{"id":3246,"date":"2025-06-01T13:46:16","date_gmt":"2025-06-01T13:46:16","guid":{"rendered":"https:\/\/diznr.com\/?p=3246"},"modified":"2025-06-01T13:46:16","modified_gmt":"2025-06-01T13:46:16","slug":"day-02part-05-operating-system-examples-based-on-first-come-first-serve-basic-algorithms-fcfs","status":"publish","type":"post","link":"https:\/\/www.reilsolar.com\/pdf\/day-02part-05-operating-system-examples-based-on-first-come-first-serve-basic-algorithms-fcfs\/","title":{"rendered":"Day 02Part 05 – Operating system- Examples based on First come first serve basic algorithms (FCFS)."},"content":{"rendered":"Day 02Part 05 – Operating system- Examples based on First come first serve basic algorithms (FCFS).<\/p>\n
[fvplayer id=”316″]<\/p>\n
First Come, First Serve (FCFS) Scheduling \u2013 Examples & Explanation<\/strong><\/h3>\n1. Introduction to FCFS Scheduling<\/strong><\/h3>\n\n- First Come, First Serve (FCFS)<\/strong> is the simplest CPU scheduling algorithm<\/strong>.<\/li>\n
- The process that arrives first<\/strong> gets executed first.<\/li>\n
- It is non-preemptive<\/strong>, meaning once a process starts, it cannot be interrupted<\/strong> until it finishes.<\/li>\n<\/ul>\n
2. Steps to Implement FCFS Scheduling<\/strong><\/h3>\nSort processes by their arrival time<\/strong> (earliest first).
Execute each process in the order of arrival<\/strong>.
Compute the completion, turnaround, and waiting times<\/strong>.<\/p>\n3. Example of FCFS Scheduling<\/strong><\/h3>\nGiven Data<\/strong><\/h4>\n\n\n\n| Process<\/th>\n | Arrival Time (AT)<\/th>\n | Burst Time (BT)<\/th>\n<\/tr>\n<\/thead>\n |
\n\n| P1<\/td>\n | 0 ms<\/td>\n | 4 ms<\/td>\n<\/tr>\n |
\n| P2<\/td>\n | 1 ms<\/td>\n | 3 ms<\/td>\n<\/tr>\n |
\n| P3<\/td>\n | 2 ms<\/td>\n | 1 ms<\/td>\n<\/tr>\n |
\n| P4<\/td>\n | 3 ms<\/td>\n | 2 ms<\/td>\n<\/tr>\n<\/tbody>\n<\/table>\nStep 1: Compute Completion Time (CT)<\/strong><\/h4>\nFCFS executes processes in the order they arrive.<\/p>\n \n\n\n| Process<\/th>\n | AT<\/th>\n | BT<\/th>\n | Start Time<\/th>\n | Completion Time (CT)<\/th>\n<\/tr>\n<\/thead>\n | \n\n| P1<\/td>\n | 0<\/td>\n | 4<\/td>\n | 0<\/td>\n | 4<\/td>\n<\/tr>\n | \n| P2<\/td>\n | 1<\/td>\n | 3<\/td>\n | 4<\/td>\n | 7<\/td>\n<\/tr>\n | \n| P3<\/td>\n | 2<\/td>\n | 1<\/td>\n | 7<\/td>\n | 8<\/td>\n<\/tr>\n | \n| P4<\/td>\n | 3<\/td>\n | 2<\/td>\n | 8<\/td>\n | 10<\/td>\n<\/tr>\n<\/tbody>\n<\/table>\nStep 2: Compute Turnaround Time (TAT)<\/strong><\/h4>\nTAT=CT\u2212ATTAT = CT – AT<\/span>T<\/span>A<\/span>T<\/span>=<\/span><\/span>CT<\/span>\u2212<\/span><\/span>A<\/span>T<\/span><\/span><\/span><\/span><\/span><\/p>\n\n\n\n| Process<\/th>\n | CT<\/th>\n | AT<\/th>\n | TAT (CT – AT)<\/th>\n<\/tr>\n<\/thead>\n | \n\n| P1<\/td>\n | 4<\/td>\n | 0<\/td>\n | 4<\/td>\n<\/tr>\n | \n| P2<\/td>\n | 7<\/td>\n | 1<\/td>\n | 6<\/td>\n<\/tr>\n | \n| P3<\/td>\n | 8<\/td>\n | 2<\/td>\n | 6<\/td>\n<\/tr>\n | \n| P4<\/td>\n | 10<\/td>\n | 3<\/td>\n | 7<\/td>\n<\/tr>\n<\/tbody>\n<\/table>\nStep 3: Compute Waiting Time (WT)<\/strong><\/h4>\nWT=TAT\u2212BTWT = TAT – BT<\/span>W<\/span>T<\/span>=<\/span><\/span>T<\/span>A<\/span>T<\/span>\u2212<\/span><\/span>BT<\/span><\/span><\/span><\/span><\/span><\/p>\n\n\n\n| Process<\/th>\n | TAT<\/th>\n | BT<\/th>\n | WT (TAT – BT)<\/th>\n<\/tr>\n<\/thead>\n | \n\n| P1<\/td>\n | 4<\/td>\n | 4<\/td>\n | 0<\/td>\n<\/tr>\n | \n| P2<\/td>\n | 6<\/td>\n | 3<\/td>\n | 3<\/td>\n<\/tr>\n | \n| P3<\/td>\n | 6<\/td>\n | 1<\/td>\n | 5<\/td>\n<\/tr>\n | \n| P4<\/td>\n | 7<\/td>\n | 2<\/td>\n | 5<\/td>\n<\/tr>\n<\/tbody>\n<\/table>\nStep 4: Calculate Average Times<\/strong><\/h4>\nAverage\u00a0Turnaround\u00a0Time=4+6+6+74=5.75\u00a0ms\\text{Average Turnaround Time} = \\frac{4 + 6 + 6 + 7}{4} = 5.75 \\text{ ms}<\/span>Average\u00a0Turnaround\u00a0Time<\/span><\/span>=<\/span><\/span>44+<\/span>6+<\/span>6+<\/span>7<\/span>\u200b<\/span><\/span><\/span><\/span><\/span>=<\/span><\/span>5.75<\/span>\u00a0ms<\/span><\/span><\/span><\/span><\/span><\/span> Average\u00a0Waiting\u00a0Time=0+3+5+54=3.25\u00a0ms\\text{Average Waiting Time} = \\frac{0 + 3 + 5 + 5}{4} = 3.25 \\text{ ms}<\/span>Average\u00a0Waiting\u00a0Time<\/span><\/span>=<\/span><\/span>40+<\/span>3+<\/span>5+<\/span>5<\/span>\u200b<\/span><\/span><\/span><\/span><\/span>=<\/span><\/span>3.25<\/span> ms<\/span><\/span><\/span><\/span><\/span><\/span><\/p>\n4. Gantt Chart Representation<\/strong><\/h3>\n | | | |