The simple computer in class has the specifications as follows: 1. Memory words are 16 bits in length 2. 2 12 = 4 K words of memory 3. Memory can be accessed in one clock cycle 4. Single Accumulator for ALU (AC) 5. Registers are fully connected

(30 pts) For the simple computer developed in class, assume it is now word-addressable (NOT byte addressable). Consider the following program starting in memory location 0 0000000000001000 0010000000000011 0010000000000100 0010000000000010 0001000000000000 Assume the contents of locations 6, 8, 10 are 5, 10, 15 (all in decimal), respectively. The content of AC is zero before the program is executed. What will be the content of AC (in hexadecimal) in two cases: (a) before the first execution of the last instruction (i.e., the last line in the above program: 0001000000000000) and (b) after the execution of the entire program? Note: You must (1) translate each memory line in the above program into an assembly instruction (e.g., "ADD 6" or ADD (6)" or "JMP 6"). (2) Show how the content of AC changes after executing each instruction. In that way, you can show how you get your answer. (30 pts) For the simple computer developed in class, assume it is now word-addressable (NOT byte addressable). Consider the following program starting in memory location 0 0000000000001000 0010000000000011 0010000000000100 0010000000000010 0001000000000000 Assume the contents of locations 6, 8, 10 are 5, 10, 15 (all in decimal), respectively. The content of AC is zero before the program is executed. What will be the content of AC (in hexadecimal) in two cases: (a) before the first execution of the last instruction (i.e., the last line in the above program: 0001000000000000) and (b) after the execution of the entire program? Note: You must (1) translate each memory line in the above program into an assembly instruction (e.g., "ADD 6" or ADD (6)" or "JMP 6"). (2) Show how the content of AC changes after executing each instruction. In that way, you can show how you get your