For a transportation problem, let \(c_{ij}\) denote the unit cost of the cell \((i, j)\). The known unit costs are shown below (a dash marks a cell whose cost is not given).
j = 1
j = 2
j = 3
i = 1
10
12
19
i = 2
11
13
-
i = 3
-
-
-
Let \(\alpha_i\) and \(\beta_j\), \(i, j = 1, 2, 3\), be the simplex multipliers associated with a basis corresponding to this unit cost table, so that \(\alpha_1 = x\), \(\alpha_2 = x + 1\), \(\beta_1 = y\) and \(\beta_2 = y + 2\). The relative cost coefficient \(d_{ij}\) is the difference between the current solution and the new improved solution.
If \(x = 4\), \(c_{13} = 19\) and \(\beta_3 = y + 5\), then which one of the following is TRUE?
Show Hint
Use the basic-cell rule \(c_{ij} = \alpha_i + \beta_j\) to first pin down \(x+y\), then use \(x=4\) to get \(y\), and finally use \(d_{13}=c_{13}-(\alpha_1+\beta_3)\) for the non-basic cell \((1,3)\).
Step 1: Understanding the Concept.
This is a u-v method (MODI method) question. For every basic cell of a transportation table, the rule $c_{ij} = u_i + v_j$ holds, where $u_i$ plays the role of $\alpha_i$ and $v_j$ plays the role of $\beta_j$. For a non-basic cell, the reduced (relative) cost is $d_{ij} = c_{ij} - (u_i+v_j)$.
Step 2: Pick one basic-cell equation and plug in x directly.
Take the cell $(2,2)$, whose cost is given as $c_{22}=13$. Since it is basic, $c_{22} = \alpha_2+\beta_2 = (x+1)+(y+2)$. With $x=4$:
\[ 13 = (4+1) + (y+2) = 5 + y + 2 = y + 7 \]
so $y = 13 - 7 = 6$.
Step 3: Cross-check with a second basic cell.
Using cell $(1,1)$: $c_{11}=\alpha_1+\beta_1 = x+y = 4+6=10$, and indeed $c_{11}=10$ as given, so the value $y=6$ is consistent across the whole table, not just from one equation.
Step 4: Work out beta_3 and the reduced cost of the empty cell (1,3).
$\beta_3 = y+5 = 6+5=11$. The cell $(1,3)$ is empty (non-basic), so its reduced cost is
\[ d_{13} = c_{13} - (\alpha_1+\beta_3) = 19 - (4+11) = 4 \]
A reduced cost of $4$ (positive) means moving material into this cell would raise the total transportation cost by $4$ per unit, confirming the current basis is not improved by bringing in that cell.
Step 5: Final Answer.
So $y=6$ and $d_{13}=4$, which is option (B); options with $y=5$ break the relation $x+y=10$, and the option with $d_{13}=7$ uses the wrong value while combining with $\beta_3$ and $c_{13}$.
\[ \boxed{y=6,\ d_{13}=4} \]
