1. Lecture 39: Review Session #1 Reminders –Final exam, Thursday 12/18/2014 @ 3:10pm Sloan 150 –Course evaluation (Blue Course Evaluation) Access through zzusis 1

2. Problem #1 Consider a system with two multiprocessors with the following configurations: –Machine A with two processors, each with local memory of 512 MB with local memory access latency of 20 cycles per word and remote memory access latency of 60 cycles per word. –Machine B with two processors, with a shared memory of 1GB with access latency of 40 cycles per word. Suppose an application has two threads running on the two processors, each of them need to access an entire array of 4096 words. –Is it possible to partition this array on the local memories of ‘A’ machine so that the application runs faster on it rather than ‘B’ machine? If so, specify the partitioning! If not, by how many more cycles should the ‘B’ memory latency be worsened for a partitioning on the ‘A’ machine to enable a faster run than the ‘B’ machine? Assume that the memory operations dominate the execution time. 2

3. Solution #1 Suppose we have ‘x’ words on one processor and (T-x) words on the other processor, where T = 4096. Execution Time on ‘A’ = max(20x + 60(T-x), 60x + 20(T-x)) = max(60T-40x, 20T+40x) The max is 40T (unit time), where x = T/2 Execution Time on ‘B’ = 40T So, we can't make ‘A’ faster than ‘B’. However, if ‘B’ access is one more cycle slower (that is, 41 cycles access latency), ‘A’ could be faster. 3

4. Problem #2 Consider a multi-core processor with heterogeneous cores: A, B, C and D where core B runs twice as fast as A, core C runs three times as fast as A and cores C and A run at the same speed (ie have the same processor frequency, micro architecture etc). Suppose an application needs to compute the square of each element in an array of 256 elements. Consider the following two divisions of labor: (a) (b) Compute (1) the total execution time taken in the two cases and (2) cumulative processor utilization (amount of total time the processors are not idle divided by the total execution time). For case (b), if you do not consider Core D in cumulative processor utilization (assuming we have another application to run on Core D), how would it change? Ignore cache effects by assuming that a perfect pre-fetcher is in operation. 4 Core A32 elements Core B128 elements Core C64 elements Core D32 elements Core A48 elements Core B128 elements Core C80 elements Core DUnused

5. Solution #2 (1) Total execution Time –Total execution time = max(32/1, 128/2, 64/3, 32/1) = 64 (unit time) –Total execution time = max(48/1, 128/2, 80/3, 0/1) = 64 (unit time) (2) Utilization: –(a) Utilization = (32/1 + 128/2 + 64/3 +32/1) / 4 * (1/64) = 0.58 –(b) Utilization = (48/1 + 128/2 + 80/3 + 0/1) / 4 * (1/64) = 0.54 –(b) Utilization (if processor D is ignored) = (48/1 + 128/2 + 80/3) / 3 * (1/64) = 0.72 5

6. Problem #3 How would you rewrite the following sequential code so that it can be run as two parallel threads on a dual-core processor ? int A[80], B[80], C[80], D[80]; for (i = 0 to 40) { A[i] = B[i] * D[2*i]; C[i] = C[i] + B[2*i]; D[i] = 2*B[2*i]; A[i+40] = C[2*i] + B[i]; } 6

7. Solution #3 The code can be written into two threads as follows: Thread 1: int A[80], B[80], C[80], D[80]; for (i = 0 to 40) { A[i] = B[i] * D[2*i]; C[i] = C[i] + B[2*i]; A[i+40] = C[2*i] + B[i]; } Thread 2: int A[80], B[80], C[80], D[80]; for (i = 0 to 40) { D[i] = 2*B[2*i]; } 7

8. 8 RAID 0 and RAID 1 RAID 0 has no additional redundancy (misnomer) – it uses an array of disks and stripes (interleaves) data across the arrays to improve parallelism and throughput RAID 1 mirrors or shadows every disk – every write happens to two disks Reads to the mirror may happen only when the primary disk fails – or, you may try to read both together and the quicker response is accepted Expensive solution: high reliability at twice the cost

9. 9 RAID 3 Data is bit-interleaved across several disks and a separate disk maintains parity information for a set of bits. On failure, use parity bits to reconstruct missing data. For example: with 8 disks, bit 0 is in disk-0, bit 1 is in disk 1, …, bit 7 is in disk-7; disk-8 maintains parity for all 8 bits For any read, 8 disks must be accessed (as we usually read more than a byte at a time) and for any write, 9 disks must be accessed as parity has to be re-calculated High throughput for a single request, low cost for redundancy (overhead: 12.5% in the above example), low task-level parallelism

10. 10 RAID 4 / RAID 5 Data is block interleaved – this allows us to get all our data from a single disk on a read – in case of a disk error, read all 9 disks Block interleaving reduces throughput for a single request (as only a single disk drive is servicing the request), but improves task-level parallelism as other disk drives are free to service other requests. On a write, we access the disk that stores the data and the parity disk – parity information can be updated simply by checking if the new data differs from the old data

11. RAID 3 vs RAID 4 11

12. 12 RAID 5 If we have a single disk for parity, multiple writes can not happen in parallel (as all writes must update parity info) RAID 5 distributes the parity block to allow simultaneous writes

13. Problem #4 Discuss why RAID 3 is not suited for transaction processing applications. What kind of applications is it suitable for and why? 13

14. Solution #4 RAID 3 is unsuited to transactional processing because each read involves activity at all disks. In RAID 4 and 5 reads only involve activity at one disk. The disadvantages of RAID 3 are mitigated when long sequential reads are common, but performance never exceeds RAID 5. For this reason, RAID 3 has been all but abandoned commercially. 14