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1. How many bits are required to address a 4M X 16 main memory if a) Main memory

ID: 3683228 • Letter: 1

Question

1. How many bits are required to address a 4M X 16 main memory if

a) Main memory is byte addressable? ______

b) Main memory is word addressable? ______

2. Suppose that a 16M X 16 main memory is built using 512K X 8 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 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.

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 0001 0111 0000

b) 1001 0000 1001 0000

c) 0011 0000 0000 1011

d) 1000 0100 0000 0000

5.List the hexadecimal code for the following program

   Hex Address                        Label               Instruction Hexadecimal Code

            100 Num, Load Addr                 

101 Store Test                   

            102 Skipcond 800                         

103 Jump Num                  

            104 Subt One                    

            105 Store Ctr                    

            106 Halt                            

            107                  Addr, Hex 130                     

            108                  Test, Hex 0                         

            109                  One, Hex 1                         

            10A                 Ctr, Hex 0                         

Explanation / Answer

Multiple Questions : Answering 1st

1)

4M X 16 = 4 x 220 x 24 = 226

a) byte addressable

226 / 8 = 223 => 23 bits

b) word addressable

226/32 = 221 => 21 bits