Consider an instruction pipeline with five stages without any branch prediction: Fetch Instruction (FI), Decode Instruction (DI), Fetch Operand (FO), Execute Instruction (EI) and Write Operand (WO). The stage delays for FI, DI, FO, EI and WO are 5 ns, 7 ns, 10 ns, 8 ns and 6 ns, respectively. There are intermediate storage buffers after each stage and the delay of each buffer is 1 ns. A program consisting of 12 instructions $\text{I}_1, \text{I}_2, \text{I}_3, \dots, \text{I}_{12}$ is executed in this pipelined processor. Instruction $\text{I}_4$ is the only branch instruction and its branch target is $\text{I}_9$. If the branch is taken during the execution of this program, the time (in ns) needed to complete the program is

1. Clock Cycle Time ($\tau$):

   $$\tau = \max(\text{Stage Delays}) + \text{Buffer Delay} = \max(5, 7, 10, 8, 6) + 1 = 10 + 1 = 11\text{ ns}$$

2. Execution Flow and Pipeline Stalls:

   • The instructions successfully executed are $\text{I}_1, \text{I}_2, \text{I}_3, \text{I}_4$ followed by the target instructions $\text{I}_9, \text{I}_{10}, \text{I}_{11}, \text{I}_{12}$ (total $8$ committed instructions).

   • Without branch prediction, the processor continues fetching sequential instructions ($\text{I}_5, \text{I}_6, \text{I}_7$) until the branch outcome is determined in the Execute Instruction (EI) stage (Stage 4).

3. Cycle-by-Cycle Pipeline Trace:

   • Cycle 1: $\text{I}_1$ in FI

   • Cycle 2: $\text{I}_1$ in DI, $\text{I}_2$ in FI

   • Cycle 3: $\text{I}_1$ in FO, $\text{I}_2$ in DI, $\text{I}_3$ in FI

   • Cycle 4: $\text{I}_1$ in EI, $\text{I}_2$ in FO, $\text{I}_3$ in DI, $\text{I}_4$ in FI

   • Cycle 5: $\text{I}_1$ in WO, $\text{I}_2$ in EI, $\text{I}_3$ in FO, $\text{I}_4$ in DI, $\text{I}_5$ in FI

   • Cycle 6: $\text{I}_2$ in WO, $\text{I}_3$ in EI, $\text{I}_4$ in FO, $\text{I}_5$ in DI, $\text{I}_6$ in FI

   • Cycle 7: $\text{I}_3$ in WO, $\text{I}_4$ in EI (branch resolved taken, $\text{I}_5, \text{I}_6, \text{I}_7$ flushed), $\text{I}_5$ in FO, $\text{I}_6$ in DI, $\text{I}_7$ in FI

   • Cycle 8: $\text{I}_4$ in WO, $\text{I}_9$ in FI (Target fetched)

   • Cycle 9: $\text{I}_9$ in DI, $\text{I}_{10}$ in FI

   • Cycle 10: $\text{I}_9$ in FO, $\text{I}_{10}$ in DI, $\text{I}_{11}$ in FI

   • Cycle 11: $\text{I}_9$ in EI, $\text{I}_{10}$ in FO, $\text{I}_{11}$ in DI, $\text{I}_{12}$ in FI

   • Cycle 12: $\text{I}_9$ in WO, $\text{I}_{10}$ in EI, $\text{I}_{11}$ in FO, $\text{I}_{12}$ in DI

   • Cycle 13: $\text{I}_{10}$ in WO, $\text{I}_{11}$ in EI, $\text{I}_{12}$ in FO

   • Cycle 14: $\text{I}_{11}$ in WO, $\text{I}_{12}$ in EI

   • Cycle 15: $\text{I}_{12}$ in WO (Program completed)

4. Total Execution Time:

   $$\text{Total Time} = 15 \text{ cycles} \times 11\text{ ns/cycle} = 165\text{ ns}$$

Hence, the correct option is (B).