{"id":67,"date":"2025-01-08T03:56:54","date_gmt":"2025-01-08T03:56:54","guid":{"rendered":"https:\/\/mitalgoswami.in\/?p=67"},"modified":"2025-07-17T06:51:02","modified_gmt":"2025-07-17T06:51:02","slug":"rr-algorithm","status":"publish","type":"post","link":"https:\/\/mitalgoswami.in\/?p=67","title":{"rendered":"RR Algorithm"},"content":{"rendered":"\n

Solving the Round Robin (RR) Scheduling Algorithm<\/strong><\/h3>\n\n\n\n

Given Data:<\/strong><\/h4>\n\n\n\n
    \n
  1. Processes: P1, P2, P3<\/li>\n\n\n\n
  2. Burst Times (BT):\n
      \n
    • P1 = 24<\/li>\n\n\n\n
    • P2 = 3<\/li>\n\n\n\n
    • P3 = 3<\/li>\n<\/ul>\n<\/li>\n\n\n\n
    • Time Quantum = 4<\/li>\n<\/ol>\n\n\n\n
      \n\n\n\n

      Step 1: Gantt Chart Construction<\/strong><\/h3>\n\n\n\n

      Using a time quantum of 4, we allocate the CPU to each process in a round-robin manner.<\/p>\n\n\n\n

        \n
      1. Time Slice 1:<\/strong>\n
          \n
        • P1 executes for 4 units (remaining = 24 – 4 = 20).<\/li>\n\n\n\n
        • P2 completes execution (remaining = 3 – 3 = 0).<\/li>\n\n\n\n
        • P3 completes execution (remaining = 3 – 3 = 0).<\/li>\n<\/ul>\n<\/li>\n\n\n\n
        • Time Slice 2:<\/strong>\n
            \n
          • P1 executes for another 4 units (remaining = 20 – 4 = 16).<\/li>\n<\/ul>\n<\/li>\n\n\n\n
          • Repeat Process:<\/strong>\n
              \n
            • P1 continues to execute until it completes.<\/li>\n<\/ul>\n<\/li>\n<\/ol>\n\n\n\n
              \n\n\n\n

              Gantt Chart<\/strong><\/h3>\n\n\n\n
              Time<\/th>Process<\/th><\/tr><\/thead>
              0 – 4<\/td>P1<\/td><\/tr>
              4 – 7<\/td>P2<\/td><\/tr>
              7 – 10<\/td>P3<\/td><\/tr>
              10 – 14<\/td>P1<\/td><\/tr>
              14 – 18<\/td>P1<\/td><\/tr>
              18 – 22<\/td>P1<\/td><\/tr>
              22 – 26<\/td>P1<\/td><\/tr>
              26 – 30<\/td>P1<\/td><\/tr><\/tbody><\/table><\/figure>\n\n\n\n
              \n\n\n\n

              Step 2: Completion Time<\/strong><\/h3>\n\n\n\n
                \n
              1. P1:<\/strong> Finishes at t=30t = 30t=30<\/li>\n\n\n\n
              2. P2:<\/strong> Finishes at t=7t = 7t=7<\/li>\n\n\n\n
              3. P3:<\/strong> Finishes at t=10t = 10t=10<\/li>\n<\/ol>\n\n\n\n
                \n\n\n\n

                Step 3: Turnaround Time (TAT)<\/strong><\/h3>\n\n\n\n

                Step 3: Turnaround Time (TAT)<\/strong><\/p>\n\n\n\n

                TAT=Completion\u00a0Time\u00a0(CT)\u2212Arrival\u00a0Time\u00a0(AT)<\/p>\n\n\n\n

                Process<\/th>Completion Time (CT)<\/th>Arrival Time (AT)<\/th>Turnaround Time (TAT)<\/th><\/tr><\/thead>
                P1<\/td>30<\/td>0<\/td>30<\/td><\/tr>
                P2<\/td>7<\/td>0<\/td>7<\/td><\/tr>
                P3<\/td>10<\/td>0<\/td>10<\/td><\/tr><\/tbody><\/table><\/figure>\n\n\n\n
                \n\n\n\n

                Step 4: Waiting Time (WT)<\/strong><\/h3>\n\n\n\n

                WT=TAT\u2212Burst\u00a0Time\u00a0(BT)<\/p>\n\n\n\n

                Process<\/th>Turnaround Time (TAT)<\/th>Burst Time (BT)<\/th>Waiting Time (WT)<\/th><\/tr><\/thead>
                P1<\/td>30<\/td>24<\/td>6<\/td><\/tr>
                P2<\/td>7<\/td>3<\/td>4<\/td><\/tr>
                P3<\/td>10<\/td>3<\/td>7<\/td><\/tr><\/tbody><\/table><\/figure>\n\n\n\n
                \n\n\n\n

                Final Results:<\/strong><\/h3>\n\n\n\n
                Process<\/th>Burst Time (BT)<\/th>Completion Time (CT)<\/th>Turnaround Time (TAT)<\/th>Waiting Time (WT)<\/th><\/tr><\/thead>
                P1<\/td>24<\/td>30<\/td>30<\/td>6<\/td><\/tr>
                P2<\/td>3<\/td>7<\/td>7<\/td>4<\/td><\/tr>
                P3<\/td>3<\/td>10<\/td>10<\/td>7<\/td><\/tr><\/tbody><\/table><\/figure>\n\n\n\n
                \n\n\n\n

                Average Times:<\/strong><\/h3>\n\n\n\n
                  \n
                1. Average Turnaround Time (TAT):<\/strong>Average\u00a0TAT=30+7+10\/3=15.67<\/li>\n\n\n\n
                2. Average Waiting Time (WT):<\/strong>Average\u00a0WT=6+4+7\/3=5.67<\/li>\n<\/ol>\n","protected":false},"excerpt":{"rendered":"

                  Solving the Round Robin (RR) Scheduling Algorithm Given Data: Step 1: Gantt Chart Construction Using a time quantum of 4, we allocate the CPU to each process in a round-robin manner. Gantt Chart Time Process 0 – 4 P1 4 – 7 P2 7 – 10 P3 10 – 14 P1 14 – 18 P1 […]<\/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-67","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\/67","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=67"}],"version-history":[{"count":1,"href":"https:\/\/mitalgoswami.in\/index.php?rest_route=\/wp\/v2\/posts\/67\/revisions"}],"predecessor-version":[{"id":68,"href":"https:\/\/mitalgoswami.in\/index.php?rest_route=\/wp\/v2\/posts\/67\/revisions\/68"}],"wp:attachment":[{"href":"https:\/\/mitalgoswami.in\/index.php?rest_route=%2Fwp%2Fv2%2Fmedia&parent=67"}],"wp:term":[{"taxonomy":"category","embeddable":true,"href":"https:\/\/mitalgoswami.in\/index.php?rest_route=%2Fwp%2Fv2%2Fcategories&post=67"},{"taxonomy":"post_tag","embeddable":true,"href":"https:\/\/mitalgoswami.in\/index.php?rest_route=%2Fwp%2Fv2%2Ftags&post=67"}],"curies":[{"name":"wp","href":"https:\/\/api.w.org\/{rel}","templated":true}]}}