The Transportation Problem. Experience the Joy! Feel the Excitement! Share in the Pleasure!

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Nickolas O’Neal’
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1 The Transportation Problem Experience the Joy! Feel the Excitement! Share in the Pleasure!

2 The Problem A company manufactures a single product at each of m factories. i has a capacity of S i per month. There are n warehouses receiving this product. The demand at warehouse j is D j. It cost factory i, c ij dollars to ship one unit to warehouse j. How many units should each factory send to each warehouse in order to minimize the total transportation costs? This is a really neat problem.

3 More of the Problem Shipping costs per unit shipped

4 The LP Formulation Let x ij = the number of units sent from factory i to warehouse j Min z = m i= subject to: n j= m i= n j= c x x S i =,,..., m x = D j =,,..., n x ij ij ij ij ij i 0 j 4

5 Initial Tableau Eq basic z no. Var x ij... z i... z m+j...rhs 0 z -c ij -M -M 0 i z i 0 s i m+j z m+j 0 d j 5

6 The Objective Function Min z = x + x + 0x + x 4 + 0x + x + 4x + 0x 4 + 4x + x + 5x + x 4 6

7 Constraints Min z = x + x + 0x + x 4 + 0x + x + 4x + 0x 4 + 4x + x + 5x + x 4 Subject to: x + x + x + x 4 = 0 x + x + x + x 4 = x + x + x + x 4 = constraints

8 Constraints Min z = x + x + 0x + x 4 + 0x + x + 4x + 0x 4 + 4x + x + 5x + x 4 Subject to: x + x + x + x 4 = 0 x + x + x + x 4 = x + x + x + x 4 = constraints x + x + x = 6 x + x + x = 5 x + x + x = x 4 + x 4 + x 4 = 8 constraints 8

9 A Redundant Constraint Min z = x + x + 0x + x 4 + 0x + x + 4x + 0x 4 + 4x + x + 5x + x 4 Subject to: x + x + x + x 4 = 0 x + x + x + x 4 = x + x + x + x 4 = constraints x + x + x = 6 x + x + x = 5 x + x + x = x 4 + x 4 + x 4 = 8 constraints

10 A Redundant Constraint Min z = x + x + 0x + x 4 + 0x + x + 4x + 0x 4 + 4x + x + 5x + x 4 Subject to: x + x + x + x 4 = 0 x + x + x + x 4 = x + x + x + x 4 = constraints x + x + x = 6 x + x + x = 5 x + x + x = x 4 + x 4 + x 4 = 8 constraints Isn t one of those constraints redundant? 0

11 The Northwest Corner Starting Solution

12 The Northwest Corner Starting Solution

13 The Northwest Corner Starting Solution

14 The Northwest Corner Starting Solution

15 The Northwest Corner Starting Solution

16 The Northwest Corner Starting Solution

17 The Northwest Corner Starting Solution Cost = $ 8

18 The Northwest Corner Starting Solution- Basic cells Cost = $ 8 8

19 The Northwest Corner Starting Solution Non-basic cells Cost = $ 8

20 0-+- = - Ring-around-the rosie-a Step : Find a non-basic cell to become basic. 0

21 Ring-around-the rosie-b Step : Find a non-basic cell to become basic.

22 Ring-around-the rosie-c = Step : Find a non-basic cell to become basic.

23 Ring-around-the rosie-d Step : Find a non-basic cell to become basic. -5

24 Ring-around-the rosie-a -0+- =

25 Ring-around-the rosie-b

26 Ring-around-the rosie-a =

27 Ring-around-the rosie-b

28 Ring-around-the rosie-c

29 Ring-around-the rosie-d

30 Ring-around-the rosie Select the most negative 0

31 Ring-around-the rosie = Step : From among the negative cells - select the one that goes to zero first.

32 Ring-around-the rosie = -5 Cost = $8-5 x 4 = $ Step : Generate the new basic solution by adding and subtracting the minimum cell quantity to each affected cell.

33 Iteration The Problem Solution demo

34 Iteration The Problem Solution demo

35 Iteration The Problem Solution demo

36 Iteration The Problem Solution demo

37 Iteration The Problem Solution demo

38 Iteration The Problem Solution demo

39 Iteration The Problem Solution

40 Iteration The Problem Solution -continued

41 Iteration The Problem Solution Cost = $08-6 x = $

42 Iteration The Problem Solution

43 Iteration The Problem Solution Cost = $0 - x 5 = $

44 Iteration 4 The Problem Solution Cost = 5 - x =

Iteration 4 The Problem Solution -6 4 0 0 + + Cost = 5 - x = 0

45 Iteration 4 The Problem Solution Cost = 5 - x = This appears to be optimal! 45

46 A Question? =00 demand =0 What happens if m i= S i n j= D j 46

47 m i= S i > n j= D j Answer -Add a Dummy Column Month Dummy M M M

48 m i= S i < n j= D j Or Add a Dummy Row Dayton New York Columbus Excess capacity DP&L First Energy Long Island Light Dummy Peak demand Power requirements in 000 KWH 48

49 A Paradox? - Less cost more? Cost = $ A B supply s 0 Cost = $5 A B supply s 0 4

50 The Transportation Problem This has been a fast paced ride through a wondrous land in which the movement of a single commodity from source to destination is optimized. 50

Linear Programming Applications. Transportation Problem

Linear Programming Applications Transportation Problem 1 Introduction Transportation problem is a special problem of its own structure. Planning model that allocates resources, machines, materials, capital