LINEAR PROGRAMMING:GRAPHICAL METHOD

Monday, January 25, 2016

LINEAR PROGRAMMING:GRAPHICAL METHOD

STRUCTURE OF LINEAR PROGRAMMING

     Structure of linear programming are commonly of the form where two or more products involved. Example of these were mixtures of different ingredients like product 1(first ingredient), and product 2 (second ingredient). Each productions has its own machine as to manufacture but with limited time, say machine A can produce product 1 for a certain time and machine B for the second product with time limit as well. Each resulting product are subject to profit. We are expressing the idea that product 1 manufactured by machine A can be sell into x amount, and product 2 can be sell y amount. Linear Algebraic equation is needed to quantify each product. We must see to it that we are minimizing the material but maximizing the profit as programming intervention shows its importance, but still it depends on the case of the problem.

Illustration

     Let two products be x and y. The set up for restrictions,profit, requirements were presented in tabular manner below. See photo provided.

Where :

1.   , this is so because we need to produce each product of x and/or y.

2. 
   , this is restriction presented in inequalities. 

    The presentation above was designed for two ingredients only. It can be more than two depending on how many ingredients were to be mixed. If ingredient 3 were added, it can be .

    The objective here is to select from x and/or y that will be the objective function (aka profit function). The equation for the objective function was found by:

       , equation of objective function

    There are some other ways to solve the linear programming problem. But the most fundamental and easiest one is the graphical method. This the most basic and convenient approach in solving such problem. 
    
    Take note also that our illustration not limits to two products, it can be three or more.

    Let us have an illustration as example how to solve a simple problem involving Linear Programming.

Sample Problem:

    How many units of x and y of the two compounds should be produced given the numbers of pounds for each of the ingredients based on the illustration below.

Illustrations:

Step 1: writing the given in-equations:

Ingredient 1: 

Ingredient 2:  

Step 2: Change the inequality symbol to equal symbol.

    Now in order to graph these, we want to change the inequalities symbol into equal symbol.

Ingredient 1: 

Ingredient 2: 

Step 3:  From ingredient A, set x=0 to get y, and y=0 to get x.This is to get the boundary as restrictions.

   In equation 2x +3y=60,@x=0 y=20, and @ y=0 x=30. With this boundaries the graph were shown below.

Graph for 2x +3y=60.

And in equation x +2y=40,@x=0 y=20, and @ y=0 x=40.

Graph for x +2y =40.

Step 4: Graphing for its Feasible region

The shaded yellow part was the feasible region.It is assumed that the students has back ground in inequalities shading.

Step 5: Solving for the objective function.(see from the table)

By solving for the value of x and y using substitution method, it was found by x=0 and y =20. Substituting these values to the objective function, we arrive to the following output;




 Answer

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SIMPLEX MAXIMIZATION PROBLEMS

Wednesday, January 13, 2016

SIMPLEX MAXIMIZATION

Problems:

Supply the simplex maximization tables. 

1. 

Cj                                           150      160     0     0     0                Prod      Qty           x1         x2       s1   s2   s3

0             S1          100           1           0         0      0      1 0             S2          200           0           1         0      1      0 0             S3          300           0           0         1      0      0

Zj              Cj-Zj

2. 

Cj                                           200      160     0     0     0                Prod      Qty           x1         x2       s1   s2   s3

0             S1          100           1           0         0      0      1 160        X2          200           0           1         0      1      0 0             S3          100           0           0         1     -1      0

Zj              Cj-Zj

3. 

Cj                                             8            10     0     0     0                Prod      Qty            x              y      s1   s2   s3

8            x               1               1           0        -1/7    3/7  -1/7 10         y                1               0           1        3/7     2/7   5/21

Zj              Cj-Zj

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DECISION MAKING IN QUANTITATIVE TECHNIQUES

Thursday, December 11, 2014

Decision Making

By definition, it is the act of drawing several answers to a certain problem and selecting the appropriate or the best option as the solution. Follow the gradual steps to do the decision making.

Steps for decision making:

       1.       Identify the problem. This to check whether the existing problem is really a problem or not. This is so because sometimes the existing problem is not a problem but a fact. 

  

       2.       Set criteria appropriate for the problem. This is to be done by giving weights (in terms of percent) in every detail of the problem identified.

       3.       Make Plan A or Plan B. This is by preparing a second option if the first solution does not work.

       4.       Assess every alternative solution. This is by rechecking if the Plan B is ideal or appropriate for the problem.

       5.       Spot the best alternative solution. This by identifying the most ideal and suitable solution.

       6.       Apply the spotted alternative solution. This is by applying the solution to the problem.

       7.       Check or consult the application of the solution for the assurance that it is appropriate and with the approval of the majority, if not all, the of 
organization.

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QUANTITATIVE TECHNIQUES IN BUSINESS

Tuesday, January 15, 2013

QUANTITATIVE TECHNIQUES IN BUSINESS:

Problems intended for graphical and Simplex method:

Problems in Graphical Method:

Problem 1:

A tailor shop makes two products, dress and pants which must be processed through assembly and Finishing department. Assembly department is available for 24 hours in every production period, while the finishing department is available for 38 hours of work. Manufacturing one dress requires 2 hours in assembly and 3 hours in the finishing department. Each pants requires 3 hours in the assembly and 5 hours in the finishing department. One dress contributes $ 150 as profit and pants $ 100. The problem is to determine the number of dress and pants to make production period in order to maximize the profit.

Problem 2:

A steel producer makes two types of steel, regular and special.A ton of regular steel requires two hours  in the open-hearth furnace and three hours in the soaking pit. A ton of special steel requires two hours in the open-hearth furnace and five hours in the soaking pit. The open-hearth furnace is available for eight hours per day and the soaking pit is fifteen hours a day. The profit on a ton of regular steel is $4000 and it is $ 6000 on a ton of special steel. Determine how many tons of each type of steel should be made to maximize the profit considering that demand on regular steel is at least one ton.


Simplex Method:

Maximization:

Steps for constructing the Simplex solution:

1.      From the problem condition, set up the constraints.

2.      Convert the inequality explicit constraints to equations by adding slack variables.

      3. Enter the numerical coefficients and variables in the Simplex Tables.

    4.Calculate the C _j and Z _j values.

      5. Determine the optimum column or entering variable by choosing the most positive value   in the C _j – Z _{j row.}

      _{6. Divide the quantity column values by the non-zero and non-negative entries in the optimum   column. The smallest quotient belongs to the pivotal row.}

     _{7. Compute the values for the replacing row by dividing all entries by the pivot. Enter the result in the next table.}

     _{8. Compute the new entries for the remaining rows by reducing the optimum column entries to zero.}

     _{9. Calculate} C _j and Z _j values.

   _{10. If there is a positive entry in the} C _{j  minus}  Z _{j  row, return to step 5. The final solution has been obtained if there is no positive value in the} C _j – Z _{j row.}

Click the reference below for additional information:

QUANTITATIVETECHNIQUESINBUSINESS

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