1. **Problem Statement:** We have four raw materials $R_1, R_2, R_3, R_4$ with constraints on their availability and usage, and costs associated with each. We want to construct a linear programming (LP) model and find an initial basic feasible solution. 2. **Define Variables:** Let $x_1, x_2, x_3, x_4$ be the quantities of raw materials $R_1, R_2, R_3, R_4$ used per day. 3. **Constraints from the problem:** - Availability of $R_1$ is thrice the required quantity: $x_1 \leq 3x_1$ is trivial, so this means $x_1$ is limited by availability, but since it says "availability is thrice the required quantity," it implies availability is $3x_1$, so no upper bound on $x_1$ from this directly. We interpret this as $x_1$ can be up to some maximum, but no explicit max given, so no constraint here. - Availability of $R_3$ is twice the required quantity: similarly, $x_3 \leq 2x_3$ is trivial, so no direct constraint. But the problem likely means availability is twice the required quantity, so $x_3$ can be up to some max, but no explicit max given. - Maximum availability of $R_4$ is 50 units: $x_4 \leq 50$ - Quality assurance conditions: - $x_2$ cannot exceed $x_1$ by more than 10 units: $x_2 - x_1 \leq 10$ - $x_2$ cannot exceed $x_3$ by more than 20 units: $x_2 - x_3 \leq 20$ - Combined storage capacity for raw materials per day is 100 units: $x_1 + x_2 + x_3 + x_4 \leq 100$ - Non-negativity: $x_i \geq 0$ for $i=1,2,3,4$ 4. **Objective function:** Minimize total cost: $$\text{Minimize } Z = 45x_1 + 25x_2 + 65x_3 + 5x_4$$ 5. **Linear Programming Model:** $$\begin{cases} \text{Minimize } Z = 45x_1 + 25x_2 + 65x_3 + 5x_4 \\ \text{subject to:} \\ x_4 \leq 50 \\ x_2 - x_1 \leq 10 \\ x_2 - x_3 \leq 20 \\ x_1 + x_2 + x_3 + x_4 \leq 100 \\ x_1, x_2, x_3, x_4 \geq 0 \end{cases}$$ 6. **Finding an Initial Basic Feasible Solution (IBFS):** We introduce slack variables $s_1, s_2, s_3, s_4$ for the inequalities: $$\begin{cases} x_4 + s_1 = 50 \\ x_2 - x_1 + s_2 = 10 \\ x_2 - x_3 + s_3 = 20 \\ x_1 + x_2 + x_3 + x_4 + s_4 = 100 \\ x_i, s_j \geq 0 \end{cases}$$ Choose basic variables as slack variables $s_1, s_2, s_3, s_4$ and set $x_1 = x_2 = x_3 = x_4 = 0$: - From $x_4 + s_1 = 50$, with $x_4=0$, $s_1=50$ - From $x_2 - x_1 + s_2 = 10$, with $x_1=0, x_2=0$, $s_2=10$ - From $x_2 - x_3 + s_3 = 20$, with $x_2=0, x_3=0$, $s_3=20$ - From $x_1 + x_2 + x_3 + x_4 + s_4 = 100$, with all $x_i=0$, $s_4=100$ All variables are non-negative, so this is a feasible solution. **Final answer:** - LP model as above. - Initial basic feasible solution: $x_1=0, x_2=0, x_3=0, x_4=0, s_1=50, s_2=10, s_3=20, s_4=100$.