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: Defining the Problem The Equipment Supply Schedule The Primal Problem Initial Tableau and variable definitions Final Tableau and comments The Dual Problem Concerns with the initial tableau Initial Tableau and variable definitions Final Tableau and comments Sensitivity Analysis Final Comments Works Cited

The Problem: A 1,000 yd 3 excavation must be dug, given a week (5 business days). Of the five different categories of rental equipment, each has a specified hourly wage (for both equipment and operator) for use, and a limited availability per 8 hour business day. The objective is to find the allocation of equipment that minimizes the cost of the excavation, while removing all 1,000 yd 3 of earth within the given week.

Given equipment: Type Cost Availability Capacity Shovel Dozer $17.50/hr 6 hrs per day 28.2 yds 3 /hr Large Backhoe $40/hr 6 hrs per day 150 yds 3 /hr Backhoe A $27.50/hr 6 hrs per day 90 yds 3 /hr Backhoe B $22/hr 8 hrs per day 60 yds 3 /hr Crane $47/hr 5.5 hrs per day 40yds 3 /hr Other constraints: Each day consists of 8 business hours The week consists of 5 business days The entire 1,000 cubic yard excavation must be complete by the end of the business week

The Primal Problem Initial Tableau 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 Final Tableau 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: With the dual problem, we have several issues to resolve. First, we must consider the fact we have an unrestricted decision variable, V, in all of the constraint equations. We must set this unrestricted value equal to two variables, v1 and v2, which are required to be greater than or equal to zero. Secondly, note that to solve the prime problem, we had to convert it to a maximization problem, which is how we then developed our dual problem. Again, we must convert this to a maximization problem by considering a dependent variable y’ as the value of our objective function. Y’ will be the negative of y, our original dual problem objective function.

The Dual Problem Initial Tableau 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 Final Tableau 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 To find how much the resource values of our primal problem could change without altering the X B vector we consider the inverse basis matrix from the initial tableau of phase one of our two-phase method and the resource column (b). The equation we used to determine individual resource value changes was: max i { -b * i / t ir | t ir > 0 } ≤ Δ b 1 ≤ min i { -b * i / t ir | < 0 }

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 Images/Links: “ Caterpillar: Home.” May 20, 2007. http://www.cat.com Procedures: Calvert, J, Voxman, W. Linear Programming . Orlando: Harcourt Brace Jovanovich, Inc, 1989.

Linear Programming: The Construction Problem

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