GATE 2018 EE Q.22: solution and answer

Official answer: 3500.0Source: Official answer key, GATE 2018 EE (IIT Guwahati)

1. The bus admittance matrix $Y_{bus}$ is $1000\times 1000$ with $8000$ non-zero elements. Split these into diagonal and off-diagonal entries.

2. Each diagonal entry $Y_{ii}$ is the sum of the admittances of everything connected to bus $i$. In a connected system every bus has at least one branch, so all $1000$ diagonal entries are non-zero.

3. The off-diagonal non-zero entries number $8000-1000=7000$. If any diagonal entry were zero, more of the $8000$ would be off-diagonal and the branch count would rise, so assuming all diagonals non-zero gives the minimum.

4. A branch between buses $i$ and $k$ gives the two off-diagonal entries $Y_{ik}=Y_{ki}=-y_{ik}$. The matrix is symmetric when there are no phase-shifting transformers.

5. So $n_{\text{branches}}\ge \frac{\text{off-diagonal non-zeros}}{2}$. Equality holds when no two branches join the same pair of buses. Parallel branches between one pair share a single pair of entries, so they would need more branches for the same $7000$ entries.

6. $$n_{\text{branches,min}}=\frac{7000}{2}=3500$$