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Msc Logistics & Supply Chain Management

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SCHOOL OF ECONOMIC SCIENCES

MSc LOGISTICS&SUPPLY CHAIN MANAGEMENT

Academic Year 2016-2017

1st Semester

Class: Operations Research

Assignment Number 1

Angelos Kottas

&

Christos Karagkounis

        


Exercise 2

Our team has chosen the Exercise 2 (named “askhsh2” in the attached PDF files sent by the class instructor).

Modeling of the Linear Problem

Let the following variables:

-x1 : the quantity of produced tables

-x2 : the quantity of produced chairs

-x3 : the quantity of produced desks

-x4 : the quantity of produced libraries

Given the fact that every produced table, chair, desk, and library requires 5, 1, 9 and 12 m3  Type 1 wood respectively, having 1500 m3 of Type 1 wood available, then the first constraint is written as follows:

5x1 + x2 + 9x3 + 12x4  ≤ 1500 (Constraint 1)

Given the fact that every produced table, chair, desk, and library requires 2, 3, 4 and 1 m3  Type 2 wood respectively, having 1000 m3 of Type 2 wood available, then the second constraint is written as follows:

2x1 + 3x2 + 4x3 + 1x4  ≤ 1000 (Constraint 2)

Given the fact that every table, chair, desk, and library requires respectively 3, 2, 5 and 10 hours of labour respectively, having 800 hours of labour available, then the third constraint is written as follows:

3x1 + 2x2 + 5x3 + 10x4  ≤ 800 (Constraint 3)

Given the fact that a minimum of 40 tables are required, then the fourth constraint is written as follows:

x1 ≥ 40 (Constraint 4)

Given the fact that a minimum of 130 chairs are required, then the fifth constraint is written as follows:

x2 ≥ 130 (Constraint 5)

Given the fact that a minimum of 30 desks are required, then the sixth constraint is written as follows:

x3 ≥ 30 (Constraint 6)

Given the fact that a maximum of 10 tables are required, then the seventh constraint is written as follows:

x4 ≤10 (Constraint 7)

As far as concerning the objective function, in our case it will represent the obtained profit. Taking into account that each produced table, chair, desk, and library provides 12, 5, 15 and 10 monetary units of profit respectively (1,000 drachmas represents a monetary unit), then the objective function is as follows:

maxz= 12x1 + 5x2 + 15x3 + 10x4  (Objective Function)

Due to the fact that we want to maximize our obtained profit, the objective function represents a maximization linear problem.

Of course, the non-negativity constraint must hold for the decision variables.

Hence, xi≥0 (i=1,2,3,4)

Solving the Modeled Linear Problem

LINDO™ Software Used

According to the LINDO™ Tutorial (accessed & downloaded the weblink http://www.lindo.com/downloads/PDF/LindoUsersManual.pdf ) the aforementioned Linear Problem command script syntax in LINDO™ software is as follows:

max 12x1+5x2+15x3+10x4

st

5x1+x2+9x3+12x4<=1500

2x1+3x2+4x3+x4<=1000

3x1+2x2+5x3+10x4<=800

x1>=40

x2>=130

x3>=30

x4<=10

end

GIN 1

GIN2

GIN 3

GIN 4

The commands “GIN1”, “GIN2”, “GIN3”, “GIN4” were used due the fact that the products that we want to produce have an “EACH” unit measurement, so any other number other than integer is not permitted.


The solution script that was obtained by LINDO™ is the following:

LP OPTIMUM FOUND AT STEP      3

 OBJECTIVE VALUE =   2660.00000

 FIX ALL VARS.(    1)  WITH RC >  0.000000E+00

 NEW INTEGER SOLUTION OF    2660.00000     AT BRANCH      0 PIVOT       3

 BOUND ON OPTIMUM:  2660.000

 ENUMERATION COMPLETE. BRANCHES=     0 PIVOTS=       3

 LAST INTEGER SOLUTION IS THE BEST FOUND

 RE-INSTALLING BEST SOLUTION...

        OBJECTIVE FUNCTION VALUE

        1)      2660.000

  VARIABLE        VALUE            REDUCED COST

        X1              130.000000            -12.000000

        X2              130.000000              -5.000000

        X3                30.000000            -15.000000

        X4                  0.000000            -10.000000

       ROW   SLACK OR SURPLUS     DUAL PRICES

        2)            450.000000                        0.000000

        3)            230.000000                        0.000000

        4)                0.000000                        0.000000

        5)              90.000000                        0.000000

        6)                0.000000                        0.000000

        7)                0.000000                        0.000000

        8)              10.000000                        0.000000

...

...

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