The Stepping Stone and the Modified Distribution Method (MODI) Stepping Stone: Procedure for finding optimal transportation tableau.

Given the s = 3 supply and d = 3 demands tableau below, first create feasible tableau by using the Northwest, VAM, Minimum cell, or Russell method. In an s x d tableau, the number of basic (allocated) cells must be s + d – 1 even if a 0 must be placed into a cell to satisfy requirement,

Initial tableau

From To A B C Supply 1 6 8 10 150 2 7 11 11 175 3 4 5 12 275

Demand 200 100 300 600 Feasible tableau using Northwest Corner Rule

From To A B C Supply 1 6 (150) 8 10 150 2 7 (50) 11 (100) 11 (25) 175 3 4 5 12 (275) 275

Demand 200 100 300 600 Notice that there are 3 + 3 – 1 = 5 allocated cells of the 9 cells; and 4 non-basic cells. Closed loops are formed starting in an empty cell; go to an allocated cell, then another and another until returning to empty cell. Never go diagonal and you may pass over cells. Never visit a cell twice. Value = 150 * 6 + 50 * 7 + 100 * 11 + 25 * 11 + 275 * 12 = 5925 Example To evaluate the unallocated cell 3A, the loop is 3A 3C 2C 2A with values

4 – 12 + 11 – 7 = –4 meaning that for every unit shifted (give and take) into 3A and 2C from 3C and 2A results in a savings of 4. When shifting, feasibility constraints of supply and demand must be maintained. Look at the giving cells 3C with 275 and 2A with 50. Shift the minimum 50 into the receiving cells 3A and 2C to create the tableau below.

From To A B C Supply

1 6 (150) 8 10 150 2 7 11 (100) 11 (75) 175 3 4 (50) 5 12 (225) 275

Demand 200 100 300 600 Value = 150 * 6 + 50 * 4 + 100 * 11 + 75 * 11 + 225 * 12 = 5725. The delta value of the tableaus is 5925 – 5725 = 200 = 50 * 4 The stepping stone evaluation is done for all empty cells and the best cell improvement is chosen for the change. Notice cell 3A entered into the basis and cell 2A left. Loops always contain the same number of + and – cells alternating as one completes the loop. In evaluating cell 3B, the loop is 3B 3C 2C 2B. Let’s evaluate cell 2A with loop 2A 2C 3C 3A with 7 – 11 + 12 – 4 = 4. Positive quantities mean no improvement. When all unallocated (non-basic) cells evaluate to positive quantities, the tableau is optimal. If the problem is one of maximization, use the same procedure except that positive quantities imply improvement. MODI Paths or loops in this method are determined mathematically. The tableau is modified with

u (row) and v (column) variables. Allocated cell costs cij = ui + vj.

Feasible tableau using Northwest procedure

va vb vc From/To A B C Supply

u1 1 6 (150) 8 10 150 u2 2 7 (50) 11 (100) 11 (25) 175 u3 3 4 5 12 (275) 275 Demand 200 100 300 600

Cell formulas for allocated or basis cells are: u1 + va = c1a;

u2 + va = c2a; u2 +vb = c2b; u2 + vc = c2c u3 + vc = c3c

The c-values are the respective cells cost. With 5 equations and 6 unknowns, we arbitrarily assign one of the unknowns, e.g., u1 = 0 (but more efficient to assign u2 to 0 since more allocated cells in the u2 row)*** Evaluate Cells 1A: u1 + va = 6 or 0 + va = 6 => va = 6 2A: u2 + va = 7 or u2 + 6 = 7 => u2 = 1; 2B: u2 + vb = 11 or 1 + vb = 11 => vb = 10; 2C: u2 + vc = 11 or 1 + vc = 11 => vc = 10; 3C: u3 + vc = 12 or u3 + 10 = 12 => u3 = 2 Evaluate empty cells with the formula Xij = cij – ui - vj

1B: X1b = c1b – u1 – vb = 8 – 0 – 10 = –2 same value using stepping stone for cell 1B 1C: X1c = c1c – u1 – vc = 10 – 0 – 10 = 0 same value using stepping stone for cell 1C 3A: X3a = c3a – u3 – va = 4 – 2 – 6 = –4 same value using stepping stone for cell 3A 3B: X3b = c3b – u3 – vb = 5 – 2 – 10 = –7 same value using stepping stone for cell 3B Re-visit stepping stone evaluations 1B: 1B 1A 2A 2B => 8 – 6 + 7 – 11 = –2 1C: 1C 1A 2A 2C =>10 – 6 + 7 – 11 = 0 3A: 3A 3C 2C 2A => 4 – 12 + 11 – 7 = –4 3B: 3B 3C 2C 2B => 5 – 12 + 11 – 11 = –7 Thus cells 3B and 1B can improve solution. The procedure is then repeated until all empty cell evaluations are positive. *** Returning to initial tableau and using a different arbitrary assignment leads to the following: Feasible tableau using Northwest procedure

va= 7 vb=11 vc=11 From/To A B C Supply

u1 = -1 1 6 (150) 8 10 150 u2 = 0 2 7 (50) 11 (100) 11 (25) 175 u3 = 1 3 4 5 12 (275) 275

Demand 200 100 300 600 Cell formulas for allocated or basis cells are: u1 + va = c1a;

u2 + va = c2a; u2 +vb = c2b; u2 + vc = c2c u3 + vc = c3c

The c-values are the respective cells cost. With 5 equations and 6 unknowns, we arbitrarily assign one of the unknowns, e.g. u2 = 0. Then u2 = 0 => va = 7; vb = 11, vc = 11, u1 + va = 6 => u1 = –1; u3 + vc = 12 => u3 = 1.

Evaluating unallocated cells 1B => X1b = 8 –7 –(–1) = 2 1C => 10 – (–1) – 11 = 0 3A => 4 – 1 – 7 = –4 3B => 5 – 1 – 11 = –7 show the same as stepping stone and the same as assigning u1 to 0. Alternate optimal solutions Notice alternate solution is available with cell 1C = 0, but with the same optimal value. Degeneracy If number of unallocated cells is less than the number of rows R + the number of columns C – 1, degeneracy has occurred (R + C – 1). The solution is to choose an empty cell and assign it a very small value. The selection is a bit tricky to ensure the evaluation of the other empty cells. For example, try evaluating any empty cell in the tableau below using the stepping stone approach. Similarly try the MODI. Note that R + C – 1 = 3 + 3 – 1 = 6 ≠ 5.

From/To A B C Supply 1 6 8 10 (225) 150 2 7 11 (100) 11 175 3 4 (200 5 12 (75) 275

Demand 200 100 300 600 Prohibited Assignments There may be times when a specified Supply is prohibited from satisfying a specified Demand. We can revert to the Big M method for the cost When Supply ≠ Demand Dummy supply or demand can be set up for the excess and whatever is assigned to the dummy demand does not get received, and to the dummy supply does not get shipped. Example Supply > Demand tableau From To A B C Dummy Supply

1 7 6 11 0 150 2 12 8 9 0 175 3 8 10 12 0 275

Demand 0 100 300 50 650 Example Demand > Supply tableau

From To A B C Supply 1 7 6 11 250 2 12 8 9 150 3 8 10 12 275

Dummy 0 0 0 100 Demand 200 175 300 675

Maximizing When maximizing rather than minimizing, similar to the assignment algorithm, subtract all cell profits from the maximum overall profit and use the same minimizing procedures. Transshipments Each transshipment locality becomes both a supply and a demand center.