Banded matrix and skyline storage

In FEM the stiffness matrix $[K]$ is symmetric, sparse and has non-zero terms only near the diagonal. Storing the full $n\times n$ matrix wastes memory.

Banded storage

Only the band is stored: the diagonal and the $b-1$ upper terms of each row (using symmetry). Half-bandwidth $b=(\text{max node difference in an element}+1)\times$ dof per node.

Example: a bar of 100 nodes (1 dof), elements connect nodes $i$ and $i+1$, so $b=2$. Full storage $=100^2=10000$ numbers; banded storage $=100\times2=200$ numbers.

 Full 5x5 (symmetric)        Banded (b=2)
 a11 a12  0   0   0          a11 a12
 a12 a22 a23  0   0          a22 a23
  0  a23 a33 a34  0   ->     a33 a34
  0   0  a34 a44 a45         a44 a45
  0   0   0  a45 a55         a55  *

Skyline (variable band / profile) storage

Each column is stored from the first non-zero entry down to the diagonal, so the band height varies with column. A pointer array stores the diagonal positions.

Example: column heights 1, 2, 2, 3, 1 give 9 stored terms against 15 for the full upper triangle of a $5\times5$ matrix.

Skyline storage stores fewer terms than a constant band when a few nodes have a large connectivity. Fill-in during elimination remains inside the skyline, so the factorisation also needs no extra storage. Node renumbering (e.g. reverse Cuthill-McKee) reduces the bandwidth and so the memory and operation count.