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I am an electrical engineer and currently I am learning some optimization for my research work. Note that this is a follow up question from my previous question at this link.

My project requires me to solve this problem and I am not sure about two things:

1/ Does case 1 easier than case 2 or does case 2 easier than case 1 ?

2/ For both cases, are there a special structure in the matrix $A$ that I can exploit ?

Expoit can mean anything, even if it means that I could solve it more faster or decompose this problem into other smaller subproblems since I have some distributed computing capability.

This is how case 1 looks:

$\begin{array}{l} \begin{array}{*{20}{c}} {{\rm{Min}}}&{{x_1} + {x_2} + {x_3} + {x_4} + {x_5} + {x_6}} \end{array}\\ \begin{array}{*{20}{c}} {}&{\underbrace {\left[ {\begin{array}{*{20}{c}} 1&1&0&{}&{}&{}\\ 0&1&1&{}&{}&{}\\ {}&{}&{}&1&1&0\\ {}&{}&{}&0&1&1 \end{array}} \right]}_A\left[ {\begin{array}{*{20}{c}} {{x_1}}\\ {{x_2}}\\ {{x_3}}\\ {{x_4}}\\ {{x_5}}\\ {{x_6}} \end{array}} \right] \ge \left[ {\begin{array}{*{20}{c}} a\\ b\\ c\\ d \end{array}} \right]}\\ {}&{a + c \le 1}\\ {}&{b + d \le 1}\\ {}&{{x_i},a,b,c,d \in \left\{ {0,1} \right\}} \end{array} \end{array}$

This is how case 2 looks:

$\begin{array}{l} \begin{array}{*{20}{c}} {{\rm{Min}}}&{{x_1} + {x_2} + {x_3} + {x_4} + {x_5} + {x_6}} \end{array}\\ \begin{array}{*{20}{c}} {}&{\underbrace {\left[ {\begin{array}{*{20}{c}} 1&1&0&{}&{}&{}\\ 0&1&1&{}&{}&{}\\ {}&{}&{}&1&1&0\\ {}&{}&{}&0&1&1 \end{array}} \right]}_A\left[ {\begin{array}{*{20}{c}} {{x_1}}\\ {{x_2}}\\ {{x_3}}\\ {{x_4}}\\ {{x_5}}\\ {{x_6}} \end{array}} \right] \ge \left[ {\begin{array}{*{20}{c}} a\\ b\\ c\\ d \end{array}} \right]}\\ {}&{a + c = 1}\\ {}&{b + d = 1}\\ {}&{{x_i},a,b,c,d \in \left\{ {0,1} \right\}} \end{array} \end{array}$

For more information, a larger scale of this problem with $x_7$, $x_8$ and $x_9$ would make the matrix $A$ into this form which is quite sparse:

$\underbrace {\left[ {\begin{array}{*{20}{c}} 1&1&0&{}&{}&{}&{}&{}&{}\\ 0&1&1&{}&{}&{}&{}&{}&{}\\ {}&{}&{}&1&1&0&{}&{}&{}\\ {}&{}&{}&0&1&1&{}&{}&{}\\ {}&{}&{}&{}&{}&{}&1&1&0\\ {}&{}&{}&{}&{}&{}&0&1&1 \end{array}} \right]}_A\left[ {\begin{array}{*{20}{c}} {{x_1}}\\ {{x_2}}\\ {{x_3}}\\ {{x_4}}\\ {{x_5}}\\ {{x_6}}\\ {{x_7}}\\ {{x_8}}\\ {{x_9}} \end{array}} \right] \ge \left[ {\begin{array}{*{20}{c}} a\\ b\\ c\\ d\\ e\\ f \end{array}} \right]$

The constraints on $a,b,c,d,e,f$ of the larger case would look like this

$\begin{array}{l} a + c + e \le 1\\ b + d + f \le 1 \end{array}$

Or this

$\begin{array}{l} a + c + e = 1\\ b + d + f = 1 \end{array}$

Would you kindly help me with this ?

Thank you for your enthusiam !

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    $\begingroup$ Are $a,b,c,d$ given values and $x_j$ decision variables? If so, what prevents you to solve the "size 1" problem by hand and then just combine the solutions of $n$ such problems into one? $\endgroup$ Commented Nov 26, 2023 at 21:31
  • $\begingroup$ A b c d are also decision variables but in some special situation, I can pre-solve them so that they become a given value. Could you elaborate more on this size 1 tactic ? $\endgroup$ Commented Nov 26, 2023 at 22:44
  • $\begingroup$ @RobPratt I have editted my question regarding $a,b,c,d,e,f$ for clarity. $a,b,c,d,e,f$ are just some control variables from a previous question that you have successfully help me. Sorry for not posting the link in the first draft of the question, I have editted my question probably. $\endgroup$ Commented Nov 27, 2023 at 2:04
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    $\begingroup$ OK, my answer below still applies. $\endgroup$ Commented Nov 27, 2023 at 2:27

1 Answer 1

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Case 1 is trivial: set all variables to $0$, yielding optimal objective value $0$.

Case 2 is almost as trivial. The zero solution is infeasible, but you can set $x_2=a=b=1$ and all other variables $0$, yielding optimal objective value $1$.

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