Effective Memory Access Time (EMAT) Problem: TLB Hit Ratios & Multi-Level Tables

Medium

Interview Question: "A system uses virtual memory with paging and a Translation Lookaside Buffer (TLB). TLB access latency is 20 ns and main memory access latency is 100 ns. Calculate the Effective Memory Access Time (EMAT) for an 80% TLB hit ratio, a 95% TLB hit ratio, and for a 2-level page table system at an 80% hit ratio. How does page fault probability change this equation?"

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Why this matters

Page tables live in RAM too. So without help, every memory access actually costs two RAM trips: one to look up where the data is (the page table entry), one to fetch the data itself — 200 ns instead of 100 ns.

The TLB is the shortcut: a small on-chip cache that, on a hit, skips the page-table trip entirely and resolves the address in 20 ns.

TLB Lookup

🔍20 ns

every access starts here

TLB Hit (prob h)

✅+ 100 ns

fetch data straight from RAM

TLB Miss (prob 1-h)

🔁+ k×100 ns, then + 100 ns

walk the page table(s), then fetch data

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Part 1: EMAT with a Single-Level Page Table (80% Hit Ratio)

Problem Parameters:

• TLB Access Time ($t_{\text{TLB}}$): 20 ns
• Main Memory Access Time ($t_{\text{mem}}$): 100 ns
• TLB Hit Ratio ($h$): 0.80 (Miss Ratio $1 - h = 0.20$)

Access Latencies:

1. On TLB Hit: Check TLB + Fetch data from memory: $\text{Cost}_{\text{hit}} = t_{\text{TLB}} + t_{\text{mem}} = 20\text{ ns} + 100\text{ ns} = 120\text{ ns}$
2. On TLB Miss: Check TLB + Read Page Table from RAM + Fetch data from RAM: $\text{Cost}_{\text{miss}} = t_{\text{TLB}} + 2 \cdot t_{\text{mem}} = 20\text{ ns} + 100\text{ ns} + 100\text{ ns} = 220\text{ ns}$

Mathematical Calculation:

$\text{EMAT} = h \cdot \text{Cost}_{\text{hit}} + (1 - h) \cdot \text{Cost}_{\text{miss}}$

$\text{EMAT} = (0.80 \times 120\text{ ns}) + (0.20 \times 220\text{ ns})$

$\text{EMAT} = 96\text{ ns} + 44\text{ ns} = \mathbf{140\text{ ns}}$

• Verdict: An 80% TLB hit ratio yields an EMAT of 140 ns, saving 60 ns compared to a TLB-less system (200 ns).

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Part 2: What Happens at a 95% Hit Ratio

Now assume software exhibits better spatial and temporal locality, raising the TLB hit ratio to $h = 0.95$ ($1 - h = 0.05$).

Mathematical Calculation:

$\text{EMAT} = (0.95 \times 120\text{ ns}) + (0.05 \times 220\text{ ns})$

$\text{EMAT} = 114\text{ ns} + 11\text{ ns} = \mathbf{125\text{ ns}}$

Sensitivity & Engineering Insights:

• A 15% increase in TLB hit ratio (80% to 95%) cuts EMAT from 140 ns to 125 ns (a 10.7% speedup).
• Because a TLB miss costs 220 ns (nearly double a hit), system throughput is hyper-sensitive to memory access patterns. Sequential array iteration and tight loop structures preserve TLB cache lines, while random pointer chasing thrashing the TLB causes CPU stalls.

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Part 3: A 2-Level Page Table, Still at 80% Hit Ratio

Modern operating systems use multi-level (hierarchical) page tables to avoid allocating megabytes of contiguous page tables for sparse address spaces. However, every additional level adds a memory access on a TLB miss.

Problem Parameters:

• Number of Page Table Levels ($k$): 2 (Outer Page Directory + Inner Page Table)
• TLB Hit Ratio ($h$): 0.80

Access Latencies:

1. On TLB Hit: Remains identical: $\text{Cost}_{\text{hit}} = t_{\text{TLB}} + t_{\text{mem}} = 20\text{ ns} + 100\text{ ns} = 120\text{ ns}$
2. On TLB Miss: Check TLB + Read Outer Directory + Read Inner Table + Fetch data: $\text{Cost}_{\text{miss}} = t_{\text{TLB}} + (k + 1) \cdot t_{\text{mem}} = 20\text{ ns} + (2 + 1) \cdot 100\text{ ns} = 320\text{ ns}$

Mathematical Calculation:

$\text{EMAT} = (0.80 \times 120\text{ ns}) + (0.20 \times 320\text{ ns})$

$\text{EMAT} = 96\text{ ns} + 64\text{ ns} = \mathbf{160\text{ ns}}$

• Verdict: In a 2-level system at an 80% hit ratio, EMAT rises to 160 ns. Without the TLB, every access would require 3 RAM accesses (300 ns).

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The General Formula, for Any Number of Levels

For an arbitrary $k$-level page table configuration with serial TLB lookup:

$\text{EMAT} = h \cdot (t_{\text{TLB}} + t_{\text{mem}}) + (1 - h) \cdot \Big[ t_{\text{TLB}} + (k + 1) \cdot t_{\text{mem}} \Big]$

Simplifying algebraically: $\text{EMAT} = t_{\text{TLB}} + t_{\text{mem}} + (1 - h) \cdot k \cdot t_{\text{mem}}$

Architecture Configuration	Levels (k)	Hit Latency	Miss Latency	EMAT at 80% Hit	EMAT at 95% Hit	EMAT at 99% Hit
1-Level Paging (32-bit flat)	1	120 ns	220 ns	140.0 ns	125.0 ns	121.0 ns
2-Level Paging (x86-32 PAE)	2	120 ns	320 ns	160.0 ns	130.0 ns	122.0 ns
4-Level Paging (x86-64 PML4)	4	120 ns	520 ns	200.0 ns	140.0 ns	124.0 ns
5-Level Paging (x86-64 PML5)	5	120 ns	620 ns	220.0 ns	145.0 ns	125.0 ns

Key Observation: In modern 64-bit systems with 4-level or 5-level paging, an un-cached memory walk costs 500 ns to 600 ns. This makes hardware TLB hit rates of 98%+ mandatory for high performance.

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The Trap: What If the Page Isn't Even in RAM?

Interviewers frequently combine TLB access with the Page Fault Rate ($p$). If the requested page is not present in physical RAM at all, a page fault interrupt is triggered, requiring disk I/O.

The Full Memory Hierarchy Equation:

$\text{EMAT}_{\text{total}} = (1 - p) \cdot \text{EMAT}_{\text{RAM}} + p \cdot t_{\text{fault}}$

Where:

• $p$ = Page Fault Probability (Miss rate in physical memory).
• $t_{\text{fault}}$ = Page Fault Service Time (Disk read + OS interrupt handling $\approx 8\text{ ms} = 8,000,000\text{ ns}$).

Numerical Proof of Sensitivity:

Assume a 1-level system with $\text{EMAT}_{\text{RAM}} = 140\text{ ns}$, and a minor page fault rate of 0.1% ($p = 0.001$):

$\text{EMAT}_{\text{total}} = (1 - 0.001) \cdot 140\text{ ns} + 0.001 \cdot 8,000,000\text{ ns}$

$\text{EMAT}_{\text{total}} \approx 140\text{ ns} + 8,000\text{ ns} = \mathbf{8,140\text{ ns}}$

• The Takeaway: A tiny 0.1% page fault rate slows memory access down by 58x (from 140 ns to 8,140 ns).
• To keep memory degradation under 10% (EMAT $\le 154$ ns): $p \cdot 8,000,000 \le 14 \implies p \le 0.00000175 \quad (p < 0.00017\%)$ Less than 1 in 570,000 accesses can be allowed to fault.

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Python: Verify the Numbers

The following standalone script calculates EMAT across variable hit ratios, page table depths, and page fault rates:

"""
Effective Memory Access Time (EMAT) Simulator
Calculates memory latencies across TLB hit rates, multi-level tables, and page faults.
"""

def calculate_emat(
    t_tlb: float,
    t_mem: float,
    hit_ratio: float,
    levels: int = 1,
    page_fault_rate: float = 0.0,
    t_fault: float = 8_000_000.0  # 8 ms in nanoseconds
) -> float:
    # Serial lookup model
    cost_hit = t_tlb + t_mem
    cost_miss_ram = t_tlb + (levels + 1) * t_mem

    emat_ram = (hit_ratio * cost_hit) + ((1.0 - hit_ratio) * cost_miss_ram)

    # Full hierarchy including disk page faults
    total_emat = (1.0 - page_fault_rate) * emat_ram + (page_fault_rate * t_fault)
    return total_emat


if __name__ == "__main__":
    t_tlb = 20.0   # ns
    t_mem = 100.0  # ns

    print("=" * 65)
    print("EFFECTIVE MEMORY ACCESS TIME (EMAT) NUMERICAL VERIFICATION")
    print("=" * 65)

    # 1. Single-level at 80% and 95%
    emat_80 = calculate_emat(t_tlb, t_mem, 0.80, levels=1)
    emat_95 = calculate_emat(t_tlb, t_mem, 0.95, levels=1)
    print(f"1-Level Page Table (80% TLB Hit): {emat_80:.2f} ns")
    print(f"1-Level Page Table (95% TLB Hit): {emat_95:.2f} ns")

    # 2. Two-level at 80%
    emat_2lvl_80 = calculate_emat(t_tlb, t_mem, 0.80, levels=2)
    print(f"2-Level Page Table (80% TLB Hit): {emat_2lvl_80:.2f} ns")

    # 3. Multi-level scaling across architectures (80% vs 98%)
    print("\n--- Architecture Scaling (k = 1, 2, 4, 5 levels) ---")
    print(f"{'Levels (k)':<12} | {'EMAT (h=0.80)':<16} | {'EMAT (h=0.98)':<16}")
    print("-" * 50)
    for k in [1, 2, 4, 5]:
        e80 = calculate_emat(t_tlb, t_mem, 0.80, levels=k)
        e98 = calculate_emat(t_tlb, t_mem, 0.98, levels=k)
        print(f"{k:<12} | {e80:8.2f} ns      | {e98:8.2f} ns")

    # 4. Page Fault Penalty Demonstration
    print("\n--- Impact of Page Fault Probability (1-Level, h=0.80) ---")
    fault_rates = [0.0, 0.00001, 0.0001, 0.001]
    for p in fault_rates:
        emat_pf = calculate_emat(t_tlb, t_mem, 0.80, levels=1, page_fault_rate=p)
        print(f"Page Fault Rate: {p*100:6.3f}% -> Total EMAT: {emat_pf:10.2f} ns")