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the best chosen various lengths for the least remainder

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    the best chosen various lengths for the least remainder

    Hi,
    there are a range of various length and numbers of pipes are required to cut out of 6 meter raw pipes.(rangeT4=6000 mm)
    I am trying to write practical vba for covering the number of cutting length with the min. waste parts.
    is there any guidance how to reach to that point?


    the method which I ve used with no desired result is that,
    first, I copy the range of length form 1st sheet and paste them into 2nd sht.
    then, I sorted them (A to Z,Ascending).
    next, I found the number of each length which could come out of one raw lenght(T4=6000mm) =.Range("C"&i).value
    after that, I,ve found the remainder of 6000/ each different length (.Range("D"&i).value=.Range("B" & i) MOD .Range("T4"))
    in If condition I told that ıf the number of parts did not reach to desired numbers and the remainder is the least one in ıts range continue the division task.
    else ....


    many thanks for any further help on this matter.

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    Re: the best chosen various lengths for the least remainder

    I would think that trying to apply solver would be a better approach than VBA. Lots of examples you can google

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    Re: the best chosen various lengths for the least remainder

    from solver you mean some online software which do this computing?

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    Re: the best chosen various lengths for the least remainder

    Solver is an Add-In to Excel, not installed by default when you install excel but can be added.

    Solver can be used for problems like cutting lengths in order to minimize wastage. Image below shows how to lay a floor with minimum wastage. You have a caton with a mix of planks with different length and these should be combined to certain length +- 3".

    So the first line shows that using 3 planks with a length of 42 you get a total length of 126 were you need a length 125 i.e. Here there would be a wastage of 1", for the second line solver uses 3 planks of length 18 and 3 planks of length 24, again this gives a total length of 126 and a wastage of 1.

    The total wastage for laying this floor is 55" and left in the carton you will have 2 planks of 36" and 3 planks of 42"

    solver_floor.jpg

    Alf

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    Re: the best chosen various lengths for the least remainder

    In cutting a stock list to different length you need to set up a relation between stock list length and the different cut you can make from it. This can be a really tedious process and you will also need to decide how excess production should be handled. Are this waste or can they be used on a latter date?

    solver_cut_list.jpg

    Alf

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    Re: the best chosen various lengths for the least remainder

    Dear Alf,
    really thank you for this straightforward way. I tried to watch a couple of tutorials. As you've found out I am completely new in solver.
    I determined:
    "set objective" as the cell which include my bar length:6000
    "changing variable cells" the range of cutting lengths
    "constrains": the cells of each length quantity = the number of that length
    "solving method":GRG Nonlinear

    But when I click solve button, it show up a message for objective cell should contain a formula! but ıt is only a length=6000

    Many thanks for ongoıng supports

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    Re: the best chosen various lengths for the least remainder

    I think the best way to help you is to upload a model of a solver cutting list file. Clear range K6 :K22 and then run solver.

    The object cell is L27 and this is set to "Min" i.e. solver shall minimize the total wastage when cutting the lists. You will also see that solver produces more lists than
    than the specified demand. They are not added to the total wastage as the could be used on a latter date.

    I also have another solver model set up that will produce just the required numbers of cut lists and add all leftovers as wastage. I've added a macro to that to run solver on a
    loop. The problem with this is solver will not find the optimal solution for all the "components" in one go find the best solution for some components, then it loops and find the
    best for some other components, and then it loops until all components are accounted for. This technique called "sub optimization" is not considered as accurate as a "normal" solver solution
    but it's quick and easy to set up.

    Could make a model like that for you, what I need to know is ho many different cut you would like to have. As you see from uploaded image this model is set for max 5
    different cuts. Model will work fine with less cuts, just put a zero value for the cut you don't need.

    You specify the cut length range C2:H2, then add in the green ranges C4;H4 the number of cuts you need and finally the length of stock in F12 and the you run the macro.

    ab_cut.jpg

    Alf
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    Re: the best chosen various lengths for the least remainder

    Dear Alf,
    Appreciate all your useful explanation AND GUİDANCE. as I mentioned before I might have a wide range of lengths , mostly more than 5 cutting lengths. The number of cutting lengths changes based on various projects.
    I try to find the logic of possibilities which is applied in your enclosed file, (range J6:J22) and also through range(D6:I22).
    Are they defined for specific project ? could be they extended for the various projects with an extensive diversity of lengths?

    I tried to cover different possibilities which crossed my mind by combining VBA codes and excel formula but unfortunately it does not cover different projects to reach the best optimized result.

    sorry for my clumsy questions which I burden you with

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    Re: the best chosen various lengths for the least remainder

    So I've set up a solver model that you have a maximum of 11 different cuts. This can of course be extended and that should not be to
    difficult using the macro I wrote as a model.

    First of all this kind of problem is a linear one and the best solver engine for linear problems is the "Simplex LP" engine.

    To solve this kind of problem I wrote a macro that runs solver in a loop. In order to be able to use this macro there must be set a reference to
    solver in Visual Basic.

    Select the "Developers" tab, click on the "Visual Basic" icon, -> "Tools" -> "References" and tick box marked "Solver". Then click
    "OK" and close "Visual Basic" window.

    Setting up the model you start by setting the different list length in range C2:M2. Here you select whatever length you like to produce. Of course the maximum length of a list must be less than the length of stock. Then you specify the number of list lengths you
    need in range C4:M4 and finally the length of stock in cell F12.

    Then you run macro "SolverLoop". You don't have to clear the range B18:O34 as macro will do this. The macro is a dynamic one and it will loop until you have produced all the pieces you specify in range C4:M4 and building a table with the result starting at row 18 where you see what pieces are produced and the wastage for that particular production.

    At the end of the table you get the total production and the total wastage.

    I mentioned that running solver in a loop may produce a sub optimal solution. As I was interested to see if this was so I also set up a solver model on the sheet "Solver_model_test". This solves the problem in one go as opposed to the macro but sadly to tell the excel solver was not really up to this as it found no solution testing for 5 minutes. Instead I tested the "OpenSolver" and got a solution in
    less than 10 seconds.

    As the object function was minimum wastage both solution are equal as they have the same total wastage. I'm inclined to say that the solver loop solution is the better one as you get bigger pieces of wastage that can be more useful than the smaller peaces of wastage produced by "OpenSolver".

    But that's my opinion, don't know what you think about this. But perhaps this indicates that the effect of sub optimization is not as bad as I thought.

    Alternative you could download the "OpenSolver" a freebie developed by the University of Auckland NZ as this is a much more powerful solver for linear problem than the solver found in excel and "OpenSolver" integrates nicely with excel.

    Alf
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    Last edited by Alf; 04-09-2018 at 04:32 PM.

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    Re: the best chosen various lengths for the least remainder

    Dear Alf,
    Im really so thankful of your solver. It works great specially for max 11 cutting lengths, of course with slight differences on demand numbers and solver result.
    there are also projects for 15 and max 20. It seems that even I revised the vb ranges, but some of solver numbers came up with high negative differences.
    I guess that it refers to the codings which are locked now, so because of no access to them I have to ask you again.
    again I appreciate all your supports.

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    Re: the best chosen various lengths for the least remainder

    OK there seems to be an inbuilt restriction in both Excel solver and OpenSolver. Testing models with solver working over to many columns is a "no go".

    Somewhere after 15 - 16 columns solver started to ignore cells to change.

    Had to rebuild the model using rows for the setup instead of columns.

    I guess that it refers to the codings which are locked now
    The coding is not locked, you will find it if you click "Visual Basic" icon -> "Module1" and "View Code"

    So hopefully this will solve your problems.

    Macro has the same name as before and you may need to set a reference to solver in "Visual Basic"

    Alf
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    Re: the best chosen various lengths for the least remainder

    how much helpful your are Alf. Really thanks for your great supports and help.

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    Re: the best chosen various lengths for the least remainder

    You are welcome and thanks for feedback

    Unfortunately things are not as simple as they first seems. As you problem was quite interesting I played around with it a bit more than I normally do and did get a better result using OpenSolver in one go. The new solution has only 5010 in waste as compared to excel solvers solution with a wastage of 11010.

    So in order to get the best (less wastage) solution if this is something you do regularly perhaps you should use OpenSolver instead? Have a look at the link
    https://opensolver.org/

    As long as you run small models / problems there should be no difference but the bigger the model the more likely it is that there may be a difference.

    Have a think about it and if you do think it's worthwhile to "upgrade" to OpenSolver do tell me and I will of course help you with model building.

    I'm also uploading an image of the OpenSolver result.

    two_solvers.jpg

    If you do have problem seeing the numbers try right click image and select "open in new tab"

    Alf
    Last edited by Alf; 04-13-2018 at 12:53 PM.

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    Re: the best chosen various lengths for the least remainder

    Dear Alf, really appreciate your kind suggestion for further help.
    I check two other projects which were done manually, and I try to solve them by this excel solver. But the differences are 10 bars for one of project and 14 bars (of 6000 mm) for another one, the diversities of demanding cutting lengths are 8 and 7 respectively.
    maybe your suggestion for outsource solver would be more practical for this kind of projects.
    many thank for your support.

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    Re: the best chosen various lengths for the least remainder

    I've set up a model that runs using "OpenSolver" as the excel solver is not powerful enough to solve problems of this size.

    If you would like to try you need to download "OpenSolver" from the link I mentioned in post #13 and install it. And since model is build using the excel solver but solution
    is found bu using "OpenSolver" a reference must be set both to Solver and "OpenSolver" in Visual Basic.

    You start by setting / updating range B3;B22 (if you wish to change length of cut, then range C3:C22 for the number of lists you wish to make and finally the stock length
    cell C28.

    Then a new cell (C30) calculates the theoretical number of stock rods you need in order to solve this problem. So you update cell P27 with a whole number bigger than the
    theoretical but close in value.

    Then you run the macro "OpenSolvMacro" and macro starts by clearing old result and then tries to solve the problem using the only the number of stock rods that was spcified
    in cell P27.

    If "OpenSolver" did not find a solution increase P27 with 1 unit do a rerun and see if "OpenSolver" finds a solution using a higher value of P27.

    If a solution is not found after 20 seconds a pop up window will tell you so (see image) as I've set a max time for 20 seconds for "OpenSolver" to solve problems.

    a_nosol.jpg

    and after clicking "OK" macro will give the cell address where cell value differ from target value.

    a_fel.jpg

    Alf
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    Last edited by Alf; 04-14-2018 at 03:17 PM.

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    Re: the best chosen various lengths for the least remainder

    Great topic handa & awesome help Alf, I've been working on something similar and have greatly enjoyed reading this topic.

    I'm looking into OpenSolver now thanks to Alf's suggestion.

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    Re: the best chosen various lengths for the least remainder

    Hi Dal

    Thanks for your kind word and rep I do think you will like the "OpenSolver" and as there is no artificial constraints imposed on it as it is on the excel solver. The limitation on the other hand is that it is only build to handle linear problems.

    Still it seems that "OpenSolver" can be used as an interface to a number of non-linear solvers something I have not tested but would be interested to do.

    I've been working on something similar
    Sounds interesting, if it's not work related / private I would like to have a look at it. Is it possible for you to post your work in the forum?

    Well whatever you are working with the best of luck to you in solving your problem.

    Regards

    Alf

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    Re: the best chosen various lengths for the least remainder

    Dear Alf,
    I enclosed one of my problem which experienced a large contradiction in numbers in both of Excel.solver and Open.solver!
    The manual calculation gave 364 Rods but by excel solver it became 10 rods more, and by open solver as pop up msg tells
    it cannot solve the problem and after that show there is distinction in demanding numbers for each cutting length.
    I guess that in excel.solver the way it works is that starting with the less remainders not combining the best chosen lengths.
    Because of the the remainders is following ascending trend and in the last rods the remainders are greater than some of cutting less.
    I know it is normal but by comparing to manual computing they are a little bit more.
    Attached Files Attached Files

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    Re: the best chosen various lengths for the least remainder

    one more thing I wanna add to my previous post(#18) is that I'm changing SolverOptions MaxTime and "I30". The matter is that if the problem was not solved manually before how can the value of I30 be estimated ? because even we consider this value as a round up of "C30" it does not give us reasonable result.
    by playing with the value of I30 all the solver values was equal to demand values except one with 33 less in numbers!
    and in your post#16 , you mentioned cell p27 but it seems not in vb code not in solver no action is done in it.
    Attached Files Attached Files
    Last edited by handa; 04-16-2018 at 06:28 AM. Reason: changing the attachement

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    Re: the best chosen various lengths for the least remainder

    Having spent some time with you last uploaded file yesterday I think I fond the right setting for you model.

    What I did was to change the constraint "D3:D22 <= C3:C22" to "D3:D22 = "C3:C22" and this made the "OpenSolver" run better and faster.

    I also changed MaxTime back to 30 seconds and this could probably be reduced further. Either program finds a solution in less than 15 - 20 seconds or there is no solution
    to be found but solver will keep on working until it reaches MaxTime or max number of iterations.

    The bad, you still need to find the optimal number of stock rods and this is a matter of trial end error. You set a value (number of rods) and if a solution is found you reduce the number. I had some idea of running this in a macro but have not succeeded.

    A last point you specify the list length down to a millimeter, how about cut size? If a solution gives 0 in wastage and you cut 3 - 4 lists from that rod how much do you
    lose in cutting?

    Checking the result of this file the maximum number of lists produced from a rod was 7 and the smallest waste was 8 cell both values found in cell JB24 so with a cut thickness
    of say 0.5 you should be fine in this case.

    Yes that reference to cell P27 was something from an earlier model so you can forget all about that.

    The manual calculation gave 364 Rods
    Well it seems "OpenSolver" did better as it managed to find a solution using 269 rods running for 17 seconds. I do think that the length of the running time is also influenced by how powerful your PC are.

    Alf
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    Last edited by Alf; 04-17-2018 at 01:06 AM.

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    Re: the best chosen various lengths for the least remainder

    This looks like a never ending story. Sorting values from lowest to highest did make it possible to solve the problem using 268 stock rods in 6 - 7 seconds so when you run your other problems it seems that one should test different approaches to find the right one in order to get the best solution.

    a_best.jpg

    Alf

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    Re: the best chosen various lengths for the least remainder

    Hi again , I became busy with another emergency project.
    As I mentiond in post#14 I had remarkable differences in amount of stock bars in a couple of projects.
    By checking them in the last file #15, the problem is that it counts numbers of stock bars (6000) without any cutting length! for ınstance , 98 stock bars consumed but 5 of them with 6000 remainder means no cutting length!

    another matter is that when checking the waste E24:XX24 most of them show "0" but actually they have remainder and also some of them are more than the length of stock bar! ı tried to modify by changng the format cell and checking the vb codes but not practical yet.

    ıt seems that opensolver need more setting for solving these kinds of problems.

    thank you again for all ur consideration.

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    Re: the best chosen various lengths for the least remainder

    What I wonder is did you test the uploaded file in #post 20? What you can do before you run it is have your list values in range B3:B10 going from the smallest value to the highest as shown in
    image in post #21.

    I've also tested the uploaded file in #post 20 and sorted the range and I get no 0 in the sum column (row 24).

    Since I can't see the problem you describe can you upload the result file if there is the same problem when you do a run with the file from #post 20.

    There are another thing that could be done but first I must know we do get the same results before it's any idea to proceed.

    Alf

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    Re: the best chosen various lengths for the least remainder

    from the attached file it could be seen that if the format of cells(above yellow line) changed into 2digits , the numbers of cutting lengths are not round.
    It means that if they would be round down , we have the wastes and also the demanding numbers and produced ones are not equal.
    handa
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    Re: the best chosen various lengths for the least remainder

    The first problem with the file was that the result range was set to the number format with 2 decimal places. Have changed that to setting to "general".

    Then the rod setting was not high enough to make it possible for solver to find a solution. I started at 102 and took it down one rod at the time to the minimum number where
    "OpenSolver" could find a solution, at a 100 stock rods.

    If "OpenSolver" can't find a solution it will tell you so and then present a decimal solution as a last resort in order to find a solution.
    Try testing a run where the number of rods are set to 99 and you will see what I mean.

    I've also made some changes to the macro. It now counts the number of list lengths specified and build the "OpenSolver" model to cover just that area as well as I've
    added a specified list of the wastage the length and the number of these so in range F34:G34 the smallest wastage produced is 5 mm and there are 3 pieces of this size.

    Take you time and familiar yourself with this model then later on we may discus changes as I'm not fully satisfied with the "OpenSolver" or perhaps my model. A proper model
    should take in account wastage whose size is among the specified list length.

    For instance you have specified 56 lists of a length of 675 but looking at the wastage list you have 3 pieces of a length of 680 and the model don't see this.
    Looking at Rod 10 there could also be made a list of 675 in addition to the production of the seven list of 760.


    Alf
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    Last edited by Alf; 04-21-2018 at 11:40 AM.

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    Re: the best chosen various lengths for the least remainder

    As it can be seen from the sheet1 of attachement, for some projects the numbers of required stock bars are close to the one solved by hand, except for two projects numbered 1 & 2 . For these two u can find remarkable distinction between (excel&open) solvers and manual solving.
    For instance, it is clearly possible to distract another length from some remainders in project 1. the first remainder 1634 could cover length 1460! or the fourth remainder,2161-2120 could give less waste. This matter may not be that much noticable for the other projects but for projects 1 n 2 the results are highly significant.

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    Re: the best chosen various lengths for the least remainder

    Had a look at your result and as your manual results are much better than both excel solver and "OpenSolver" there can be no doubt that you should stick to your manual method.

    So the conclusion is that you do have the best method for calculating list cutting with a minimum of loss. Care to post your setup?

    I did a bit of fiddling with your last uploaded file. I started by setting list of length 856 and 1460 to 0 and found a solution for this setup using 367 rods.

    The wastage list after that run looked like this:

    awast_1.jpg

    I then wrote a macro like this in order to add the the 16 lists with a length of 1460.

    Please Login or Register  to view this content.
    This added the required 16 lists in cell D4, so the wastage looked like this after an update.

    awast_2.jpg

    For the final macro making 160 lists of a length 856 I wrote this macro.

    Please Login or Register  to view this content.
    Problem with this macro it picks the waste where it finds it as it works its way along the row, could of course have made to macros. One for 154 pieces in the range 990 to 1634 and finally one for 6 pieces of the 2147 waste.

    But still with 367 rods "OpenSolver" only managed to come second best as 364 is the winning number!!!

    awast_3.jpg

    So this should be the end of this thread unless you care to post you manual setup (and much better method superior to solver)

    Alf
    Last edited by Alf; 04-25-2018 at 09:58 AM.

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    Re: the best chosen various lengths for the least remainder

    As I do dislike not being able to solve a problem properly and with input from the OP that he had found a better solution than I had managed to find with the models I build
    I decided to have another go at this problem.

    This time I build a model based on my comments in #post 5 and the uploaded file in #post 7.

    To cover all possibilities in the cutting pattern I needed 51 different cutting pattern. This model runs as a stand alone no macro required and both OpenSolver as well as
    the excel solver works with this setup producing identical result using the same number of rods to produce the required number of lists with the same total wastage.

    The only difference I found was that two solvers do not choose the same cutting pattern showing that there are a number of possible solutions using the same number of rods
    for the lists production with the same amount of total wastage.

    The problem with a setup like this is when list length is changed one could be forced to rebuild the cutting pattern. In this model with 7 different list length I needed
    51 lines with cut pattern. The uploaded model in #post 7 I had 6 different list length and to cover all possible cuts I needed 17 cut lines.

    So the number of cut lines may wary very much dependent of list length, stock rod length and number of different lists length.

    When the popup message appears about "time limit" for OpenSolver click "OK", for excel solver click "Stop" and "OK".

    To get rid of theses messages I could do so by deleting 27 of the unused cut rows for OpenSolver or 37 when running the excel solver but hardly worth the effort. Better then
    to do this after a run to make the end result more easily readable if the number of cuts extends over a page.

    Alf
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    Last edited by Alf; 04-28-2018 at 01:12 PM.

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