COD hw 4

Xinglu Wang
3140102282
2016-12-27 21:28:01

5.3

5.3.1

One block has 32 8-bits data, express into words: cache block size is${2^{5 - 2}} = 8$ words

5.3.2

$2^{9-5+1}=32$ entries

5.3.3

Data storage bits:$2^5 \times 8 \times 2^5$
Cache implementation:$2^5 \times (22+1+8 \times 2^5)$

The ratio is 1.086

5.3.4

Thus, 4 blocks are replaced.

5.3.5

Hit ratio is $4/12=1/3$

5.3.6

$$\begin{align*} <& index, tag, data> \\<& 0, 3, Mem[3072] \tilde{\quad} Mem[3103] > \\<& 4, 2, Mem[2176] \tilde{\quad} Mem[2207]>\\<& 5, 0, Mem[160] \tilde{\quad} Mem[191]>\\<& 7, 0, Mem[224] \tilde{\quad} Mem[255]> \end{align*}$$

5.4

5.4.1

It is needed to design a buffer between L1 and L2 cache:

When L1 cache write miss, the data need to write to L2 cache directly, but it is recommended to store this data into buffer first, and write to L2 cache when buffer overflow. It is because L2 cache spend more times to write data in.

It is also needed to design a buffer between L2 cache and mem:

When L2 write hit, there will be a dirty data on L2 cache, which is not consist with Mem.

When L2 write miss, we need to allocate on L2 cache, if L2 cache is full, we have to choose an old and dirty data to be replaced, and this old data has to be written back to mem. Store this old data in buffer first, can reduce penalty time.

5.4.2

解决L1 写失败的过程：
直接写通到L2，在L2上写分以下两种情况：

• L2上写中：替换对应数据，更改标志位：有效，dirty

• L2上写失败，写分配，分以下两种情况：

• • L2 cache上有空闲空间，直接分配
• • L2 cache上没有空闲空间，根据选择策略选择old/dirty的数据替换，而这个dirty的数据也同时被放入buffer。

5.4.3

L1 写失败：数据直接写入L2，L1上不留副本

之后如果读取同一地址数据，导致L1读失败：从L2读出该数据，L2标志位置为无效，L2中该数据写入Mem和L2间的buffer

5.4.4 ?????

2 CPI $\Rightarrow$ 0.5 instrution/cycle

• Instruction bandwidth = $0.5 \times 0.3\% 64 bytes=0.096 bytes$
• Data read bandwidth = $0.5 \times \frac{250}{1000} \times 2\% 64 bytes=0.16bytes$
• Total read bandwidth = $0.256bytes/cycle$
• Data write bandwidth = $0.5 \times \frac{100}{1000} \times 4bytes=0.2bytes$ , write through write direct to Mem 1 words per instruction.
• Total write bandwidth = $0.2bytes/cycle$

5.4.5

• Total read bandwidth = $0.256bytes/cycle$
• Data write bandwidth = $0.5 \times \frac{100}{1000} \times 30\% \times 64bytes=0.96bytes$
• Total write bandwidth = $0.96bytes/cycle$
  
  

5.4.6

5.6

5.6.1

• P1 Cycle Time = L1 Hit Time = $0.66ns$, P1 Clock Rate = $1/0.66ns=1.52GHz$
• P2 Cycle Time = L2 Hit Time = $0.90ns$, P2 Clock Rate = $1/0.90ns=1.11GHz$

5.6.2

• P1 AMAT in ns = $0.66ns+8\% \times 70 ns=6.26ns$
• P2 AMAT in ns = $0.90ns+6\% \times 70 ns=5.1ns$

5.6.3

First, express in cycles:

• P1 AMAT in Cycles =$ 6.26/0.66 = 9.56 Cycles$
• P2 AMAT in Cycles =$ 5.1/0.90 = 5.68 Cycles$
  Then, calculate CPI:
• P1 CPI = $9.56+(9.56-1) \times 0.36 = 12.64 CPI$
• P2 CPI = $5.68+(5.68-1) \times 0.36 = 7.36 CPI$
  Campare in ns:
• P1 time per inst: 8.34 ns/inst
• P2 time per inst: 6.63 ns/inst
  $\Rightarrow $ P2 is faster than P1

5.6.4

P1 AMAT = $0.66+8\% \times 70 = 6.26ns$
P1 with L2 AMAT = $0.66+8\%(5.62+95\% \times 70 )=6.43ns$
$ \Rightarrow $ P1 with L2 becomes worser!

5.6.5

P1 with L2 AMAT in cycles: $\frac{0.66+8\%(5.62+95\% \times 70)}{0.66} =9.7418 Cycles $
P1 with L2 CPI: $9.74+(9.74-1) \times 0.36=12.88$

5.6.6

P1 with L2 AMAT = $0.66+8\%(5.62+95\% \times 70 )=6.43ns$
P2 without L2 AMAT = $0.9+6\% \times 70=5.1ns$

$\Rightarrow $ P1 should be improved to match P2:
$$5.1=0.66+MR \times (5.62+95\% \times 70 ) $$
$\Rightarrow MR=6.156\%$

5.7

5.7.1

5.7.2

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