
A problem that benefits from the side model construction
is the multiperiod problem. Here the problem is to plan the
operations of some system over several periods during a timeplanning
horizon. Some decisions are made at the beginning of the horizon,
while others are made in each period. Certain variables link
the decisions from one period to the next.
We illustrate this case with the aggregate scheduling problem
described in the modeling
section. A company wants a high level, aggregate production
plan for the next 6 months. Projected orders for the company's
product are listed in the table. Over the 6month period, units
may be produced in one month and stored in inventory to meet
some later month's demand. Because of seasonal factors, the
demand for the product and the cost of production is not constant,
as shown in the table.
Aggregate planning data 

Demand 
Production 
Month 
(units) 
cost ($/unit) 
1 
1300 
100 
2 
1400 
105 
3 
1000 
110 
4 
800 
115 
5 
1700 
110 
6 
1900 
110 
The cost of holding an item in inventory for 1 month is $4/unitmo.
Items produced and sold in the same month are not put in inventory.
The maximum number of units that can be held in inventory is
250. The inventory level at the beginning of the planning horizon
is 200 units; the inventory level at the end of the planning
horizon is to be 100. The problem is to determine the optimal
amount to produce in each month so that demand is met while
minimizing the total cost of production and inventory. Shortages
are not permitted. The general model is below.
We construct the model with the Math Programming addin and
solve it with the Excel solver to obtain the solution below. 

Using the Side Model 

In the formulation above one notes
that the constraint matrix is sparse, that is, it contains
many 0's. This is usually true in a multiperiod model because
most of the variables appear in the constraints associated
with only a single period. For this simple example there is
only one constraint per period. Each production variable is
present in only one constraint while each inventory variable
appears in two. We will construct a more compact model using
the side model feature. For this purpose we identify
the inventory variables as the linking variables in the master
problem and the production variables as the side variables
in the subproblems. First we construct the master problem shown
below.
Only the inventory variables are included with the two constraints
describing the beginning and ending inventory. The solution
to the model has no meaning since important information
regarding demand and production cost is not included.
To create the subproblems. choose Side Model under
the Math Programming addin entries in the ORMM menu
item. The dialog below contains the parameters of the side
model.
The cell field indicates the location
on the worksheet for the side model. Replications is
the number of rows, or individual subproblems to construct. Parameters indicates
the number of columns to include with model related parameters.
Transfer is the number of linking variables. Variables is
the number of side variables in each of the subproblem rows.
The Obj. box indicates that a column is to provided
for objective coefficients of the side variables. Constraints indicates
the number of constraints for each row of the side model. The RHS box
indicates that a row is to be included for the Righthandside values. Clicking
OK results in the side model below. The formulas shown in row
19 link the data in the parameter columns to the objective
and RHS columns. 


The T and W matrices are composed
entirely of 1's. The D rows determine the contribution of the
master transfer variables to the constraints.The constraints
described by the side form are:
z0  z1 + x1 = 1300
z1  z2 + x2 = 1400
z2  z3 + x3 = 1000
z3  z4 + x4 = 800
z4  z5 + x5 = 1700
z5  z6 + x6 = 1900
The model form includes a Change button that allows
columns of the side model to be added or deleted. Addition
of columns is possible only if the associated region currently
has at least two columns of that type. In the example, the
numbers of parameters and transfer variables can be increased,
but the number of side variables and side constraints cannot.
When we click the Solve button at the top of the
page. The solution to the combination of the main model and
the side model is obtained. The solution is the same as when
all the variable and constraints were included in the same
model. 

Adding Fixed Charges 

For such a simple problem, there
does not seem to be a very large savings when using the side
model, rather than including the entire problem in the master
problem. More savings are apparent when the side model grows
in size. For an example we change the problem to add a fixed
charge whenever nonzero production occurs. The fixed charge
might represent the setup cost for producing a product.
The master problem shown below is the same as
the original except we have increased the maximum inventory
to allow a more interesting solution. The figure shows the
optimum solution 


The side model is shown below with
the optimum solution to the subproblems. We have included the
parameters directly as the RHS of the period constraints and
the objective coefficients. The setup costs result in fewer production
runs over the time horizon. 


For this example, the addition of
setup costs increased the number of variables and constraints
by 6. If the side model was not used the constraint matrix would
have increased from 8*13 = 104 cells to 14*19 = 266 cells plus
additional cells in the variable and constraint limits. The change
more than doubles the number of cells required. Using the side
model, the change required only 49 additional cells for the side
model. The value of the side model increases as the complexity
of the individual subproblems. increases. The structure of the
subproblems is easy to check and the parameters are easy to apply.
The side model formulation does not change the model. It simply
makes it easier to represent the model on the worksheet. 

