i) a) Draw the process state diagram indicating clearly the transitions between states. ` +-------+ +-------+ +---------+ | New | --> | Ready | <-> | Running | --> | Terminated | +-------+ +-------+ +---------+ ^ | | v +-------------+ | Blocked | +---------+ ` New: The process is being created. Ready: The process is waiting to be assigned to a processor. Running: Instructions are being executed. Blocked (Waiting): The process is waiting for some event to occur (e.g., I/O completion, signal). Terminated: The process has finished execution. b) Describe the transition from the Running state to the Blocked (Waiting) state. Give an example of an event that would cause this transition. A process transitions from the Running state to the Blocked (Waiting) state when it requires an event to occur that it cannot immediately satisfy. This typically involves waiting for an I/O operation to complete or for a resource to become available. The CPU is then relinquished by the process. Example: A process requests to read data from a hard disk. Since disk I/O is much slower than CPU operations, the process enters the Blocked state until the data is read and available. ii) Given processes arriving at time 0: | Process | Burst Time (ms) | Priority | | :------ | :-------------- | :------- | | P1 | 10 | 3 | | P2 | 1 | 1 | | P3 | 2 | 4 | | P4 | 1 | 5 | | P5 | 5 | 2 | A smaller priority number implies a higher priority. a) Draw a Gantt chart illustrating the execution of these processes using a Preemptive Priority scheduling algorithm. At time $t=0$, all processes arrive. We sort them by priority (1 is highest, 5 is lowest): 1. P2 (Priority 1, Burst 1) 2. P5 (Priority 2, Burst 5) 3. P1 (Priority 3, Burst 10) 4. P3 (Priority 4, Burst 2) 5. P4 (Priority 5, Burst 1) Since all processes arrive at $t=0$ and no new processes arrive later, the preemptive priority scheduling behaves like non-preemptive in this specific scenario. Gantt Chart: $$ \begin{array}{|c|c|c|c|c|} \hline \text{P2} & \text{P5} & \text{P1} & \text{P3} & \text{P4} \\ \hline \end{array} $$ $$ \begin{array}{cccccccccc} 0 & 1 & & 6 & & 16 & & 18 & & 19 \\ \end{array} $$ b) Calculate the average waiting time for the processes using the scheduling algorithm in part (a). First, calculate the completion time for each process: P2 completes at $1 \text{ ms}$ P5 completes at $6 \text{ ms}$ P1 completes at $16 \text{ ms}$ P3 completes at $18 \text{ ms}$ P4 completes at $19 \text{ ms}$ Waiting Time = Completion Time - Burst Time - Arrival Time Since all arrival times are $0 \text{ ms}$: Waiting Time = Completion Time - Burst Time P2 Waiting Time: $1 \text{ ms} - 1 \text{ ms} - 0 \text{ ms} = 0 \text{ ms}$ P5 Waiting Time: $6 \text{ ms} - 5 \text{ ms} - 0 \text{ ms} = 1 \text{ ms}$ P1 Waiting Time: $16 \text{ ms} - 10 \text{ ms} - 0 \text{ ms} = 6 \text{ ms}$ P3 Waiting Time: $18 \text{ ms} - 2 \text{ ms} - 0 \text{ ms} = 16 \text{ ms}$ P4 Waiting Time: $19 \text{ ms} - 1 \text{ ms} - 0 \text{ ms} = 18 \text{ ms}$ Total Waiting Time = $0 + 1 + 6 + 16 + 18 = 41 \text{ ms}$ Number of processes = $5$ Average Waiting Time = $\frac{\text{Total Waiting Time}}{\text{Number of Processes}}$ Average Waiting Time = $\frac{41 \text{ ms}}{5} = 8.2 \text{ ms}$ The average waiting time for the processes is $\boxed{\text{8.2 ms}}$. c) Explain one disadvantage of priority-based scheduling. One significant disadvantage of priority-based scheduling is starvation (or indefinite blocking). Low-priority processes may never get to execute if there is a continuous stream of higher-priority processes entering the ready queue. This can lead to a situation where a process waits indefinitely for the CPU.