{"id":342,"date":"2025-11-13T03:26:46","date_gmt":"2025-11-13T03:26:46","guid":{"rendered":"https:\/\/mitalgoswami.in\/?p=342"},"modified":"2025-12-19T04:30:53","modified_gmt":"2025-12-19T04:30:53","slug":"round-robin-rr-scheduling-algorithm","status":"publish","type":"post","link":"https:\/\/mitalgoswami.in\/?p=342","title":{"rendered":"Round Robin (RR) Scheduling Algorithm"},"content":{"rendered":"\n

Definition:<\/strong>
Round Robin is a preemptive CPU scheduling algorithm<\/strong> where each process is assigned a fixed time slice or quantum<\/strong>. The CPU executes each process for this fixed amount of time in a cyclic order. If a process does not finish in its time quantum, it is moved to the end of the ready queue.<\/p>\n\n\n\n

\n\n\n\n

Key Features:<\/strong><\/h3>\n\n\n\n
    \n
  • Type:<\/strong> Preemptive scheduling<\/li>\n\n\n\n
  • Basis:<\/strong> Time-sharing system<\/li>\n\n\n\n
  • Queue type:<\/strong> FIFO (First In First Out)<\/li>\n\n\n\n
  • Preemption:<\/strong> Occurs after a fixed time quantum expires<\/li>\n\n\n\n
  • Fairness:<\/strong> Every process gets equal CPU time turn by turn<\/li>\n<\/ul>\n\n\n\n
    \n\n\n\n

    Working of Round Robin:<\/strong><\/h3>\n\n\n\n
      \n
    1. All processes are kept in a ready queue<\/strong>.<\/li>\n\n\n\n
    2. Each process gets the CPU for a fixed time quantum (say 4 ms)<\/strong>.<\/li>\n\n\n\n
    3. If a process completes within the time quantum, it leaves the CPU.<\/li>\n\n\n\n
    4. If not, it is preempted<\/strong> and moved to the end of the ready queue<\/strong>.<\/li>\n\n\n\n
    5. The CPU then schedules the next process in the queue.<\/li>\n<\/ol>\n\n\n\n
      \n\n\n\n

      Example:<\/strong><\/h3>\n\n\n\n
      Process<\/th>Burst Time (ms)<\/th><\/tr><\/thead>
      P1<\/td>10<\/td><\/tr>
      P2<\/td>5<\/td><\/tr>
      P3<\/td>8<\/td><\/tr><\/tbody><\/table><\/figure>\n\n\n\n

      Time Quantum = 4 ms<\/strong><\/p>\n\n\n\n

      Execution Order:<\/strong><\/p>\n\n\n\n

      P1 (4) \u2192 P2 (4) \u2192 P3 (4) \u2192 P1 (4) \u2192 P2(1) \u2192 P3 (4) \u2192 p1 (2)\n<\/code><\/pre>\n\n\n\n
      Gantt Chart:<\/strong><\/p>\n\n\n\n
      | P1 | P2 | P3 | P1 | P2 | P3 | p1\n0    4    8   12   16    17   21   23\n<\/code><\/pre>\n\n\n\n
      \n\n\n\n
      Advantages:<\/strong><\/h3>\n\n\n\n
        \n
      • Fairness:<\/strong> Every process gets equal CPU time.<\/li>\n\n\n\n
      • Good for time-sharing systems.<\/strong><\/li>\n\n\n\n
      • No starvation:<\/strong> All processes get a chance to execute.<\/li>\n<\/ul>\n\n\n\n
        \n\n\n\n
        Disadvantages:<\/strong><\/h3>\n\n\n\n
          \n
        • Performance depends on the time quantum:<\/strong>\n
            \n
          • If too large<\/strong>, RR \u2192 FCFS (First Come First Serve).<\/li>\n\n\n\n
          • If too small<\/strong>, too many context switches<\/strong> (inefficient).<\/li>\n<\/ul>\n<\/li>\n\n\n\n
          • Higher overhead<\/strong> due to frequent context switching.<\/li>\n<\/ul>\n\n\n\n
            \n\n\n\n
            Applications:<\/strong><\/h3>\n\n\n\n
              \n
            • Widely used in time-sharing<\/strong> and interactive systems<\/strong> (e.g., modern operating systems).<\/li>\n<\/ul>\n\n\n\n
              his is a CPU Scheduling \u2013 Round Robin (RR)<\/strong> problem.<\/p>\n\n\n\n
              Given<\/h3>\n\n\n\n
                \n
              • Time Quantum (TQ) = 3<\/strong><\/li>\n<\/ul>\n\n\n\n
                Process<\/th>Arrival Time<\/th>Burst Time<\/th><\/tr><\/thead>
                P1<\/td>0<\/td>5<\/td><\/tr>
                P2<\/td>2<\/td>3<\/td><\/tr>
                P3<\/td>3<\/td>2<\/td><\/tr>
                P4<\/td>1<\/td>3<\/td><\/tr><\/tbody><\/table><\/figure>\n\n\n\n
                \n\n\n\n
                Step 1: Order by Arrival Time<\/h2>\n\n\n\n
                Arrival sequence:<\/p>\n\n\n\n
                  \n
                • Time 0 \u2192 P1<\/strong><\/li>\n\n\n\n
                • Time 1 \u2192 P4<\/strong><\/li>\n\n\n\n
                • Time 2 \u2192 P2<\/strong><\/li>\n\n\n\n
                • Time 3 \u2192 P3<\/strong><\/li>\n<\/ul>\n\n\n\n
                  \n\n\n\n
                  Step 2: Draw the Gantt Chart (Round Robin, TQ = 3)<\/h2>\n\n\n\n
                  Execution Flow<\/h3>\n\n\n\n
                    \n
                  1. Time 0\u20133<\/strong> \u2192 P1 executes (remaining burst = 2)<\/li>\n\n\n\n
                  2. Time 3\u20136<\/strong> \u2192 P4 executes (burst finished)<\/li>\n\n\n\n
                  3. Time 6\u20139<\/strong> \u2192 P2 executes (burst finished)<\/li>\n\n\n\n
                  4. Time 9\u201311<\/strong> \u2192 P1 executes (burst finished)<\/li>\n\n\n\n
                  5. Time 11\u201313<\/strong> \u2192 P3 executes (burst finished)<\/li>\n<\/ol>\n\n\n\n
                    Gantt Chart<\/h3>\n\n\n\n
                    0   3   6   9   11  13\n| P1 | P4 | P2 | P1 | P3 |\n<\/code><\/pre>\n\n\n\n
                    \n\n\n\n
                    Step 3: Completion Time (CT)<\/h2>\n\n\n\n
                    Process<\/th>Completion Time<\/th><\/tr><\/thead>
                    P1<\/td>11<\/td><\/tr>
                    P2<\/td>9<\/td><\/tr>
                    P3<\/td>13<\/td><\/tr>
                    P4<\/td>6<\/td><\/tr><\/tbody><\/table><\/figure>\n\n\n\n
                    \n\n\n\n
                    Step 4: Turnaround Time (TAT)<\/h2>\n\n\n\n
                    TAT = Completion Time \u2212 Arrival Time<\/strong><\/p>\n\n\n\n
                    Process<\/th>TAT<\/th><\/tr><\/thead>
                    P1<\/td>11 \u2212 0 = 11<\/strong><\/td><\/tr>
                    P2<\/td>9 \u2212 2 = 7<\/strong><\/td><\/tr>
                    P3<\/td>13 \u2212 3 = 10<\/strong><\/td><\/tr>
                    P4<\/td>6 \u2212 1 = 5<\/strong><\/td><\/tr><\/tbody><\/table><\/figure>\n\n\n\n
                    \n\n\n\n
                    Step 5: Waiting Time (WT)<\/h2>\n\n\n\n
                    WT = Turnaround Time \u2212 Burst Time<\/strong><\/p>\n\n\n\n
                    Process<\/th>WT<\/th><\/tr><\/thead>
                    P1<\/td>11 \u2212 5 = 6<\/strong><\/td><\/tr>
                    P2<\/td>7 \u2212 3 = 4<\/strong><\/td><\/tr>
                    P3<\/td>10 \u2212 2 = 8<\/strong><\/td><\/tr>
                    P4<\/td>5 \u2212 3 = 2<\/strong><\/td><\/tr><\/tbody><\/table><\/figure>\n\n\n\n
                    \n\n\n\n
                    Final Answer Summary<\/h2>\n\n\n\n
                    ✅ Gantt Chart<\/h3>\n\n\n\n
                    0 | P1 | 3 | P4 | 6 | P2 | 9 | P1 | 11 | P3 | 13\n<\/code><\/pre>\n\n\n\n
                    ✅ Average Times<\/h3>\n\n\n\n
                      \n
                    • Average Waiting Time<\/strong>
                      = (6 + 4 + 8 + 2) \/ 4 = 5<\/strong><\/li>\n\n\n\n
                    • Average Turnaround Time<\/strong>
                      = (11 + 7 + 10 + 5) \/ 4 = 8.25<\/strong><\/li>\n<\/ul>\n\n\n\n
                      \n\n\n\n
                      If you want, I can also:<\/p>\n\n\n\n
                        \n
                      • Explain this step-by-step in exam writing format<\/strong><\/li>\n\n\n\n
                      • Solve it using FCFS \/ SJF \/ Priority<\/strong> for comparison<\/li>\n<\/ul>\n","protected":false},"excerpt":{"rendered":"
                        Definition:Round Robin is a preemptive CPU scheduling algorithm where each process is assigned a fixed time slice or quantum. The CPU executes each process for this fixed amount of time in a cyclic order. If a process does not finish in its time quantum, it is moved to the end of the ready queue. Key […]<\/p>\n","protected":false},"author":1,"featured_media":0,"comment_status":"open","ping_status":"open","sticky":false,"template":"","format":"standard","meta":{"footnotes":""},"categories":[5],"tags":[],"class_list":["post-342","post","type-post","status-publish","format-standard","hentry","category-unix"],"aioseo_notices":[],"_links":{"self":[{"href":"https:\/\/mitalgoswami.in\/index.php?rest_route=\/wp\/v2\/posts\/342","targetHints":{"allow":["GET"]}}],"collection":[{"href":"https:\/\/mitalgoswami.in\/index.php?rest_route=\/wp\/v2\/posts"}],"about":[{"href":"https:\/\/mitalgoswami.in\/index.php?rest_route=\/wp\/v2\/types\/post"}],"author":[{"embeddable":true,"href":"https:\/\/mitalgoswami.in\/index.php?rest_route=\/wp\/v2\/users\/1"}],"replies":[{"embeddable":true,"href":"https:\/\/mitalgoswami.in\/index.php?rest_route=%2Fwp%2Fv2%2Fcomments&post=342"}],"version-history":[{"count":5,"href":"https:\/\/mitalgoswami.in\/index.php?rest_route=\/wp\/v2\/posts\/342\/revisions"}],"predecessor-version":[{"id":367,"href":"https:\/\/mitalgoswami.in\/index.php?rest_route=\/wp\/v2\/posts\/342\/revisions\/367"}],"wp:attachment":[{"href":"https:\/\/mitalgoswami.in\/index.php?rest_route=%2Fwp%2Fv2%2Fmedia&parent=342"}],"wp:term":[{"taxonomy":"category","embeddable":true,"href":"https:\/\/mitalgoswami.in\/index.php?rest_route=%2Fwp%2Fv2%2Fcategories&post=342"},{"taxonomy":"post_tag","embeddable":true,"href":"https:\/\/mitalgoswami.in\/index.php?rest_route=%2Fwp%2Fv2%2Ftags&post=342"}],"curies":[{"name":"wp","href":"https:\/\/api.w.org\/{rel}","templated":true}]}}