Question

1. In a computer structure format, the instruction length is 12 bits and the size of an address field is 4 bits. Assume a computer architect has already designed 14 two-address and 36 zero-address instructions. What is the maximum number of one-address instructions that can be added to the instruction set?

2. Marie Code:

A.   Load D
       Skipcond 800
       Jump C
B.    Load D
       Add D
       Store D
       Subt F
       Store E
C.    Load D
       Add E
       Store F
       Halt
D.    Dec 3
E.    Dec 4
F.    Dec 5

a. Write the above MARIE code in high level language such as JAVA.
b. What are the values of D, E and F after execution?

3. Using CRC polynomial 10011, compute the CRC code word for the information
10110011001

CRC code word =____

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1. Answer:
The 2-address instructions have the opcode of 4 bits and 8 bits for the two addresses (4 bits for each).
The architect can represent the 2-address instructions between:
0000 xxxx xxxx – 1111 xxxx xxxx (since he represented only 14 out of 16 – maximum for 4 bits opcode) => 0000 xxxx xxxx – 1101 xxxx xxxx are 2-address instructions.

Then we know he had represented 36 0-address instructions (no bits are for addresses)....

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