A computer has a 256K word addressable memory Module with 16 bits per word. Theinstruction set consists of 166 different instructions. All instructions have an operation codepart (opcode) and an address part (allowing for only one address). Each instruction isstored 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 theassigned size of the opcode part? Justify.

Dear Student, Given that size of main memory = 256K bits = 256*1024 bits, Now bits in the main…

Answered: A computer has a 256K word addressable…

Computer Networking: A Top-Down Approach (7th Edition)

7th Edition

ISBN:9780133594140

Author:James Kurose, Keith Ross

Publisher:James Kurose, Keith Ross

Chapter1: Computer Networks And The Internet

Section: Chapter Questions

Problem R1RQ: What is the difference between a host and an end system? List several different types of end...

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Write down the binary and the hexadecimal values of the following and MIPS instruction. Consider the correct values of opcode table and Register File from the tables. The registers size is 5-bit.
Please answer all the following questions. 1. In a hypothetical computer, the processor has four registers: an 8-bit Program Counter (PC), a 16-bit Accumulator (AC), a 16-bit Counter (CTR), and a 12-bit Pointer (PTR). The memory is divided into words each of which is 16-bit long. Each word can hold either an instruction or a piece of data. For each instruction X, the four most significant bits (denoted by X15-12) represent an opcode. The rest of the instruction (denoted by X11-0) can be either an address or a value of an operand. The table below explains some of the instructions supported by the processor. Opcode (binary) 0000 0001 0010 0011 0100 0101 0110 0111 1000 1001 1010 1011 Operation Load AC from a memory location whose address is X11-0. Store AC into a memory location whose address is X11-0. Add to AC the contents of a memory location whose address is the value of PTR. The result is saved to AC. 1100 Subtract from AC the contents of a memory location whose address is the value…
A hypothetical machine is given as follows. The partial list of opcodes (bits 0-3) for the machine are as follows. 0001 = Load AC from Memory 0010 = Store AC to Memory 0011 = Add to AC from Memory 0100 = Load AC from I/O 0101 = Store AC to I/O 0110 = Load Register from Memory 0111 = Store Register to Memory 1000 = Add to AC from Register The list of registers (bits 4-5) for the machine are as follows. 00 Register 1 01 Register 2 10 Register 3 11 Register 4 In the following cases, the remaining 10 bits identify a particular I/O device or memory location. Show the program execution (contents of IR, PC, AC) for the following program where the following instructions are executed from memory locations 300, 301, 302, 303, and 304 respectively. 1.Load Register 1 from Device 5. Assume that the next value retrieved from device 5 is 30 2.Add content of Register 2. (Assume that Register 2 contains a value of 20.) 3.Store AC to device 6.
DO YOU THINK IT'S TRUE OR FALSE?The advantage of zero-address instruction computers is that their programs are short; the disadvantage is that the instructions need a large number of bits, resulting in programs that are quite long.
Computer Architecture (Already submit this question, but I think I got wrong solution) Consider a computer that has a number of registers such that the three registers R0 =1500, R1 = 4500, and R2 = 1000 Show the effective address of memory and the registers’ contents in each of the following instructions 1. ADD (R1), R2 2. MOVE 500(R0), R2 3. ADD (40), R1 4. SUBTRACT (5000), R2 5. ADD #30, R2
Suppose the RAM for a certain computer has 4M words, where each word is 16 bits long. What is the capacity of this memory expressed in bytes? If this RAM is byte addressable, how many bits must an address contain? If this RAM is word addressable, how many bits must an address contain?
Fill in the following table showing the results of the indicated instructions, using the register values shown. Also, show the OF and CF values AFTER the instruction executes. Show all results in hexadecimal. Show OF and CF as 0 or 1.
Convert the given hex value to its R-format MIPS instruction by completing the following table. The instruction and letters for registers must be in lower case. DO NOT TYPE DOLLAR SIGNS WITH REGISTERS. Hex Opcode (decimal) rs (decimal) rt (decimal) rd (decimal) funct (decimal) Instruction 02488822 O 18 16 17 34 sub s4 $$1 " $ to " LA
In this implementation, each instruction is provided with its own data independently of how other instructions acquire theirs. We insert a(n): A Multiple Input/Output B The Case of Many Pieces of Information or Directions C Unique Information Only One D Several Insts From One

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