Consider the following linear problem

$$\max tx_1+x_2\\ s.t. 4x_1+3x_2\le12 \\ 3x_1+4x_2\le12\\ x_1,x_2\ge0$$

where the parameter $t$ grows exponentially $t\in[1,\infty).$

Find the sequence of basic optimal solutions.


First we should convert to standar form

$$\max tx_1+x_2\\ s.t.\ 4x_1+3x_2+x_3=12 \\ 3x_1+4x_2+x_4=12\\ x_1,x_2\ge0$$

Then, I am not sure how should I proceed.

I've solved exercise where you have say $3$ instead of $t$ and then you change $3$ by $4$ and then analyze how the final simplex tableau changes, as it is a coefficient of a basic variable the optimality and factibility might both be affected.

But this could be fixed by using again simplex method or dual-simplex in the final tableau.

My question is how should I proceed in this case where there is a $t$ instead of a number?

Could someone help me please?

I would really appreciate any help you're willing to provide. Thank you.

  • $\begingroup$ Just iterate through a couple of times, and see how the process depends on $t$, and where you end up $\endgroup$ – gt6989b Dec 5 '18 at 19:45

You can work geometrically. The domain is given in the picture below :

enter image description here

Now, the gradient of $f(x_1,x_2) = tx_1+x_2$ equals $\pmatrix{t\\1}$.

If $t=1$, the optimum lies at the intersection of $3x_1+4x_2=12$ and $4x_1+3x_2=12$, i.e. $(\frac{12}{7},\frac{12}{7})$.

If $t=\frac{4}{3}$, the gradient is orthogonal to $4x_1+3x_2=12$, which means that all points on this line are optimal, in particular $(\frac{12}{7},\frac{12}{7})$ and $(3,0)$.

So for any $t\in [1,\frac{4}{3}[$, $(\frac{12}{7},\frac{12}{7})$ is the optimal point.

Then, as $t$ increases, the gradient points towards $(3,0)$.

Note that this matches gt6989b's answer : solving $3t = \frac{12(t+1)}{7}$ for $t$ yields $t=\frac{4}{3}$.

  • $\begingroup$ Thank you. What would be the sequence of basic optimal solutions? $\endgroup$ – user441848 Dec 5 '18 at 21:06
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    $\begingroup$ $(12/7,12/7)$, $(3,0)$, $(3,0)$, $(3,0)$,....... $\endgroup$ – Kuifje Dec 5 '18 at 21:09
  • $\begingroup$ ok. Could you explain this part if $t=1$ the optimum lies at the intersection of $3x_1+4x_2=12$ and $3x_1+4x_2=12$ ? $\endgroup$ – user441848 Dec 5 '18 at 21:29
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    $\begingroup$ If $t=1$, the gradient is $(1,1)$, so it points in the direction of the line $x_2=x_1$. Since $(12/7,12/7)$ lies on this line, the gradient points towards $(12/7,12/7)$. $\endgroup$ – Kuifje Dec 5 '18 at 22:25
  • $\begingroup$ ok I understand that part, I don't see how the gradient points towards $(3,0)$ when $t$ increases; there is a constant $1$, how can you know that it will point towards $(3,0)$ ? $\endgroup$ – user441848 Dec 5 '18 at 22:33

Here is one high-level approach. Your region looks like this:

LP Plot

Easy to see $(0,0)$ produces the optimization value 0, so only 3 points are in play: $(3,0),(0,3)$ and $(12/7,12/7)$.

Which one will get chosen and how will the optimization converge?


You are maximizing $f(x_1,x_2) = tx_1 + x_2$. It produces the following values at the vertices:

$$ f(0,3) = 3, f(3,0) = 3t, f(12/7,12/7) = (t+1)12/7 $$

As $t$ evolves, which of the values is larger?

  • $\begingroup$ $(3,0)$ will get chosen, I don't know how the optimization will converge. How? $\endgroup$ – user441848 Dec 5 '18 at 20:07
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    $\begingroup$ Nice picture ! ^^ $\endgroup$ – Kuifje Dec 5 '18 at 20:15
  • $\begingroup$ @Alt. see update, not so simple $\endgroup$ – gt6989b Dec 5 '18 at 20:18
  • $\begingroup$ $f(3,0)=3t$ is the larger value $\endgroup$ – user441848 Dec 5 '18 at 20:26
  • $\begingroup$ Am I correct? ${}$ $\endgroup$ – user441848 Dec 5 '18 at 20:28

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