O slideshow foi denunciado.
Utilizamos seu perfil e dados de atividades no LinkedIn para personalizar e exibir anúncios mais relevantes. Altere suas preferências de anúncios quando desejar.

Lp simplex method example in construction managment

1.151 visualizações

Publicada em

using linear programing in solve example in construction mangment

Publicada em: Engenharia

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

×