Transportation model

Optimization Technique in Construction Management...
mainly Transportation Model

1. Presentation on Transportation Model Methods of Optimization Techniques BY SATISH AWARE ( M.Tech Civil)

2. INTRODUCTION It is a one Method of Optimization Techniques Introduce about transportation model methods of solving transportation problem to its optimistic solution stage. 2

3. What Is Optimization ? Transportation Model Applications of Transportation Model… Phases of Solution Obtains the Initial Basic Feasible Solution Obtains the Optimal Basic Solution Actual model Conclusion Use of this technique CONTENTS

4. What Is Optimization ? Optimization problem : Maximizing or minimizing some function Common applications : Minimal cost, maximal profit, minimal error, optimal design, optimal management, variation principles

5. Transportation Model The transportation problem is a special type of LPP The objective is to minimize the cost of distributing a product Because of its special structure the usual simplex method is not suitable require special method of solution. Aim : find out optimum transportation schedule keeping in mind cost of transportation to be minimized

6. Assumptions in the transportation Model : Total qty of the item available at different sources is equal to the total requirement at different destinations. Item can be transported conveniently from all sources to destinations. The unit transportation cost of the item from all sources to destinations is certainly & precisely known. The transportation cost on given route is directly proportional to the no. of units shipped on that route. The objective is to minimize the total transportation cost for the organization as a whole & not for individual supply & distribution centre.

7. Application of Transportation Model Minimize shipping costs Determine low cost location Find minimum cost production schedule Military distribution system

8. Phases of Solution of Transportation Problem :- Phase I- obtains the initial basic feasible solution North West Corner Method (NWC) Row Minima Method Column Minima Method Least Cost Method Vogel's Approximation Method (VAM) Phase II-obtains the optimal basic solution Stepping Stone Method Modified Distribution Method a.k.a. MODI Method

9. Example. Factory takes 16 Rs/km transport cost per truck for supply of bricks. Following is requirement of bricks on various sites with respective distances(km) from warehouses Available Single truck contains 7000 bricks Sites  warehouses  Karad Saidapur Masur Agashiv Nagar Available Bricks Tasavde 2 3 11 7 42,000 Tembhu 1 6 1 7,000 Koparde 5 8 15 9 70,000 Required Bricks 49,000 35,000 21,000 14,000 1,19,000

10. Solution  Unit cost of single tuck transport =17x travelling distance(km) Sites  warehouses  Karad Saidapur Masur Agashiv Nagar Available Bricks (trucks) Tasavde 2 3 11 7 6 Tembhu 1 6 1 1 Koparde 5 8 15 9 10 Required Bricks (trucks) 7 5 3 2 17*

11. NWC: 2 3 11 7 6 1 6 1 1 5 8 15 9 10 7 1 5 3 2 17* Transport distance =(2x6)+(1x1)+(8x5)+(15x3)+(9x2)=116 km Transport Cost =116x16=1856 Rs.

12. 2) LCM : Bricks Factory Karad Saidapur Masur Agashiv Nagar Available Bricks (trucks) Tasavde 2 3 11 7 6 Tembhu 1 6 1 1 Koparde 5 8 15 9 10 Required Bricks (trucks) 7 1 5 4 3 2 17* Transport distance =(0x1)+(2x6)+(5x1)+(8x4)+(15x3)+(9x2)=112 km Transportation Cost = 112x16 = 1792 Rs.

13. 3) VAM : Bricks Factory Karad Saidapur Masur Agashiv Nagar Available Bricks (trucks) penalties Tasavde 2 3 11 7 6 1 1 5  Tembhu 1 6 1 1 . 1 . Koparde 5 8 15 9 10 . 3 4 Required Bricks (trucks) 7 6 5 3 2 1 17* penalties 1 3 3 5  5 4 6  2 Transport distance =(1x1)+(3x5)+(2x1)+(5x6)+(15x3)+(9x1)=102 km Transport COST= 102 kmx16=1632 Rs.

14. IBFS by VAM & check optimality: 2 3 11 7 1 5 1 6 1 1 5 8 15 9 6 3 1 Modified Distribution Method :- 2 -3 5 0 1 10 4 A B C P Q R S Cell Values =Cell Amount –U-V AR= 11-2-10= -1 AS= 7-2-4 =1 BP= 1-(-3)-0 =4 BQ= 0- (-3)-1 =2 BR= 6- (-3)-10 = -1 CQ= 8-5-1 =2 1 ST TRIAL :- Allocated cells =Nc+Nr-1 Here 4+3-1=6_____we can check optimality

15. 1 -3 5 0 1 10 4 A B C P Q R S Cell Values =Cell Amount –U-V AP= 2-1-0= 1 AS= 7-1-4 =2 BP= 1-(-3)-0 =4 BQ= 0- (-3)-1 =2 BR= 6- (-3)-10 = -1 CQ= 8-5-1 =2 2 nd TRIAL :- 2 3 11 7 5 1 1 6 1 1 5 8 15 9 7 2 1

16. 2 3 11 7 5 1 1 6 1 1 5 8 15 9 7 1 2 Cell Values =Cell Amount –U-V AP= 2-1-0= 1 AS= 7-1-4 =2 BP= 1-(-4)-0 =5 BQ= 0- (-4)-2 =2 BS= 1- (-4)-4 = CQ= 8-5-2 =1 (NO ANY – ve CELL VALUE , IT IS OPTIMAL STAGE OF MODEL) 3 nd TRIAL :- 1 -4 5 0 2 10 4 A B C P Q R S TRANSPORTATION DISTANCE =(3x5)+(11x1)+(6x1)+(5x7)+(15x1)+(9x2) = 100 km TRANSPORTATION COST =100 x16=1600 Rs.

17. Conclusion :

18. Use of This Technique :- This model can be used for a wide variety of situations such as scheduling, production, investment, mix problems & many other, so that the model is really not confined to transportation or distribution only. The objective is to minimize the cost of transportation while meeting the requirements at the destinations.

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