"I am your cool tool from the motor pool…we have 2-by's, 4-by'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 (Troops Available)						Destination (Troops Needed)
Fort Bragg (14,000)	To Ada --> To Dha --> To Whl -->				-> 7 Days From FtB -> 10 Days From FtH	Adana (5,000)
					-> 8 Days From FtB -> 7 Days From FtH	Dhahran (10,000)
Fort Hood (6,000)	To Ada --> To Dha --> To Whl -->				-> 8 Days from FtB -> 5 Days From FtH	Wheelus (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

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

Origin (Troops Available)						Destination (Troops Needed)
Fort Bragg (14,000)	To Ada --> To Dha --> To Whl --> To Frf ---> To Toj -->				-> 7 Days from FtB -> 10 Days from FtH -> 4 Days from Frf -> 3 Days from Toj	Adana (5,000)
	-> 3 Days from FtB		Frankfurt
	-> 5 Days from FtH
					-> 8 Days from FtB -> 7 Days from FtH -> 4 Days from Frf -> 1 Day from Toj	Dhahran (10,000)
	-> 6 Days from FtB
	-> 3 Days from FtH		Torrejon
Fort Hood (6,000)	To Ada --> To Dha --> To Whl --> To Frf ---> To Toj -->				-> 8 Days from FtB -> 5 Days from FtH -> 5 Days from Frf -> 2 Days from Toj	Wheelus (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 out-of-state 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_Whl

FtB_Frf + FtB_Toj < 14,000

Fort 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 = 5,000

Dhahran Demand: FtB_Dha + FtH_Dha

+ Frf_Dha + Toj_Dha = 10,000

Wheelus 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_Frf-1Frf_Ada-1Frf_Dha-1Frf_Whl+1FtH_Frf=0

7) 1FtB_Toj+1FtH_Toj-1Toj_Ada-1Toj_Dha-1Toj_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 trade-off. 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.