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# Lp simplex method example in construction managment

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using linear programing in solve example in construction mangment

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### Lp simplex method example in construction managment

1. 1. 1 Application example for using linear programing in construction management By : Nagham nawwar Abbas 1
2. 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. 3 Fig.1 (detail of area ) 3
4. 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 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 6 Solution
7. 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. 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. 9 𝑏 𝑐 b Variables BasisIteration 𝑋5𝑋4𝑋3𝑋2𝑋1 _0000-12-30Z 1 76.12560900001600800𝑋3 40480001060120𝑋4 33.3320000000100200000600000𝑋5 9 ……………………………..continued to next slide
10. 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. 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. 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. 13. 13Distribution of building 13
14. 14. 14 Solving by software 1- excel (solver) 2- matlab 14
15. 15. 15 1- excel (solver ) insert decision variables , objective function X1 and x2 any intuitive value))and constrains as shown in picture 15
16. 16. 16Insert solver parameters 16
17. 17. 17Finally get the results 17
18. 18. 18 2-matlab (optimization tools) open matlab start toolboxes optimization optimization tool (optimtool) 18
19. 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. 20 Run solver and view results Maximum value of function = 1080 X1 = 20 , x2=40 20
21. 21. 21 Thank you 21