A system contains three programs and each requires three tape units for its operation. The minimum number of tape units which the system must have such that deadlocks never arise is _________.

To ensure that deadlocks never arise in this system, we need to follow the Banker's Algorithm. This algorithm avoids deadlock by denying or postponing the request when there is a possibility of future deadlock.

In this case, each program requires three tape units for its operation. So, if a program requests for a tape unit, the system must have enough units left for the remaining programs.

Let's calculate:

• If the first program takes 3 units, there should be at least 3*2 = 6 units left for the other two programs.
• If the second program also takes 3 units, there should be at least 3*1 = 3 units left for the last program.
• The last program will also need 3 units.

So, the minimum number of tape units the system must have to ensure that deadlocks never arise is 3 (for the first program) + 6 (for the second and third programs) = 9 units.