A computer employs RAM chips of 512 x 16 and ROM chips of 1024 x 8. The computer system needs 4K bytes of RAM and 2K bytes of ROM along with interface unit of 128 registers each. A memory mapped I/O configuration is used. The two higher order bits are assigned for RAM, ROM and interface as 00, 01 and 10 respectively. How many RAM and ROM chips are needed
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- A(n) ________________ instruction always alters the instruction execution sequence. A(n) ______________ instruction alters the instruction execution sequence only if a specified Condition is true.A computer employs RAM chips of 512 x 16 and ROM chips of 1024 x 8. The computer system needs 4K bytes of RAM and 2K bytes of ROM along with interface unit of 128 registers each. A memory mapped I/O configuration is used. The two higher order bits are assigned for RAM, ROM and interface as 00, 01 and 10 respectively. Show the chip layout for the above designA computer employs RAM chips of 512 x 16 and ROM chips of 1024 x 8. The computer system needs 4K bytes of RAM and 2K bytes of ROM along with interface unit of 128 registers each. A memory mapped I/O configuration is used. The two higher order bits are assigned for RAM, ROM and interface as 00, 01 and 10 respectively.
- A computer employs RAM chips of 128 x 8 and ROM chips of 512 x 8. The computer system needs 1K bytes of RAM, 2K bytes of ROM, and two interface units, each with two registers. A memory-mapped 1/0 configuration is used. The two highest-order bits of the 16-bit address bus are assigned 11 for RAM, 10 for ROM, and 01 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 interfaceA 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 with a 32 bit word uses an instruction format that includes direct and indirect addressing of 8 megabytes and on of 16 registers. The highest order bits are used for the opcode, followed by the bits indicating the register, followed by indirect/direct bit, followed by the bits indicating memory address Draw the instruction word's format, showing how many bits are used for all four fields
- A computer has a 256K word addressable memory Module with 16 bits per word. The instruction set consists of 166 different instructions. 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. Based on the above, answer the following questions: a. How many bits are there in the main memory? (Represent it in power of 2) b. How many bits are needed for the opcode? c. How many bits are left for the address part of the instruction? d. How many additional instructions can be added to the existing 166 without affecting the assigned size of the opcode part? Justify.The memory unit of a computer has 256K words of 32 bits each. The computer has an instruction format with 4 fields: an opcode field; a mode field to specify 1 of 7 addressing modes; a register address field to specify one of 60 registers; and a memory address field. Assume an instruction is 32 bits long. Answer the following:a. How large must the mode field be?b. How large must the register field be?c. How large must the address field be?d. How large is the opcode field?A digital computer has a memory unit with 24 bits per word. The instruction set consists of 150 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.Q.) What is the maximum allowable size for memory?