This document presents a linear programming problem to determine the optimal number and type of buildings to construct on a plot of land based on budget, area, and population constraints. There are two types of buildings that can be built: five-story buildings or two-story buildings. The objective is to maximize the total population housed while staying within budget and area limits. The problem is solved using the simplex method, with the optimal solution being 20 five-story buildings and 40 two-story buildings, housing a total population of 1080.

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

21. 21
Thank you
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