Final — Spring 2024

Final — Spring 2024

CSEE 4340 · Spring 2024

Historically numbered as CSEE 4824.

Question 1(a) 4 pts

What kind of dependences does register renaming eliminate?

Register renaming eliminates write-after-write (WAW) and write-after-read (WAR) dependences.

Verified correct

Correct. WAW and WAR are name dependences (a.k.a. false dependences): they arise only because two instructions happen to reference the same architectural register name, not because data actually flows between them. Mapping each architectural write to a fresh physical register removes both. RAW is a true (flow) dependence and cannot be eliminated by renaming — the consumer must still wait for the producer’s value.

Question 1(b) 6 pts

In the P6 microarchitecture, an instruction’s source operand value may come from three places before being placed in the reservation station. Name all three.

1. Register file
2. Common data bus (CDB)
3. Reorder buffer (ROB)

Verified correct

Correct. In a P6-style machine, when an instruction is dispatched to a reservation station, each source operand is sourced from (1) the architectural register file if the value has already committed, (2) the ROB entry of an in-flight producer that has executed but not yet retired, or (3) a tag waiting on the CDB when the producer has not yet executed.

Question 1(c) 6 pts

What are the main advantages/disadvantages of the R10k in comparison to the P6-style architecture?

• P6 style: easier rollback / easier implementation, but more data movement.
• R10k style: faster implementations and less data movement, but harder implementation / harder rollback.

(Both rely on register renaming and a ROB for precise interrupts; the difference is where operand values live.)

Verified correct

Correct at the level expected. The key distinction is that R10k-style designs use a unified physical register file so values stay in place and only tags move, avoiding the P6’s repeated copying of operands through ROB entries and reservation-station fields. The cost is a more complex free-list / map-table recovery scheme on misspeculation, since the ROB no longer holds the architectural value.

Question 1(d) 3 pts

Name a technique that can eliminate compulsory cache misses.

Prefetching.

Verified correct

Correct. A compulsory (cold) miss happens the first time a block is referenced. If a hardware or software prefetcher brings the block into the cache before the demand reference, the demand access becomes a hit, so the compulsory miss is avoided. Larger block size also reduces compulsory misses by amortizing one miss across spatially adjacent words.

Question 1(e) 4 pts

What is the main advantage of using a virtually-addressed cache over a physically-addressed one?

A virtually addressed cache can be accessed before virtual-to-physical translation completes, so cache lookup latency can be lower (the TLB lookup is removed from the critical path of a hit).

Verified with caveat

Correct, but worth noting the cost: a virtually addressed cache must handle synonyms (two virtual addresses mapping to the same physical address) and homonyms (the same virtual address meaning different physical addresses across processes), typically by flushing on context switch or by adding ASIDs. Modern designs often compromise with virtually-indexed, physically-tagged (VIPT) caches that overlap TLB translation with set lookup but compare physical tags.

Question 2(a) 6 pts

A repeating branch pattern of T, T, T, T, T, N, N is predicted with a PC-indexed 2-bit saturating-counter predictor. What is the steady-state prediction accuracy?

The steady-state accuracy is:

4 / 7 = 0.5714 = 57.14%

Reasoning in steady state:

Actual outcome	Predictor state before outcome	Prediction	Result	State after outcome
T1	strongly taken	T	correct	strongly taken
T2	strongly taken	T	correct	strongly taken
T3	strongly taken	T	correct	strongly taken
T4	strongly taken	T	correct	strongly taken
T5	strongly taken	T	correct	strongly taken
N1	strongly taken	T	wrong	weakly taken
N2	weakly taken	T	wrong	weakly not taken
next T1	weakly not taken	N	wrong	weakly taken

Per 7-outcome period, there are 3 mispredictions and 4 correct predictions.

Verified correct

Correct. In steady state the counter reaches strongly-taken during the run of five Ts, gets demoted by the two Ns, and is in weakly-not-taken when the next pattern begins — so the very first T of the next period is also mispredicted. That gives 3 mispredictions per 7 outcomes, i.e. 4/7 accuracy. A 2-bit counter is provably worse than a 1-bit predictor on this pattern (a 1-bit predictor would mispredict the same 3 transitions but always produce 4/7 correct as well — accuracy is limited by the pattern’s runs, not the counter width).

Question 2(b) 8 pts

CPI is 1.2; ideal CPI is 1; 20% of instructions are branches; branch-prediction accuracy is 90%. How many cycles does each misprediction cost? What accuracy is needed to reduce CPI to 1.01 only by improving branch prediction?

Current CPI overhead:

1.2 - 1 = 0.2 cycles/instruction

Branch misprediction rate per instruction:

0.20 branch/instruction * (1 - 0.90) = 0.02 mispredictions/instruction

Misprediction cost:

0.2 / 0.02 = 10 cycles per misprediction

For CPI = 1.01:

1.01 = 1 + 0.20 * (1 - accuracy) * 10
0.01 = 2 * (1 - accuracy)
1 - accuracy = 0.005
accuracy = 0.995 = 99.5%

Verified correct

Correct. The arithmetic is the standard $\text{CPI} = \text{CPI}_{\text{ideal}} + \text{branch frequency} \times \text{miss rate} \times \text{penalty}$. With penalty fixed at 10 cycles, dropping the CPI overhead from 0.2 to 0.01 is a 20× reduction, which requires the misprediction rate to drop by 20× (from 10% to 0.5%), i.e. accuracy of 99.5%.

Question 2(c) 6 pts

A two-level/correlated branch predictor uses 8 PC bits to select a BHT entry; each BHT entry holds an 8-bit branch history; the PHT is indexed by that 8-bit history and stores 2-bit saturating counters. How many storage bits are needed for the BHT and the PHT?

BHT:

2^8 entries * 8 bits/entry = 256 * 8 = 2048 bits

PHT:

2^8 entries * 2 bits/entry = 256 * 2 = 512 bits

Total predictor table storage:

2048 + 512 = 2560 bits

Verified correct

Correct for the stated design where the 8-bit history alone indexes a single shared PHT (a global-history-indexed style, like the second level of a two-level predictor). If instead the design concatenated PC and history to index the PHT, the PHT would have $2^{16}$ entries — but that is not what was specified.

Question 3(a) 8 pts

Compute the cache index bits for L1i, L1d, L2, and L3 caches with the following parameters. The block size is 64 B for all caches.

Cache	Size	Associativity	Block size
L1 instruction	32 KB	8-way	64 B
L1 data	32 KB	8-way	64 B
L2	256 KB	8-way	64 B
L3	20 MB	20-way	64 B

Formula:

# sets = cache size / (block size * associativity)
index bits = log2(# sets)

Answers:

Cache	Sets	Index bits
L1 instruction	32 KB / (64 B * 8) = 64	6
L1 data	32 KB / (64 B * 8) = 64	6
L2	256 KB / (64 B * 8) = 512	9
L3	20 MB / (64 B * 20) = 16384	14

Verified correct

Correct. The L3 calculation is the only one that requires care: $20 \text{ MB} / (64 \text{ B} \times 20) = (20 \times 2^{20}) / (64 \times 20) = 2^{20} / 64 = 2^{14} = 16{,}384$ sets, which is exactly $\log_2$-clean. If the L3 were not 20-way (e.g., 16-way), the set count would not be a power of two — a tell that real L3s use unusual ways exactly to make this divide come out clean.

Question 3(b) 9 pts

Compute the average access time (AAT) for each associativity given the per-associativity hit access time, a 10 ns miss access time, 0.3 data references per instruction, and the table’s MPKI values. Which associativity performs best?

Miss rate conversion:

miss rate = misses / accesses
          = (MPKI / 1000) / 0.3

Average access time:

AAT = hit_rate * hit_access_time + miss_rate * miss_access_time
    = (1 - miss_rate) * hit_access_time + miss_rate * 10 ns

Associativity	Hit access time	MPKI	Miss rate	Average access time
Direct-mapped	0.86 ns	6.64	0.02213	~1.061 ns
2-way	1.12 ns	3.66	0.01220	~1.228 ns
4-way	1.37 ns	0.987	0.00329	~1.398 ns
8-way	2.03 ns	0.266	0.000887	~2.037 ns

The direct-mapped cache performs best by average access time.

Verified with caveat

Arithmetic is correct given the model. One subtle point: this AAT formula treats the 10 ns miss access time as the total time including the hit-time component (i.e., it is the miss penalty plus hit time, used as the weighted alternative outcome). An equally common formulation is $\text{AAT} = \text{hit time} + \text{miss rate} \times \text{miss penalty}$, in which case the 10 ns would be miss penalty alone. Both produce the same ordering of the four designs here, so the conclusion that the direct-mapped cache wins still holds — its low hit time (0.86 ns) dominates the 0.022 miss rate it pays for the lack of associativity.