How to Download and Install the Trial version of What’sBest! 11.1 in Excel 2010

What’sBest! is an excellent add-in to Excel that allows us to solve linear, non linear, integer and probabilistic (stochastic) optimization models through an easy and intuitive interface. This program is highly recommended both for students as for professors in the field of Operations Research and is available in a free trial version. The following tutorial shows, step by step, how to download and install the trial version of What’sBest! 11.1 if you are user of Excel 2010 (if you have another operating system or version of Excel this tutorial will also help you for sure).

How to Download and Install the Trial version of What’sBest!

Step 1: Verify the operating system you use and the number of bits associated. What’sBest! It is compatible with Windows 2000, XP, Vista, Windows 7 and Windows 8. In this case we will show how to activate the add-in on a computer that uses Windows 7 Home Premium with a 64-bit operating system. To verify this setting go to My PC (or Computer) and then Properties of the System.

Properties of the System

In the system information, you can find the number of bits associated with your operating system as shown in the following image:

operating system bits

Step 2: Log in the download section of What’sBest! in the website of Lindo, a company located in Chicago, United States, with more than 22 years of experience in the development of software and applications for optimization and decision support. After getting to the download link, you must select the version of the program that is compatible with your operating system and your version of Excel.


Step 3: Fill out the form to get the file with the program. The fields with an asterisk (*) are mandatory.

form whatsbest

After completing the above correctly and pressing Submit, you will get a message indicating that a download link for the version of What’sBest! has been sent to the email address that you have provided.


Step 4: Log in your email address (the one provided when filling in the form). You should have received an email from LINDO Systems Inc with the link to download the program as shown below. (Confidential information has been hidden with black stripes and download link with red). Select the download link and will begin to download the program that comes on a file compressed in ZIP format to your computer.


Step 5: Once the download is complete (by default the file will be saved in the Downloads section of your computer) open the ZIP file and then run the setup.exe file inside as shown in the following image:


This will make the installation application that will guide you in the process of software activation.


licence agreement whatsbest

Step 6: The installation is complete. What’sBest! will be available at the right of the Add-Ins menu. The program is ready to be used and solve your optimization models.


Now that What’sBest! is installed on your computer you are ready to solve an optimization model.

Problem of Cutting, Assembling and Production of Chairs solved with Excel Solver

“The Old Chairs”, a rustic company manufactures, among many other products, three types of chairs A, B and C, which are sold at price of 11, 13 and 12 dollars each respectively. The chairs go through three processes, Cutting, Assembly and Painting, for which up to 17, 13 and 15 hours a week respectively are dedicated to these operations for these products. The type A chair takes 3 hours for cutting, 1 hour for assembling and 3 hours for painting. The type B chair takes 1 hour for cutting, 4 hours for assembling and 3 hours for painting. And finally the type C chair takes 5 hours for cutting, 2 hours for assembling and 2 hours for painting. According to the above information:

a) Solve the problem with continuous variables and point out the results for each variable.

Decision Variables: the level of weekly production was set for each of the varieties of chair as detailed below:


Objective Function: Maximize weekly earnings associated with the production and sale of the chairs.


Constraints: In the process of cutting, assembling and painting the availability of weekly hours must be respected. Additionally the constraints of non-negativity must be met.


The computational implementation of the above problem with Excel Solver can achieve the following results:


Where the optimal solution is A=1,914286, B=1,828571 y C=1,885714 with optimal value V(P)=67,45714.

b) Modify the conditions of the variables and select them whole (integer) and observe the difference between the response from point a) and this new one found.

When defining the decision variables as integers we are facing an Integer Programming model (being the first stage a problem in Linear Programming). The results are:


The optimal solution is A=1B=2 y C=2 with optimal value V(PE)=61.

c) Reach to a conclusion about what happened between continuous variables and integer variables.

It is important to note that the domain of feasible solutions of the problem (part b) is a subset of the domain of feasible solutions of the linear problem (part a). Therefore it is natural that, without obtaining a solution with integer values for the decision variables in the first problem, the optimal value necessarily will decrease in whole variant of this maximization (V(PE)<V(P)). It can also be noted that the integer solution, not necessarily is achieved by approximating the fractional results of a solution of a linear problem to the nearest lower or higher integer. Consequently, to address the resolution of a model which takes integer values for the decision variables efficiently, it requires a specific algorithmic alternative such as the Branch and Bound Method.

Below is a link to download the Excel file used for solving the problem of cutting, assembling and production of chairs. Two sheets corresponding to part a) and b) the proposed problem are included in the file: Production of Chairs.

What is a Basic Feasible Solution in Linear Programming

In Linear Programming (LP) a basic feasible solution is one that also belong to the feasible region or problem area can be represented by a feasible solution in implementing the Simplex Method satisfying nonnegative conditions. In this context, a basic solution corresponds to one of the vertices whose coordinate feasibility domain or solution can be represented by a set of active constraints for the model. To develop the above concept consider the following mathematical optimization problem (linear):

linear programming example

The graphical solution of the above problem using Geogebra is presented in the following graph:


The hatched area corresponds to the feasibility of the problem domain, in particular 5 identifying vertices have arbitrarily called A, B, C, D and E. The optimal solution of the linear model is reached in the vertex C where X=100 and Y=350 with optimal value V(P)=3.100. Note that this solution can be obtained by solving a system of equations with the constraints 1 and 3 (R1 and R3) in equality.

Consequently the vertex C besides being a basic solution is an optimal basic solution. As for the vertices A, B, D and E are feasible basic solutions (not optimal) because the application of the simplex method at least one non-basic variable have negative reduced cost (which will improve the current value of the function target). The table below is obtained by taking the problem into standard form, adding S1, S2 and S3 as slack variables for the constraints 1, 2 and 3, respectively (R1, R2 and R3).


Both non-basic variables (initial) X and Y have negative reduced cost (-3 and -8) for both X=0 and Y=0 if it is a feasible basic solution ( vertex A) is not optimal solution. To continue the show will make an iteration of the simplex method incorporating the variable Y to the base (reduced cost criterion “more negative”) and where the minimum quotient Min {1600/4;1700/2; 350/1} = 350 ==> S3 leaves the base:


The basic solution is now X=0 and Y=350 ( point B ), however, the reduced cost of the variable X is still negative and thus we are not yet at the optimum. Consequently X enters the base and get the minimum quotient: Min {200/2; 1000/6} = 100 ==> S1  leaves the base:


Finally the optimal solution (optimal is reached basic feasible solution) with X=100 and Y=350 ( point C) where all non-basic variables (S1 and S3) have reduced greater or equal to zero cost, fulfilling the optimality criterion .

What happens to the vertices D and E? also are (suboptimal) feasible basic solutions could be found for example by incorporating in the first instance (initial table) to the variable X to the base. In this way, the vertex E should then one iteration and the vertex D in a second iteration. Note that there are other (non-basic) feasible solutions such as X=100 and Y=100 belonging to the domain of feasible solutions but can not be represented by solving a system of equations.