using linear programing in solve example in construction mangment

1. 1
Application example for
using linear programing in
construction management
By : Nagham nawwar Abbas
1

2. 2
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
:
2

3. 3
Fig.1 (detail of area )
3

4. 4
4
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
600,00012080030five
200,0006060012two

5. 5
5
1- The total budget not
exceed 20,000,000 $ .
2- Working hours not
exceed 4800 hours .
3- total available area
60900 m2 .

6. 6
6
Solution

7. 7
By simplex method
When x1= no. of five stores building
When x2= no. of two stores building
max Z = 30X1 + 12X2
when :
800X1 + 600X2 ≤ 60900
120X1 + 60X2 ≤ 4800
600000X1 + 200000X2 ≤ 20000000
X1 , X2 > 0
7

8. 8
Z – 30𝑋1 -12X2 + 0X3 + 0X4 +0X5 = 0
800X1 +600X2 + X3 + 0X4 + 0X5 = 60900
120X1 + 60X2 + 0X3 + X4 + 0X5 = 4800
600000X1 +200000X2 + 0X3 +0X4 + X5 = 20000000
8

9. 9
𝑏
𝑐
b
Variables
BasisIteration
𝑋5𝑋4𝑋3𝑋2𝑋1
_0000-12-30Z
1
76.12560900001600800𝑋3
40480001060120𝑋4
33.3320000000100200000600000𝑋5
9
……………………………..continued to next slide

10. 10
𝑏
𝑐
b
Variables
BasisIteration
𝑋5𝑋4𝑋3𝑋2𝑋1
_10001
20000
00-20Z
2
1027
10
102700
3
−1
750
011000
3
0𝑋3
40800−1
5000
10200𝑋4
_100
3
1
600000
001
3
1𝑋1
10
……………………………..continued to next slide

11. 11
𝑏
𝑐
b
Variables
BasisIteration
𝑋5𝑋4𝑋3𝑋2𝑋1
_10803
100000
1
10
000Z
3
_209001
500
−50
30
100𝑋3
_40−1
100000
1
20
010𝑋2
_200−1
60
001𝑋1
11
Optimum value of Z = 1080
Value of X1 = 20 , Value of X2 = 40

12. 12
total 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 ?
12

13. 13Distribution of building 13

14. 14
Solving by software
1- excel (solver)
2- matlab
14

15. 15
1- excel (solver )
insert decision variables , objective function
X1 and x2 any intuitive value))and constrains as shown in picture
15

16. 16Insert solver parameters 16

17. 17Finally get the results
17

18. 18
2-matlab (optimization tools)
open matlab start toolboxes
optimization optimization tool (optimtool)
18

19. 19
max Z = 30X1 + 12X2
when :
800X1 + 600X2 ≤ 60900
120X1 + 60X2 ≤ 4800
600000X1 + 200000X2 ≤
20000000
Convert the LP into MATLAB format
F =-
30
12
A =
800 600
120 60
600000 200000
B =
60900
4800
20000000
19

20. 20
Run solver and view results
Maximum value of function = 1080
X1 = 20 , x2=40
20

21. 21
Thank you
21