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Organization and Architecture

   

Added on  2019-10-30

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Running head: ORGANIZATION AND ARCHITECTUREOrganization and ArchitectureName of the Student:Name of the University:Author Note

1ORGANIZATION AND ARCHITECTURE 1.Suppose that a 256M * 128 Memory built using 1024 K *32 RAM Chips and memory is word addressable. [10 marks]a.How many RAM Chips are necessary?We need 256 Kbytes, i.e., 256 x 1024 x 128 bits.We have RAM chips of capacity 32 Kbits = 32 x 1024 bits.Number of Ram chips required are (256 * 1024 * 128)/(32 * 1024) = 1024b.How many RAM Chips are there per memory word?We need 256 Kbytes, i.e., 256 x 1024 x 8 bits.We have RAM chips of capacity 32 Kbits = 32 x 1024 bits.Number of Ram chips required are (256 * 1024 * 8)/(32 * 1024) = 64c.How many address bits are needed for each RAM Chip?Number of RAM chips required are 1024 = 2^30.Hence address bits needed for each RAM chip required are 30.d.How many banks will this memory have?The number bits present in the RAM = 1024Hence the number of memory banks required for this system are: 256/32 = 8e.How many address bits are needed for all memory? Address bits required for all memory = 1024/32 = 32

2ORGANIZATION AND ARCHITECTURE 2. A digital computer has a memory unit with 48 bits per word. The instruction set consists of 240 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. [2+2+2 = 6 marks].a.How many bits are needed for the opcode? 16 bits would be required for the opcode part of the word.b.How many bits are needed to specify the register? How many bits are left for the address part of the instruction?32 bits would be left for addressing and the register parts. This can be manipulated according to the requirements of the users.c.What is the largest unsigned binary number that can be accommodated in one word of memory?The largest unsigned binary number that can be accommodated in one word of the memory are: 2^48 -1.

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