I'm having difficulties with the logic with the last part of the reformulation.


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asked 08 Apr '16, 06:06

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BCLC
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edited 07 May '16, 07:23


You can omit #3 and #4 in your final constraints because the logical implications you want to model are only one direction.

Also, make sure you use smallish values (separate for each constraint) of \(M\), based on upper bounds on \(x_i\). You can derive such upper bounds from the other constraints in the problem.

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answered 08 Apr '16, 11:00

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Rob Pratt
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Are you sure? What if \( y_3 = 0 \) ?

(08 Apr '16, 11:26) BCLC

Yes. If \(y_3 = 0\), then constraint #1 does nothing and constraint #2 forces \(x_1 \le 3\), just like your logical implications specify. You do not need to model the converse implications.

(08 Apr '16, 13:49) Rob Pratt

Rob Pratt, what about \( x_2 \ge 4 \)? I think that is in where, for example, constraint 3 would come

(08 Apr '16, 14:00) BCLC

Constraint #3 models the converse, if \(y_3 = 0\) then \(x_2 \ge 4\), but you do not need that. Statement b.iii in the problem is only a one-way if-then logical implication, not a two-way if-and-only-if.

(08 Apr '16, 14:29) Rob Pratt

Only constraints #1 and #2: If \(y_3 = 1\), then \(x_2 \le 3\). If \(y_3 = 0\), then \(x_2 \le 3 + M\) and \(x_1 \le 3\)...right? Is \(x_2 \le 3 + M\) supposed to be the same as \(x_2 \ge 4\)?

(08 Apr '16, 14:36) BCLC

Let \(x_i\) be the the number of ships of type \(i\) to purchase.


For 4a:

\[\min \ z = 20,000x_1 + 1,000x_2\]

s.t.

  1. \[2,000x_1 + 1,000x_2 \le 7,500\]

  2. $$12,000x_1 + 7,000x_2 \le 55,000$$

  3. $$250x_1 + 100x_2 \le 900$$

  4. $$x_1, x_2 \in {0,1,2,...}$$


For 4b:

\[\min \ z = 20,000x_1 + 1,000x_2 \color{red}{+ 2,000y_1 + 1,000y_2}\]

s.t.

  1. same

  2. same

  3. same

  4. same

  5. $$x_2 \le My_1, y_1 \in {0,1}$$

  6. $$x_2 - 2 \le My_2, y_2 \in {0,1}$$

  7. ...

I think we have to say, linearly, that:

  1. if \(x_2 \ge 4\), then \(y_3 = 0\).

  2. if \(y_3 = 0\), then \(x_1 \le 3\).

then exactly one of the following (also linearly):

  1. converse of 'if \(x_2 \ge 4\), then \(y_3 = 0\)'

  2. converse of 'if \(y_3 = 0\), then \(x_1 \le 3\)'

and also

  1. \(y_3\) is binary.

I believe these translate to:

  1. $$x_2 - 3 \le M(1-y_3)$$

  2. $$x_1 - 3 \le M(y_3)$$

  3. $$4 - x_2 \le My_3$$

  4. $$4 - x_1 \le M(1-y_3)$$

  5. $$y_3 \in {0,1}$$

Is that right? I think if I use 4 instead of 3, I'm going to get the contrapositive of \((iii)\).

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answered 07 May '16, 07:23

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Asked: 08 Apr '16, 06:06

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Last updated: 07 May '16, 07:23

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