QuestionQuestion

1. How many bits are required to address a 16M x 32 main memory if
a) Main memory is byte addressable?   
b) Main memory is word addressable?

2. Suppose that a 64M x 64 main memory is built using 8M x 32 RAM chips and memory is word addressable.
a) How many RAM chips are necessary?
b) How many RAM chips are needed for each memory word?
c) How many address bits are needed for each RAM chip?
d) How many address bits are needed for all memory?

3. A digital computer has a memory unit with 30 bits per word. The instruction set consists of 200 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.
a) How many bits are needed for the opcode?
b) How many bits are left for the address part of the instruction?
c) What is the maximum allowable size for memory?

4. Write the following MARIE assembly language equivalent of the following machine language instructions
a) 0010 0000 1000 1111
b) 1001 1000 1011 0000
c) 0100 0011 0010 0001
d) 0101 0000 0000 0000      

5. List the hexadecimal code for the following program.
Hex Address Label Instruction
100 Load A
101 Add One
102 Jump S1
103 S2, Add One
104 Store A
105 Halt
106 S1, Add A
107 Jump S2
108 A, HEX 0103
109 One, HEX 0001

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4.
We are provided the Binary Content of Memory Address. In MARIE language, the bits 0-11 represent the address of the instruction and the bits 12-15 show the opcode of the instruction.
In part a), the opcode 0010 corresponds to STORE instruction. The remaining part, namely 0000 1000 1111 corresponds to 08B. So the instruction stores the content of AC at address 8B.
In part b), the opcode 1001 corresponds to JUMP instruction. The remaining part, namely 1000 1011 0000 corresponds to 8B0; the instruction loads the value of address 8B0 into PC....

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