Difference: CokeSupplyChain (18 vs. 19)

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Case Study: The Coke Supply Chain Problem

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FORM FIELD ComputationalModel	ComputationalModel	The computational model...
FORM FIELD Results	Results	The results...
FORM FIELD Conclusions	Conclusions	In conclusion...

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|*FORM FIELD ExtraForExperts*|ExtraForExperts|This problem is similar to the Cosmic Computers Problem, but the facility location is a little different. Rather than face a decision about whether to build a plant or not, we need to determine what type of plant to build. This gives rise to the following variable:

$y_{ps} = 1 \text{ if plant $p$ is built with size $s$, 0 otherwise}$

While solving the Coke Supply Chain Problem, the expression

$\sum_{s \in \text{sizes}} y_{ps}$

will often be fractional. This means that the capacity of the plant

$\sum_{s \in \text{sizes}} C_s y_{ps}$

where

is the capacity of a plant of size

will be between the capacities of two different size plants, e.g., 280 kT/yr is between 225 kT/yr plant and a 300 kT/yr plant. We could use constraints to remove the fractionality, e.g.,

$\begin{array}{r@{\,}l@{\quad\quad}l} \displaystyle \sum_{s \in \text{sizes}} C_s y_{ps} &\leq 225 & \text{(branching down)} \\ \displaystyle \sum_{s \in \text{sizes}} C_s y_{ps} &\geq 300 & \text{(branching up)} \end{array}$

These branches will ensure plants have the right capacities, but will still not resolve fractionalities in the $y_{ps}$ . However, if we want the capacity of plant to be $\leq$ 225, then we can remove all the $y_{ps}$ that allow for the capacity to be larger than 225:

$\sum_{s \in \text{sizes} | C_s > 225} y_{ps} = 0$

Conversely, if we want to branch up, i.e., the capacity is $\geq$ 300, then we can remove all the $y_{ps}$ that allow the capacity to be smaller than 300:

$\sum_{s \in \text{sizes} | C_s < 300} y_{ps} = 0$

Note For these branches to work properly, the possibility of building no plant must be modeled as building a plant with no capacity (i.e., = 0) for no cost. |

>
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$y_{ps} = 1 \text{ if plant $p$ is built with size $s$, 0 otherwise}$

While solving the Coke Supply Chain Problem, the expression

$\sum_{s \in \text{sizes}} y_{ps}$

will often be fractional. This means that the capacity of the plant

$\sum_{s \in \text{sizes}} C_s y_{ps}$

where

is the capacity of a plant of size

will be between the capacities of two different size plants, e.g., 280 kT/yr is between 225 kT/yr plant and a 300 kT/yr plant. We could use constraints to remove the fractionality, e.g.,

$\sum_{s \in \text{sizes} | C_s > 225} y_{ps} = 0$

Conversely, if we want to branch up, i.e., the capacity is $\geq$ 300, then we can remove all the $y_{ps}$ that allow the capacity to be smaller than 300:

$\sum_{s \in \text{sizes} | C_s < 300} y_{ps} = 0$

Note For these branches to work properly, the possibility of building no plant must be modelled as building a plant with no capacity (i.e., = 0) for no cost. |

|*FORM FIELD StudentTasks*|StudentTasks|

Write AMPL model, data and script files (coke.mod, coke.dat and coke.run respectively) to solve the Coke Production Problem. Write a management summary for your solution.
What to hand in Your 3 AMPL files. Your management summary.
Experts Only Solve the LP relaxation of the Coke Supply Chain Problem integer programme. Using the branches described in Extra for Experts, use AMPL to perform branching on at least 4 nodes and draw the resulting branch-and-bound tree.
Hint See Extra for Experts in the Cosmic ComputersProblem for an example of constraint branching in AMPL.
What to hand in Your AMPL code for implementing the branches. Your drawing of your branch-and-bound tree.

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