"Transportation and Transshipment Problems" 
Index to Module Seven Notes 
"I am your cool tool from the motor pool…we have 2by's, 4by's and those big things that bend in the middle and go chhhh… chhhhhh!"Overhead telephone response from Motor Pool Dispatcher, Cam Ranh Bay Air Base Motor Pool, 1967
Module 7.1: Structure of the
Transportation Problem
Introduction
Ah… finally after 14 weeks…my favorite of all
quantitative method applications… the transportation problem.
For seventeen years in the Air Force, most of my career was spent
figuring out how to efficiently move troops and their equipment,
weapons and weapon systems, communication equipment, and medical
supplies from point A to point B. Then, when I moved to the Pentagon
to finish my career as Chief of Air Force Transportation Programs, I
worked for three years on how to effectively deploy troops and their
equipment, weapons and weapon systems, communication equipment, and
medical suppliers from many ports of embarkation to multiple ports of
debarkation.
After I retired from the Air Force and joined the faculty at the
University of South Florida, my initial applied research was working
with supply chain managers at Johnson & Johnson and 3M to figure
out how to effectively move J&J medical products and 3M consumer
products through their supply chains. For 3M, this included using
quantitative methods to help decide where intermediate distribution
centers should be located, especially to meet European expansion in
the early 1990's.
Whether in military or commercial applications, the transportation
problem is that problem which addresses what origin
point should ship to what final destination point over
which route so as to minimize transportation
costs while meeting the problem constraints. The problem
constraints are staying within the available supply at the
origins, and meeting customer demand at the
destinations.
The problem can become quickly complex by adding intermediate
transshipment points, such as distribution centers,
which add constraints such as whatever is shipped into transshipment
points must be shipped out. The transshipment problem
is the subject of Module 7.2 Notes. Complexity is also added by
placing capacity constraints on the routes or the transshipment
points.
You will see that the transportation problem is simply another
application of linear programming, but it is such a widespread
application that this and other quantitative texts devote a chapter
just to this application. Companies like 3M that have multiple
manufacturing sites, many distribution centers and warehouses, and
multiple consumer demand locations find the transportation problem to
be so complex that linear programming applications are in common
use.
Illustration of the General Transportation Problem
Let's start this module with a simplified example of the problem
facing logistics staffs at the Pentagon. How many troops should be
deployed from aerial and water ports of embarkation to aerial and
water ports of debarkation so as to minimize total deployment time.
For example, we could use Fort Bragg, North Carolina and Fort Hood,
Texas as aerial ports of embarkation and Adana, Turkey; Dhahran,
Saudi Arabia; and Wheelus, Libya as aerial ports of debarkation.
Please understand that this is not an actual deployment scenario; but
rather an example for illustration.
The constraints include "supply" and "demand." Supply constraints are
used to ensure that the number of troops deployed do not exceed the
number available at the two ports of embarkation. Demand constraints
are used to ensure that the number of troops deployed meet the need
for troops at the ports of debarkation.
Before continuing, please understand that the two origins could be
Detroit and Memphis automobile production plants. The three
destinations could be customers (automobile dealerships) in
Philadelphia, Washington DC and Miami. Instead of moving 20,000
troops, we may be moving 20,000 new cars out of Detroit and Memphis
to meet dealer demand at the three destinations.
The following table presents a picture of this transportation
problem. The table includes the nodes (origins and destinations) and
the arcs (deployment or shipping routes).
Table 7.1.1
Origin Destination Fort Bragg To Ada > > 7 Days From FtB Adana > 8 Days From FtB Dhahran Fort Hood To Ada > > 8 Days from FtB Wheelus
(Troops
Available)
(Troops
Needed)
(14,000)
To Dha >
To Whl >
> 10 Days From FtH
(5,000)
> 7 Days From FtH
(10,000)
(6,000)
To Dha >
To Whl >
> 5 Days From FtH
(5,000)
The table shows that 14,000 troops are
available for deployment out of Fort Bragg. It takes 7 days to deploy
troops from Fort Bragg to Adana, which has a demand for 5,000 troops.
So, one alternative is to deploy 5,000 troops out of Fort Bragg to
Adana, leaving 9,000 troops available at Fort Bragg for Dhahran or
Wheelus. The "cost" of this move would be 5,000 troops times 7 days
giving 35,000 troop deployment days.
Another way of satisfying the demand at Adana is to deploy 5,000
troops from Fort Hood. Note it takes 10 days to deploy troops from
Fort Hood to Adana, which would give 5,000 times 10 days or 50,000
troop deployment days. We could continue to try different
combinations of origins and destinations to minimize total troop
deployment days while meeting demand and staying within troop
availability constraints. For a small problem, it would not be too
difficult to try all of the combinations of solutions to find the
optimal solution. But if there were many origins, such as 10 or more,
and many destinations, such as 15 or more, the problem would be too
cumbersome to work "by hand."
I should note here that in the parallel automobile example, the 7, 10
and other deployment day coefficients might be $700, $1000 and other
freight charges. The objective for the commercial application would
be to minimize transportation costs while meeting demand and staying
within the available supply constraints.
Linear programming is an excellent quantitative method for
application to the transportation problem. Recall from Module 6, that
to formulate a linear program we need to decide on the decision
variables, create the objective function as a linear equation, and
then formulate the constraints as linear equations.
For the transportation problem, the decision variables
are:
FtB_Ada = Nbr of troops to deploy from Fort Bragg to Adana
FtB_Dha = Nbr of troops to deploy from Fort Bragg to Dhahran
FtB_Whl = Nbr of troops to deploy from Fort Bragg to Wheelus
FtH_Ada = Nbr of troops to deploy from Fort Hood to Adana
FtH_Dha = Nbr of troops to deploy from Fort Hood to Dhahran
FtH_Whl = Nbr of troops to deploy from Fort Hood to Wheelus
Note that the number of variables for the
standard transportation problem is the number of origins times the
number of destinations. For this problem, there are two origins and
three destinations which gives 2 times 3 or 6 decision variables. The
decision variables then represent the units shipped over the
deployment or shipping routes. Also note that I used a code for
naming the variables. The text uses X12 to represent the number of
units shipped from origin one to destination 2. I like to use a more
descriptive code to represent the route, and abbreviate names to keep
within the 8 character variable name restrictions of The
Management Scientist.
The objective function is to minimize troop deployment days, where
days ("costs") are the coefficients multiplied times the number of
troops routing decision variables.
Minimize Z = 7 FtB_Ada + 8 FtB_Dha + 8 FtB_Whl+10 FtH_Ada + 7 FtH_Dha + 5 FtH_Whl
To show how this equation works, let's compute the deployment days for the solution FtB_Ada = 5,000; FtB_Dha = 5,000; Fth_Whl = 4,000; Fth_Dha = 5,000 and FtH_Whl = 1,000.
Minimize Z = 7 (5,000) + 8 (5,000) + 8 (4,000) + 7 (5,000)+ 5 (1,000) = 147,000 troop deployment days
This is a feasible solution but we do not know if it is optimal until we try all combinations (or finish the constraint set and let the computer software run the combinations). Now for the constraints which include staying within the available supply at each origin node.
Fort Bragg Supply: FtB_Ada + FtB_Dha + FtB_Whl < 14,000
Fort Hood Supply: FtH_Ada + FtH_Dha + FtH_Whl < 6,000
The constraints also include meeting the customer demand:
Adana Demand: FtB_Ada + FtH_Ada = 5,000
Dhahran Demand: FtB_Dha + FtH_Dha = 10,000
Wheelus Demand: FtB_Whl + FbH_Whl = 5,000
Note that I made the demand constraints strict
equalities, but allowed for slack in the supply constraints. This
general formulation works as long as supply = demand or supply is
greater than demand. If supply is greater than demand, the slack
constraint will indicate which origin should have the excess supply.
I will talk about how to handle the special case of demand being
greater than supply after we look at the solution.
Printout 7.1.1 provides The Management Scientist printout of
the linear programming formulation and solution.
Printout 7.1.1
LINEAR PROGRAMMING PROBLEM
MIN 7FtB_Ada+8FtB_Dha+8FtB_Whl+10FtH_Ada+7FtH_Dha+5FtH_Whl
S.T.
1) 1FtB_Ada+1FtB_Dha+1FtB_Whl<14000
2) 1FtH_Ada+1FtH_Dha+1FtH_Whl<6000
3) 1FtB_Ada+1FtH_Ada=5000
4) 1FtB_Dha+1FtH_Dha=10000
5) 1FtB_Whl+1FtH_Whl=5000
OPTIMAL SOLUTION
Objective Function Value = 139000.000
Variable Value Reduced Costs
  
FtB_Ada 5000.000 0.000
FtB_Dha 9000.000 0.000
FtB_Whl 0.000 2.000
FtH_Ada 0.000 4.000
FtH_Dha 1000.000 0.000
FtH_Whl 5000.000 0.000
Constraint Slack/Surplus Dual Prices
  
1 0.000 0.000
2 0.000 1.000
3 0.000 7.000
4 0.000 8.000
5 0.000 6.000
OBJECTIVE COEFFICIENT RANGES
Variable Lower Limit Current Value Upper Limit
   
FtB_Ada No Lower Limit 7.000 11.000
FtB_Dha 7.000 8.000 10.000
FtB_Whl 6.000 8.000 No Upper Limit
FtH_Ada 6.000 10.000 No Upper Limit
FtH_Dha 5.000 7.000 8.000
FtH_Whl No Lower Limit 5.000 7.000
RIGHT HAND SIDE RANGES
Constraint Lower Limit Current Value Upper Limit
   
1 14000.000 14000.000 No Upper Limit
2 6000.000 6000.000 15000.000
3 0.000 5000.000 5000.000
4 1000.000 10000.000 10000.000
5 0.000 5000.000 5000.000
The optimal solution is to deploy 5,000 troops from Fort Bragg to
Adana; 9,000 troops from Fort Bragg to Dhahran; 1,000 troops from
Fort Hood to Dhahran; and 5,000 troops from Fort Hood to Wheelus. The
total deployment days value of the objective function is 139,000. The
reduced costs of 2 for FtB_Whl indicates that the solution would get
worse (increase since this is a minimization problem) by 2 for every
troop deployed over that route.
The dual price of 1 for the second constraint means that if the right
hand side of the Fort Hood supply constraint was raised from 6,000 to
6,001, the total troop deployment days would decrease by 1 (change
sign for minimization problem). The dual price of 7 for the Adana
demand constraint means that if the demand of 5,000 was increased to
5,001, the troop deployment days would increase by 7.
The objective function coefficient ranges provide valuable
sensitivity analysis as discussed in Module 6 Notes. For example,
this solution (values of the decision variables) remains the optimal
deployment solution as long as the deployment days from Fort Bragg to
Adana remain at or under 11. Likewise, the ranges on the right hand
side constraints provide the allowable changes for continued
interpretation of the dual prices.
Earlier I mentioned that the general form of the transportation
problem works when supply = demand, or when supply > demand. Let's
look at this problem if Fort Bragg and Fort Hood both have 14,000
troops to deploy, but the total demand stays at 20,000.
Printout 7.1.2
LINEAR PROGRAMMING PROBLEM
MIN 7FtB_Ada+8FtB_Dha+8FtB_Whl+10FtH_Ada+7FtH_Dha+5FtH_Whl
S.T.
1) 1FtB_Ada+1FtB_Dha+1FtB_Whl<14000
2) 1FtH_Ada+1FtH_Dha+1FtH_Whl<14000
3) 1FtB_Ada+1FtH_Ada=5000
4) 1FtB_Dha+1FtH_Dha=10000
5) 1FtB_Whl+1FtH_Whl=5000
OPTIMAL SOLUTION
Objective Function Value = 131000.000
Variable Value Reduced Costs
  
FtB_Ada 5000.000 0.000
FtB_Dha 1000.000 0.000
FtB_Whl 0.000 2.000
FtH_Ada 0.000 4.000
FtH_Dha 9000.000 0.000
FtH_Whl 5000.000 0.000
Constraint Slack/Surplus Dual Prices
  
1 8000.000 0.000
2 0.000 1.000
3 0.000 7.000
4 0.000 8.000
5 0.000 6.000
In this solution, Fort Hood deploys the
majority of the troops to Dhahran taking advantage of the shorter
deployment time. As a result, the total deployment troops day
solution is 131,000; a savings of 8,000 troop days. The slack of
8,000 is located at Fort Bragg. These troops would be in reserve in
case they are needed elsewhere.
There are some special cases to the general transportation problem
that are discussed in the next section: demand exceeds supply and
capacity constraints on the routes.
Special Cases
The first special case occurs when demand is greater than supply. In
a commercial application, this would be a stock out, lost sale, or
back order at one or more of the destinations. In this military
deployment scenario, one or more of the deployment destinations would
be without a full complement of troops. For a linear programming
formulation, we have to create dummy variables for each of the demand
destinations to determine which destination will not have its demand
met.
Dum_Ada: Number of troops short at Adana
Dum_ Dha: Number of troops short at Dhahran
Dum_Whl: Number of troops short at Wheelus
Suppose in our original example that the demand
is for 6,000 troops at Adana rather than 5,000. The other demands
remain at the 10,000 for Dhahran and 5,000 for Wheelus. The supply is
14,000 at Fort Bragg and 6,000 at Fort Hood. So the total demand is
21,000 versus a supply of 20,000, meaning one of the demand sites
will be short 1,000 troops.
Here is the computer formulation and solution. Note in the linear
programming problem section of the output, the dummy variables do not
show up in the objective function since their coefficients (costs)
are zero. However, note that the dummy variable for Adana is included
in the Adana constraint (Constraint Number 3), likewise for the
Dhahran and Wheelus demand constraints. Finally, a new constraint is
added to ensure that the total of all the dummy variables stays at
1,000, the amount by which demand exceeds supply.
Printout 7.1.3
LINEAR PROGRAMMING
PROBLEM
MIN 7FtB_Ada+8FtB_Dha+8FtB_Whl+10FtH_Ada+7FtH_Dha+5FtH_Whl
S.T.
1) 1FtB_Ada+1FtB_Dha+1FtB_Whl<14000
2) 1FtH_Ada+1FtH_Dha+1FtH_Whl<6000
3) 1FtB_Ada+1FtH_Ada+1Dum_Ada=6000
4) 1FtB_Dha+1FtH_Dha+1Dum_Dha=10000
5) 1FtB_Whl+1FtH_Whl+1Dum_Whl=5000
6) 1Dum_Ada+1Dum_Dha+1Dum_Whl=1000
OPTIMAL SOLUTION
Objective Function Value = 138000.000
Variable Value Reduced Costs
  
FtB_Ada 6000.000 0.000
FtB_Dha 8000.000 0.000
FtB_Whl 0.000 2.000
FtH_Ada 0.000 4.000
FtH_Dha 1000.000 0.000
FtH_Whl 5000.000 0.000
Dum_Ada 0.000 1.000
Dum_Dha 1000.000 0.000
Dum_Whl 0.000 2.000
Constraint Slack/Surplus Dual Prices
  
1 0.000 0.000
2 0.000 1.000
3 0.000 7.000
4 0.000 8.000
5 0.000 6.000
6 0.000 8.000
This is an interesting solution, with lower
total deployment troop days that the original problem (138,000 versus
139,000). The reason for the improvement is that relatively more
troops could be sent over a relatively shorter deployment route, with
the longest deployment route (Dhahran) having the shortage. If that
is not an acceptable solution, a new constraint could always be
added. That is actually the subject of the next special case.
The next special case concerns capacity limitations on the routes of
distribution. For example, what if in the original problem we could
only deploy 3,000 troops over the route from Fort Hood to Wheelus. We
simply add the constraint:
FtB_Ada < 3000
The solution shows an increase in total troop
deployment days to 149000 reflecting the shift of traffic from a "low
cost" to a "high cost" route.
Printout 7.1.4
LINEAR PROGRAMMING PROBLEM
MIN 7FtB_Ada+8FtB_Dha+8FtB_Whl+10FtH_Ada+7FtH_Dha+5FtH_Whl
S.T.
1) 1FtB_Ada+1FtB_Dha+1FtB_Whl<14000
2) 1FtH_Ada+1FtH_Dha+1FtH_Whl<6000
3) 1FtB_Ada+1FtH_Ada=5000
4) 1FtB_Dha+1FtH_Dha=10000
5) 1FtB_Whl+1FtH_Whl=5000
6) 1FtB_Ada<3000
OPTIMAL SOLUTION
Objective Function Value = 149000.000
Variable Value Reduced Costs
  
FtB_Ada 3000.000 0.000
FtB_Dha 10000.000 0.000
FtB_Whl 1000.000 0.000
FtH_Ada 2000.000 0.000
FtH_Dha 0.000 2.000
FtH_Whl 4000.000 0.000
Constraint Slack/Surplus Dual Prices
  
1 0.000 0.000
2 0.000 3.000
3 0.000 13.000
4 0.000 8.000
5 0.000 8.000
6 0.000 6.000
Origin Destination Fort Bragg To Ada > > 7 Days from FtB Adana > 3 Days from FtB Frankfurt > 5 Days from FtH > 8 Days from FtB Dhahran > 6 Days from FtB > 3 Days from FtH Torrejon Fort Hood To Ada > > 8 Days from FtB Wheelus
The last subject with the transportation
problem concerns adding intermediate nodes between the origin and
destination, such as the addition of distribution centers,
warehouses, intermediate aerial and sea port, and so forth.
7.2: Transshipment Problem
The more realistic transportation
models are those that model intermediate nodes in the supply chain or
distribution system. Suppose we placed intermediate stopping points
at Frankfurt (Frf), Germany and Madrid (Torrejon  Toj), Spain.
Again, this deployment scenario is not an actual situation, but
rather used for illustrative purposes.
In the automobile example scenario, this would be like placing a
large auto mart in Atlanta that would be an intermediate distribution
site between Detroit and Memphis origins and the Washington DC and
Miami destinations.
The primary reason for these intermediate nodes called transshipment
points is transportation cost savings (sometimes truck load can be
used into distribution sites and less than truck load out of the
sites), although sometimes there could be inventory savings as well
(distribution center warehouses might offer cheaper storage than
storing items at production plants). In the military example, we may
use transshipment points to get troops closer to battle theaters.
That is, we may think a future contingency will occur in the Middle
East but we do not know where. In preparation, the troops might be
"staged" in Spain or Germany. The ultimate staging is when we place
Marine brigades in lighters that are placed on larger LASH vessels
(lighter aboard ship vessels).
Back to the example problem. Table 7.2.1 illustrates the nodes
(origins, transshipment points, and destinations) along with the arcs
(deployment routes).
Table 7.2.1
(Troops
Available)
(Troops
Needed)
(14,000)
To Dha >
To Whl >
To Frf >
To Toj >
> 10 Days from FtH
> 4 Days from Frf
> 3 Days from Toj
(5,000)
> 7 Days from FtH
> 4 Days from Frf
> 1 Day from Toj
(10,000)
(6,000)
To Dha >
To Whl >
To Frf >
To Toj >
> 5 Days from FtH
> 5 Days from Frf
> 2 Days from Toj
(5,000)
Now, note that the origins can ship troops directly to the
destinations and/or through the two transshipment points at
Frankfurt, Germany (Frf) and Madrid (Torrejon  Toj) Spain. Also note
that the transshipment staging bases ship to either of the three
destinations.
Some of the deployment times may seem unusual to you, like the 5 days
from Frankfurt to Wheelus which is the same as 5 days from Fort Hood
to Wheelus. There are a number of factors to consider in developing
these times, such as time to stage the troops at the origin aerial
ports, type of aircraft (fast strategic for long hauls, slower
tactical for shorter hauls), and country over flight restrictions 
sadly, the US isn't an ally with all nations in that part of the
world and several nations do not allow over flight resulting in
longer deployment times). Also note that I am not using actual
deployment scenarios for obvious reasons.
Side note: for commercial transshipment models, some transportation
costs also seem unusual, such as relatively high instate rates
compared to outofstate transportation rates for the same type of
transportation over the same miles. While transportation costs
generally do reflect distance and type of transportation, the market
place (under transportation deregulation) can sometimes play funny
tricks with tariffs.
To formulate the transshipment problem as a linear program, we need
to create decision variables for the arcs into and out of the two
transshipment points. The new decision variables are:
FtB_Frf = Nbr of troops to deploy from Fort Bragg to Frankfurt
FtB_Toj = Nbr of troops to deploy from Fort Bragg to Torrejon
Frf_Ada = Nbr of troops to deploy from Frankfurt to Adana
Frf_Dha = Nbr of troops to deploy from Frankfurt to Dhahran
Frf_Whl = Nbr of troops to deploy from Frankfurt to Wheelus
FtH_Frf = Nbr of troops to deploy from Fort Hood to Frankfurt
FtH_Toj = Nbr of troops to deploy from Fort Hood to Torrejon
Toj_Ada = Nbr of troops to deploy from Torrejon to Adana
Toj_Dha = Nbr of troops to deploy from Torrejon to Dhahran
Toj_Whl = Nbr of troops to deploy from Torrejon to Wheelus
The objective function now incorporates all of the decision variables
representing all of the deployment routes (shipment arcs) and their
deployment time ("costs").
Minimize Z = 7 FtB_Ada + 8 FtB_Dha + 8 FtB_Whl+10 FtH_Ada + 7 FtH_Dha + 5 FtH_Whl
+ 3 FtB_Frf + 6 FtB_Toj
+ 4 Frf_Ada + 4 Frf_Dha + 5 Frf_Whl
+ 5 FtH_Frf + 3 FtH_Toj
+ 3 Toj_Ada + 1 Toj_Dha + 2 Toj_Whl
We use the original two supply constraints that require we stay within the available supply at each origin node, but we add the transshipment routes:
Fort Bragg Supply: FtB_Ada + FtB_Dha + FtB_WhlFtB_Frf + FtB_Toj < 14,000Fort Hood Supply: FtH_Ada + FtH_Dha + FtH_Whl
FtH_Frf + FtH_Toj < 6,000
The constraints also include the original customer demand constraints but we have to add the possibility of getting demand satisfied from the transshipment points:
Adana Demand: FtB_Ada + FtH_Ada+ Frf_Ada + Toj_Ada = 5,000Dhahran Demand: FtB_Dha + FtH_Dha
+ Frf_Dha + Toj_Dha = 10,000Wheelus Demand: FtB_Whl + FbH_Whl
+ Frf_Whl + Toj_Whl = 5,000
Note that I made the demand constraints strict
equalities, but allowed for slack in the supply constraints. This
general formulation works as long as supply = demand or supply is
greater than demand. If supply is greater than demand, the slack
constraint will indicate which origin should have the excess supply.
If demand would be greater than supply, we would have to handle this
with dummy variables as described in the special case of Section
7.1.
We are almost finished except with transshipment problems, we have to
ensure that everything going into the transshipment points come out
of the transshipment points. We do this with strict equalities for
each transshipment node:
FtB_Frf + FtH_Frf = Frf_Ada + Frf_Dha + Frf_Whl
which gives
FtB_Frf + FtH_Frf  Frf_Ada  Frf_Dha  Frf_Whl = 0
FtB_Toj + FtH_Toj = Toj_Ada + Toj_Dha + Toj_Whl
which gives
FtB_Toj + FtH_Toj  Toj_Ada  Toj_Dha  Toj_Whl = 0
Printout 7.2.1 provides The Management Scientist printout of
the linear programming formulation and solution.
Printout 7.2.1
LINEAR PROGRAMMING
PROBLEM
MIN 7FtB_Ada+8FtB_Dha+8FtB_Whl+10FtH_Ada+7FtH_Dha+5FtH_Whl+3FtB_Frf+6FtB_Toj+4Fr
f_Ada+4Frf_Dha+5Frf_Whl+5FtH_Frf+3FtH_Toj+3Toj_Ada+1Toj_Dha+2Toj_Whl
S.T.
1) 1FtB_Ada+1FtB_Dha+1FtB_Whl+1FtB_Frf+1FtB_Toj<14000
2) 1FtH_Ada+1FtH_Dha+1FtH_Whl+1FtH_Frf+1FtH_Toj<6000
3) 1FtB_Ada+1FtH_Ada+1Frf_Ada+1Toj_Ada=5000
4) 1FtB_Dha+1FtH_Dha+1Frf_Dha+1Toj_Dha=10000
5) 1FtB_Whl+1FtH_Whl+1Frf_Whl+1Toj_Whl=5000
6) 1FtB_Frf1Frf_Ada1Frf_Dha1Frf_Whl+1FtH_Frf=0
7) 1FtB_Toj+1FtH_Toj1Toj_Ada1Toj_Dha1Toj_Whl=0
OPTIMAL SOLUTION
Objective Function Value = 127000.000
Variable Value Reduced Costs
  
FtB_Ada 0.000 0.000
FtB_Dha 0.000 1.000
FtB_Whl 0.000 0.000
FtH_Ada 0.000 6.000
FtH_Dha 0.000 3.000
FtH_Whl 0.000 0.000
FtB_Frf 14000.000 0.000
FtB_Toj 0.000 0.000
Frf_Ada 5000.000 0.000
Frf_Dha 9000.000 0.000
Frf_Whl 0.000 0.000
FtH_Frf 0.000 5.000
FtH_Toj 6000.000 0.000
Toj_Ada 0.000 2.000
Toj_Dha 1000.000 0.000
Toj_Whl 5000.000 0.000
Constraint Slack/Surplus Dual Prices
  
1 0.000 0.000
2 0.000 3.000
3 0.000 7.000
4 0.000 7.000
5 0.000 8.000
6 0.000 3.000
7 0.000 6.000
OBJECTIVE COEFFICIENT RANGES
Variable Lower Limit Current Value Upper Limit
   
FtB_Ada 7.000 7.000 No Upper Limit
FtB_Dha 9.000 8.000 No Upper Limit
FtB_Whl 8.000 8.000 No Upper Limit
FtH_Ada 16.000 10.000 No Upper Limit
FtH_Dha 10.000 7.000 No Upper Limit
FtH_Whl 5.000 5.000 No Upper Limit
FtB_Frf No Lower Limit 3.000 3.000
FtB_Toj 6.000 6.000 No Upper Limit
Frf_Ada No Lower Limit 4.000 4.000
Frf_Dha 4.000 4.000 1.000
Frf_Whl 5.000 5.000 No Upper Limit
FtH_Frf 10.000 5.000 No Upper Limit
FtH_Toj No Lower Limit 3.000 3.000
Toj_Ada 5.000 3.000 No Upper Limit
Toj_Dha No Lower Limit 1.000 0.000
Toj_Whl 4.000 2.000 2.000
RIGHT HAND SIDE RANGES
Constraint Lower Limit Current Value Upper Limit
   
1 No Lower Limit 14000.000 No Upper Limit
2 No Lower Limit 6000.000 No Upper Limit
3 No Lower Limit 5000.000 No Upper Limit
4 No Lower Limit 10000.000 No Upper Limit
5 No Lower Limit 5000.000 No Upper Limit
6 No Lower Limit 0.000 No Upper Limit
7 No Lower Limit 0.000 No Upper Limit
This solution shows the general result of
utilizing the transshipment points to save costs (deployment time in
this case), when comparing this solution to Printout 7.1.1.
Side note for the careful reader. This is a small problem which you
could play around with pencil and paper. You might even find that
instead of deploying 14,000 troops from Fort Bragg to Frankfurt, for
transhipment on (5,000 from Frankfurt to Adana and 9,000 from
Frankfurt to Dhahran as in this solution) you could deploy 5,000
troops from Fort Bragg to Adana and 9,000 troops from Fort Bragg to
Frankfurt for onward movement to Dhahran. This solution also results
in a total deployment x days of 127,000. This is a special case in
linear programming called Alternate optimal solutions.
Different values for decision variables give the same total value for
the objective function. You can tell if a solution has alternative
optimal solution values by looking at the reduced costs for the
decision variables in the computer output. Note that the reduced cost
for FtB_Ada is 0 even though that variable is not in the solution.
That means that the amount the objective function reduces by is 0 if
that variable comes into solution. Thus, there is a tradeoff. We
could bring FtB_Ada into solution, reduce the deployment amount from
FtB_Frf, and eliminate the deployment from Frf_Ada. After these
adjustments, we get the same optimal solution value of 127,000.
That's it! You should be ready to tackle Case 8. I have a hint on
Case 8. Normally, transportation and transshipment problems involve
minimization of shipping or distribution costs. However, when more
than one production plant are used as origins, and when the
manufacturing costs are different, then we have to minimize
production + distribution costs over routes coming out of the origins
(production costs are not included in routes coming out of
transshipment points since no additional production takes place at
the transshipment point).
It just occurred to me that I did not include decision variables from
one transshipment point to another in my example (nor will you
include decision variables showing shipments between Ft. Worth, Sante
Fe and Las Vegas distribution centers in the case). In my experience,
the only time we ship between transshipment points is when somebody
screwed up! That is, product should always move forward to the
customer  no value would be added to ship between warehouses.
I take that back. I remember shipping the same cargo back and forth
between Cam Ranh Bay and Saigon distribution centers before a Tet
Offensive in 1968. I did this so we could show aircraft utilization
in the II Corps area, to keep the headquarters from pulling aircraft
out of the theater during a lull in operations. Thus, when activity
picked up, we did not have to wait a day or two to get aircraft
redeployed. ...... ..end of Module 7.


