lp simplex method example in construction managment

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Application example for using linear programing in

construction management

By : Nagham nawwar Abbas

22 On a particular area with 60900 m2 (detail in fig.1)

we would like to build several buildings and we would like

some of the floors of these buildings are five- stores and

some two-stores , how it should be the number of the first

type of these buildings and how many should be number

of other type to accommodate the largest number of

populations , knowing that the data set out table follow :

33

Fig.1 (detail of area )

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Cost for each building $

Required working hours for

each building

Required area for

each building

m2

No. of population in each building

 No. of storey

600,000 120 800 30 five

200,000 60 600 12 two

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5

1- The total budget not

exceed 20,000,000 $ .

2- Working hours not exceed

4800 hours .

3- total available area 60900

m2 .

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Solution

77By simplex method

When x1= no. of five stores buildingWhen x2= no. of two stores building

max Z = 30X1 + 12X2when :800X1 + 600X2 ≤ 60900120X1 + 60X2 ≤ 4800600000X1 + 200000X2 ≤ 20000000X1 , X2 > 0

88

Z – 30 -12X2 + 0X3 + 0X4 +0X5 = 0

800X1 +600X2 + X3 + 0X4 + 0X5 = 60900

120X1 + 60X2 + 0X3 + X4 + 0X5 = 4800

600000X1 +200000X2 + 0X3 +0X4 + X5 = 20000000

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b

Variables Basis Iteration

_ 0 0 0 0 -12 -30 Z

176.125 60900 0 0 1 600 800

40 4800 0 1 0 60 120

33.33 20000000 1 0 0 200000 600000..……………………………continued to next slide

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b

Variables Basis Iteration

_ 1000 0 0 -2 0 Z

20 1 0

40 800 1 0 20 0

_ 0 0 1..……………………………continued to next slide

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b

Variables Basis Iteration

_ 1080 0 0 0 Z

3_ 20900 1 0 0

_ 40 0 1 0

_ 20 0 0 0 1Optimum value of Z = 1080Value of X1 = 20 , Value of X2 = 40

1212total area = 60900m2

total building area=800*40+600*20=40000m2

The remaining area=60900-40000=20900m2

how to distribute the buildings

and the remaining area?

1313Distribution of building

1414

Solving by software

1- excel (solver)

2- matlab

15151 -excel (solver )

insert decision variables , objective function X1 and x2 any intuitive value) )and constrains as shown in picture

1616Insert solver parameters

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Finally get the results

18182-matlab (optimization tools)

open matlab start toolboxes optimization optimization tool (optimtool)

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max Z = 30X1 + 12X2when :800X1 + 600X2 ≤ 60900120X1 + 60X2 ≤ 4800600000X1 + 200000X2 ≤ 20000000

Convert the LP into MATLAB format

F -=

A =

B =

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Run solver and view results

Maximum value of function = 1080 X1 = 20 , x2=40

2121

Thank you