If you still have a modelling question (e.g., how do I express this requirement using the rules for a linear program), please ask it on. (Thanks to Stu Mitchell of PuLP fame for pointing out this resource.) The NEOS guide is also a great resource. The AIMMS folks have a great online chapter describing how to take non-linear requirements ( local copy of pdf) and model them using linear programming. (OpenSolver uses CBC by default to solve these linear programmes.) Check out this Wiki page on linear programming. However, most of the time you want to build these “linear programming” models (and avoid non-linear models) because these are easier and more reliable to solve using packages such as OpenSolver. Writing optimisation models that only use linear mathematical equations and inequalities is not easy. However, if you still have a question, please post it below. When we run this code, you would expect a type mismatch error, but see what happens when we run this code.We hope that most of your questions will be answered on the Using OpenSolver page.
In the above code, we have assigned a value of 4556, which is not as per the data type values of TRUE or FALSE. read more that can hold the value of either TRUE or FALSE. The declaration of this data type is similar to all the other data types. The value this data type holds is either TRUE or FALSE and is used for logical comparison.
We have declared the variable “x” as Boolean.īoolean is a data type Boolean Is A Data Type Boolean is an inbuilt data type in VBA used for logical references or logical variables. Look at the below code.Ĭode: Sub Type_MisMatch_Example2() Dim x As Boolean x = 4556 MsgBox x End Sub Now, look at one more example with a different data type. The data type cannot hold a text’s value, so the Type Mismatch Error comes. But in the next line, we have assigned the value for the variable “k” as “Hiii.” The variable “k” can hold values from 0 to 255. We have declared the variable “k” as Byte. Sub Type_MisMatch_Example1() Dim k As Byte k = "Hiii" MsgBox k End Sub
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