The following program is stored in the memory unit of the basic computer. Give…

Solution:

1. Execute CLA at location 010: accumulator becomes 0000.

2. Execute ADD 016 at location 011: add the word at 016 (C1A5) to accumulator: accumulator = 0000 + C1A5 = C1A5.

3. Execute BUN 014 at location 012: branch to location 014.

4. Execute AND 017 at location 014: perform bitwise AND of C1A5 and 93C6.

   Compute nibblewise: C & 9 = 8, 1 & 3 = 1, A & C = 8, 5 & 6 = 4.

   Result of AND is 8184, so accumulator = 8184.

5. Execute BUN 013 at location 015 to jump to 013, then execute HLT at 013 to stop.

6. Final accumulator (hex): 8184