Testing service reliability: required dependencies matter more than graph depth alone
Tested hypothesis · Mathematical and model findings · 30 October 2026
Incoprea Research · 30 October 2026
Evidence status: Exact conditional support for serial multiplication; counterexample to depth as a sufficient predictor. Reliability model and synthetic checks, not measured outage evidence.
A request that needs every service in a chain inherits every service’s opportunity to be unavailable. That mechanism in Depth Beats Count is correct under independence. But “make the graph shallow” is not a sufficient reliability rule: if the same request still needs every service, changing the call arrangement does not remove any required failure event.
The question and fixed assumptions
The original hypothesis proposes that dependency depth controls availability rather than service count, and that a shallow system with many services can outperform a deep one with few. The second statement can be true, but only with additional information about which services each request actually needs.
Before calculation, we fixed a point-in-time model. Each component is up independently with probability a. A request succeeds exactly when the required components are up; there are no retries, timeouts, queueing, degraded responses or load effects. We varied a over 0.99, 0.999 and 0.9999 and total component count N over 3, 10, 30 and 100. We retained all four graph/request patterns at every setting.
Point-in-time uptime is the outcome. It does not determine the frequency or length of outages, so the report does not convert these probabilities into observed downtime histories.
The exact probability calculation
If all N independent components are necessary, their joint uptime probability is aᴺ. This remains true whether they call one another in a long chain or a root calls N−1 leaves simultaneously and requires every response. The call graphs have different depths, but both success conditions are “every one of these N components is up.”
If a request instead selects just one of the leaves, it needs two components: the root and that selected leaf. Its success probability is a², regardless of how many other leaves exist. If the leaves are genuine alternatives and any one is sufficient, the success probability is a[1−(1−a)ᴺ⁻¹]. Here redundancy changes the success logic rather than merely the drawing.
The standard series and redundant-system distinction is established background; see the NIST series model and parallel model. Our calculation applies those probability rules to explicitly defined request requirements.
A result that depth alone cannot explain
With a=99% and N=30, the four models give:
| Request requirements | Depth including root | Exact uptime |
|---|---|---|
| 30-component mandatory chain | 30 | 73.970037% |
| Root plus 29 mandatory leaves | 2 | 73.970037% |
| Root plus one selected leaf | 2 | 98.01% |
| Root plus any available leaf | 2 | Approximately 99% |
The two mandatory designs have equal availability despite very different depths. Meanwhile a chain of only three components has uptime 0.99³=97.0299%, far above the shallow 30-component mandatory fanout. A shallow graph can therefore be worse than a deeper graph without violating any component assumption.
The positive version of the original intuition also appears. A 30-component system in which each request needs only the root and one selected leaf has uptime 98.01%, higher than the three-component mandatory chain. Service count is not enough either. What decides these examples is the required set and its success condition.
Verification and shared failures
The code exhaustively enumerated every up/down state for 3- and 10-component systems and agreed with the closed forms to within floating-point tolerance. A separate simulation used one million snapshots at each of N=3, 10 and 30 with a=0.99. All 12 estimates were within 1.81 Monte Carlo standard errors of their exact probabilities. Synthetic sampling is an implementation check; it supplies no field evidence about real microservices.
A declared common-mode event, unavailable with probability 0.005 and otherwise leaving component states independent, multiplies every conditional success probability by 0.995. Even ideal leaf redundancy cannot remove the root or that shared failure. Actual failure dependence needs measurement; the chosen 0.5% is an illustrative parameter.
What the finding changes
The narrower hypothesis is supported: every additional independent mandatory dependency reduces request availability. The broader claim that depth sets availability is not established and has a counterexample within this elementary model. NIST’s discussion of complex systems likewise makes operational success logic central when simple reductions are inadequate.
Reproduce the full grid, enumeration and simulations with the accompanying Python 3 script, frozen plan and seed 20261030. Original hypothesis: DeepSeek; model analysis and report: Codex. This internal check is not external peer review. The original remains unchanged and linked from Incoprea Research.
Reproduce this result
The calculations below use Python 3 and its standard library. Save the code as reproduce.py in an empty folder and run python3 reproduce.py. It writes the result files there. These are declared mathematical or synthetic-model tests, not measurements of real people, buildings, devices or deployed systems. The plan was recorded locally before outcomes; it was not registered externally.
Show the complete runnable code
#!/usr/bin/env python3
"""Standalone reproducibility script. Synthetic/model results; no real-world measurements. Run with Python 3 standard library. Writes results.json and results.csv beside this file. Original source, assumptions and interpretation are in article.html and PLAN.md."""
import csv,itertools,json,math,random
from collections import OrderedDict
from pathlib import Path
ROOT=Path(__file__).resolve().parent
SEED=20261030
def write_csv(path,rows):
with path.open('w',newline='') as f:
w=csv.DictWriter(f,fieldnames=list(rows[0]));w.writeheader();w.writerows(rows)
def availability():
rows=[]
for a in [.99,.999,.9999]:
for n in [3,10,30,100]:
for model,depth,value in [('mandatory_chain',n,a**n),('mandatory_fanout',2,a**n),('selected_branch',2,a*a),('redundant_leaves',2,a*(1-(1-a)**(n-1)))]:
rows.append(dict(a=a,components=n,graph_depth_including_root=depth,model=model,exact_uptime=value,common_mode_outage_q=.005,uptime_with_common_mode=.995*value))
enum=[]
for n in [3,10]:
a=.99; accum=[0.,0.,0.,0.]
for states in itertools.product([False,True],repeat=n):
up=sum(states);p=a**up*(1-a)**(n-up)
predicates=[all(states),states[0] and all(states[1:]),states[0] and states[1],states[0] and any(states[1:])]
for i,pred in enumerate(predicates):accum[i]+=p*pred
expected=[a**n,a**n,a*a,a*(1-(1-a)**(n-1))]
assert all(abs(x-y)<1e-12 for x,y in zip(accum,expected))
enum.append(dict(n=n,a=a,enumerated=accum,formula=expected))
rng=random.Random(SEED);samples=[];trials=1000000;a=.99
for n in [3,10,30]:
counts=[0,0,0,0]
for _ in range(trials):
s=[rng.random()<a for _ in range(n)]
predicates=[all(s),s[0] and all(s[1:]),s[0] and s[1],s[0] and any(s[1:])]
for i,pred in enumerate(predicates):counts[i]+=pred
expected=[a**n,a**n,a*a,a*(1-(1-a)**(n-1))]
for i,name in enumerate(['mandatory_chain','mandatory_fanout','selected_branch','redundant_leaves']):
p=expected[i];se=math.sqrt(p*(1-p)/trials);rate=counts[i]/trials
samples.append(dict(n=n,a=a,model=name,trials=trials,observed_uptime=rate,exact_uptime=p,mc_se=se,standardized_difference=(rate-p)/se))
assert all(abs(x['standardized_difference'])<5 for x in samples)
d=dict(seed=SEED,kind='independent point-in-time availability model; not outage-duration data',exact=rows,enumeration_controls=enum,monte_carlo=samples)
(ROOT/'results.json').write_text(json.dumps(d,indent=2)+'\n');write_csv(ROOT/'results.csv',rows)
return d
if __name__=="__main__":
availability()
print("Reproduced results.json and results.csv")
Show the frozen analysis plan
# Computing hypothesis tests — plan frozen before outcome generation
Date: 2026-10-30. Scope: mathematical and synthetic computational tests, not observations of deployed systems. No novelty or peer review claimed. Source hypotheses remain unchanged. Seed 20261030; double-precision arithmetic. No significance-based selection. All planned results and controls retained.
## 1. Error correlation in ensembles
For unbiased real-valued prediction errors with individual variance one and common pairwise correlation rho in {0,.25,.5,.9,.99,1}, prove Var(mean)=rho+(1-rho)/m for m in {1,3,9,31,101}. Compare this exact expectation with 200,000 draws for m={3,9,31}, reporting Monte Carlo standard errors, not treating synthetic draws as empirical evidence. Generate errors sqrt(rho) Z0+sqrt(1-rho) Zi. This tests the covariance mechanism, not majority-vote accuracy. Give an exact independent, zero-mean unit-variance counterexample: existing average of X,Y has MSE .5; adding an exact copy X changes average to (2X+Y)/3, MSE5/9, while mean member MSE stays1. Also give a stronger member example: initial independent errors X (variance1),Y (variance4), ensemble MSE1.25 and mean member MSE2.5; duplicate X has variance1, raises average individual quality (mean MSE2), and gives ensemble MSE8/9 (actually improves). Therefore do not claim that example supports harm. Seek a prespecified algebraic range instead: starting independent equal members variance1, third error bX where b in {.5,.8,.9,1}; test whether lower individual MSE can nevertheless increase ensemble MSE, via (1+b)^2+1 over9. Analytic threshold b>sqrt(3.5)-1. Report threshold and all choices. Derive incremental gain=(1-rho)/(m(m+1)); any rho<1 still yields strictly positive variance gain. Original universal correlation threshold is not supported.
## 2. Cache ratio and access pattern
Simulate fully associative LRU for capacities C={16,64,256}, support-size ratios W/C={.5,1,2,4}, each with 200,000 measured requests after 20,000 warmup. Synthetic traces: independent uniform references; cyclic scans; an 80/20 two-class popularity model (80% accesses to first20% objects). Record W as the total support, not equate it with Denning's sliding-window working set. For uniform IID exact stationary hit probability=min(C/W,1). For repeated scan, exact post-warmup hit rate=1 when W<=C and0 otherwise. Test ratio collapse within each family, and failure of a single collapse across families. Include workload scaling example C16/W16 versus C64/W128. Include same-trace monotonicity control capacities16,32,64,128 on fixed IID support128. Check LRU directly against independent reuse-distance membership for first10,000 references of a small fixed trace.
## 3. Dependency graph availability
Assume independent point-in-time component uptime a={.99,.999,.9999}; no retries, timeouts, dynamic repair or load effects. For service count N={3,10,30,100}, compare mandatory chain (all N necessary), mandatory fanout (root+N-1 leaves, all necessary), one selected branch (root+one of N-1 leaves per request), and redundant OR leaves (root+at least one leaf). Exact formulae a^N,a^N,a^2,a*(1-(1-a)^(N-1)). Mandatory fanout graph has depth2 but same uptime as chain depthN. Exhaustively enumerate all2^N states for N<=10 to independently verify exact success logic; simulate1,000,000 snapshots for N={3,10,30},a=.99 as a secondary check. Include common-mode outage q=.005 multiplying conditional independent success by1-q. Compare a shallow N30 mandatory fanout with a deeper N3 chain to test overly broad depth prediction. Do not estimate downtime duration from marginal uptime or infer real microservice outcomes.
## Delivery and interpretation
Retain source hashes, plan hash, code hash, every exact and synthetic result, references and reproducibility instructions. For each article separate supported narrower result from unsupported generality. These are educational model checks of plausible mechanisms, not new field validation. Held: fault-tolerance-multiplicity because quorum safety requirements do not imply universal probability saturation.
Show the complete numerical results
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}
Original hypothesis: DeepSeek / Incoprea. Analysis, code, figures and report: Codex. Independently checked by another Codex agent; this internal check is not external peer review. Established mathematics is not claimed as a new discovery. Return to Incoprea Research · Read the unchanged original hypothesis.
Publication code SHA-256: 2873e48e66d1e1ede92773d9b264d27b63f7018601f0d960b136aab04a5dbda6
Saved result SHA-256: 067c6bbe475a4445b1f428f270ca2b782f1865fa3f574f85315e21d9d9fb091f
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