FIFO Aging Simulation












2














I am trying to simulate a backlog of cases by age.



Assumptions:





  1. FIFO: (First In First Out): This means we will always prioritize cases in the 4+ bucket then move to the bucket to it's left and so on. This means current day Inflow has last priority.


  2. Productivity = 1,000 = Constant: The most that can be worked in a single day is 1,000

  3. Before the simulation begins we have 1'000 aged 1 day, 750 aged 2 days, 500 aged 3 days, & 250 aged 4 days




Given this information, I believe it is possible to input any Inflow and have the buckets complete themselves. I cannot figure out what equations to use to populate the desired result though.



I have tried a series of SUM, SUMIF, IF, and checking if the differences between a bucket and productivity is negative (all unsuccessful).





In Photo, the buckets for each day represent the end of day age of everything. Towards the end, when backlog is cleared, all buckets show 0 since the inflow is less then the productivity I.E. cleared that day and never gets to actually age. The inflow column essentially represents 0 days old



Gray columns are locked



enter image description here



How can I correctly populate the buckets and produce no negatives?










share|improve this question






















  • I'm not clear on your process. On any given day are you "working" the 1000 oldest units?. Do those get removed from the table or do you tote those up somewhere? For the units that don't get worked, do they get carried down one row (to the next day's work) and one column to the right (to give them one more day of aging)?
    – xidgel
    Nov 20 at 23:31












  • That all seems correct. On the first day they will work 1000 starting from the right bucket and keep moving left until they have reached the max productivity. The next day the remainder of the work to be done less productivity (1000) should be distributed to the appropriate buckets. Each day it is not cleared it will move up in age by 1 day. Once they are worked they fall off. This is only tracking backlog (things not worked) @xidgel
    – urdearboy
    Nov 20 at 23:33












  • It seems logical to start working equations from 4+ bucket then 3, 2, 1 but just can’t get there
    – urdearboy
    Nov 20 at 23:37
















2














I am trying to simulate a backlog of cases by age.



Assumptions:





  1. FIFO: (First In First Out): This means we will always prioritize cases in the 4+ bucket then move to the bucket to it's left and so on. This means current day Inflow has last priority.


  2. Productivity = 1,000 = Constant: The most that can be worked in a single day is 1,000

  3. Before the simulation begins we have 1'000 aged 1 day, 750 aged 2 days, 500 aged 3 days, & 250 aged 4 days




Given this information, I believe it is possible to input any Inflow and have the buckets complete themselves. I cannot figure out what equations to use to populate the desired result though.



I have tried a series of SUM, SUMIF, IF, and checking if the differences between a bucket and productivity is negative (all unsuccessful).





In Photo, the buckets for each day represent the end of day age of everything. Towards the end, when backlog is cleared, all buckets show 0 since the inflow is less then the productivity I.E. cleared that day and never gets to actually age. The inflow column essentially represents 0 days old



Gray columns are locked



enter image description here



How can I correctly populate the buckets and produce no negatives?










share|improve this question






















  • I'm not clear on your process. On any given day are you "working" the 1000 oldest units?. Do those get removed from the table or do you tote those up somewhere? For the units that don't get worked, do they get carried down one row (to the next day's work) and one column to the right (to give them one more day of aging)?
    – xidgel
    Nov 20 at 23:31












  • That all seems correct. On the first day they will work 1000 starting from the right bucket and keep moving left until they have reached the max productivity. The next day the remainder of the work to be done less productivity (1000) should be distributed to the appropriate buckets. Each day it is not cleared it will move up in age by 1 day. Once they are worked they fall off. This is only tracking backlog (things not worked) @xidgel
    – urdearboy
    Nov 20 at 23:33












  • It seems logical to start working equations from 4+ bucket then 3, 2, 1 but just can’t get there
    – urdearboy
    Nov 20 at 23:37














2












2








2







I am trying to simulate a backlog of cases by age.



Assumptions:





  1. FIFO: (First In First Out): This means we will always prioritize cases in the 4+ bucket then move to the bucket to it's left and so on. This means current day Inflow has last priority.


  2. Productivity = 1,000 = Constant: The most that can be worked in a single day is 1,000

  3. Before the simulation begins we have 1'000 aged 1 day, 750 aged 2 days, 500 aged 3 days, & 250 aged 4 days




Given this information, I believe it is possible to input any Inflow and have the buckets complete themselves. I cannot figure out what equations to use to populate the desired result though.



I have tried a series of SUM, SUMIF, IF, and checking if the differences between a bucket and productivity is negative (all unsuccessful).





In Photo, the buckets for each day represent the end of day age of everything. Towards the end, when backlog is cleared, all buckets show 0 since the inflow is less then the productivity I.E. cleared that day and never gets to actually age. The inflow column essentially represents 0 days old



Gray columns are locked



enter image description here



How can I correctly populate the buckets and produce no negatives?










share|improve this question













I am trying to simulate a backlog of cases by age.



Assumptions:





  1. FIFO: (First In First Out): This means we will always prioritize cases in the 4+ bucket then move to the bucket to it's left and so on. This means current day Inflow has last priority.


  2. Productivity = 1,000 = Constant: The most that can be worked in a single day is 1,000

  3. Before the simulation begins we have 1'000 aged 1 day, 750 aged 2 days, 500 aged 3 days, & 250 aged 4 days




Given this information, I believe it is possible to input any Inflow and have the buckets complete themselves. I cannot figure out what equations to use to populate the desired result though.



I have tried a series of SUM, SUMIF, IF, and checking if the differences between a bucket and productivity is negative (all unsuccessful).





In Photo, the buckets for each day represent the end of day age of everything. Towards the end, when backlog is cleared, all buckets show 0 since the inflow is less then the productivity I.E. cleared that day and never gets to actually age. The inflow column essentially represents 0 days old



Gray columns are locked



enter image description here



How can I correctly populate the buckets and produce no negatives?







excel excel-formula






share|improve this question













share|improve this question











share|improve this question




share|improve this question










asked Nov 20 at 23:15









urdearboy

5,9832725




5,9832725












  • I'm not clear on your process. On any given day are you "working" the 1000 oldest units?. Do those get removed from the table or do you tote those up somewhere? For the units that don't get worked, do they get carried down one row (to the next day's work) and one column to the right (to give them one more day of aging)?
    – xidgel
    Nov 20 at 23:31












  • That all seems correct. On the first day they will work 1000 starting from the right bucket and keep moving left until they have reached the max productivity. The next day the remainder of the work to be done less productivity (1000) should be distributed to the appropriate buckets. Each day it is not cleared it will move up in age by 1 day. Once they are worked they fall off. This is only tracking backlog (things not worked) @xidgel
    – urdearboy
    Nov 20 at 23:33












  • It seems logical to start working equations from 4+ bucket then 3, 2, 1 but just can’t get there
    – urdearboy
    Nov 20 at 23:37


















  • I'm not clear on your process. On any given day are you "working" the 1000 oldest units?. Do those get removed from the table or do you tote those up somewhere? For the units that don't get worked, do they get carried down one row (to the next day's work) and one column to the right (to give them one more day of aging)?
    – xidgel
    Nov 20 at 23:31












  • That all seems correct. On the first day they will work 1000 starting from the right bucket and keep moving left until they have reached the max productivity. The next day the remainder of the work to be done less productivity (1000) should be distributed to the appropriate buckets. Each day it is not cleared it will move up in age by 1 day. Once they are worked they fall off. This is only tracking backlog (things not worked) @xidgel
    – urdearboy
    Nov 20 at 23:33












  • It seems logical to start working equations from 4+ bucket then 3, 2, 1 but just can’t get there
    – urdearboy
    Nov 20 at 23:37
















I'm not clear on your process. On any given day are you "working" the 1000 oldest units?. Do those get removed from the table or do you tote those up somewhere? For the units that don't get worked, do they get carried down one row (to the next day's work) and one column to the right (to give them one more day of aging)?
– xidgel
Nov 20 at 23:31






I'm not clear on your process. On any given day are you "working" the 1000 oldest units?. Do those get removed from the table or do you tote those up somewhere? For the units that don't get worked, do they get carried down one row (to the next day's work) and one column to the right (to give them one more day of aging)?
– xidgel
Nov 20 at 23:31














That all seems correct. On the first day they will work 1000 starting from the right bucket and keep moving left until they have reached the max productivity. The next day the remainder of the work to be done less productivity (1000) should be distributed to the appropriate buckets. Each day it is not cleared it will move up in age by 1 day. Once they are worked they fall off. This is only tracking backlog (things not worked) @xidgel
– urdearboy
Nov 20 at 23:33






That all seems correct. On the first day they will work 1000 starting from the right bucket and keep moving left until they have reached the max productivity. The next day the remainder of the work to be done less productivity (1000) should be distributed to the appropriate buckets. Each day it is not cleared it will move up in age by 1 day. Once they are worked they fall off. This is only tracking backlog (things not worked) @xidgel
– urdearboy
Nov 20 at 23:33














It seems logical to start working equations from 4+ bucket then 3, 2, 1 but just can’t get there
– urdearboy
Nov 20 at 23:37




It seems logical to start working equations from 4+ bucket then 3, 2, 1 but just can’t get there
– urdearboy
Nov 20 at 23:37












2 Answers
2






active

oldest

votes


















2














I think this is what you're after.



In H4: =MAX(SUM(G3:H3-$A$3),0)
In G4: =MAX(SUM(F3:H3)-$A$3-SUM(H4),0)
In F4: =MAX(SUM(E3:H3)-$A$3-SUM(G4:H4),0)
In E4: =MAX(SUM(E3:H3)+D4-$A$3-SUM(F4:H4),0)


Then fill down.
In those formulas $A$3 is the productivity. Hope that helps.






share|improve this answer





















  • very elegant answer
    – Bharat Anand
    Nov 21 at 1:12



















1














Step1: Guiding principle for each bucket:



If a bucket has more than the remaining productivity for that bucket for that day, then (BUCKETVALUE - REMAININGUNITS), else 0.



We will deploy this IF logic in steps 3 - 6.



Step 3 defines what is REMAININGUNITS for each bucket, which is again used in subsequent steps 3 - 6



Step2:
Now let's define REMAININGUNITS as the amount of units that can be cleared for the day for a bucket



So, REMAININGUNITS for Bucket 4 is PRODUCTIVITY, i.e. potentially all of A3 can be used to clear Bucket 4                               
=$A$3
Remaining units for bucket 3 = PRODUCTIVITY - Actual Reduction in Bucket 4
=($A$3-(H3-H4))
Remaining units for bucket 2 = PRODUCTIVITY - Actual Reduction in Bucket 4 - Actual Reduction in Bucket 3
=($A$3-(H3-H4)-(G3-G4))
Remaining units for bucket 1 = PRODUCTIVITY - Actual Reduction in Bucket 4 - Actual Reduction in Bucket 3 - Actual Reduction in Bucket 2
=($A$3-(H3-H4)-(G3-G4)-(F3-F4))


Step3:
Clear as much of 4th bucket as possible,



H4=IF(H3>$A$3,H3-$A$3,0)



Step4:
Clear as much of 3rd bucket as possible,



G4=IF(G3>($A$3-(H3-H4)),G3-($A$3-(H3-H4)),0)



Step5:
Clear as much of 2nd bucket as possible,



F4=IF(F3>($A$3-(H3-H4)-(G3-G4)),F3-($A$3-(H3-H4)-(G3-G4)),0)



Step6:
Clear as much of (1st bucket + INFLOW) as possible,



E4=IF(E3+D3>($A$3-(H3-H4)-(G3-G4)-(F3-F4)),E3+D3-($A$3-(H3-H4)-(G3-G4)-(F3-F4)),0)



Step7:
Fill the formulaes from Row 4 to the remanining rows






share|improve this answer























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    2 Answers
    2






    active

    oldest

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    2 Answers
    2






    active

    oldest

    votes









    active

    oldest

    votes






    active

    oldest

    votes









    2














    I think this is what you're after.



    In H4: =MAX(SUM(G3:H3-$A$3),0)
    In G4: =MAX(SUM(F3:H3)-$A$3-SUM(H4),0)
    In F4: =MAX(SUM(E3:H3)-$A$3-SUM(G4:H4),0)
    In E4: =MAX(SUM(E3:H3)+D4-$A$3-SUM(F4:H4),0)


    Then fill down.
    In those formulas $A$3 is the productivity. Hope that helps.






    share|improve this answer





















    • very elegant answer
      – Bharat Anand
      Nov 21 at 1:12
















    2














    I think this is what you're after.



    In H4: =MAX(SUM(G3:H3-$A$3),0)
    In G4: =MAX(SUM(F3:H3)-$A$3-SUM(H4),0)
    In F4: =MAX(SUM(E3:H3)-$A$3-SUM(G4:H4),0)
    In E4: =MAX(SUM(E3:H3)+D4-$A$3-SUM(F4:H4),0)


    Then fill down.
    In those formulas $A$3 is the productivity. Hope that helps.






    share|improve this answer





















    • very elegant answer
      – Bharat Anand
      Nov 21 at 1:12














    2












    2








    2






    I think this is what you're after.



    In H4: =MAX(SUM(G3:H3-$A$3),0)
    In G4: =MAX(SUM(F3:H3)-$A$3-SUM(H4),0)
    In F4: =MAX(SUM(E3:H3)-$A$3-SUM(G4:H4),0)
    In E4: =MAX(SUM(E3:H3)+D4-$A$3-SUM(F4:H4),0)


    Then fill down.
    In those formulas $A$3 is the productivity. Hope that helps.






    share|improve this answer












    I think this is what you're after.



    In H4: =MAX(SUM(G3:H3-$A$3),0)
    In G4: =MAX(SUM(F3:H3)-$A$3-SUM(H4),0)
    In F4: =MAX(SUM(E3:H3)-$A$3-SUM(G4:H4),0)
    In E4: =MAX(SUM(E3:H3)+D4-$A$3-SUM(F4:H4),0)


    Then fill down.
    In those formulas $A$3 is the productivity. Hope that helps.







    share|improve this answer












    share|improve this answer



    share|improve this answer










    answered Nov 21 at 0:05









    xidgel

    2,1272717




    2,1272717












    • very elegant answer
      – Bharat Anand
      Nov 21 at 1:12


















    • very elegant answer
      – Bharat Anand
      Nov 21 at 1:12
















    very elegant answer
    – Bharat Anand
    Nov 21 at 1:12




    very elegant answer
    – Bharat Anand
    Nov 21 at 1:12













    1














    Step1: Guiding principle for each bucket:



    If a bucket has more than the remaining productivity for that bucket for that day, then (BUCKETVALUE - REMAININGUNITS), else 0.



    We will deploy this IF logic in steps 3 - 6.



    Step 3 defines what is REMAININGUNITS for each bucket, which is again used in subsequent steps 3 - 6



    Step2:
    Now let's define REMAININGUNITS as the amount of units that can be cleared for the day for a bucket



    So, REMAININGUNITS for Bucket 4 is PRODUCTIVITY, i.e. potentially all of A3 can be used to clear Bucket 4                               
    =$A$3
    Remaining units for bucket 3 = PRODUCTIVITY - Actual Reduction in Bucket 4
    =($A$3-(H3-H4))
    Remaining units for bucket 2 = PRODUCTIVITY - Actual Reduction in Bucket 4 - Actual Reduction in Bucket 3
    =($A$3-(H3-H4)-(G3-G4))
    Remaining units for bucket 1 = PRODUCTIVITY - Actual Reduction in Bucket 4 - Actual Reduction in Bucket 3 - Actual Reduction in Bucket 2
    =($A$3-(H3-H4)-(G3-G4)-(F3-F4))


    Step3:
    Clear as much of 4th bucket as possible,



    H4=IF(H3>$A$3,H3-$A$3,0)



    Step4:
    Clear as much of 3rd bucket as possible,



    G4=IF(G3>($A$3-(H3-H4)),G3-($A$3-(H3-H4)),0)



    Step5:
    Clear as much of 2nd bucket as possible,



    F4=IF(F3>($A$3-(H3-H4)-(G3-G4)),F3-($A$3-(H3-H4)-(G3-G4)),0)



    Step6:
    Clear as much of (1st bucket + INFLOW) as possible,



    E4=IF(E3+D3>($A$3-(H3-H4)-(G3-G4)-(F3-F4)),E3+D3-($A$3-(H3-H4)-(G3-G4)-(F3-F4)),0)



    Step7:
    Fill the formulaes from Row 4 to the remanining rows






    share|improve this answer




























      1














      Step1: Guiding principle for each bucket:



      If a bucket has more than the remaining productivity for that bucket for that day, then (BUCKETVALUE - REMAININGUNITS), else 0.



      We will deploy this IF logic in steps 3 - 6.



      Step 3 defines what is REMAININGUNITS for each bucket, which is again used in subsequent steps 3 - 6



      Step2:
      Now let's define REMAININGUNITS as the amount of units that can be cleared for the day for a bucket



      So, REMAININGUNITS for Bucket 4 is PRODUCTIVITY, i.e. potentially all of A3 can be used to clear Bucket 4                               
      =$A$3
      Remaining units for bucket 3 = PRODUCTIVITY - Actual Reduction in Bucket 4
      =($A$3-(H3-H4))
      Remaining units for bucket 2 = PRODUCTIVITY - Actual Reduction in Bucket 4 - Actual Reduction in Bucket 3
      =($A$3-(H3-H4)-(G3-G4))
      Remaining units for bucket 1 = PRODUCTIVITY - Actual Reduction in Bucket 4 - Actual Reduction in Bucket 3 - Actual Reduction in Bucket 2
      =($A$3-(H3-H4)-(G3-G4)-(F3-F4))


      Step3:
      Clear as much of 4th bucket as possible,



      H4=IF(H3>$A$3,H3-$A$3,0)



      Step4:
      Clear as much of 3rd bucket as possible,



      G4=IF(G3>($A$3-(H3-H4)),G3-($A$3-(H3-H4)),0)



      Step5:
      Clear as much of 2nd bucket as possible,



      F4=IF(F3>($A$3-(H3-H4)-(G3-G4)),F3-($A$3-(H3-H4)-(G3-G4)),0)



      Step6:
      Clear as much of (1st bucket + INFLOW) as possible,



      E4=IF(E3+D3>($A$3-(H3-H4)-(G3-G4)-(F3-F4)),E3+D3-($A$3-(H3-H4)-(G3-G4)-(F3-F4)),0)



      Step7:
      Fill the formulaes from Row 4 to the remanining rows






      share|improve this answer


























        1












        1








        1






        Step1: Guiding principle for each bucket:



        If a bucket has more than the remaining productivity for that bucket for that day, then (BUCKETVALUE - REMAININGUNITS), else 0.



        We will deploy this IF logic in steps 3 - 6.



        Step 3 defines what is REMAININGUNITS for each bucket, which is again used in subsequent steps 3 - 6



        Step2:
        Now let's define REMAININGUNITS as the amount of units that can be cleared for the day for a bucket



        So, REMAININGUNITS for Bucket 4 is PRODUCTIVITY, i.e. potentially all of A3 can be used to clear Bucket 4                               
        =$A$3
        Remaining units for bucket 3 = PRODUCTIVITY - Actual Reduction in Bucket 4
        =($A$3-(H3-H4))
        Remaining units for bucket 2 = PRODUCTIVITY - Actual Reduction in Bucket 4 - Actual Reduction in Bucket 3
        =($A$3-(H3-H4)-(G3-G4))
        Remaining units for bucket 1 = PRODUCTIVITY - Actual Reduction in Bucket 4 - Actual Reduction in Bucket 3 - Actual Reduction in Bucket 2
        =($A$3-(H3-H4)-(G3-G4)-(F3-F4))


        Step3:
        Clear as much of 4th bucket as possible,



        H4=IF(H3>$A$3,H3-$A$3,0)



        Step4:
        Clear as much of 3rd bucket as possible,



        G4=IF(G3>($A$3-(H3-H4)),G3-($A$3-(H3-H4)),0)



        Step5:
        Clear as much of 2nd bucket as possible,



        F4=IF(F3>($A$3-(H3-H4)-(G3-G4)),F3-($A$3-(H3-H4)-(G3-G4)),0)



        Step6:
        Clear as much of (1st bucket + INFLOW) as possible,



        E4=IF(E3+D3>($A$3-(H3-H4)-(G3-G4)-(F3-F4)),E3+D3-($A$3-(H3-H4)-(G3-G4)-(F3-F4)),0)



        Step7:
        Fill the formulaes from Row 4 to the remanining rows






        share|improve this answer














        Step1: Guiding principle for each bucket:



        If a bucket has more than the remaining productivity for that bucket for that day, then (BUCKETVALUE - REMAININGUNITS), else 0.



        We will deploy this IF logic in steps 3 - 6.



        Step 3 defines what is REMAININGUNITS for each bucket, which is again used in subsequent steps 3 - 6



        Step2:
        Now let's define REMAININGUNITS as the amount of units that can be cleared for the day for a bucket



        So, REMAININGUNITS for Bucket 4 is PRODUCTIVITY, i.e. potentially all of A3 can be used to clear Bucket 4                               
        =$A$3
        Remaining units for bucket 3 = PRODUCTIVITY - Actual Reduction in Bucket 4
        =($A$3-(H3-H4))
        Remaining units for bucket 2 = PRODUCTIVITY - Actual Reduction in Bucket 4 - Actual Reduction in Bucket 3
        =($A$3-(H3-H4)-(G3-G4))
        Remaining units for bucket 1 = PRODUCTIVITY - Actual Reduction in Bucket 4 - Actual Reduction in Bucket 3 - Actual Reduction in Bucket 2
        =($A$3-(H3-H4)-(G3-G4)-(F3-F4))


        Step3:
        Clear as much of 4th bucket as possible,



        H4=IF(H3>$A$3,H3-$A$3,0)



        Step4:
        Clear as much of 3rd bucket as possible,



        G4=IF(G3>($A$3-(H3-H4)),G3-($A$3-(H3-H4)),0)



        Step5:
        Clear as much of 2nd bucket as possible,



        F4=IF(F3>($A$3-(H3-H4)-(G3-G4)),F3-($A$3-(H3-H4)-(G3-G4)),0)



        Step6:
        Clear as much of (1st bucket + INFLOW) as possible,



        E4=IF(E3+D3>($A$3-(H3-H4)-(G3-G4)-(F3-F4)),E3+D3-($A$3-(H3-H4)-(G3-G4)-(F3-F4)),0)



        Step7:
        Fill the formulaes from Row 4 to the remanining rows







        share|improve this answer














        share|improve this answer



        share|improve this answer








        edited Nov 21 at 0:50

























        answered Nov 21 at 0:43









        Bharat Anand

        325410




        325410






























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