The

Grain Elevator 


Farm 
3. K ANSAS C ITY 
4. O MAHA 
5. D ES M OINES 
1. Nebraska 
$16 
$10 
$12 
2. Colorado 
15 
14 
17 
The basic structure of this model is shown in the graphical network in Figure 6.3.
Figure 6.3. Network of transshipment routes
As with the transportation problem, a linear programming model is developed with supply and demand constraints. The available supply constraints for the farms in Nebraska and Colorado are
x _{ 13 } + x _{ 14 } + x _{ 15 } = 300
x _{ 23 } + x _{ 24 } + x _{ 25 } = 300
The demand constraints at the Chicago, St. Louis, and Cincinnati mills are
x _{ 36 } + x _{ 46 } + x _{ 56 } = 200
x _{ 37 } + x _{ 47 } + x _{ 57 } = 100
x _{ 38 } + x _{ 48 } + x _{ 58 } = 300
Next we must develop constraints for the grain elevators (i.e., transshipment points) at Kansas City, Omaha, and Des Moines. To develop these constraints we follow the principle that at each transshipment point, the amount of grain shipped in must also be shipped out . For example, the amount of grain shipped into Kansas City is
x _{ 13 } + x _{ 23 }
and the amount shipped out is
x _{ 36 } + x _{ 37 } + x _{ 38 }
Thus, because whatever is shipped in must also be shipped out, these two amounts must equal each other:
x _{ 13 } + x _{ 23 } = x _{ 36 } + x _{ 37 } + x _{ 38 }
or
x _{ 13 } + x _{ 23 } x _{ 36 } x _{ 37 } x _{ 38 } = 0
The transshipment constraints for Omaha and Des Moines are
x _{ 14 } + x _{ 24 } x _{ 46 } x _{ 47 } x _{ 48 } = 0
x _{ 15 } + x _{ 25 } x _{ 56 } x _{ 57 } x _{ 58 } = 0
The complete linear programming model, including the objective function, is summarized as follows:
minimize Z 
= 
$16 x _{ 13 } + 10 x _{ 14 } + 12 x _{ 15 } + 15 x _{ 23 } + 14 x _{ 24 } + 17 x _{ 25 } + 6 x _{ 36 } + 8 x _{ 37 } + 10 x _{ 38 } 
+ 7 x _{ 46 } + 11 x _{ 47 } + 11x _{ 48 } + 4x _{ 56 } + 5 x _{ 57 } + 12 x _{ 58 } 

subject to 


Computer Solution with Excel
Because the transshipment model is formulated as a linear programming model, it can be
Exhibit 6.10 shows the spreadsheet solution and Exhibit 6.11 the solver for our wheat shipping transshipment example. A network diagram of the optimal solution is shown in Figure 6.4. The spreadsheet is similar to the original spreadsheet for the regular transportation problem in Exhibit 6.1 except that there are two tables of variablesone for shipping from the farms to the grain elevators and one for shipping grain from the elevators to the mills. Thus, the decision
Exhibit 6.10.
Exhibit 6.11.
Two cost arrays have been developed for the shipping costs in cells I6:K7 and cells J13:L15 , which are then multiplied by the variables in cells B6:D7 and C13:E15 and added together. The objective function, =SUMPRODUCT(B6:D7,I6:K7)+SUMPRODUCT(C13:E15,J13:L15) , is shown on the toolbar at the top of Exhibit 6.10. Constructing the objective function with cost arrays like this is a little easier than typing in all the variables and costs in a single objective function when there are a lot of variables and costs.
Figure 6.4. Transshipment network solution for wheat shipping example