A boutique brewery has two warehouses from which it distributes beer to five carefully chosen bars. At the start of every week, each bar sends an order to the brewery’s head office for so many crates of beer, which is then dispatched from the appropriate warehouse to the bar. The brewery would like to have an interactive computer program which they can run week by week to tell them which warehouse should supply which bar so as to minimize the costs of the whole operation.
For example, suppose that at the start of a given week the brewery has 1000 cases at warehouse A, and 4000 cases at warehouse B, and that the bars require 500, 900, 1800, 200, and 700 cases respectively.
The transportation costs are given in Table 1.
Table 1 Transportation costs from warehouses to bars
Cost ($/crate) | Warehouse A | Warehouse B |
---|---|---|
Bar 1 | 2 | 3 |
Bar 2 | 4 | 1 |
Bar 3 | 5 | 3 |
Bar 4 | 2 | 2 |
Bar 5 | 1 | 3 |
Which warehouse should supply which bar?
Since crates of beer are being shipped from the warehouses to the bars, this problem is a bipartite network and can be modelled as a transportation problem.
The supply nodes are the warehouses, the demand nodes are the bars. The total supply is 1,000 + 4,000 = 5,000 and the total demand is 500 + 900 + 1,800 + 200 + 700 = 4,100, so the problem is unbalanced. We can either alter the balanced formulation to incorporate this or add a dummy demand node to receive the extra supply.
Since this is a transportation problem we can use the AMPL model file for transportation problems.
The data file will have the supply nodes as the warehouses:
set SUPPLY_NODES := A B ;and the demand nodes as the bars:
set DEMAND_NODES := 1 2 3 4 5 ;
The supply and demand is given and can be specified:
param Supply := A 1000 B 4000 ; param Demand := 1 500 2 900 3 1800 4 200 5 700 ;
Finally, the transportation costs are easier to specify using a transposed table:
param Cost (tr) : A B := 1 2 3 2 4 1 3 5 3 4 2 2 5 1 3 ;
When solving The Beer Distribution Problem there are some things happening "behind the scenes" in AMPL. First, note that the attributes given to the parameters, e.g.,
\begin{verbatim} # The supply at the supply nodes param Supply {SUPPLY_NODES} >= 0, integer; # The demand at the demand nodes param Demand {DEMAND_NODES} >= 0, integer; \end{verbatim}
ensure that both {\tt Supply} and {\tt Demand} are integer. This requirement has a huge effect on the solution process as we will see shortly. If you require data to have specific properties before solving you can ensure this by setting parameter attributes or by using a {\tt check} statement in your model file.
Notice that solving The Beer Distribution Problem did not require CPLEX to use any branch-and-bound nodes (which it would use if the optimal linear programming solution was not integer).
Thus, the linear programme solution was also an integer solution. Were we just lucky (as in The Surfboard Production Problem)? In fact, as long as {\tt Supply} and {\tt Demand} are integer, the linear programming solution will always be integer. Read about naturally integer solutions for more details.
Since we have guaranteed the {\tt Supply} and {\tt Demand} are integer, we know that the solution to the linear programme will be integer, so we don't need to check the integrality of the solution.
In The Beer Distribution Problem, the total supply is 5000 cases of beer, but the total demand is only for 4100 cases. The extra supply can be sent to an dummy demand node. The cost of flow going to the dummy demand node is then the storage cost at each of the supply nodes.
Now that total supply = total demand, i.e., the problem is balanced, we can create {\tt balanced_transportation.mod} from {\tt transportation.mod} by checking the problem is balanced and changing the {\tt Enough Supply} constraint to be an {\tt =} constraint.
\begin{verbatim} . . . # Make sure the problem is balanced check : sum {s in SUPPLY_NODES} Supply[s] = sum {d in DEMAND_NODES} Demand[d];
# Flow must not exceed supply subject to EnoughSupply {s in SUPPLY_NODES}: sum {t in DEMAND_NODES} Flow[s, t] = Supply[s]; . . . \end{verbatim}
The data file can now incorporate the storage costs (usually known as holding costs). The storage cost at Warehouse A is 50c per crate of beer, but at Warehouse B refrigerator space is at a premium, so holding a crate of beer costs $1.
\begin{verbatim} data; set SUPPLY_NODES := A B ; set DEMAND_NODES := 1 2 3 4 5 Dummy ; # Added the extra (dummy) demand node
param Cost:
param Supply := A 1000 B 4000 ;
param Demand := 1 500 2 900 3 1800 4 200 5 700 Dummy 900 ; # There are 900 cases of beer left to store \end{verbatim}
If a transportation problem has more demand than supply, we can balance the problem using a dummy supply node. For example, consider The Beer Distribution Problem, and assume there has been a production problem and only 4000 cases of beer could be produced. Since the total demand is 4100, we need to get extra cases of beer from the dummy supply node.
We can still use the balanced transportation model {\tt balanced_transportation.mod}, but our data file will change to reflect the extra cost of getting cases of beer from an outside supplier at a cost of $5 per crate.
\begin{verbatim} data; set SUPPLY_NODES := A B Dummy ; # Extra supply from a competitor set DEMAND_NODES := 1 2 3 4 5 ;
param Cost:
param Supply := A 1000 B 4000 Dummy 100 ; # Need 100 extra crates of beer
param Demand := 1 500 2 900 3 1800 4 200 5 700 ;
\end{verbatim}
\begin{verbatim} # brewery.run reset;
model transportation.mod; data brewery.dat;
option solver cplex;
option presolve 0; # Turn off the AMPL presolve option cplex_options 'presolve 0 sensitivity'; # Turn off the CPLEX presolve, # Turn on the sensitivity analysis solve;
display Flow;
display _varname, _var, _var.rc, _var.current, _var.up, _var.down; display _conname, _con.body, _con.slack, _con.dual, _con.current, _con.up, _con.down; \end{verbatim}
The sensitivity analysis shows that the current solution is largely unaffected by changes in the transportation costs. The two exceptions are the transportation costs from Warehouse A to Bar 1 and from Warehouse B to Bar 1. The transportation cost from Warehouse A to Bar 1 can only increase from 2 to 3 before a solution change will occur. Similarly, the cost of transportation from Warehouse B to Bar 1 can only decrease from 3 to 2 before causing a solution change .
The sensitivity analysis also shows that a change in any of the constraints will cause the solution to change, except for the supply at Warehouse B. This supply can decrease to 3100 before any change in the solution happens. However, changing the supply at Warehouse B will not change the objective function value as {\tt EnoughSupply['B']} has a dual value (i.e., shadow price) of 0.
There are many ways to present the solution to The Beer Distribution Problem: as a list of shipments, in a table, etc. You can use AMPL's {\tt display}, {\tt print} and {\tt printf} commands to create formatted output that can be easily used in your management summary. The following file was produced by an AMPL script file (the actual numbers have been removed):
\begin{verbatim} TRANSPORTATION SOLUTION -- Non-zero shipments TotalCost = __
Ship _ crates of beer from warehouse A to pub 1 Ship _ crates of beer from warehouse A to pub 5 Ship _ crates of beer from warehouse B to pub 1 Ship _ crates of beer from warehouse B to pub 2 Ship __ crates of beer from warehouse B to pub 3 Ship _ crates of beer from warehouse B to pub 4 \end{verbatim}
This information gives rise to the following management summary:
The brewery has two warehouses (A and B respectively) and 5 bars (1, 2, 3, 4 and 5).
The supply of crates of beer at the warehouses is:
______
The forecasted demand (in crates of beer) at the bars is:
______
The cost of transporting 1 crate of beer from a warehouse to a bar is given in the following table:
______
To minimise the transportation cost the brewery should make the following shipments:
Ship _ crates of beer from warehouse A to pub 1 Ship _ crates of beer from warehouse A to pub 5 Ship _ crates of beer from warehouse B to pub 1 Ship _ crates of beer from warehouse B to pub 2 Ship __ crates of beer from warehouse B to pub 3 Ship _ crates of beer from warehouse B to pub 4
The total transportation cost of this shipping schedule is $_____.
The current solution is robust under changes to all the costs except for the following transportation costs:
______ to _____ which can only ______ to _____; and,
__________ to _____ which can only ______ to _____.
However, any change in the amount of supply at the warehouses or the demand at the bars will cause the solution to change with the following exception:
The ______ at ______ can ______ to ______ .
If this ______ occurs then the objective function value ______ by ______ per unit ______ .
Ongoing Monitoring may take the form of:
You can use AMPL as a scripting language to generalise your models. For example, you can solve The Beer Distribution Problem for differing supplies and demands using the {\tt balanced_transportation.mod} model file, an extra model file {\tt balance.mod} and a script file {\tt balance.run}. The {\tt balance.mod} file declares three new parameters:
\begin{verbatim} # The following parameters are needed to use balance.run param difference; # The difference between total supply and total demand
param dummyDemandCost {SUPPLY_NODES}; # The cost of sending supply to a dummy demand node param dummySupplyCost {DEMAND_NODES}; # The cost of getting supply from a dummy supply node \end{verbatim}
and the {\tt balance.run} file automatically balances the data:
\begin{verbatim} let difference := (sum {s in SUPPLY_NODES} Supply[s]) - (sum {d in DEMAND_NODES} Demand[d]);
if difference > 0 then { let DEMAND_NODES := DEMAND_NODES union {'Dummy'}; let Demand['Dummy'] := difference; let {s in SUPPLY_NODES} Cost[s, 'Dummy'] := dummyDemandCost[s]; } else if difference < 0 then { let SUPPLY_NODES := SUPPLY_NODES union {'Dummy'}; let Supply['Dummy'] := - difference; let {d in DEMAND_NODES} Cost['Dummy', d] := dummySupplyCost[d]; }; # else the problem is balanced \end{verbatim}
*Note* The {\tt ;} at the end of the file is necessary because the second {\tt if} statement ({\tt if difference < 0 then}) is the only statement within the {\tt else} part of the first {\tt if} statement.
Using these files and the {\tt reset data} command we can solve both of the balanced transportation examples with a single script file:
\begin{verbatim} reset;
model balanced_transportation.mod; data brewery.dat; # Unbalanced brewery model
# Define parameters to be used with balance.run model balance.mod;
let dummyDemandCost['A'] := 0.5; let dummyDemandCost['B'] := 1;
include balance.run;
option solver cplex; solve; option display_1col 5; display Flow;
reset data;
data brewery.dat; let Supply['B'] := Supply['B'] - 1000;
let {d in DEMAND_NODES} dummySupplyCost[d] := 5;
include balance.run;
solve; display Flow; \end{verbatim}