Analysis

$$T(n) = c_1 n + c_2(n-1) + c_3(n-1) + c_4 \sum_{j=2}^n t_j + ... ...sum_{j=2}^n (t_j - 1) + c_6 \sum_{j=2}^n (t_j - 1) + c_7(n-1)$$

Best Case:

Array already sorted, t_j = 1 for all j

$$\begin{eqnarray*}T(n) & = & c_1 n + c_2(n-1) + c_3(n-1) + c_4(n-1) + c_5(0) + c_... ...+ c_7) \\ & = & an + b \;\;\; ({\rm linear} \; {\rm in} \; n) \end{eqnarray*}$$

Worst Case:

Array in reverse order, t_j = j for all j
Note that

$$\begin{eqnarray*}T(n) & = & c_1 n + c_2(n-1) + c_3(n-1) + c_4(\frac{n(n+1)}{2} -... ... & = & an^2 + bn + c \;\;\; ({\rm quadratic} \; {\rm in} \; n) \end{eqnarray*}$$

Average Case:

Check half of array on average, t_j = j/2 for all j
T(n) = an² + bn + c