Computational Methods for Management and Economics Carla Gomes

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Computational Methods forManagement and
EconomicsCarla Gomes

• Module 8b
• The transportation simplex method

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The transportation and assignment problems

• Special types of linear programming problems.
• The structure of these problems leads to
  algorithms streamlined versions of the simplex
  method - more efficient than the standard simplex
  method.

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The transportation problem
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Setting up the tableau for the general simplex
method

• Convert minimization problem into maximization
  problem
• Big M or 2 phase method to introduce m n
  artificial variables (equality constraints)
• Find an initial BF solution

Insert table 8.13
Noteall entries not shown are 0.
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Iterations in the general simplex method

• Optimality test and step 1 of iteration
• Select an entering basic variable ? knowing
  current row 0 (which is obtained by subtracting a
  certain multiple of another row from the
  preceding row 0).
• Step 2 (leaving basic variable) ? identifies
  which basic variable reaches zero as the entering
  variable is increased (which is done by comparing
  the current coefficients of the entering basic
  variable and the corresponding rhs).
• Step 3 (determine new BF solution) ? subtracting
  certain multiples of one row from other rows

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Iterations in the general simplex method
applied to transportation problem

• General form of row 0
• ui multiple of the original row i that has been
  subtracted (directly or indirectly) from original
  row 0.
• vj multiple of the original row i that has been
  subtracted (directly or indirectly) from original
  row 0.
• cij ui vj 0 if it is a basic variable
  otherwise it is interpreted as the rate at which
  Z will change as xij is increased.

Insert table 814
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Transportation simplex method
Does the transportation simplex method do all
the general simplex method operations?

• NO!
• No artificial variables are needed ? simple
  procedure is used to identify the initial BF
  solution.
• Row 0 can be obtained without using any other row
  a simple procedure allows us to calculate the
  values of ui and vi directly.
• Basically by solving a simple procedure that
  allows us to find the solution to the set of
  equations for the basic variables that have a
  coefficient 0 in row 0, i.e., cij ui vj0.
• The leaving variable can also be identified in a
  simple way due to the special structure of the
  problem, without explicitly using the
  coefficients of the new entering basic variable.
• The new BF solution can also be identified very
  easily.

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Transportation simplex method

• Much faster than standard simplex method
• standard simplex method
• mn1 rows
• (m1)(n1) mn m n 1 columns
• transportation simplex method
• m 2 rows
• n 2 columns

Assume m 10 n 100 and see the difference!
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Tableau for Transportation simplex method
Insert table 8.15
() Because there are mn equations a BF solution
in this case only has mn-1 basic variables
the other variable can be obtained as a function
of the mn-1 variables
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Transportation simplex method

• Initialization finding the initial BF solution
• Optimality test
• Iteration

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General procedure for construction of an initial
basic feasible solution for TP

1. From the rows and columns still under
   consideration select the next basic variable
   (allocation) according to some criterion
2. Make that allocation the largest possible (the
   smallest of the remaining supply in its row vs.
   remaining demand in its column)
3. Eliminate the row or column that zeroed the
   corresponding supply/demand (in case of tie,
   randomly pick between the row vs. column this is
   a case of a degenerate basic solution, i.e, a
   basic variable with value 0).
4. If only one row or column remains under
   consideration, then the procedure is complete by
   selecting every remaining variable associated
   with that row or column to be basic with the only
   feasible allocation. Otherwise return to step 1.

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Criteria for selecting next basic variable

• There are several methods. Here are three
• Northwest Corner Method
• Minimum Cost Method
• Vogels Method

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1. Northwest Corner Method

• To find the bfs by the NWC method
• Begin in the upper left (northwest) corner of the
  transportation tableau and set x11 as large as
  possible (here the limitations for setting x11 to
  a larger number, will be the demand of demand
  point 1 and the supply of supply point 1. Your
  x11 value can not be greater than minimum of this
  2 values).

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According to the explanations in the previous
slide we can set x113 (meaning demand of demand
point 1 is satisfied by supply point 1).
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After we check the east and south cells, we saw
that we can go east (meaning supply point 1 still
has capacity to fulfill some demand).
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After applying the same procedure, we saw that we
can go south this time (meaning demand point 2
needs more supply by supply point 2).
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Finally, we will have the following bfs, which
is x113, x122, x223, x232, x241, x342
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2. Minimum Cost Method

• The Northwest Corner Method dos not utilize
  shipping costs. It can yield an initial bfs
  easily but the total shipping cost may be very
  high. The minimum cost method uses shipping costs
  in order come up with a bfs that has a lower
  cost. To begin the minimum cost method, first we
  find the decision variable with the smallest
  shipping cost (Xij). Then assign Xij its largest
  possible value, which is the minimum of si and dj

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• After that, as in the Northwest Corner Method we
  should cross out row i and column j and reduce
  the supply or demand of the noncrossed-out row or
  column by the value of Xij. Then we will choose
  the cell with the minimum cost of shipping from
  the cells that do not lie in a crossed-out row or
  column and we will repeat the procedure.

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An example for Minimum Cost MethodStep 1 Select
the cell with minimum cost.
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Step 2 Cross-out column 2
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Step 3 Find the new cell with minimum shipping
cost and cross-out row 2
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Step 4 Find the new cell with minimum shipping
cost and cross-out row 1
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Step 5 Find the new cell with minimum shipping
cost and cross-out column 1
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Step 6 Find the new cell with minimum shipping
cost and cross-out column 3
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Step 7 Finally assign 6 to last cell. The bfs is
found as X115, X212, X228, X315, X334 and
X346
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3. Vogels Method

1. Begin with computing each row and column a
   penalty. The penalty will be equal to the
   difference between the two smallest shipping
   costs in the row or column.
2. Identify the row or column with the largest
   penalty. Find the first basic variable which has
   the smallest shipping cost in that row or column.
   Assign the highest possible value to that
   variable, and cross-out the row or column as in
   the previous methods.
3. Compute new penalties and use the same procedure.

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An example for Vogels MethodStep 1 Compute the
penalties.
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Step 2 Identify the largest penalty and assign
the highest possible value to the variable.
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Step 3 Identify the largest penalty and assign
the highest possible value to the variable.
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Step 4 Identify the largest penalty and assign
the highest possible value to the variable.
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Step 5 Finally the bfs is found as X110, X125,
X135, and X2115
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Russells method

1. For each row and column remaining under
   consideration determine its ui the largest cij
   still in that row, and its vj -- the largest cij
   still in that column.
2. For each variable xij not previous selected in
   these rows and columns calculate ?ij cij - ui -
   vj.
3. Select the variable xij with the largest (in
   absolute terms) negative ?ij (ties my be broken
   randomly).

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An example for Russells MethodStep 1 Compute
uis and vjs for each row and column .
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An example for Russells MethodStep 2 For each
variable xij not previous selected in these rows
and columns calculate ?ij cij - ui - vj.
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An example for Russells MethodStep 2 For each
variable xij not previous selected in these rows
and columns calculate ?ij cij - ui - vj.
Note v1 and v3 changed
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An example for Russells MethodStep 2 For each
variable xij not previous selected in these rows
and columns calculate ?ij cij - ui - vj.
?11 16 - 22 19 - 25 ?21 14 - 19 19 -
24 ?31 19 - M 19 M
?12 16 - 22 19 25 ?22 14 - 19 19
24 ?32 19 - M 19 - M
?13 13 - 22 20 29 ?23 13 - 19 20
26 ?33 20 - M 20 M
?14 22 - 22 23 23 ?24 19 - 19 23
23 ?34 23 - M 23 M
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Russells method Interactive Solver
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Initial Basic Feasible Solution
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Comparison of different methods for finding
initial basic feasible solution

• NW corner rule very simple to implement and
  therefore fast. However it doesnt factor in the
  costs.
• The minimum cost method and Vogels method take
  into consideration the costs and they are simple
  to compute
• Russells method also takes into consideration
  the costs and it closely patterns the decision
  criterion used by the transportation simplex
  method to pick the entering variable.()
• Nevertheless it is not clear which method
  performs better on average, which may depend on
  the particular class of problem

() In particular ?ij cij - ui vj estimate
the relative values of cij - ui vj
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Transportation simplex method

• Initialization finding the initial BF solution
• Optimality test how do we check if it is the
  optimal solution?
• Iteration

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Transportation simplex methodOptimality Test

• Very similar to the standard simplex method we
  check if all the cij - ui vj are non-negative.

Optimality test A BF solution is optimal iff
cij - ui vj? 0 for every (i,j) such that xij
is non-basic .
Note the cij - ui vj correspond to the
coefficients in row 0.
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Transportation simplex methodOptimality Test

• So what do we have to do?
• calculate the uis and vjs for the current BF
  solution
• Use the values obtained in 1 to calculate the cij
  - ui vj for the non-basic variables.

Optimality test A BF solution is optimal iff
cij - ui vj? 0 for every (i,j) such that xij
is non-basic .
Note the cij - ui vj correspond to the
coefficients in row 0.
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Calculation of the uis and vjs for the
current BF solution

• We know that when a variable is basic its
  corresponding row 0 coefficient is zero
• So cij - ui vj 0 ?? cij ui vj for all the
  basic variables
• Given that there are m n 1 variables, there
  are m n 1 of these equations
• Since the number of unknowns (uis and vjs ) is
  mn, we can assign an arbitrary value to one of
  them a convenient choice is to assign to the ui
  that has the largest number of allocations the
  value 0.
• It is then very easy to solve the remaining
  equations.

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Calculation of the uis and vjs for the
current BF solution
x31 19 u3 v1 set u30 ? v1 19 x32 19
u3 v2 ? v2 19 x34 19 u3 v4 ? v4
23 x21 14 u2 v1 v1 19 ? u2
-5 x23 13 u2 v3 u2 -5 ? v3
18 x13 13 u1 v3 v3 18 ? u1
-5 x15 17 u1 v5 u1 -5 ? v5
22 x45 0 u4 v5 v5 22 ? u4 -22
Now we can compute the cij - ui vj for the
non-basic variables.
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x11 16 5 - 19 2 x12 16 5 19 2
x14 22 5 23 4 x22 14 5 19 0
v1 19 x24 19 5 23 1 v3 x25
15 5 22 -2 x33 20 0 18 2 x41 M
22 19 M 3 x42 0 22 19 3 x43
M 22 18 M 4 x43 0 22 23 -1
Which non-basic variable will enter the basis?
Is it optimal?
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Transportation simplex method

• Initialization finding the initial BF solution
• Optimality test how do we check if it is the
  optimal solution?
• Iteration

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How to Pivot a Transportation Problem

• Select as the entering variable the variable with
  the largest (in absolute value) negative cij - ui
  vj value.
• Select as the leaving variable. Increasing the
  entering variable from zero sets off a chain
  reaction of compensating changes in other basic
  variables. The first variable to be decreased to
  zero then becomes the leaving basic variable.
• Find the chain involving the entering variable
  and some of the basic variables.
• Counting the cells in the chain, label them
  alternating as () and () cells, starting with
  the entering variable with label ()

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• 3. Find the (-) cells whose variable assumes the
  smallest value. Call this value ?. The variable
  corresponding to this (-) cell will leave the
  basis.
• 4. To perform the pivot, decrease the value of
  each (-) cell by ? and increase the value of
  each cell by ?. The variables that are not in
  the loop remain unchanged. The pivot is now
  complete.
• ( If ? 0, the entering variable will equal 0,
  and a - variable that has a current value of 0
  will leave the basis. In this case a degenerate
  bfs existed before and will result after the
  pivot. If more than one - cell in the loop
  equals ?, you may arbitrarily choose one of these
  odd cells to leave the basis again a degenerate
  bfs will result)

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Illustration of pivoting procedure. We want to
find the bfs that would result if x14 were
entered into the basis.

• Example of the chain

Ent. Variable
-
-


-
What is the value of the entering variable?
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New bfs after x14 is pivoted into basis. Since
There is no loop involving the cells (1,1),
(1,4), (2,1), (2,2), (3,3) and (3, 4) the new
solution is a bfs.
After the pivot the new bfs is x1115, x1420,
x2130, X2220, X3330 and X3410.
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Two important points!

• In the pivoting procedure
• Since each row has as many 20s as 20s, the new
  solution will satisfy each supply and demand
  constraint.
• By choosing the smallest (-) variable (X23) to
  leave the basis, we ensured that all variables
  will remain nonnegative.

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• The 1st initial BF solution is degenerate because
  x310. But it is not a problem since this cell
  becomes a () cell.
• Another degenerate solution arises in the 3rd
  tableau (two (-) cells tie for the smallest
  value - 30). We selected (2,1) to be the leaving
  variable if we selected (3,4), (2,1) would
  become the degenerate basic variable. Cell (3,4)
  is a (-) cell but at the level 0. So the entering
  variable has the value 0.
• Because none of the cij - ui vj is negative in
  the 4th tableau, it means that the 3rd tableau is
  also optimal.