Transcript Slides from 11/5 (Transportation Simplex)

Operations Research
TRANSPORTATION PROBLEM
Consider…
One of the main products of the P&T Company is canned peas.
The peas are prepared at three canneries (Bellingham, WA;
Eugene, OR; Albert Lea, MN) and then shipped by truck to
four distributing warehouses in the western US (Sacramento,
Salt Lake City, Rapid City, Albuquerque). Because shipping
costs are a major expense, management is initiating a study to
reduce them as much as possible. For the upcoming season,
an estimate has been made of the output from each cannery,
and each warehouse has been allocated a certain amount from
the total supply of peas. This information (in units of
truckloads, along with the shipping cost per truckload for each
cannery-warehouse combination is provided. There are a total
of 300 truckloads to be shipped. Determine which plan for
assigning these shipments to the various cannery-warehouse
combinations would minimize shipping costs.
What does the problem formulation look like?
Minimize Z = 464x11 + 513x12 + 654x13 + 867x14 +
352x21 + 416x22 + 690x23 + 791x24 +
995x31 + 682x32 + 388x33 + 685x34
Subject To:
x11 + x12 + x13 + x14 = 75
x21 + x22 + x23 + x24 = 125
x31 + x32 + x33 + x34 = 100
x11 + x21 + x31 = 80
x12 + x22 + x32 = 65
x13 + x23 + x33 = 70
x14 + x24 + x34 = 85
With:
xij >= 0, (i = 1, 2, 3; j = 1, 2, 3, 4)
Exercise!
The Move-It Company has two plants producing forklift trucks that
then are shipped to three distribution centers. The production costs
are the same as the two plants, and the cost of shipping for each
truck is shown for each combination of plant and distribution
center:
Distribution Center
1
2
3
Plant A
$800
$700
$400
Plant B
$600
$800
$500
A total of 60 forklift trucks are produced and shipped per week. Each
plant can produce and ship any amount up to a maximum of 50
trucks per week, so there is considerable flexibility how to divide the
total production between the two plants so as to reduce shipping
costs. However, each distribution center must receive exactly 20
trucks per week.
Initialization
1.) From the rows and columns still under consideration,
select the next basic variable (allocation) using one of the
criteria noted next slide.
2.) Make that allocation large enough to exactly use up the
remaining supply in its row or the remaining demand in
its column (whichever is smaller).
3.) Eliminate that row or column (whichever went to 0). If
tied, choose the row.
4.) If only one row or only one column remains under
consideration, then the procedure is completed by
selecting every remaining variable associated with that
row to be basic with the only feasible allocation.
Otherwise, return to step 1.
Criterion
 Northwest Corner Rule: Begin by selecting x11. Thereafter, if xij was
the last basic variable selected, then next select xij+1 if source I has
any supply remaining. Otherwise, select xi+1j
 Vogel’s Approximation: For each row and column remaining under
consideration, calculate the arithmetic difference between the
smallest and next-to-the-smallest unit cost still remaining in that
row or column. In that row or column having the largest difference,
select the variable having the smallest remaining unit cost.
 Russell’s Approximation: For each source row i remaining under
consideration determine the largest unit cost cij (call this x) still
remaining in that row. For each destination column j remaining
under consideration determine the largest unit cost cij still
remaining in that column (call this y). For each variable xij not
previously selected in these rows and columns calculate z = cij – x –
y. Select the variable having the largest (in absolute terms) negative
value of z.
Optimality
 Derive ui, vj by selecting the row having the largest
number of allocations, setting its ui = 0, and then
solving the set of equations cij = ui + vj for each (i,j)
such that xij is basic. If cij – ui – vj >= 0 for every (i,j)
such that xij is nonbasic, then the current solution is
optimal and stop. Otherwise, go to an iteration.
Iteration
1.) Determine the entering basic variable: select the nonbasic
variable xij having the largest (in absolute terms) negative
value of cij – ui – vj.
2.) Determine the leaving basic variable: Identify the chain
reaction required to retain feasibility. Basically, increase the
value of the entering variable and see what happens to the
other basic variables when you attempt to retain feasibility.
The basic variable that gets driven to 0 first becomes the
leaving basic variable. From this you can determine the
“donor” cells (ones that decrease) and “recipient” cells (ones
that increase, including the entering variable).
3.) Determine the new feasible solution: add the value of the
leaving basic variable to the allocation of recipient cells and
subtract from donor cells.