The memory access time is 1 nanosecond for a read operation with a hit in cache, 5 nanoseconds for a read operation with a miss in cache, 2 nanoseconds for a write operation with a hit in cache and 10 nanoseconds for a write operation with a miss in cache. Execution of a sequence of instructions involves 100 instruction fetch operations, 60 memory operand read operations and 40 memory operand write operations. The cache hit-ratio is 0.9. The average memory access time (in nanoseconds) in executing the sequence of instructions is ___________.

1. Categorize the operations:

   • Instruction fetch is a read operation: $100$ operations.

   • Memory operand read is a read operation: $60$ operations.

   • Total read operations $= 100 + 60 = 160$.

   • Memory operand write is a write operation: $40$ operations.

   • Total memory operations $= 160 + 40 = 200$.

2. Calculate average access time for each type of operation:

   • Cache hit-ratio $h = 0.9$, miss-ratio $m = 1 - 0.9 = 0.1$.

   • Average time for a read operation:

     $$T_{\text{read}} = (0.9 \times 1\text{ ns}) + (0.1 \times 5\text{ ns}) = 0.9 + 0.5 = 1.4\text{ ns}$$

   • Average time for a write operation:

     $$T_{\text{write}} = (0.9 \times 2\text{ ns}) + (0.1 \times 10\text{ ns}) = 1.8 + 1.0 = 2.8\text{ ns}$$

3. Calculate total memory access time:

   $$T_{\text{total}} = (160 \times 1.4) + (40 \times 2.8) = 224 + 112 = 336\text{ ns}$$

4. Calculate average memory access time per operation:

   $$\text{AMAT} = \frac{T_{\text{total}}}{\text{Total Operations}} = \frac{336}{200} = 1.68\text{ ns}$$

Hence, the average memory access time is 1.68 ns.