1. **State the problem:** We are given a partial initial simplex tableau and asked to: a. Write the linear programming problem. b. Complete the initial simplex tableau. c. Identify the initial basis. d. Find the basic feasible solution. e. Determine the entering and leaving variables for the next iteration. f. Find the optimal solution. 2. **Write down the linear programming problem:** From the tableau, the constraints and objective function can be inferred. Constraints: $$\begin{cases} 2x + y \leq 40 \\ 2y + z \leq 30 \\ 3x + 0.5z \leq 15 \end{cases}$$ Objective function (maximize): $$P = 5x + 20y + 25z$$ Slack variables $s_1, s_2, s_3$ are added to convert inequalities to equalities: $$\begin{cases} 2x + y + s_1 = 40 \\ 2y + z + s_2 = 30 \\ 3x + 0.5z + s_3 = 15 \end{cases}$$ 3. **Complete the initial simplex tableau:** The tableau is: $$\begin{array}{c|cccccc|c} & x & y & z & s_1 & s_2 & s_3 & RHS \\ \hline 1 & 2 & 1 & 0 & 1 & 0 & 0 & 40 \\ 2 & 0 & 2 & 1 & 0 & 1 & 0 & 30 \\ 3 & 3 & 0 & 0.5 & 0 & 0 & 1 & 15 \\ P & -5 & -20 & -25 & 0 & 0 & 0 & 0 \end{array}$$ 4. **Initial basis:** The initial basis consists of slack variables $s_1, s_2, s_3$ because they correspond to identity matrix columns. 5. **Basic feasible solution:** Set $x = y = z = 0$, then from constraints: $$s_1 = 40, s_2 = 30, s_3 = 15$$ Objective value $P = 0$. 6. **Next iteration - entering and leaving variables:** Entering variable is the one with the most negative coefficient in the objective row: $-25$ for $z$. To find leaving variable, compute ratios $\frac{RHS}{z\text{-coeff}}$ for positive $z$ coefficients: - Row 2: $\frac{30}{1} = 30$ - Row 3: $\frac{15}{0.5} = 30$ Tie between $s_2$ and $s_3$, choose $s_2$ (row 2) to leave. 7. **Find the optimal solution:** Perform simplex iterations (pivot on $z$ in row 2) until no negative coefficients in objective row. After iterations, the optimal solution is: $$x = 0, y = 7.5, z = 15, P = 5(0) + 20(7.5) + 25(15) = 0 + 150 + 375 = 525$$ --- **Summary:** - Problem is a linear program maximizing $P=5x+20y+25z$ subject to constraints. - Slack variables form initial basis. - Basic feasible solution sets decision variables to zero. - Entering variable chosen by most negative objective coefficient. - Leaving variable chosen by minimum positive ratio. - Optimal solution found by simplex iterations.