Simplex Method |Part 11| Two Phase Method (Feasible Optimal Solution)

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  • เผยแพร่เมื่อ 18 ธ.ค. 2024

ความคิดเห็น • 14

  • @beulahdivyakannan9355
    @beulahdivyakannan9355 10 หลายเดือนก่อน +2

    So for the second problem at 10:09, since it is the final table in phase 1 for a minimization problem , shouldn't all the values of Zj-Cj be 0,0,1,1,2,0. I'm confused. Could you please explain?

  • @girikgarg1268
    @girikgarg1268 ปีที่แล้ว +3

    Calculation mistake at 6:02, Zj-Cj should be -11 for a2.

  • @PunmasterSTP
    @PunmasterSTP 8 หลายเดือนก่อน

    I'm still a bit unsure about all of the nuances, such as when you can drop rows like at 18:05, but your examples are definitely helping me to understand!

  • @shahzaibnaveed5397
    @shahzaibnaveed5397 ปีที่แล้ว +1

    Mam fantastic ☺️☺️☺️

  • @yasharora4114
    @yasharora4114 4 ปีที่แล้ว

    is there any difference between the cost of artifcial avriables in maximization case and in minimization case

  • @varunsakalkale
    @varunsakalkale 2 ปีที่แล้ว

    At 17.32 shouldnt min () be either 2/2=1?
    And the pivot element corresponding to it?

    • @MathPod
      @MathPod  2 ปีที่แล้ว

      At this time, i am explaining Transition from Phase 1 to Phase 2. Phase 1 table is completed so usual minimum ratio rule is not applied, rather i am looking which variable we can take in phase 2.

  • @MathPod
    @MathPod  4 ปีที่แล้ว +4

    If you using Two phase method, then we start by writing auxiliary LPP, and here objective function is Minimization=sum of artificial variables. So cost is +1. But if in auxiliary LPP you keep objective function as Maximization then it is Maximization=- (sum of artificial variables), so here you can take -1 as cost of artificial variables.
    So depending upon what we are considering objective function in auxiliary LPP accordingly you can choose cost of artificial variables.
    So selection of maximum/ minimum objective is irrelevant, you can select any.
    Similarly in BigM method, if objective function is Maximization then cost is -M, and for Minimization the cost is +M for artificial variables.

    • @yasharora4114
      @yasharora4114 3 ปีที่แล้ว

      Thank you for the explanation, been struggling for a while with these topics. The best explanation video for these topics I've seen.

    • @bikkykumar6312
      @bikkykumar6312 ปีที่แล้ว

      yes you are right i stuck again and again but seeing your cooment i understand that thanks to share your thoughts.
      moreover earlier they said when minimum ratio tie go with artificial variable but here she go with x_{2}. however answer in unaltered.

  • @simranjitsingh6047
    @simranjitsingh6047 4 ปีที่แล้ว +1

    Please make video on matlab code

    • @MathPod
      @MathPod  4 ปีที่แล้ว +1

      I do hope to make matlab code in future. Thank you for watching videos.

  • @girish1052
    @girish1052 2 ปีที่แล้ว +1

    Screenshot: drive.google.com/file/d/1YNMVQtz98Hmim7uGctYeqxOdJBK0zW16/view?usp=drivesdk
    For optimality the vector with most positive c - z value (in this case x1) should be entered into the solution (Table 4.15), but author stressed that if x1 is allowed it will lead to infeasibility w/o any explanation, instead chose to go with s2. Can you please give a hint why he said so?