Using result of a decisionvariable for further computations

回答済み

2020年04月03日 14:05 編集済み

Hello,

I have the following problem (I am working with Python):

I have a quadratic constraint with binary variables x and y, which computes the sum of a product of x*y*some value of a dictionary, where only one x and one y is 1. The result of the sum is between -5 and 5. J1 and J2 have about 100 entries each. It looks like

model.addConstr(z == quicksum(y[j1] * x[j2] * dictionary1(j1,j2) for j1 in J1 for j2 in J2))

Now I need to add another value, which depends on this result. It is a lookup in some second dictionary and would simplified look like:

model.addConstr(a == dictionary2[z])) s.t. the endresult will look like:

model.addConstr(result == z+a)

As I am a beginner in Gurobi, I am not sure, whether it is the same problem as this one and how I need to change my model to implement it correctly:

https://support.gurobi.com/hc/en-us/community/posts/360055448071-Is-Gurobi-able-to-handle-a-decision-variable-in-the-index-as-a-boundary-in-a-sum-

Or whether it is not that similar and there is a trick or something that I do not see.

Thanks in advance

Simon

6件のコメント

正式なコメント
Simranjit Kaur

Gurobi Staff

2025年06月23日 03:15 編集済み
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Eli Towle

Gurobi Staff

2020年04月03日 22:50
Hi Simon,

The current formulation won't work because you cannot use model variables to index other variables/parameters. However, we can model this as follows. Let $ c $ denote the first dictionary, $ d $ the second dictionary, and $ v $ the variable you call "result" in your code. We add the following constraint to the model:

$$\begin{align*}v &= \sum_{i \in J_1} \sum_{j \in J_2} y_i x_j(c_{ij} + d_{c_{ij}}). \end{align*}$$

Because exactly one $ y_i $ and one $ x_j $ are $ 1 $, exactly one $ y_i x_j $ term in the summation is equal to $ 1 $ (the rest are $ 0 $). Thus, the equation simplifies to $ v = c_{ij} + d_{c_{ij}} $, where $ y_i = x_j = 1 $.

Translated to Python code, this is:
```
model.addConstr(result == quicksum(y[i] * x[j] * (dictionary1[i,j] + dictionary2[dictionary1[i,j]]) for i in J1 for j in J2))
```
Could you try this out? Thanks!

Eli
0
Simon Lackmann

2020年04月04日 00:24
Hey Eli,

Thank you for your reply. Your answer would perfectly fit my simplified model. Unfortunately, my simplified model is too loose, because I made a mistake simplifying it and so the suggested solution wouldn't work for me. Sorry for that!

I'll give it a more precise try, please have a look at the following:

I have an array, let's say 'membersArray' and I need to calculate the sum for each member. If this array has 3 entries,

I would get three constraints, which I can sum up to: result == z[firstMember]+z[secondMember]+z[thirdMember].

In my python code, it looks like:
```
for member in membersArray:
    model.addConstr(z[member] == quicksum((y[j1]*x[member,j2] * dictionary1[j1,j2] for j1 in J1 for j2 in J2)
```
This 'result' value, which I could define as:
```
model.addConstr(result == quicksum(z[member] for member in membersArray)
```
is what I need to look up in the second dictionary (for the only y which is 1, so I don't need the j2 for this).

The second dictionary look-up would look like:
```
dictionary2[result]
```
As I cannot use model variables to index other variables or parameters, this wouldn't be feasible, would it?

Simon
0
Simon Lackmann

2020年04月04日 12:43
As the result is some Integer value between -5 and 5, I thought about something like an implication. But the only way I could imagine this is as follows (with an example value of 3):
```
 model.addConstr(result==3 <= u_3)
```
where u is some binary variable for each of the 11 outcomes. But of course this would also not work, as 'result' has no bool value. Is there any workaround?

Simon
0
Eli Towle

Gurobi Staff

2020年04月06日 17:30
Hi Simon,

This should still be possible with a different formulation. Again, let $ v $ be the variable you call "result," and let $ d $ be the second dictionary. In addition to the constraints you have written, we introduce variables $ u \in \{0,1\}^{11} $ (there are 11 possible values for $ v $). Then, we add the following constraints:

$$\begin{align*}\sum_{i = 1}^{11} (i-6) u_i &= v \\ \sum_{i = 1}^{11} u_i &= 1. \end{align*}$$

The idea is that exactly one the $ u_i $ variables will be equal to $ 1 $, and this index $ i $ will satisfy $ v = i - 6 $. Finally, we introduce a new variable $ w $ and set it equal to the value of "dictionary2[result]" with the constraint:

$$w = \sum_{i = 1}^{11} d_{i - 6} u_i.$$

In Python, we can equivalently write this as:
```
# Integers from -5 to 5
M = range(-5,6)

# Add auxiliary variables
u = model.addVars(M, vtype=GRB.BINARY, name="u")
w = model.addVar(lb=-GRB.INFINITY, name="w")

# Exactly one u_i is 1, such that result = i
model.addConstr(u.sum() == 1)
model.addConstr(result == quicksum(i*u[i] for i in M))

# Set w equal to dictionary2[result]
model.addConstr(w == u.prod(dictionary2))
```
Does this work? Thanks!

Eli
0
Simon Lackmann

2020年04月07日 22:15
Hello Eli,

Thanks again for your reply. Your approach helped me to understand my problem and I was able to solve it :)

Simon
0

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