© Copyright Andrew Hall, 2002 FOMGT 353 Introduction to Management Science Lecture 17v Slide 1 Network Models Lecture 17 (Part v.) Vogel’s Approximation Method

© Copyright Andrew Hall, 2002 FOMGT 353 Introduction to Management Science Lecture 17v Slide 2 Vogel’s Approximation Starting Procedure 1.Calculate Row and Column Penalties by taking the absolute value of the two lowest “available” costs in the row or column deducted from one another. 2.Select the Maximum of the Row and Column Penalties to identify the Row or Column into which flow will go. 3.In the Row selected find the lowest “available” cost to identify the Column. OR in the Column selected find the lowest “available” cost to identify the Row. We have identified the Xij to be increased!

© Copyright Andrew Hall, 2002 FOMGT 353 Introduction to Management Science Lecture 17v Slide 3 Vogel’s Approximation Starting Procedure cont… 4.Fill in as much flow for that Decision Variable, Xij, as possible. 5.Adjust the Border Demands and Supplies to take into account the Flow allocated. 6.If there is still a positive amount of Border Demand, goto 1, otherwise stop.

© Copyright Andrew Hall, 2002 FOMGT 353 Introduction to Management Science Lecture 17v Slide 4 Vogel’s Approximation 1 2 Lowest Costs in Column 1 are 4 and 5 so Column Penalty is 1 2 Lowest Costs in Column 2 are 10 and 11 so Column Penalty is 1 2 Lowest Costs in Row 1 are 12 and 6 so Row Penalty is 6 2 Lowest Costs in Row 2 are 11 and 4 so Row Penalty is 7 2 Lowest Costs in Row 3 are 10 and 5 so Row Penalty is 5 In Row 2 the Column with the Lowest cost is Column 1, so insert flow into x21 The Maximum Penalty is 7 [max(1,1,5,7,6) ] So Row 2 will get some flow!

© Copyright Andrew Hall, 2002 FOMGT 353 Introduction to Management Science Lecture 17v Slide 5 Vogel’s Approximation 2 15,000 0 In x21, insert the min of S2 and D1 Subtract x21 from S2 and from D1.

© Copyright Andrew Hall, 2002 FOMGT 353 Introduction to Management Science Lecture 17v Slide 6 Vogel’s Approximation 3 2 Lowest Costs in Row 2 are 11 and 11 so Row Penalty is 0 Row 1 Penalty unchanged at 6 Row 3 Penalty unchanged at 5 2 Lowest Costs in Column 1 are 5 and 6 so Column Penalty is 1 2 Lowest Costs in Column 2 are 10 and 12 so Column Penalty is 2 The Maximum Penalty is 6 [max(1,2,5,0,6) ] So Row 1 will get some flow! In Row 1 the Column with the Lowest cost is Column 1, so insert flow into x11

© Copyright Andrew Hall, 2002 FOMGT 353 Introduction to Management Science Lecture 17v Slide 7 Vogel’s Approximation 4 5,000 0 In x11, insert the min of S1 and D1 Subtract x11 from S1 and from D1.

© Copyright Andrew Hall, 2002 FOMGT 353 Introduction to Management Science Lecture 17v Slide 8 Vogel’s Approximation 5 2 Lowest Costs in Row 2 are 11 and 11 so Row Penalty is 0 2 Lowest Costs in Column 2 are 10 and 12 so Column Penalty is 2 2 Lowest Costs in Row 3 are 10 and 10 so Row Penalty is 0 2 Lowest Costs in Row 1 are 12 and 12 so Row Penalty is 0 The Maximum Penalty is 2 [max(0,2,0,0,0) ] So Col 2 will get some flow! In Column 2 the Row with the Lowest cost is Row 3, so insert flow into x32 2 Lowest Costs in Column 1 are 5 and 5 so Column Penalty is 0 Note Column 1 has zero Border Demand, so “Green Out” Column 1!

© Copyright Andrew Hall, 2002 FOMGT 353 Introduction to Management Science Lecture 17v Slide 9 Vogel’s Approximation 6 5,000 0 In x32, insert the min of S3 and D2 Subtract x32 from S3 and from D2.

© Copyright Andrew Hall, 2002 FOMGT 353 Introduction to Management Science Lecture 17v Slide 10 Vogel’s Approximation Initial Feasible Solution