Download presentation

Presentation is loading. Please wait.

Published byHailey Byrne Modified over 5 years ago

1
Optimizing the Cost of Excavation MTH-363-0016/1/2007 Joseph Jess Wesley Allen Parker Christopher Bullard Construction LP Problem #3

2
Table of Contents: Defining the ProblemDefining the Problem The Equipment Supply ScheduleThe Equipment Supply Schedule The Primal ProblemThe Primal Problem –Initial Tableau and variable definitions –Final Tableau and comments The Dual ProblemThe Dual Problem –Concerns with the initial tableau –Initial Tableau and variable definitions –Final Tableau and comments Sensitivity AnalysisSensitivity Analysis Final CommentsFinal Comments Works CitedWorks Cited

3
The Problem:

4
Given equipment: Type CostAvailabilityCapacity Shovel Dozer Shovel Dozer $17.50/hr6 hrs per day28.2 yds 3 /hr Shovel Dozer Large Backhoe Large Backhoe $40/hr6 hrs per day150 yds 3 /hr Large Backhoe ABackhoe A$27.50/hr6 hrs per day90 yds 3 /hr Backhoe A Backhoe BBackhoe B$22/hr8 hrs per day60 yds 3 /hr Backhoe B Crane$47/hr5.5 hrs per day40yds 3 /hr Other constraints: Each day consists of 8 business hoursEach day consists of 8 business hours The week consists of 5 business daysThe week consists of 5 business days The entire 1,000 cubic yard excavation must be complete by the end of the business weekThe entire 1,000 cubic yard excavation must be complete by the end of the business week

5
The Primal Problem Initial Tableau Tableau 1 Equationx1x2x3x4x5s1s2s3s4s5y1B 480/17150.0090.0060.0040.000.00 1.001000.00 s11.000.00 1.000.00 30.00 s20.001.000.00 1.000.00 30.00 s30.00 1.000.00 1.000.00 30.00 s40.00 1.000.00 1.000.00 40.00 s50.00 1.000.00 1.000.0055/2 W-480/17-150.00-90.00-60.00-40.000.00 -1000.00 Z35/240.0055/222.0047.000.00 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

6
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. Tableau 2 Equationx1x2x3x4x5s1s2s3s4s5y1B x216/851.003/52/54/150.00 1/15020/3 s11.000.00 1.000.00 30.00 s2-16/850.00-3/5-2/5-4/150.001.000.00 -1/15070/3 s30.00 1.000.00 1.000.00 30.00 s40.00 1.000.00 1.000.00 40.00 s50.00 1.000.00 1.000.0055/2 W0.00 10 Z339/340.007/26.00109/30.00 -4/15-800/3

7
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.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.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.

8
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. Tableau 1 v1v2u1u2u3u4u5s1s2s3s4s5B s1-480/17480/170.00 1.000.00 35/2 s2-150.00150.000.000.00 1.000.00 40.00 s3-90.0090.000.00 0.00 1.000.00 55/2 s4-60.0060.000.00 0.00 1.000.0022.00 s5-40.0040.000.00 0.00 1.0047.00 Z1000.00-1000.0030.00 40.0055/20.00

9
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. v1v2u1u2u3u4u5s1s2s3s4s5B s10016/850.00 1.00-16/850.00 339/4 v210.00-1/1500.00 1/1500.00 4/15 s3000.003/50.00 -3/51.000.00 7/2 s4000.002/50.000.00 -2/50.001.000.006 s5000.004/150.00 0.00-4/150.00 1.00109/3 Z0030.0070/330.0040.0055/20.0020/30.00 800/3

10
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).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 } Δ b 1 min i { -b * i / t ir | < 0 }

11
Sensitivity Data B-1 Matrix 0.006666670.00 1.000.00 -0.00666670.001.000.00 1.000.00 1.000.00 1.00 b vector 6.67 30 23.33 30 40 27.5

12
Sensitivity Results Sensitivity AnalysisSingle b i change -1000.00Δb13500.00 -30.00Δb2Infinite -23.33Δb3Infinite -30.00Δb4Infinite -40.00Δb5Infinite -27.50Δb6Infinite Resulting b i for Simultaneous Change Δb1>-1000.00 Δb2>-30.00 Δb3>-23.33 Δb4>-30.00 Δb5>-40.00 Δb6>-27.50 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

13
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.

Similar presentations

OK

Fractions Simplify: 36/48 = 36/48 = ¾ 125/225 = 125/225 = 25/45 = 5/9

Fractions Simplify: 36/48 = 36/48 = ¾ 125/225 = 125/225 = 25/45 = 5/9

© 2019 SlidePlayer.com Inc.

All rights reserved.

To make this website work, we log user data and share it with processors. To use this website, you must agree to our Privacy Policy, including cookie policy.

Ads by Google