Question 41

1. Let $n = 3$ be the number of processes (programs).

2. Each process has a maximum resource requirement $m = 3$ tape units.

3. Worst-Case Allocation (Deadlock State):

   • In the worst possible state without deadlock resolution, every process holds one unit less than its maximum requirement ($m - 1$ units):
     $$\text{Units per process} = 3 - 1 = 2.$$
   • Total units allocated across all $3$ processes in this state:
     $$n \times (m - 1) = 3 \times 2 = 6\text{ units}.$$
   • If only 6 tape units are available, all 3 processes are waiting for one additional unit, resulting in a deadlock.

4. Deadlock-Free Threshold:

   • If the system has at least 1 additional unit:
     $$R_{\min} = n(m - 1) + 1 = 3(3 - 1) + 1 = 6 + 1 = 7\text{ units}.$$
   • At least one process is guaranteed to receive its third unit, finish execution, and release all 3 units to satisfy other processes.

Hence, the minimum number of tape units required is 7.