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© Copyright Andrew Hall, 2002 FOMGT 353 Introduction to Management Science Lecture 18 Slide 1 Network Models Lecture 18 The Transportation Algorithm II.

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Presentation on theme: "© Copyright Andrew Hall, 2002 FOMGT 353 Introduction to Management Science Lecture 18 Slide 1 Network Models Lecture 18 The Transportation Algorithm II."— Presentation transcript:

1 © Copyright Andrew Hall, 2002 FOMGT 353 Introduction to Management Science Lecture 18 Slide 1 Network Models Lecture 18 The Transportation Algorithm II

2 © Copyright Andrew Hall, 2002 FOMGT 353 Introduction to Management Science Lecture 18 Slide 2 The Initialized Transportation Simplex Tableau Recall at the end of the last lecture we had initialized the Transportation Simplex Tableau starting from the results of the Least Cost Starting Procedure. LCSP

3 © Copyright Andrew Hall, 2002 FOMGT 353 Introduction to Management Science Lecture 18 Slide 3 Transportation Simplex cont…

4 © Copyright Andrew Hall, 2002 FOMGT 353 Introduction to Management Science Lecture 18 Slide 4 The Transportation Simplex Method – Identify Entering Variable Recall from the Generalized Simplex Method that a variable is an Entering Variable in a Minimization problem if its Cij-Zij is the most negative of the Cij-Zij values of the non-basic variables. So we need to calculate Cij-Zij values for all the non-basic variables. So first we need to calculate Ui and Vj values for the Basic Variables using the relationship:  Cij – Ui + Vj = 0

5 © Copyright Andrew Hall, 2002 FOMGT 353 Introduction to Management Science Lecture 18 Slide 5 Transportation Simplex cont…

6 © Copyright Andrew Hall, 2002 FOMGT 353 Introduction to Management Science Lecture 18 Slide 6 The Transportation Simplex Method – Identify Entering Variable We use the relationship Cij – Ui + Vj = 0 for x12 to calculate V2. C12 – U1 + V2 = 0 => 12 – 0 + V2 = 0 => V2 = -12 LCSP

7 © Copyright Andrew Hall, 2002 FOMGT 353 Introduction to Management Science Lecture 18 Slide 7 The Transportation Simplex Method – Identify Entering Variable We use the relationship Cij – Ui + Vj = 0 for x13 to calculate V3. C13 – U1 + V3 = 0 => 100 – 0 + V3 = 0 => V3 = -100 LCSP

8 © Copyright Andrew Hall, 2002 FOMGT 353 Introduction to Management Science Lecture 18 Slide 8 The Transportation Simplex Method – Identify Entering Variable We use the relationship Cij – Ui + Vj = 0 for x32 to calculate U3. C32 – U3 + V2 = 0 => 10 – U3 + (-12) = 0 => U3 = -2 LCSP

9 © Copyright Andrew Hall, 2002 FOMGT 353 Introduction to Management Science Lecture 18 Slide 9 The Transportation Simplex Method – Identify Entering Variable We use the relationship Cij – Ui + Vj = 0 for x31 to calculate V1. C31 – U3 + V1 = 0 => 5 – (-2) + V1 = 0 => V1 = -7 LCSP

10 © Copyright Andrew Hall, 2002 FOMGT 353 Introduction to Management Science Lecture 18 Slide 10 The Transportation Simplex Method – Identify Entering Variable We use the relationship Cij – Ui + Vj = 0 for x21 to calculate U2. C21 – U2 + V1 = 0 => 4 – U2 + (-7) = 0 => -3 – U2 = 0 => -3 = U2 LCSP

11 © Copyright Andrew Hall, 2002 FOMGT 353 Introduction to Management Science Lecture 18 Slide 11 The Next Steps..

12 © Copyright Andrew Hall, 2002 FOMGT 353 Introduction to Management Science Lecture 18 Slide 12 We use the relationship Cij – Zij = Cij – Ui + Vj for x11 to calculate C11 – Z11. We use the relationship Cij – Zij = Cij – Ui + Vj for x22 to calculate C22 – Z22. C22 – Z22 = C22 – U2 + V2 C22 – Z22 = 11 – (-3) + (-12) = 2 The Transportation Simplex Method – Identify Entering Variable C11 – Z11 = C11 – U1 + V1 C11 – Z11 = 6 – 0 + (-7) = -1 We use the relationship Cij – Zij = Cij – Ui + Vj for x33 to calculate C33 – Z33. We use the relationship Cij – Zij = Cij – Ui + Vj for x23 to calculate C23 – Z23. C23 – Z23 = C23 – U2 + V3 C23 – Z23 = 100 – (-3) + (-100) = 3 C33 – Z33 = C33 – U3 + V3 C33 – Z33 = 100 – (-2) + (-100) = 2 The Most –ve Cij – Zij is “–1” so x11 is the Entering Variable! LCSP

13 © Copyright Andrew Hall, 2002 FOMGT 353 Introduction to Management Science Lecture 18 Slide 13 The Pivot

14 © Copyright Andrew Hall, 2002 FOMGT 353 Introduction to Management Science Lecture 18 Slide 14 Label the Entering Variable with a “+” and the other variables around the Cycle “-”, “+” and “-”. We look for a “Cycle” of at least 3 Basic Variables and the Entering Variable. The Transportation Simplex Pivot x11 is the Entering Variable! LCSP

15 © Copyright Andrew Hall, 2002 FOMGT 353 Introduction to Management Science Lecture 18 Slide 15 The Transportation Simplex Pivot Then work out the max x11 can be increased by without decreasing x12 or x31 below 0. i.e. Min(x12, x31) = LCSP

16 © Copyright Andrew Hall, 2002 FOMGT 353 Introduction to Management Science Lecture 18 Slide 16 The Transportation Simplex Pivot Increase x11 and x32 by 10,000. Decrease x12 and x31 by 10,000. LCSP

17 © Copyright Andrew Hall, 2002 FOMGT 353 Introduction to Management Science Lecture 18 Slide 17 A Reminder…

18 © Copyright Andrew Hall, 2002 FOMGT 353 Introduction to Management Science Lecture 18 Slide 18 The Transportation Simplex Now, we start again calculating the revised values for Ui and Vj. LCSP

19 © Copyright Andrew Hall, 2002 FOMGT 353 Introduction to Management Science Lecture 18 Slide 19 The Transportation Simplex We use the relationship Cij – Ui + Vj = 0 for x11 to calculate V1. C11 – U1 + V1 = 0 => 6 – 0 + V1 = 0 => V1 = -6 LCSP

20 © Copyright Andrew Hall, 2002 FOMGT 353 Introduction to Management Science Lecture 18 Slide 20 The Transportation Simplex We use the relationship Cij – Ui + Vj = 0 for x13 to calculate V3. C13 – U1 + V3 = 0 => 100 – 0 + V3 = 0 => V3 = -100 LCSP

21 © Copyright Andrew Hall, 2002 FOMGT 353 Introduction to Management Science Lecture 18 Slide 21 The Transportation Simplex We use the relationship Cij – Ui + Vj = 0 for x21 to calculate U2. C21 – U2 + V1 = 0 => 4 – U2 + (-6) = 0 => -2 – U2 = 0 => -2 = U2 LCSP

22 © Copyright Andrew Hall, 2002 FOMGT 353 Introduction to Management Science Lecture 18 Slide 22 The Transportation Simplex We use the relationship Cij – Ui + Vj = 0 for x31 to calculate U3. C31 – U3 + V1 = 0 => 5 – U3 + (-6) = 0 => U3 = -1 LCSP

23 © Copyright Andrew Hall, 2002 FOMGT 353 Introduction to Management Science Lecture 18 Slide 23 The Transportation Simplex We use the relationship Cij – Ui + Vj = 0 for x32 to calculate V2. C32 – U3 + V2 = 0 => 10 – (-1) + V2 = 0 => 11 + V2 = 0 => V2 = -11 LCSP

24 © Copyright Andrew Hall, 2002 FOMGT 353 Introduction to Management Science Lecture 18 Slide 24 Another Reminder…

25 © Copyright Andrew Hall, 2002 FOMGT 353 Introduction to Management Science Lecture 18 Slide 25 C22 – Z22 = C22 – U2 + V2 C22 – Z22 = 11 – (-2) + (-11) = 2 We use the relationship Cij – Zij = Cij – Ui + Vj for x22 to calculate C22 – Z22. We use the relationship Cij – Zij = Cij – Ui + Vj for x12 to calculate C12 – Z12. C12 – Z12 = C12 – U1 + V2 C12 – Z12 = 12 – 0 + (-11) = 1 The Transportation Simplex We use the relationship Cij – Zij = Cij – Ui + Vj for x33 to calculate C33 – Z33. We use the relationship Cij – Zij = Cij – Ui + Vj for x23 to calculate C23 – Z23. C23 – Z23 = C23 – U2 + V3 C23 – Z23 = 100 – (-2) + (-100) = 2 C33 – Z33 = C33 – U3 + V3 C33 – Z33 = 100 – (-1) + (-100) = 1 LCSP

26 © Copyright Andrew Hall, 2002 FOMGT 353 Introduction to Management Science Lecture 18 Slide 26 The Transportation Simplex There are no negative Cij – Zij values so stop. This is an optimal solution! We ship 10,000 from Boston to New York, with 10,000 slack capacity in Boston. We ship 10,000 from Hartford to New York, 5,000 from Worcester to New York and 20,000 from Worcester to Montreal. LCSP

27 © Copyright Andrew Hall, 2002 FOMGT 353 Introduction to Management Science Lecture 18 Slide 27 Another Example Starting from the Initial Feasible Solution generated using the Vogel’s Approximation Starting Procedure

28 © Copyright Andrew Hall, 2002 FOMGT 353 Introduction to Management Science Lecture 18 Slide 28 The Transportation Simplex Method – Identify Entering Variable We use the relationship Cij – Ui + Vj = 0 for x11 to calculate V1. C11 – U1 + V1 = 0 => 6 – 0 + V1 = 0 => V1 = -6 VAM

29 © Copyright Andrew Hall, 2002 FOMGT 353 Introduction to Management Science Lecture 18 Slide 29 The Transportation Simplex Method – Identify Entering Variable We use the relationship Cij – Ui + Vj = 0 for x21 to calculate U2. C21 – U2 + V1 = 0 => 4 – U2 + (-6) = 0 => U2 = -2 VAM

30 © Copyright Andrew Hall, 2002 FOMGT 353 Introduction to Management Science Lecture 18 Slide 30 The Transportation Simplex Method – Identify Entering Variable We use the relationship Cij – Ui + Vj = 0 for x13 to calculate V3. C12 – U1 + V3 = 0 => 100 – 0 + V3 = 0 => V3 = -100 VAM

31 © Copyright Andrew Hall, 2002 FOMGT 353 Introduction to Management Science Lecture 18 Slide 31 The Transportation Simplex Method – Identify Entering Variable We use the relationship Cij – Ui + Vj = 0 for x33 to calculate U3. C33 – U3 + V3 = 0 => 100 – U3 + (-100) = 0 => U3 = 0 VAM

32 © Copyright Andrew Hall, 2002 FOMGT 353 Introduction to Management Science Lecture 18 Slide 32 The Transportation Simplex Method – Identify Entering Variable We use the relationship Cij – Ui + Vj = 0 for x32 to calculate V2. C32 – U3 + V2 = 0 => 10 – 0 + V2 = 0 => 10 – 0 + V2 = 0 => -10 = V2 VAM

33 © Copyright Andrew Hall, 2002 FOMGT 353 Introduction to Management Science Lecture 18 Slide 33 We use the relationship Cij – Zij = Cij – Ui + Vj for x12 to calculate C12 – Z12. We use the relationship Cij – Zij = Cij – Ui + Vj for x22 to calculate C22 – Z22. C22 – Z22 = C22 – U2 + V2 C22 – Z22 = 11 – (-2) + (-10) = 3 The Transportation Simplex Method – Identify Entering Variable C12 – Z12 = C12 – U1 + V2 C12 – Z12 = 12 – 0 + (-10) = 2 We use the relationship Cij – Zij = Cij – Ui + Vj for x31 to calculate C31 – Z31. We use the relationship Cij – Zij = Cij – Ui + Vj for x23 to calculate C23 – Z23. C23 – Z23 = C23 – U2 + V3 C23 – Z23 = 100 – (-2) + (-100) = 2 C31 – Z31 = C31 – U3 + V1 C31 – Z31 = 5 – 0 + (-6) = -1 x31 is the Entering Variable! VAM

34 © Copyright Andrew Hall, 2002 FOMGT 353 Introduction to Management Science Lecture 18 Slide 34 Label the Entering Variable with a “+” and the other variables around the Cycle “-”, “+” and “-”. We look for a “Cycle” of at least 3 Basic Variables and the Entering Variable. The Transportation Simplex Pivot x31 is the Entering Variable! VAM

35 © Copyright Andrew Hall, 2002 FOMGT 353 Introduction to Management Science Lecture 18 Slide 35 The Transportation Simplex Pivot Then work out the max x31 can be increased by without decreasing x11 or x33 below 0. i.e. Min(x11, x33) = VAM

36 © Copyright Andrew Hall, 2002 FOMGT 353 Introduction to Management Science Lecture 18 Slide 36 The Transportation Simplex Pivot Increase x31 and x13 by 5,000. Decrease x11 and x33 by 5,000. VAM

37 © Copyright Andrew Hall, 2002 FOMGT 353 Introduction to Management Science Lecture 18 Slide 37 The Transportation Simplex We have seen this before! It is the optimal solution!! We ship 10,000 from Boston to New York, with 10,000 slack capacity in Boston. We ship 10,000 from Hartford to New York, 5,000 from Worcester to New York and 20,000 from Worcester to Montreal. VAM

38 © Copyright Andrew Hall, 2002 FOMGT 353 Introduction to Management Science Lecture 18 Slide 38 If we look at the result of applying the North West Corner method to derive a Basic Feasible Solution. We will see in the following slides, that we need to undertake a more complex pivot involving a cycle which has more than four elements… A More Complex Pivot

39 © Copyright Andrew Hall, 2002 FOMGT 353 Introduction to Management Science Lecture 18 Slide 39 We look for a “Cycle” of at least 3 Basic Variables and the Entering Variable. A cycle which includes the four corners looks good, but if we use this we get 6 Basic Variables! A More Complex Pivot x13 is the Entering Variable! NWC

40 © Copyright Andrew Hall, 2002 FOMGT 353 Introduction to Management Science Lecture 18 Slide 40 So we need to find a “Cycle” of at least 5 Basic Variables and the Entering Variable. A More Complex Pivot We then need to find the minimum leaving variable… So, add 5000 to x13, x32, x21 and deduct 5000 from x33, x22 and x11 to complete the pivot. Remembe r x13 is the Entering Variable! NWC

41 © Copyright Andrew Hall, 2002 FOMGT 353 Introduction to Management Science Lecture 18 Slide 41 This time we can find a simpler pivot involving 3 Basic Variables and the Entering Variable. A More Complex Pivot NWC

42 © Copyright Andrew Hall, 2002 FOMGT 353 Introduction to Management Science Lecture 18 Slide 42 The Transportation Simplex We have seen this before! It is the optimal solution!! We ship 10,000 from Boston to New York, with 10,000 slack capacity in Boston. We ship 10,000 from Hartford to New York, 5,000 from Worcester to New York and 20,000 from Worcester to Montreal. NWC

43 © Copyright Andrew Hall, 2002 FOMGT 353 Introduction to Management Science Lecture 18 Slide 43 Reading and Homework. Read Network Model Algorithms Supplement 5 Section III Read the Assignment Problem Handout.


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