The Cache-Fronthaul Tradeoff

Two Resources, One Latency

The NDT formula Δ=max⁡(1,K(1−μ)/(NENC)+K(1−μ)/(NENLEN))\Delta = \max(1, K(1-\mu)/(N_\text{EN} C) + K(1-\mu)/(N_\text{EN} L_\text{EN})) reveals a fundamental substitution: cache and fronthaul capacity can be traded off to achieve the same delivery time. An operator can choose a cache-heavy architecture (cheap cache at each EN, modest fronthaul) or a cloud-heavy one (lean ENs, fat fronthaul). Both can deliver the same NDT.

The Pareto frontier — the set of (M,C)(M, C) pairs achieving a given NDT target — defines the architectural design space. This section characterizes this frontier and relates it to deployment costs.

Theorem: Iso-NDT Contours

For fixed K,NEN,LENK, N_\text{EN}, L_\text{EN} and target NDT Δ∗\Delta^*, the set of (M/N,CF)(M/N, C_F) pairs achieving Δ≤Δ∗\Delta \leq \Delta^* is {(μ,C):K(1−μ)NENC+K(1−μ)NENLEN≤Δ∗−1if Δ∗>1}.\left\{(\mu, C) : \frac{K(1-\mu)}{N_\text{EN} C} + \frac{K(1-\mu)}{N_\text{EN} L_\text{EN}} \leq \Delta^* - 1_\text{if} \, \Delta^* > 1\right\}. The iso-NDT contours in the (μ,C)(\mu, C) plane are hyperbolae: smooth tradeoff between cache and fronthaul at any given latency target.

Doubling the cache fraction μ\mu halves the "1 - μ\mu" factor, doubling the available slack. Doubling fronthaul CC halves the fronthaul contribution. Either works to meet the NDT budget.

Proof
Equation

Setting Δ(μ,C)=Δ∗\Delta(\mu, C) = \Delta^* and solving: K(1−μ)(1/(NENC)+1/(NENLEN))=Δ∗−1K(1-\mu)(1/(N_\text{EN} C) + 1/(N_\text{EN} L_\text{EN})) = \Delta^* - 1. Rearranging: 1/C=(Δ∗−1)⋅NEN/(K(1−μ))−1/LEN1/C = (\Delta^* - 1) \cdot N_\text{EN}/(K(1-\mu)) - 1/L_\text{EN}.

Feasibility

Both μ\mu and CC must be non-negative; μ≤1\mu \leq 1; C>0C > 0. These bound the iso-NDT contour, yielding the feasible region.

Shape

In the (μ,C)(\mu, C) plane, the contour is a hyperbola. Cache- cheap regime: large CC, small μ\mu. Fronthaul-cheap regime: large μ\mu, small CC.

NDT vs Memory Ratio

NDT as a function of memory ratio μ=M/N\mu = M/N, for varying fronthaul capacities CC. At high CC, even small cache is enough to hit Δ=1\Delta = 1. At low CC, need large cache to approach NDT = 1. This is the "vertical" view of the cache-fronthaul tradeoff: hold CC fixed, vary cache.

Parameters
4
4
1

NDT Surface Δ(M/N,C)\Delta(M/N, C)

3D surface plot of the NDT as a function of both cache ratio and fronthaul capacity. The surface is monotonically non-increasing in both variables; iso-NDT curves are the level sets. The Pareto frontier for any target Δ∗\Delta^* is an iso-NDT contour on this surface.

Parameters
4
4

NDT Cache-Fronthaul Tradeoff Curves

NDT Δ\Delta as a function of memory ratio M/NM/N, for four fronthaul capacities C∈{0.5,1,2,5}C \in \{0.5, 1, 2, 5\} files/use. Smaller CC shifts the curves up; larger cache lowers NDT toward the baseline Δ=1\Delta = 1. The curves are iso-CC "slices" of the full NDT surface.

Example: Architectural Choice: Cloud-Heavy vs Cache-Heavy

A 5G deployment has K=100K = 100, NEN=10N_\text{EN} = 10, LEN=4L_\text{EN} = 4. Target NDT ≤2\leq 2. Two architectures: (A) Cloud-heavy: μ=0.05\mu = 0.05, find minimum CC. (B) Cache-heavy: C=1C = 1, find minimum μ\mu.

Solution
(A) Cloud-heavy

K(1−μ)/(NENC)+K(1−μ)/(NENLEN)≤1K(1-\mu)/(N_\text{EN} C) + K(1-\mu)/(N_\text{EN} L_\text{EN}) \leq 1. 100⋅0.95/(10C)+100⋅0.95/(10⋅4)≤1100 \cdot 0.95/(10 C) + 100 \cdot 0.95/(10 \cdot 4) \leq 1. 9.5/C+2.375≤19.5/C + 2.375 \leq 1. 9.5/C≤−1.3759.5/C \leq -1.375: infeasible!

NDT = 2 is too aggressive for μ=0.05\mu = 0.05; the downlink alone contributes 2.375 already. Relax target: for Δ≤4\Delta \leq 4, 9.5/C≤1.6259.5/C \leq 1.625, C≥5.85C \geq 5.85 files/use. That works.

(B) Cache-heavy

C=1C = 1: 100(1−μ)/10+100(1−μ)/40=10(1−μ)+2.5(1−μ)=12.5(1−μ)≤1100(1-\mu)/10 + 100(1-\mu)/40 = 10(1-\mu) + 2.5(1-\mu) = 12.5(1-\mu) \leq 1. 1−μ≤0.081 - \mu \leq 0.08, so μ≥0.92\mu \geq 0.92. Need 92% cache fraction — cache the entire library almost.

Comparison

Neither extreme meets NDT ≤ 2 comfortably. A balanced point — say μ=0.5\mu = 0.5, C=2C = 2 — gives 100⋅0.5/(10⋅2)+100⋅0.5/40=2.5+1.25=3.75100 \cdot 0.5/(10 \cdot 2) + 100 \cdot 0.5/40 = 2.5 + 1.25 = 3.75, still above 2. The "sweet spot" depends on deployment constraints and cost ratios.

Cost consideration

Cache at RU costs ~10/GB;fronthaulcosts 10/GB; fronthaul costs ~500/Gbps/month. The economical choice depends on the content-to-bandwidth ratio. Video-heavy (large libraries) favors cache; general traffic favors fronthaul.

Definition:

Marginal Substitution Rate

At a given operating point (μ0,C0)(\mu_0, C_0) on an iso-NDT contour, the marginal substitution rate −∂C/∂μ-\partial C/\partial \mu measures how much fronthaul can be saved by adding one unit of cache. From the NDT formula: −∂C∂μ∣Δ∗  =  NENC02K⋅1(1−μ0)2.-\frac{\partial C}{\partial \mu}\Bigg|_{\Delta^*} \;=\; \frac{N_\text{EN} C_0^2}{K} \cdot \frac{1}{(1 - \mu_0)^2}. This rate is steep near the cache-limited corner (small μ\mu, need lots of cache growth for small CC reduction) and shallow near the fronthaul-limited corner.

In deployment economics: buy cache if cM<cC⋅∣∂C/∂μ∣Δ∗c_M < c_C \cdot |\partial C/\partial \mu|_{\Delta^*}, where cM,cCc_M, c_C are unit costs. This determines the optimal architecture along the iso-NDT curve.

Pareto Frontier in Operator Terms

The NDT iso-contour is the Pareto frontier for architectural decisions: any point on the contour is Pareto-optimal (no other point dominates it in both μ\mu and CC). Deployment decisions slide along this frontier based on:

  1. Cost structure. Cache hardware (storage + controllers) vs. fronthaul bandwidth lease/installation.
  2. Content catalog size. Larger library needs more aggregate cache; fixed μ\mu scales with NN.
  3. Traffic pattern. Cacheable (video) benefits from cache; mixed traffic has JLEC separation concerns (Ch 6).
  4. Access latency. Cache at RU gives ∼0.1\sim 0.1 ms access; cloud fronthaul adds ∼1-10\sim 1\text{-}10 ms.

A typical 5G deployment sits in the "balanced" region of the Pareto frontier, with moderate cache (μ∼0.1-0.3\mu \sim 0.1\text{-}0.3) and moderate fronthaul (C∼1-5C \sim 1\text{-}5 files/use at high SNR).

Common Mistake: NDT Is Not a Rate

Mistake:

Confusing NDT with a throughput or capacity value.

Correction:

NDT is a dimensionless latency ratio. Δ=2\Delta = 2 means "delivery takes twice as long as the baseline MU-MIMO." It does not mean "rate is half" — the relation to rate involves the SNR and the reference baseline. For rate conversions: Rper-user=log⁡2SNR⋅K/(T⋅Nusers)=log⁡2SNR/ΔR_\text{per-user} = \log_2 \text{SNR} \cdot K/(T \cdot N_\text{users}) = \log_2 \text{SNR} / \Delta at high SNR.

NDT = 1 corresponds to per-user rate = log⁡2SNR/1\log_2 \text{SNR}/1 — the asymptotic single-user capacity.

Normalized Delivery Time (NDT)NDT Upper and Lower Bounds