Add transportation problem specifics to the solution

TRANSPORTATION PROBLEM (semicolon or tab separated)1 ^{st} row = supply 1^{st} column = demand All other cells = transport cost per quantity |

In this case, demand should be met by supply against minimal transportation costs. The table below shows demand, supply and the transportation cost per quantity for each supply-demand connection. Note that total demand equals total supply (no requirement as such). With r referring to row, and c to column, and X

Minimize Σ quantity

Σ 1 × X

_{rc}to the transport quantity from Supply_{r}to Demand_{c}, the LP problem can be formulated as:Minimize Σ quantity

_{rc}× cost_{rc}, subject toΣ 1 × X

_{rc}≥ Demand_{c}∀ cc

Σ 1 × X_{rc}≤ Supply_{r}∀ rr

## Demo data

COST PERQUANTITY | SUPPLY | ||||||||||

14 | 15 | 16 | 18 | 9 | 4 | 7 | 15 | 20 | 232 | ||

DEMAND | 40 | 334 | 356 | 336 | 18 | 45 | 795 | 59 | 291 | 89 | 585 |

14 | 285 | 234 | 805 | 283 | 440 | 325 | 957 | 330 | 796 | 600 | |

26 | 431 | 965 | 600 | 241 | 190 | 417 | 117 | 916 | 631 | 436 | |

18 | 763 | 719 | 550 | 396 | 200 | 619 | 689 | 514 | 108 | 874 | |

10 | 484 | 933 | 792 | 607 | 322 | 548 | 851 | 715 | 802 | 272 | |

30 | 816 | 611 | 225 | 79 | 851 | 425 | 841 | 481 | 500 | 903 | |

18 | 899 | 391 | 165 | 155 | 657 | 231 | 757 | 730 | 48 | 282 | |

8 | 922 | 320 | 661 | 38 | 726 | 760 | 150 | 891 | 483 | 407 | |

4 | 316 | 504 | 123 | 587 | 938 | 706 | 382 | 729 | 547 | 310 | |

6 | 751 | 37 | 825 | 754 | 140 | 323 | 229 | 592 | 571 | 914 | |

6 | 117 | 706 | 414 | 674 | 275 | 703 | 395 | 315 | 115 | 891 | |

12 | 498 | 898 | 858 | 498 | 187 | 411 | 717 | 353 | 560 | 368 | |

18 | 169 | 470 | 643 | 433 | 132 | 652 | 898 | 719 | 555 | 444 | |

2 | 840 | 336 | 860 | 577 | 112 | 292 | 422 | 294 | 411 | 474 | |

20 | 935 | 569 | 443 | 639 | 523 | 549 | 60 | 415 | 409 | 791 | |

32 | 753 | 614 | 610 | 426 | 136 | 578 | 118 | 335 | 751 | 494 | |

20 | 83 | 572 | 108 | 570 | 241 | 946 | 651 | 762 | 790 | 40 | |

36 | 167 | 838 | 963 | 299 | 362 | 90 | 843 | 39 | 50 | 45 | |

28 | 517 | 657 | 914 | 79 | 678 | 811 | 173 | 364 | 930 | 593 | |

2 | 746 | 440 | 747 | 392 | 946 | 456 | 560 | 462 | 798 | 462 |

## Optimal quantities: total costs are 102587

OPTIMALQUANTITIES | SUPPLY | ||||||||||

14 | 15 | 16 | 18 | 9 | 4 | 7 | 15 | 20 | 232 | ||

DEMAND | 40 | 4 | 9 | 2 | 2 | 23 | |||||

14 | 1 | 9 | 4 | ||||||||

26 | 26 | ||||||||||

18 | 18 | ||||||||||

10 | 10 | ||||||||||

30 | 16 | 14 | |||||||||

18 | 18 | ||||||||||

8 | 8 | ||||||||||

4 | 4 | ||||||||||

6 | 6 | ||||||||||

6 | 6 | ||||||||||

12 | 12 | ||||||||||

18 | 7 | 11 | |||||||||

2 | 2 | ||||||||||

20 | 7 | 13 | |||||||||

32 | 32 | ||||||||||

20 | 20 | ||||||||||

36 | 36 | ||||||||||

28 | 28 | ||||||||||

2 | 2 |

With this larger test set you will notice run time difference between the two different solver routines (1A versus 1B), approx. 6 seconds versus 0.2 seconds.

Add network problem specifics to the solution

NETWORK FLOW PROBLEM (semicolon or tab separated) | |

NODES | ARCS |

NodeID; Supply_Min*; Supply_Max*; Costs_In; Costs_Out; Throughput_Min; Throughput_Max | ArcID; From_NodeID; To_NodeID; Costs; Flow_Min; Flow_Max |

## Intro

You can define a netwok consisting of nodes and arcs. The decision variables in this type of problem are the arc quantities. Each**arc has a cost per quantity**(e.g. transportation costs), a

**lowerbound (minimum flow quantity)**and

**upperbound (maximum flow quantity)**. Each

**node has supply quantity**(demand = negative supply).

## Demo case

The graph below shows a minimum cost flow problem. The arcs are labeled with pairs of numbers: the first number is the capacity and the second number is the cost. The number in parentheses next to a node represent supply or demand. Node 0 is a supply node that supplies 20, while nodes 3 and 4 are demand nodes (or: sink nodes) that demand 5 and 15 respectively (demand = negative supply).## Advanced node modeling options

For modeling convenience this web app offers advanced node modeling options:- Each node has a
**cost per incoming and outgoing quantity**(e.g. inbound and outbound warehouse handling costs). - Each node has
**flexible supply/demand**(supply upper and lowerbound), so - for example - a node may supply between 0 and 1000, or have a demand between 6 and 10 (entered as a negative supply between -10 and -6). If demand/supply is not flexible, then enter the same value for both Min and Max. - Each node has a
**throughput upper and lowerbound**. Throughput could be either based on incoming or outgoing quantity. We have (arbitrarily) chosen incoming quantity as the basis.

## General network modeling remarks

- By adding dummy nodes and arcs you can create completely balanced systems, and make a model 'better readible'.
- Other network flow software may not offer advanced node modeling options.
- You can create node upper and lowerbounds by splitting a node in two nodes, adding a zero costs arc in between, and setting upper and lowerbound at this arc (connecting all original incoming arcs to the first node, all original outgoing arcs to the second).
- You can create a node with flexible demand/supply by adding a dummy node that 'eats' any non-used over-supply at zero costs (creating a balanced system in which total supply equals total demand).
- Incoming/outgoing node costs can be transferred/added to all incoming/outgoing arcs of the node.
- A network model can be used to
**model transitions in time**, with an arc representing a transition from time to time+1.

**Remark**: Min(imum) and Max(imum) are called Lowerbound (LB) and Upperbound (UB) in LP terminology.

LP PROBLEM (semicolon or tab separated) |

## Standard LP problem formulation

Maximize ΣC

You can also minimize and/or use ≥ and = as restriction type as shown by the demo cases. See also: wikipedia: Linear Programming_{i}× X_{i}, subject to Σ (A_{ij}× X_{i}) ≤ Bj ∀ j , and X_{i}≥ 0 ∀ ii

i

## Simplex tableau legend

- Top row contains the cost factors.
- 1
^{st}cell = MIN (minimize) or MAX (maximize). MIN is default in logistics. Though the standard LP problem formulation is about maximizing profit (MAX), logistics problems are usually about minimizing costs (MIN). - 2
^{nd}cell = initial goal value = always 0. - 3
^{rd}and next cells = costs factors C_{i}. Each of these column 'belongs to' one decision variable (X_1, X_2,...). - All other rows formulate the constraints.
- 1
^{st}cell = constraint type <= or >= or = - 2
^{nd}cell = constraint value B_{j} - 3
^{rd}and next cells = values of A_{ij}(optional: can be left empty if zero). Each of these columns 'belongs to' one decision variable (X_1, X_2,...)

*Normally you need to setup - so called - slack variables as well (read: dummy variables required for the Simplex method to work).*

Do

Do

__not__enter slack variables! They are created automatically.**The tableau is prefilled with the transportation problem demo case.**Putting the demo case table next to this Simplex tableau you will recognize the numbers. Note that instead of X

_{rc}(20 rows × 10 columns) all has been translated into X

_{j}(200 variables sitting on a single row), so X

_{r=6,c=3}has become X

_{53}that has a value of 16 in the optimal solution. If you run "Solve network minimum cost flow problem" the tableau is filled with the translation of the network flow problem demo case.

The solution will appear here, once run.