--- title: "Banded matrix and skyline storage" subject: "Computational Techniques in Civil Engineering" chapter: 2 key: "semester_8/computational_techniques_in_civil_engineering:banded-skyline-storage" source: https://www.bctengineer.com/q/semester_8/computational_techniques_in_civil_engineering/banded-skyline-storage site: BCT Engineering --- # Banded matrix and skyline storage Computational Techniques in Civil Engineering · Chapter 2: Solutions of linear equations - Asked in: 2077 Chaitra, 3 marks - Video: [Bandwidth of Stiffness matrix | Effect of node numbering on Bandwidth | feaClass](https://www.youtube.com/watch?v=jk2LILvTzsI) — Msquare Analysis Projects - Answer checked by Claude Opus 5.5 (high effort) on 2026-10-14 - Page: https://www.bctengineer.com/q/semester_8/computational_techniques_in_civil_engineering/banded-skyline-storage - Whole chapter: https://www.bctengineer.com/md/semester_8/computational_techniques_in_civil_engineering/chapter-2 ## Question Explain with relevant examples how banded matrix and skyline storage scheme optimize the memory. ## Short answer Banded storage stores only the diagonal and upper b−1 terms per row using symmetry, reducing memory to n×b. Example: 100-node bar with b=2 needs only 200 numbers instead of 10000. Skyline storage stores each column from first nonzero to diagonal, saving more memory when bandwidth varies; renumbering minimizes profile. ## Answer 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. ```text 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. ## Check yourself The answer's key points as multiple-choice questions, with the correct option and why. 1. How does skyline storage store each column? A. Only the diagonal B. The full column C. Only non-zeros with index arrays D. From the first non-zero entry down to the diagonal Answer: D. Column heights vary, and a pointer array locates the diagonals. (Key point: Skyline) 2. Bar of 100 nodes, 1 dof, b = 2: banded storage? A. 100 B. 10000 C. 200 numbers D. 400 Answer: C. n x b = 100 x 2. (Key point: Banded example) 3. Why is no extra storage needed for fill-in in skyline? A. No fill-in occurs B. Fill-in during elimination stays inside the skyline C. Elimination is not used D. Fill-in is discarded Answer: B. The profile contains all fill-in. (Key point: Fill-in) ## Related in Computational Techniques in Civil Engineering - [Sparse, banded matrix and memory optimization](https://www.bctengineer.com/question.md?key=semester_8%2Fcomputational_techniques_in_civil_engineering%3Asparse-banded-memory) — Computational Techniques in Civil Engineering, chapter 2 ## Related in other subjects - [File system vs virtual memory; contiguous allocation](https://www.bctengineer.com/question.md?key=semester_6%2Foperating_system%3Afs-vs-virtual-memory-contiguous-tape) — Operating System, chapter 5