How did $s_2$ suddenly become negative? What did I do wrong?
You've chosen the wrong entering variable. This give rise to an unfeasible solution.
To set up the initial tableau, multiply the second equation of the system
$$
\left\{
\begin{array}{c}
x - s_1 + a_1=1\\
-2x +y - s_2=0,
\end{array}
\right.
$$
by $-1$, so that the coefficient of the current basic variable $s_2$ becomes $1$. This makes the tableau proper.
\begin{array}{c|ccccc|cc}
&x&y&s_1&s_2&a_1&A&\text{ratio}\\
\hline
a_1&1&0&-1&0&1&1 & 1/1 = 1\\
s_2&\fbox{2}&-1&0&1&0&0 & \fbox{0/2 = 0}\\
\hline
p&-1&0&1&0&0&-1
\end{array}
For feasibility reason, one has to pivot around $(1,2)$.
\begin{array}{c|ccccc|c}
&x&y&s_1&s_2&a_1&A\\
\hline
a_1&0&\fbox{1/2}&-1&-1/2&1&\fbox1\\
x&1&-1/2&0&1/2&0&0\\
\hline
p&0&-1/2&1&1/2&0&-1
\end{array}
\begin{array}{c|ccccc|c}
&x&y&s_1&s_2&a_1&A\\
\hline
y&0&1&-2&-1&2&2\\
x&1&0&-1&0&1&1\\
\hline
p&0&0&0&0&1&0
\end{array}
This gives us a feasible nondegenerate solution $(x,y) = (1,2)$.
To start phase II, we discard the column representing artificial variable $a_1$, calculate the objective function row, and reuse the rest of the above tableau.
$$\min Z = y \iff \max z = -y$$
\begin{array}{c|cccc|c}
&x&y&s_1&s_2&A\\
\hline
y&0&1&-2&-1&2\\
x&1&0&-1&0&1\\
\hline
z&0&1&0&0&0
\end{array}
To make it a simplex tableau, make the entries representing the current basis zero at the objective function row.
\begin{array}{c|cccc|c}
&x&y&s_1&s_2&A\\
\hline
y&0&1&-2&-1&2\\
x&1&0&-1&0&1\\
\hline
z&0&0&2&1&-2
\end{array}
Therefore, the solution is $(x,y) = (1,2)$ with $Z = -(-2) = 2$. Note that it is nondegenerate and unique because we have strictly positive entries on the RHS and entries representing nonbasic variables at the objective function row.
A quicker solution using dual simplex method.
Observe that that $\max z = -y$ has nonpositve coefficients. This gives rise to a nonnegative objective function row, so optimality is satisfied. This allows us to use dual simplex method.
Rewrite the LPP as
$\max z = -y$ where $-x \le -1$ and $2x-y \le 0, x,y \ge 0$.
Note that $y \ge 2x \ge 2(1) \ge 1 > 0$, so adding $x,y \ge 0$ doesn't change the problem.
\begin{array}{c|cccc|c}
&x&y&s_1&s_2&A\\
\hline
s_1&\fbox{-1}&0&1&0&\fbox{-1}\\
s_2&2&-1&0&1&0\\
\hline
z&0&1&0&0&0 \\
\text{ratio} & \fbox0 & -
\end{array}
\begin{array}{c|cccc|c}
&x&y&s_1&s_2&A\\
\hline
x&1&0&-1&0&1\\
s_2&0 & -1 & 2 & 1 & \fbox{-2}\\
\hline
z&0&1&0&0&0 \\
\text{ratio} & & \fbox{-1} & -
\end{array}
\begin{array}{c|cccc|c}
&x&y&s_1&s_2&A\\
\hline
x&1&0&-1&0&1\\
y&0 & 1 & -2 & -1 & 2 \\
\hline
z&0&0&2&1&-2
\end{array}
Observe that we arrived at the same tableau much quicker than the usual two phase method.
Your answers are basically correct.
Even if the primal is not optimal, do we just adapt the (negative) coefficient of objective function from the primal.
Yes, the current dual solution can be always obtained in this way.
And how can I check the feasibility of dual solution?
The dual is feasible since all dual variables are non-negative.
[In item c)] how to analyze of dual problem with only this given information?
You can't continue since the dual is unbounded. Thus, by weak duality the primal is infeasible.
Best Answer
There are two options in the case of minimizing problem for choosing a pivot column:
You multiply the objective function $z$ by $(-1) $ and maximize $-z$, which means that the coefiffients of the objective function have the opposite sign in the table, $+z$. Then you choose the pivot column by the following rule: Choose the colomn as pivot colomn with the most negative value at the row of the objective function. You stop pivoting if the row of the objective function constains non-negative values only.
You minimize $z$, which means that the coefiffients of the objective function have the opposite sign in the table, $-z$. Then you choose the pivot column by the following rule: Choose the colomn as pivot colomn with the most positive value at the row of the objective function. You stop pivoting if the row of the objective function constains only non-positive values. Then your optimal value for the objective function is the negative value of the RHS.
You have the second case here. So your next pivot column is $s_2$. But you cannot choose it since the ratios are negative. So there exists no solution for the LP if you have mode no flaws in the the steps before.