Lecture Topics. Why Predict Branches? ECE 486/586. Computer Architecture. Lecture # 17. Basic Branch Prediction. Branch Prediction

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1 Lecture Topics ECE 486/586 Computer Architecture Branch Prediction Reference: Chapter : Section. Lecture # 17 Spring 2015 Portland State University Why Predict Branches? The decision about control flow (where to fetch the next instruction from?) is made in the fetchstage The branch penalty is non-zero because when the processor computes the branch outcome (in decodestage), a useless instruction may have already been fetched and needs to be discarded To prevent the fetching of useless instruction, the processor needs to know about the branch outcome in the fetchstage This involves the following steps: Anticipating that the instruction being fetched is a branch instruction Predicting whether the branch instruction will be taken or not taken Predicting the branch target address (for a taken branch) Basic Branch Prediction Branch prediction buffer (branch history table) Memory indexed by low order bits of branch instruction address Stores previous branch outcomes to predict next outcome Memory is not tagged (unlike cache) Consequence: entry may reflect a different branch (aliasing) PC PC[11:2] = 1K entries

2 Static Branch Prediction In static branch prediction, the prediction made for a conditional branch remains constant (static) throughout the execution of a program Example 1: Always-predict-not-taken Simplest form of prediction, always fetch next instruction in the sequential order In case of a misprediction, the incorrectly fetched instruction is discarded and branch penalty is incurred Low prediction accuracy because many branches in the program are taken Typically, branch outcomes are not completely random In a loop with many iterations, forward branches (beginning of loop) are mostly not taken and backward branches (end of loop) are mostly taken Example 2:Predict not-takenfor forward branches and takenfor backward branches Improves prediction accuracy as compared to the always-not-taken prediction Mispredictions still happen during the last loop iteration Dynamic Branch Prediction Outcomes for a branch instruction often change during program execution Static prediction may result in high misprediction accuracy But, outcomes for a particular branch often follow a predictable pattern Key idea behind dynamic branch prediction: Track the past outcomes for a branch instruction to make predictions about future outcomes In its simplest form, a dynamic prediction algorithm can use the result of the most recent execution of a branch instruction This result can be captured in a single bit (e.g., 0 if the branch was taken and 1 if the branch was not taken) The processor assumes that the next time, the branch instruction is executed, its outcome is the same as the last time 1-bit Branch Prediction Example The algorithm is implemented by a 2-state state machine: LT -- Branch is likely to be taken LNT -- Branch is likely not to be taken The prediction for a branch is based on the current stateof the state machine The state transitions are based on the actual outcome computed after the branch has been executed

3 LT T T LT LT T T LT LT T NT LNT 4 LNT NT T LT 5 LT T T LT 6 LT T NT LNT 2 LT T T LT LT T NT LNT 4 LNT NT T LT 5 LT T T LT 6 LT T NT LNT Prediction Accuracy = 2/6 Mispredictionshappen during both the first and last iterations of the loop => one bit of state not enough to capture the branch outcome pattern accurately

4 2-bit Branch Prediction Example Branch predicted as Not takenin these two states N = Not taken. Assume that the 2-bit branch predictor starts in the LT ST: Strongly likely to be taken LT: Likely to be taken LNT: Likely not to be taken SNT: Strongly likely not to be taken Branch predicted as Takenin these two states 1 LT T T ST 2 ST T T ST ST T NT LT 4 LT T T ST 5 ST T T ST 6 ST T NT LT Prediction Accuracy of 4K 2-bit Predictor N = Not taken. Assume that the 2-bit branch predictor starts in the LT 1 LT T T ST 2 ST T T ST ST T NT LT 4 LT T T ST 5 ST T T ST 6 ST T NT LT Prediction Accuracy = 4/6 Mispredictions happen only during the last iteration of the loop => less mispredictions that 1-bit prediction

5 Having more Entries Isn t the Solution Correlating Branch Predictors Simple 2-bit prediction schemes use branch history of single branch to predict future behavior of that branch. This is called a local branch prediction Behavior of other branches may have impact on the current branch Outcomes of different branches often correlated Example: If (a == 2) a = 0; If (b == 2) b = 0; If ( a == b) { DADDi R, R1, -2 BNEZ R, L1 ; a!= -2 DADD R1, R0, R0 L1: DADDI R, R2, -2 BNEZ R, L2 ; b!= -2 DADD R2, R0, R0 L2: DSUB R, R1, R2 BEQZ R, L ; a== b } If the first two branches are not taken, then the third one is taken. Local branch prediction cannot capture this behavior Correlating Branch Predictor with 2- bit Global History Register Correlating Branch Predictor with m-bit Global History Register Correlating (or 2-level) Predictors use the behavior of other branches (global branch history) to make branch predictions Can extend branch history as m-bits recording history of last m branches Requires 2 m tables of length 2 (branch address bits used) Global branch history implemented as a m-bit shift register where each bit records whether a branch was taken or not taken Branch Address 01 2-bit per-branch predictors Prediction = 11 (m,n) correlating predictor uses behavior of last mbranches to choose from 2 m branch predictors, each of which is an n- bit predictor Total number of bits = 2 m * n * Number of entries in each prediction table = 2 m * n * 2 (branch address bits used) For a predictor that does not use any global history, m = 0, e.g., a (0,2) is a 2-bit predictor with no global history Branch Address 10.1 n-bit per-branch predictors bit global branch history m-bit global branch history

6 Correlating Predictor Examples Comparison of 2-bit Predictors Question:How many bits are in the (0,2) branch predictor with 4K entries? How many entries are in a (2,2) predictor with the same number of bits? Solution: Number of bits = 2 m * n * 2 (branch address bits used) For the (0,2) predictor: Number of bits = 2 0 * 2 * 4K = 8K bits For the (2,2) predictor: Number of bits = 8K 8K = 2 2 * 2 * Number of predictor entries => Number of predictor entries = 1K