look once.
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A: Given: Given that SS=2400, SP=8631H, AX=4FA6H, and DX=8C3FH. What the contents are of registers AH,…
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A: The solution for the above questions are:
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A: We should start with knowing some general information about Virtual Memory, Pages, Blocks,…
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A: We are going to calculate tag, line id and word id for given memory address.
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A: Given: Memory address= 32 bits Cache Size=64KB = 216 bytes Block size=32 Bytes = 25 bytes 4 way set…
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A: Introduction: Data is updated to cache and memory at the same time in write-through. This method is…
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A: The question is to find state at the top of the stack after the given sequence of operations.
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A: The answer for the tag and index fields respectively in the addresses generated by the processor is
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A: Each block has 32 words, so the block offset = 25 words. So Block offset field can be represented by…
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A: This type of mapping has an improved and enhanced form of direct mapping which helps us to remove…
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A: Your solution is given below with an explanation.
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A: First find number of blocks = total blocks in cache/ block size. = 128/8 = 16 Assuming each word is…
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A: Following information has been given: Main memory size = 4 Gbytes(GB) = 210x 210x 210x 22 bytes…
Q: A certain processor uses a fully associative cache of size 16 kB, The cache block size is 16 bytes.…
A:
Q: Given The Memory Map and the register values correspond to the state of an 8086 microprocessor. Ss=…
A: Given : Value of SS = 7A20 H Value of SP = 0125 H Value of Stack : 35H 3DH 2BH 67H 5AH
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A: It is defined as the data or contents of the main memory that are used frequently by CPU are stored…
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A: Given :-A 4-way set-associative cache memory unit with a capacity of 16KB is built using a block…
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A: 80851 instruction byte for different call of instruction
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A: Please find the answer below
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A: Answer is given below-
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A: Here have to determine about physically addressed cache might be integrated with virtual memory.
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A: Actually, 1 byte =8 bits. cache memory is a fast access memory.
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A: Given: leftmost 3 bits of the 14-bit address address bus contains 2FBAH find the selected device ID
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A: Blank#1 : instruction cache Blank#2 : data cache
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A: According to the information given:- We have to use realtively Simple CPU include 32byte, 2 way set…
Q: A computer system using the Relatively Simple CPU includes a 32-byte, 2-way set associative cache…
A: Number of bytes in cache = 32. Eache cache line has 2 bytes. So number of cache lines = 32/2 = 16…
Q: A 2-way set-associative cache consists of four sets. Main memory contains 2K blocks of 8 bytes each…
A: Given: No of lines in a set=2No of set in Cache = 4Block size=8 bytes Therefore, Size of each cache…
Q: in a certain microcomputer system, the microprocessor is connected to an external memory of size…
A: In a certain microcomputer system, the microprocessor is connected to an external memory of size…
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A: According to the question In computing a physical deal with additionally real deal with, or binary…
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A: Here in this question we have given main memory size of 1G words Block size = 32 words Find =…
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A: Introduction Given, 32 bit address. 32 KB ,4 way set associative cache. Block size = two 32 bit…
Q: A virtual memory has a page size of 16 words. There are 8 pages and 4 blocks. The associative memory…
A: Virtual Memory have:- Page size = 16 words Page number = 8 Block number = 4
Q: Suppose a computer using fully associative cache has 2^24 bytes of byte-addressable main memory and…
A: Full associative cache size=224 bytes.Size of each block=64 bytesBlocks of main memory=cache…
Q: The 16-bits data addressing registers and their functions Registers that can do division The flags…
A: Actually, registers are used to stores data/information:
Q: A cache has been designed such that it has 512 lines, with each line or block containing 8 words.…
A: As per given information:- Address size = 20 bit No of lines in cache = 512 lines = 29 lines So, 9…
Q: So, what should happen when a processor sends a request that fails to be fulfilled in the cache…
A: The request is retried on the block that is currently in the write buffer. The block is moved to the…
Q: if the RAM has two byte data word and address bits are grouped for direct mapping with a cache…
A: According to the Question below the Solution:
Q: Consider a cache with 128 blocks. The block size is 32 bytes. Find the number of tag bits, index…
A: Given, Cache memory size = 128 block Block size = 32 byte Memory address = 32 bit
Q: When a block is being returned to main memory from the write buffer, what should happen if a…
A: If requested data is contained in the cache (cache hit), this request can be served by simply…
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- physcal addresses are 4s ng 4 Ame dat in a cetain compe, te addresses can be translaled without y TLB entries At most how many ditina vid the address translation peh has 12 vld The Translation Look aside Bulfer (TLB)i sine is kB and the word size iby The memory is word addresible. The pe virtual addresses are 64 bea long d th sine is miss?In assembly 68k write the: Initialize the supervisor stack pointer to $8000 Initialize the program counter to $1000 Reserve locations at $2000 to hold your first name only (Joshua) as text and call it NAM Reserve locations at $3000 to store 9 longwords, call it MEM, and write your family name (Kimmich) as a comment Reserve locations at $12400 to store 32-bit unsigned integer variable called RES (to be computed)9. Assuming that SP=2236H, AX=FFD4H, BX=8875H, and CX39812H, show the contents of the stack as each of the following instructions is executed: PUSH AX PUSH BX PUSH CX a. SP= 2233H Ob. SP= 2230H Oc. PS= 2242H
- What should happen if the processor submits a request that is denied in the cache while the write buffer is writing a block back to main memory?3. The table below represents a stack stored in a contiguous block of memory cells (as discussed in the text). If the base of the stack is at address 0x10 and the stack pointer contains the value 0x12, what value is retrieved by a pop instruction? What value is in the stack pointer after the pop operation? Address 0x10 0x11 0x12 0x13 0x14 Contents 'F' 'C' 'A' 'B' 'E'A request that cannot be satisfied in the cache is issued by the processor while the write buffer returns a block to main memory. What should happen in this case?
- The STACK is a dynamic data structure. The 80x86 computer controls its stack via stack pointer ESP. Whenever you POP data out of the stack segment memory using POP BX, the 80x86 will transfer data by: Oa. a. Increasing the stack pointer ESP by 4. Ob. Decreasing the stack pointer ESP by 2. Oc. c. Decreasing the stack pointer ESP by 4. Od. Increasing the stack pointer ESP by 2.5. The stack memory is addressed by a combination of the segment plus_ offset. 6. The PUSH and POP instructions always transfer -bit number between the stack and a register or memory location in the 8086 microprocessors. 7. For string instructions, DI always addresses data in the segment. 8. The 8086 LOOP instruction decrements register and tests it for a 0 to decide if a jump occurs 9. The last executable instruction in a procedure must be 10.A bus cycle is equal to clocking. periods. 11. If A0 is a logic 0, then the memory bank is selected. 12.the 8086 processor is partitioned into two logical units------------ and C. 13. ------- is the value of the control word register (CWR) of PPI 8255 if all ports are input 14.Fixed address for I/O instructions bytes while Variable address for --- --- ---bytes I/O instructions is 15. Write bus cycle need -clocks to complete one write of data to memorySelect common examples of when an assembly programmer would want to use the stack: to pass arguments to save return address for CALL local variables temporary save area for registers applications which have FIFO nature, such as customers waiting in a bank queue
- The STACK is a dynamic data structure. The 80x86 computer controls its stack via stack pointer ESP. Whenever you POP data out of the stack segment memory using POP BX, the 80x86 will transfer data by: Increasing the stack pointer ESP by 2. а. O b. Increasing the stack pointer ESP by 4. O . Decreasing the stack pointer ESP by 2. O d. Decreasing the stack pointer ESP by 4.What should happen if the processor provides a request that is not satisfied in the cache while a block is being transferred back to main memory from the write buffer?A 2-way set associative cache consists of four sets. Main memory contains 2K blocks of eight words each. Show the main memory address format that allows us to map addresses from main memory to cache. Be sure to include the fields as well as their sizes. Compute the hit ratio for a program that loops 6 times from locations 8 to 51 in main memory. You may leave the hit ratio in terms of a fraction. Please show details how you obtain the result.