Math

Variables

• A boolean variable $\text{open}_i$ per warehouse i is needed, set to true if the corresponding warehouse is open, false otherwise.

  $$\forall i \in [1,5],\, \text{open}_i = \{0,1\}$$

• An integer variable $\text{supplier}_j$ per store j is needed, it indicates which warehouse supplies it.

  $$\forall j \in [1,10],\, \text{supplier}_j = [\![1,5]\!]$$

• An integer variable $\text{cost}_j$ per store j is needed too, it stores the cost of being supplied by a warehouse (the range is deduced from the matrix P).

  $$\forall j \in [1,10],\, \text{cost}_j = [\![1, 96]\!]$$

• An integer variable $tot_cost$ totals all costs:

  $$tot_{cost} = [\![1, {+\infty})$$

Constraints

• if a warehouse i supplies a store j, then, the warehouse is open:

  $$\forall j \in [1,10], \text{open}_{\text{supplier}_j} = 1$$

  Here $\text{supplier}_j$ defines the index the array $\text{open}$ to be valuated to 1. This is a encoded with an element constraint.

• if a warehouse i supplies a store j, it is related to a specific cost:

  $$\forall j \in [1,10], P_{j,\text{supplier}_{j}} = \text{cost}_j$$

  Here again, an element constraint is used to bind the supplier and the supply cost matrix to the cost of supplying a store.

• the maximum number of stores a warehouse i can supply is limited to $K_i$:

  $$\forall i \in [1,5], \sum_{j = 1}^{10} (\text{supplier}_j == i) = \text{occ}_i$$ $$\forall i \in [1,5], \text{occ}_i \leq K_i$$ $$\forall i \in [1,5], \text{occ}_i \geq \text{open}_i$$

  The first constraint counts the number of occurrences of the value i in the array supplier and stores the result $\text{occ}_i$ variable. This variable is then constrained to be less than or equal to $K_i$, to ensure the capacity is satisfied, but also to be greater or equal to $\text{open}_i$ to better propagation.

• the assignment cost has then to be maintained, including fixed costs and supplying costs:

  $$tot_{cost} = \sum_{i = 1}^{5} 30 \cdot \text{open}_i + \sum_{j = 1}^{10} \text{cost}_j$$

Objective

The objective is not to simply find a solution but one that minimizes $tot_{cost}$.