Linear Programming: The Construction Problem

Optimizing the Cost of Excavation MTH-363-001 6/1/2007 Joseph Jess Wesley Allen Parker Christopher Bullard Construction LP Problem #3

Table of Contents: ,[object Object],[object Object],[object Object],[object Object],[object Object],[object Object],[object Object],[object Object],[object Object],[object Object],[object Object],[object Object]

The Problem: ,[object Object],[object Object]

Given equipment: ,[object Object],[object Object],[object Object],[object Object],[object Object],[object Object],[object Object],[object Object],[object Object],[object Object]

The Primal Problem ,[object Object],x i = the number of hours used by machine “i” where 1 < i < 5 s i = slack variables for the unused hours of machine “i” where 1 < i < 5 y 1 = artificial variable to resolve the LP problem with the Two-Phase Method Note the pivot location (circled above) does not rely on the Simplex Method 0.00 0.00 0.00 0.00 0.00 0.00 0.00 47.00 22.00 55/2 40.00 35/2 Z -1000.00 0.00 0.00 0.00 0.00 0.00 0.00 -40.00 -60.00 -90.00 -150.00 -480/17 W 55/2 0.00 1.00 0.00 0.00 0.00 0.00 1.00 0.00 0.00 0.00 0.00 s5 40.00 0.00 0.00 1.00 0.00 0.00 0.00 0.00 1.00 0.00 0.00 0.00 s4 30.00 0.00 0.00 0.00 1.00 0.00 0.00 0.00 0.00 1.00 0.00 0.00 s3 30.00 0.00 0.00 0.00 0.00 1.00 0.00 0.00 0.00 0.00 1.00 0.00 s2 30.00 0.00 0.00 0.00 0.00 0.00 1.00 0.00 0.00 0.00 0.00 1.00 s1 1000.00 1.00 0.00 0.00 0.00 0.00 0.00 40.00 60.00 90.00 150.00 480/17 y1 B y1 s5 s4 s3 s2 s1 x5 x4 x3 x2 x1 Equation Tableau 1

The Primal Problem ,[object Object],x i = the number of hours used by machine “i” where 1 < i < 5 s i = slack variables for the unused hours of machine “i” where 1 < i < 5 y 1 = artificial variable to resolve the LP problem with the Two-Phase Method Minimum: z=$266.67 for 6 hours, 40 minutes of labor with the large backhoe. -800/3 -4/15 0.00 0.00 0.00 0.00 0.00 109/3 6.00 7/2 0.00 339/34 Z 0 1 0.00 0.00 0.00 0.00 0.00 0.00 0.00 0.00 0.00 0.00 W 55/2 0.00 1.00 0.00 0.00 0.00 0.00 1.00 0.00 0.00 0.00 0.00 s5 40.00 0.00 0.00 1.00 0.00 0.00 0.00 0.00 1.00 0.00 0.00 0.00 s4 30.00 0.00 0.00 0.00 1.00 0.00 0.00 0.00 0.00 1.00 0.00 0.00 s3 70/3 -1/150 0.00 0.00 0.00 1.00 0.00 -4/15 -2/5 -3/5 0.00 -16/85 s2 30.00 0.00 0.00 0.00 0.00 0.00 1.00 0.00 0.00 0.00 0.00 1.00 s1 20/3 1/150 0.00 0.00 0.00 0.00 0.00 4/15 2/5 3/5 1.00 16/85 x2 B y1 s5 s4 s3 s2 s1 x5 x4 x3 x2 x1 Equation Tableau 2

Concerns with the Dual Problem: ,[object Object],[object Object],[object Object]

The Dual Problem ,[object Object],v i = the hourly wage for the shovel dozer where 1 < i < 2 u j = the hourly wage for machines “i+1”, where 1 < i < 5 s k = slack variables, where 1 < k < 5 Note the Simplex Method will work in this revised form. 0.00 0.00 0.00 0.00 0.00 0.00 55/2 40.00 30.00 30.00 30.00 -1000.00 1000.00 Z 47.00 1.00 0.00 0.00 0.00 0.00 -1.00 0.00 0.00 0.00 0.00 40.00 -40.00 s5 22.00 0.00 1.00 0.00 0.00 0.00 0.00 -1.00 0.00 0.00 0.00 60.00 -60.00 s4 55/2 0.00 0.00 1.00 0.00 0.00 0.00 0.00 -1.00 0.00 0.00 90.00 -90.00 s3 40.00 0.00 0.00 0.00 1.00 0.00 0.00 0.00 0.00 -1.00 0.00 150.00 -150.00 s2 35/2 0.00 0.00 0.00 0.00 1.00 0.00 0.00 0.00 0.00 -1.00 480/17 -480/17 s1 B s5 s4 s3 s2 s1 u5 u4 u3 u2 u1 v2 v1 Tableau 1

The Dual Problem ,[object Object],Finally, note we have achieved the same optimum value for y’ as we did z. The solution we find is (0,4/15, 0, 0, 0, 0, 0) for our decision variables, with (339/4, 0, 7/2, 6, 109/3) for our slack variables. Considering the fundamental principle of duality, we confirm our dual problem has decision vectors u* and V* which correspond to the –c* row vector of our prime problem. 800/3 0.00 0.00 0.00 20/3 0.00 55/2 40.00 30.00 70/3 30.00 0 0 Z 109/3 1.00 0.00 0.00 -4/15 0.00 -1.00 0.00 0.00 4/15 0.00 0 0 s5 6 0.00 1.00 0.00 -2/5 0.00 0.00 -1.00 0.00 2/5 0.00 0 0 s4 7/2 0.00 0.00 1.00 -3/5 0.00 0.00 0.00 -1.00 3/5 0.00 0 0 s3 4/15 0.00 0.00 0.00 1/150 0.00 0.00 0.00 0.00 -1/150 0.00 1 -1 v2 339/4 0.00 0.00 0.00 -16/85 1.00 0.00 0.00 0.00 16/85 -1.00 0 0 s1 B s5 s4 s3 s2 s1 u5 u4 u3 u2 u1 v2 v1

Sensitivity Analysis ,[object Object],[object Object]

Sensitivity Data b vector 1.00 0.00 0.00 0.00 0.00 0.00 0.00 1.00 0.00 0.00 0.00 0.00 0.00 0.00 1.00 0.00 0.00 0.00 0.00 0.00 0.00 1.00 0.00 -0.0066667 0.00 0.00 0.00 0.00 1.00 0.00 0.00 0.00 0.00 0.00 0.00 0.00666667 B-1 Matrix 27.5 40 30 23.33 30 6.67

Sensitivity Results From this information we can find how much perturbation each variable can undergo without changing the basic variables in the basic solution. The only change that ends up being important to us is b 1 , which can take values from -993.33 to 3506.67 Infinite ≤ Δb6≤ -27.50 Infinite ≤ Δb5≤ -40.00 Infinite ≤ Δb4≤ -30.00 Infinite ≤ Δb3≤ -23.33 Infinite ≤ Δb2≤ -30.00 3500.00 ≤ Δb1≤ -1000.00 Single b i change Sensitivity Analysis -27.50 Δb6> -40.00 Δb5> -30.00 Δb4> -23.33 Δb3> -30.00 Δb2> -1000.00 Δb1> Resulting b i for Simultaneous Change

Works Cited ,[object Object],[object Object],[object Object],[object Object],[object Object]

Linear Programming: The Construction Problem

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