Pallet Rebalancing With a Parameterised Matrix
A wholesale distributor supplies three retail clusters with three products. A weekly rebalancing plan is encoded by Mx = b, where x = (x1, x2, x3) is the number of pallets sent to clusters 1, 2, 3, b = (10, 9, 10) is the net stock shortage at the three clusters in pallets, and M = [[2, 1, k], [1, 2, 1], [k, 1, 2]]. The parameter k is a shared-dock coupling factor set by a trial run. For k = 1 the operations team wants a unique plan. For k = 2 a dock constraint links the first and third cluster equations. The manager is uneasy: two different k values cannot both be handled by simply 'solving the matrix' if the coefficient matrix changes character. Before releasing pallets, the team must solve the k = 1 case exactly, explain what happens at k = 2, and state what minimal change in the shortage data would make the k = 2 system inconsistent rather than underdetermined.
1. For k = 1, show that the coefficient matrix is invertible and solve Mx = b.
Read the worked answer
For k = 1,
M = [[2, 1, 1], [1, 2, 1], [1, 1, 2]].
Its determinant is
det M = 4 + 2(1) - 2(1)² = 4.
Since det M ≠ 0, the system has a unique solution. The equations are:
2x1 + x2 + x3 = 10 x1 + 2x2 + x3 = 9 x1 + x2 + 2x3 = 10.
Subtract the second equation from the first: x1 - x2 = 1, so x1 = x2 + 1. Subtract the third from the second: x2 - x3 = -1, so x3 = x2 + 1. Thus x1 = x3.
Substitute into the first equation: 2(x2 + 1) + x2 + (x2 + 1) = 10 ⇒ 4x2 + 3 = 10 ⇒ x2 = 7/4.
Hence x1 = 11/4, x2 = 7/4, x3 = 11/4. The plan is unique.
2. For k = 2, determine whether the system is consistent, and describe the solution set.
Read the worked answer
For k = 2,
M = [[2, 1, 2], [1, 2, 1], [2, 1, 2]].
The first and third rows are identical, so det M = 0. The first and third equations are both 2x1 + x2 + 2x3 = 10; the data b1 = b3 = 10, so there is no immediate contradiction.
Let S = x1 + x3. The first two equations become: 2S + x2 = 10 S + 2x2 = 9.
Solving gives x2 = 8/3 and S = 11/3. Thus x1 + x3 = 11/3, with x3 free. The solution set is
x2 = 8/3, x1 = 11/3 - t, x3 = t, t arbitrary.
The system is consistent but underdetermined: one free variable. The dock constraint has removed the unique plan; extra operational information is needed to choose t.
3. What is the smallest change to b that would make the k = 2 system inconsistent? Justify using row structure.
Read the worked answer
At k = 2, rows 1 and 3 of M are identical: both are (2, 1, 2). A consistent system must therefore have equal right-hand entries b1 and b3.
Currently b1 = b3 = 10. Changing only b3 from 10 to any value other than 10, for example b3 = 11, makes the first and third equations require the same left-hand expression to equal two different numbers. That is impossible.
So the minimal change is a one-entry change: set b3 ≠ 10 while keeping b1 = 10. Then the k = 2 system has no solution. This is a row-consistency check, not a determinant-only check: det M = 0 tells us uniqueness fails, but comparing the dependent rows with b decides consistency.