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Estructura de Computadores (2025-26, Grado en Ingeniería del Software. Plan 2023 Grupo B, Grado en Ingeniería Informática. Plan 2023 Grupo B y Grado en Matemáticas + Ingeniería Informática. Plan 2023 Grupo B)

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Assume a full associative cache which can allocate up to 4 blocks. The next sequence of block references are generated by the processor:

Block references  -->381, 1053, 570, 683, 2122, 1053, 570, 1501, 1053, 168

After the fourth reference, the state of the cache is: 

cache block      content

    block 0 --> block 381 of MM

    block 1 --> block 1053 of MM

    block 2 --> block 570 of MM

    block 3 --> block 683 of MM

What block of MM will be  in the block 3 of the cache after the last reference (168) if the processor uses the LRU algorithm for replacement?

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Assume a full associative cache which can allocate up to 4 blocks. The next sequence of block references are generated by the processor:

Block references  -->314, 1080, 566, 659, 2052, 1080, 566, 1503, 1080, 154

After the fourth reference, the state of the cache is: 

cache block      content

    block 0 --> block 314 of MM

    block 1 --> block 1080 of MM

    block 2 --> block 566 of MM

    block 3 --> block 659 of MM

What block of MM will be  in the block 2 of the cache after the last reference (154) if the processor uses the LRU algorithm for replacement?

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Assume a full associative cache which can allocate up to 4 blocks. The next sequence of block references are generated by the processor:

Block references  -->314, 1004, 530, 665, 2220, 1004, 530, 1546, 1004, 125

After the fourth reference, the state of the cache is: 

cache block      content

    block 0 --> block 314 of MM

    block 1 --> block 1004 of MM

    block 2 --> block 530 of MM

    block 3 --> block 665 of MM

What block of MM will be  in the block 1 of the cache after the last reference (125) if the processor uses the LRU algorithm for replacement?

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Consider a MIPS processor with a main meory of 16 Mbytes, with a cache of 4 Kbytes, with write-back policy, direct mapped and blocks of 64 words. The miss penalty is 27 cycles. Nevertheless, if a block has to be copied from Cache to Main Mermory, the miss penatly is doubled. Calculate the number of cycles of the memmory stage (MEM stage) when the instruction lw $5, 120($15)  is executed taking into account that this instruction accesses to Main Memory address 0x75B53C to get the data. In the nex table you have some information about the control area of the data cache (the valid bit of the cache blocks in the table is 1; in the rest of blocks the valid bit is 0).

Block

DTAG
0xA0

 75B

0x5

0

 55B
0xD1 CF8

Block=num. of block of Cache, D=Dirty bit

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Assume a full associative cache which can allocate up to 4 blocks. The next sequence of block references are generated by the processor:

Block references  -->344, 1028, 518, 631, 2115, 1028, 518, 1508, 1028, 153

After the fourth reference, the state of the cache is: 

cache block      content

    block 0 --> block 344 of MM

    block 1 --> block 1028 of MM

    block 2 --> block 518 of MM

    block 3 --> block 631 of MM

What block of MM will be  in the block 0 of the cache after the last reference (153) if the processor uses the LRU algorithm for replacement?

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Consider a MIPS processor with the configuration (1). The miss penalty is 17 cycles. Calculate the number of cycles of the memmory stage (MEM stage) when the instruction sw $3, 240($7) is executed taking into account that memory position (data) of this instruction is not found in the cache memory. We also know that there is a write buffer and it is not full.

Configurations:

(0) Write through, no-write allocate

(1) Write through, write allocate

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A processor with a main memory of 21 Mbytes has a cache memory of 28Kbytes, and blocks of 27 bytes (words of 1 byte). Assume a 4-way set associative organization. Calculate the size (number of bits) of the field "index" in an physical address.

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Consider a cache with 16 bytes per block (words of 1 byte). Calculate the TAG for a direct mapped organization for the address 3883832 (decimal), assuming that the cache has 27 blocks. Note: the solution have to be written in decimal (advise: transform the address to binary, work in binary and transform the final result to decimal). Note: for systems with word of 1 byte, the byte_offset field does not exist

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Consider a cache system with blocks of 23 words, and words of 22 bytes. Calculate the block number of the main memory for the address  115419 (decimal). Note: The anwer has to be provided in decimal (advise: convert 115419 to binary, work in binary and trasform the final solution from binary to decimal).

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Calculate the physical address of main memory of a data of one byte wich is currently in cache memory, assuming a 2-way set associative organization. The data is the word 1 of a block in memory which is located in the set 23 of the cache. The cache has a size of 26 Kbytes, blocks of 24 bytes (consider words of 1 byte) and the assocciate TAG is 65 (decimal).

Note: provide the result in decimal

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