Date post: | 22-Jan-2017 |
Category: |
Engineering |
Upload: | naghamnagham |
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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 :
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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
5
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 =
20
20
Run solver and view results
Maximum value of function = 1080 X1 = 20 , x2=40
2121
Thank you