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Lp simplex method example in construction managment
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.
In this document
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Introduction to the application of linear programming in construction management focusing on building plans.
Cost, working hours, and area requirements for building types provided, outlining project constraints.
Setting up a linear programming model using simplex method to maximize population accommodation.
Iterative calculations for optimization highlighting variable values leading to optimum population capacity.
Distribution of buildings based on calculated area, discussing allocation of remaining space.
Methods to solve the linear programming problem using Excel Solver tool.
Implementing optimization methods in MATLAB, converting LP format and achieving optimal solutions.
Wrap-up and gratitude for the audience's attention.
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Lp simplex method example in construction managment
- 1. 1 Application example for usinglinear programing in construction management By : Nagham nawwar Abbas 1
- 2. 2 On a particulararea 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 ofarea ) 3
- 4. 4 4 Cost for each building $ Required working hoursfor 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 totalbudget not exceed 20,000,000 $ . 2- Working hours not exceed 4800 hours . 3- total available area 60900 m2 .
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- 7. 7 By simplex method Whenx1= 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
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- 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 ofZ = 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
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- 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
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- 17. 17Finally get theresults 17
- 18. 18 2-matlab (optimization tools) openmatlab 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 andview results Maximum value of function = 1080 X1 = 20 , x2=40 20
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