Question

A company must allocate oil produced from their two plants to meet all the demands of two markets. The cost per litre of allocation from plant \( i \in \{1,2\} \) to market \( j \in \{1,2\} \) is denoted by \( C_{ij} \). The market demand \( D_j \), plant production capacity \( K_i \) and \( C_{ij} \) values are given in the table.

	Market 1 (j = 1)	Market 2 (j = 2)	Capacity \( K_i \) (litres)
Plant 1 (i = 1)	\( C_{11} = 250 \)	\( C_{12} = 280 \)	\( K_1 = 500 \)
Plant 2 (i = 2)	\( C_{21} = 150 \)	\( C_{22} = 180 \)	\( K_2 = 600 \)
Demand \( D_j \) (litres)	\( D_1 = 300 \)	\( D_2 = 400 \)

The company hired an intern to formulate an optimization model to decide on the quantity (\( X_{ij} \)) to be allocated from plant \( i \) to market \( j \), and the formulation is given below:

Minimize \( Z = \sum_{i=1}^{2} \sum_{j=1}^{2} C_{ij} X_{ij} \)

Subject to:
\( \sum_{j=1}^{2} X_{ij} \le K_i \quad \forall i \in \{1,2\} \)
\( \sum_{i=1}^{2} X_{ij} \le D_j \quad \forall j \in \{1,2\} \)
\( X_{ij} \ge 0 \)

The optimal value of the objective function of the linear programming problem formulated by the intern is ______ (in integer).

Hint:

Check the direction of the demand constraint carefully. A "less than or equal to" sign on demand does not force any oil to actually be delivered.

Answer 1

Correct Answer: 0

Step 1: Understanding the Concept:
Before running any optimization, always check whether the constraints actually describe the problem the way you would expect.
Here the intern's model looks like a transportation problem, but a small sign error changes its meaning completely.

Step 2: Key Formula or Approach:
In a proper transportation problem the demand side must be written as an equality or a "greater than or equal to" constraint, $\sum_i X_{ij} \ge D_j$ or $= D_j$, to force the markets to actually receive their oil.
The intern instead wrote $\sum_i X_{ij} \le D_j$, an upper limit.
Since the objective is Minimize $Z = \sum C_{ij} X_{ij}$ with all $C_{ij} > 0$ and $X_{ij} \ge 0$, the smallest possible value any variable can take is zero, and zero never violates a "less than or equal to" bound.

Step 3: Detailed Explanation:
Because every cost is positive, $Z$ is a non-negative combination of the decision variables, so $Z \ge 0$ for any feasible point, meaning zero is a lower bound on the objective.
This lower bound is actually reachable, because setting all four variables to zero, $X_{11}=X_{12}=X_{21}=X_{22}=0$, does not violate any constraint: the capacities 500 and 600 are not exceeded, the demand upper bounds 300 and 400 are not exceeded, and non-negativity holds trivially.
Since the lower bound $Z \ge 0$ is attained by a feasible point, it must be the optimal value:
\[ Z^{*} = \min\big(250 X_{11} + 280 X_{12} + 150 X_{21} + 180 X_{22}\big) = 0 \]

Final Answer:
Because the demand constraint only caps supply instead of requiring it, the true optimum of the flawed model is 0.
\[ \boxed{Z^{*} = 0} \]