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Chapter 3 Transportation Problems

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QEM - Chapter 3 Design of Transport Networks Situation: Location of warehouses and customers are given Supply and demand given Example: 3 warehouses and 4 customers Transportation cost per unit from i to j Total demand must be = total supply Factory orcustomer warehouseV1V2V3V4Production F F F Demand Transportation Problem: Model & LP (c) Prof. Richard F. Hartl Kapitel 3 / 2

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(c) Prof. Richard F. HartlQEM - Chapter 3 Kapitel 3/3 Solution in the Uncapacitated case No capacity constraint solution using column minimum procedure customer FactoryV1V2V3V4Capacity used F F F Demand = = 50 Total cost =15 * * * * 4 = 295

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(c) Prof. Richard F. HartlQEM - Chapter 3 Kapitel 3/4 Capacity constraints Not all customers will be delivered from their closest factory! Solution as a special LP problem - starting (basic feasible) solution - iteration (modi, stepping stone)

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(c) Prof. Richard F. HartlQEM - Chapter 3 Kapitel 3/5 Capacity constraints Total capacity = total demand Customer FactoryV1V2V3V4capacity F F F Demand

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(c) Prof. Richard F. HartlQEM - Chapter 3 Kapitel 3/6 Constructive Heuristics (Basic Feasible Solution) Heuristics: Simple to apply, Simple to understand Reasonably good solutions Optimality not guaranteed Examples: Column minimum (Greedy) Vogel approximation (Regret)

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(c) Prof. Richard F. HartlQEM - Chapter 3 Kapitel 3/7 Column Minimum Algorithm 0. Initialization: empty transport tableau. No rows or columns are deleted 1.Proceed with the first (not yet deleted) column (from left to right) 2.In choose the cell with the smallest unit cost c ij and choose transportation quantity x ij maximal. 3.If column resource depleted delete column j, OR if row resource depleted delete row i. 4. Just one row or columd not deleted all cells in this row or colum get maximal transportation quantity Otherwise continue with 1. EXAMPLE

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(c) Prof. Richard F. Hartl QEM - Chapter 3Kapitel 3/8 Column Minimum Algorithm i / j 1234Capacity Demand Total cost = 15 * * * * * * 8 = 470 Just one colums not deleted ALGORITHM

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SS 2011EK Produktion & Logistik Kapitel 3/9 Vogel Approximation 0.Initialization: empty transport tableau. No rows or columns are deleted 1. In each row and column (not yet deleted) compute opportunity cost = difference between the smallest and the second smallest c ij (not yet deleted). 2. Where this opportunity cost is largest choose smallest c ij and make x ij maximal. 3. If column resource depleted delete column j, OR if row resource depleted delete row i. 4. Just one row or columd not deleted all cells in this row or colum get maximal transportation quantity Otherwise continue with 1.

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SS 2011 EK Produktion & LogistikKapitel 3/10 Vogel Approximation i / j 1234capacity Nachfrage Total cost = 15 * * * * * * 8 = 445 Column deleted recompute row differences Row deleted recompute column differences Jut one row not deleted (besser als 470 zuvor)

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QEM - Chapter 3 General formulation: m supplyers with supply s i, i = 1, …, m n customers with demand d j, j = 1, …, n Transportation cost c ij per unit from i to j, i = 1, …, m; j = 1, …, n variable: Transportation quantity x ij from i to j LP-Formulation: Transportation cost Supply Demand Non negativity i = 1, …, m j = 1, …, n i = 1, …, m; j = 1, …, n Data must satisfyTotal demand = Total supply (c) Prof. Richard F. Hartl Kapitel 3 / 11

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QEM - Chapter 3 In example: Supply: x 11 + x 12 + x 13 + x 14 = 25 (i=1) x 21 + x 22 + x 23 + x 24 = 25 (i=2) x 31 + x x 33 + x 34 = 50 (i=3) : Damand: x 11 + x 21 + x 31 = 15 (j=1) x 12 + x 22 + x 32 = 20 (j=2) x 13 + x 23 + x 33 = 30 (j=3) x 14 + x 24 + x 34 = 35 (j=4) Non negativity: x ij 0 für i = 1, …, 3; j = 1, …, 4 K = (10x 11 +5x 12 +6x x 14 ) + (x 21 +2x 22 +7x 23 +4x 24 ) + (9x 31 +x 32 +4x 33 +8x 34 ) min (c) Prof. Richard F. Hartl Kapitel 3 / 12

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QEM - Chapter 3 Solution: As LP (o.k. but less efficient) In each column exactly 2 of the m + n elements are 0 Make use of special structure: Transportation simplex or network simplex Starting solution Iteration (stepping stone) (c) Prof. Richard F. Hartl Kapitel 3 / 13

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QEM - Chapter 3 Exakt Algorithm: MODI, Stepping Stone Equivalent to simplex method Start with basic feasible solution ( m+n-1 basic variables) Iteration step: Initalize the tableau i\j12…nsisi uiui 1 c 11 c 12 … c 1n s1s1 u1u1 2 c 21 c 22 … c 2n s2s2 u2u2 ………… …… m c m1 c m2 … c mn smsm umum djdj d1d1 d2d2 …dndn vjvj v1v1 v2v2 …vnvn (c) Prof. Richard F. Hartl Kapitel 3 / 14

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QEM - Chapter 3 For all non basic variables compute c ij – u i – v j (reduced cost coefficient in objective function). New (entering) BV where this coefficient is most negative. If all coefficients non-negative optimal solution. Compute dual varables u i und v j [MODI] c ij = u i + v j if x ij is basic variable Since u i and v j are not unique, normalize one of these dual varables = 0. (Choose row or colums with most basic varaibles easier computation of u i and v j ) Increase new BV and perform chain reaction (donor cell, recipient cell). Transportation quantities in each row and column must remain same. The BV, which first becomes zero, leaves the basis. [stepping stone] Compute new basic solution (perform chain reaction). (c) Prof. Richard F. Hartl Kapitel 3 / 15

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QEM - Chapter 3 Example i\j1234sisi uiui djdj vjvj For didactical raesons we choose poor starting solution, because after Vogel only few steps (if any) can be demonstrated (c) Prof. Richard F. Hartl 16

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QEM - Chapter 3 Total cost = 10*15 + 5*10 + 2*10 + 7*15 + 4*15 + 8*35 = 665. Check equality of primal and dual objective (optional): The most negative cost ceofficient c ij – u i – v j of a NBV is -7 for x 24 new basic variable x 24. Chain rection: increase new BV (starting from 0) by investigate consequences for other BV (in some row and some column must be subtracted) The BV, that constrains most is the leaving BV. K = 25*0 + 25*(-3) + 50*(-6) + 15* *5 + 30* *14 = 665 (c) Prof. Richard F. Hartl Kapitel 3 / 17

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QEM - Chapter 3 Chain reaction: i\j1234sisi uiui djdj vjvj New BV x 24 increases from 0 to. Add + odor - for some other BV. If x 24 increases by, x 23 and x 34 will decrease by, and x 33 increases by. For = 15 BV x 23 becomes 0 BV x 23 leaves. + K = 665 – 7 * = 665 – 7*15 = 560 (c) Prof. Richard F. Hartl Kapitel 3 / 18

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QEM - Chapter 3 New BV x 24 gets value = 15 chain reaction x 34 = = 20 x 33 = = 30 x 23 is no BV anymore, all other BV remain same i\j1234sisi uiui djdj vjvj Next iteration: K = 560 – 6 * = 560 – 6*10 = 500 (c) Prof. Richard F. Hartl Kapitel 3 / 19

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QEM - Chapter 3 Next iteration: i\j1234sisi uiui djdj vjvj K = 500 – 3 * = 500 – 3*5 = 485 (c) Prof. Richard F. Hartl Kapitel 3 / 20

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QEM - Chapter 3 Next iteration: i\j1234sisi uiui djdj vjvj K = 485 – 2 * = 485 – 2*20 = 445 (c) Prof. Richard F. Hartl Kapitel 3 / 21

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QEM - Chapter 3 Next iteration: i\j1234sisi uiui djdj vjvj All coefficients non- negative optimal solution found. Basic variables: x 13 = 25 x 21 = 15 x 24 = 10 x 32 = 20 x 33 = 5 x 34 = 25 Total cost: K = 445 (c) Prof. Richard F. Hartl Kapitel 3 / 22

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QEM - Chapter 3 TP is a LP-Problem with equality constraints dual variables u i azw. v j can also be negative (free variables). Sensitivity analysis If small enough, then the dual variables u i and v j do not change cost change by (u i + v j ): From duality theory follows Clearly some s i AND soem d j must change at the same time; Otherwise total demand total capacity s i s i + for some i and d j d j + for some j K K + (u i + v j ) (c) Prof. Richard F. Hartl Kapitel 3 / 23

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QEM - Chapter 3 Example: i\j1234sisi uiui djdj vjvj Change of data? s 1 s 1 + and d 2 d 2 + Optimal cost change to K = (u 1 + v 2 ) = ; i.e. cost decrease! (this paradoxon can occur if some u i and/or v j are negative. In most cases cost will increase.) (c) Prof. Richard F. Hartl Kapitel 3 / 24

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QEM - Chapter 3 How large can be before basis changes: similar to stepping stone i\j1234sisi uiui djdj vjvj K = Clearly x 33 first becomes 0, if increases. Hence upper bound 10. Similar for negative. Here x 34 first becomes 0, if decreases. Hence lower bound -6. Check: K = 1*10+2*(12+ ) + 1*(13+ ) +6*(11- ) +3*(10- ) +5*(6 + ) = (c) Prof. Richard F. Hartl Kapitel 3 / 25

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