An output program resides in memory starting from address 2300. It is executed after the computer recognizes an interrupt when FGO becomes a 1 (while IEN = 1).

a. What instruction must be placed at address 1?

b. What must be the last two instructions of the output program?

answer to a). The instruction at address 1 will be 0 BUN^(2300)₁₀

Reason: the branch instruction will make the output transfer to the I/O service(I/O operation stands for input and output service which means to read or write operation). This will make sure the status of the flag is checked and will accordingly transfer essential I/O programs.

answer to b). The last two instructions of the program will be:

• ION
• BUN 0 1,indirect branch with address 0,

Afterward, the ION instruction will execute and will set IEN to 1, then the branch indirect instruction which has the address 0 will return the computer to its original position in the main program.