## APMonitor Summation with Vectors

Variable or object arrays are defined by square brackets with a range of integers and separated by a colon as *variable[index 1:index 2]*. Arrays may be used to define multiple equations or connections on one line. Any line with an array is processed sequentially from the lowest to the highest index. The model parser creates and processes the arrays as if they were written sequential in non-array form as shown in the example.

### Higher dimensional arrays

Arrays with more than one dimension are allowed. The array indices are separated by brackets as *var[i][j][k]*. For operations on matrices, the precedence of operations is determined by the number of colons separating the vector indices. Matrix elements with fewer colon separators are executed first. For example, a set of 24 intermediate variables posed as:

- x[1:2][1::3][1:::4] = 1

results in the following set of equations:

- x[1][1][1] = 1
- x[2][1][1] = 1
- x[1][2][1] = 1
- x[2][2][1] = 1
- x[1][3][1] = 1
- x[2][3][1] = 1
- x[1][1][2] = 1
- etc...

### Array Index Consistency

When processing the arrays, the parser checks for array size consistency. An error with an appropriate message is returned if the vector indeces are of different dimension.

### Example

! Summation with arrays Model array Constants n = 5 End Constants Parameters p[1:n] = 1 End Parameters Variables sum End Variables Intermediates z[1] = p[1] z[2:n] = z[1:n-1] + p[2:n] End Intermediates Equations sum = z[n] End Equations End Model |

! Summation without arrays Model array Parameters p[1] = 1 p[2] = 1 p[3] = 1 p[4] = 1 p[5] = 1 End Parameters Variables sum End Variables Intermediates z[1] = p[1] z[2] = z[1] + p[2] z[3] = z[2] + p[3] z[4] = z[3] + p[4] z[5] = z[4] + p[5] End Intermediates Equations sum = z[5] End Equations End Model |

! Matrix Summation Model Parameters p[1:10][1::5] = 1 End Parameters Variables x End Variables Intermediates ! sum the rows n[0][1:5] = 0 n[1:10][1::5] = n[0:9][1::5] + p[1:10][1::5] ! sum the columns that are summation of rows m[0] = 0 m[1:5] = m[0:4] + n[10][1:5] End Intermediates Equations ! solution = 50 x = m[5] End Equations End Model |