Resource Augmentation
Beyond Worst-case Analysis Reading Group Lecture 3
Victor Sanches Portella
Computer Science @ UBC
May, 2020
Online Paging Problem
Online Paging (Sledor & Tarjan '85)
Main/Slow Memory with \(N\) pages
Fast Memory (Cache) with \(k\) pages
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1
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Page requests arrive online
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0 cost if in cache
1 cost if not in cache
Move to cache, may evict another page
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Deciding which pages to evict
Goal: Algorithm that evicts pages in a way to minimize cache misses
First In, First Out (FIFO)
Least Recently Used (LRU)
Furthest In the Future (FIF)
Both perform equally bad in the worst case analysis
Idea: Compare with algorithms that know the entire sequence before-hand
Optimal Offline Algorithm (OPT)
Competitive Analysis
Competitive Ratio of \(A\):
\mathrm{cost}(A, \sigma) =
# cache misses of \(A\) in the sequence of requests \(\sigma\)
\displaystyle \sup_{\sigma } \frac{\mathrm{cost}(A, \sigma)}{\mathrm{cost}(\text{OPT}, \sigma)}
\geq 1
\alpha
\mathrm{cost}(A, \sigma) \leq \alpha \;\mathrm{cost}(\text{OPT}, \sigma)
For any sequence \(\sigma\) and any algorithm \(B\)
\leq \alpha \;\mathrm{cost}(B, \sigma)
\(A\) is \(\alpha\)-instance optimal
... this is really strong
Competitive Analysis for Online Paging
An Optimal Offline Algorithm
We want to compare the online algorithms with a optimal offline algorithm. Do we know any optimal offline algorithm?
Theorem 2.1 (Belady's Theorem) The Furthest In the Future (FIF) algorithm is offline optimal
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\sigma
A Lower Bound on the Competitive Ratio
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2
...
k
k+1
Proof:
Theorem 4.1 Every deterministic algorithm has competitive ratio at least the cache size \(k\).
Cache size \(k\). Main memory size \(N = k+1\).
Online algorithm \(A\).
Idea: Construct (inductively) a seq. where \(A\) misses all the time
\sigma
\(A\) evicts:
p_1
p_1
p_2
p_2
p_3
p_3
p_4
...
\text{cost}(A, \sigma) = |\sigma|
Cost of FIF?
Warm-up
A Lower Bound on the Competitive Ratio
Theorem 4.1 Every deterministic algorithm has competitive ratio at least the cache size \(k\).
Proof:
Cache size \(k\). Main memory size \(N = k+1\).
Online algorithm \(A\).
Idea: Construct (inductively) a seq. where \(A\) misses all the time
\text{cost}(A, \sigma) = |\sigma|
Cost of FIF?
\(N-1 = k\) distinct of \(q\)
q
q
p
Cache of size \(k\)
\implies
Only \(q\) is out of cache during this time (post warm-up)
\text{cost}(\text{FIF}, \sigma) \leq k + \frac{|\sigma|}{k}
Length \(\geq k\)
A Lower Bound on the Competitive Ratio
Theorem 4.1 Every deterministic algorithm has competitive ratio at least the cache size \(k\).
Proof:
Cache size \(k\). Main memory size \(N = k+1\).
Online algorithm \(A\).
Idea: Construct (inductively) a seq. where \(A\) misses all the time
\text{cost}(A, \sigma) = |\sigma|
\text{cost}(\text{FIF}, \sigma) \leq k + \frac{|\sigma|}{k}
\displaystyle \frac{\text{cost}(A, \sigma)}{\text{cost}(\text{FIF}, \sigma)}
\text{cost}(A, \sigma) = |\sigma|
\displaystyle \geq \frac{|\sigma|}{k + \frac{|\sigma|}{k}}
\displaystyle \longrightarrow k
\displaystyle \text{as}~|\sigma| \to \infty
Discussing the Lower bound
Every deterministic algorithm performs at least \(k\) times worse than OPT
In practice, algorithms such as LRU do not suffer from this problem
Horrible performance
This lower bound holds even for algorithms that can look a finite amount into the future.
Maybe competitive ratio doesn't reflect real-world performance
OPT is incredibly powerful
Is competitive ratio useful to compare algorithms?
LRU is online optimal
Theorem 4.3 LRU has competitive ratio \(k\)
Proof:
Let \(\sigma\) be a rquest sequence.
\sigma
Divide \(\sigma\) in block \(\sigma_1, \dotsc, \sigma_b\), where \(\sigma_i\) is maximal prefix with \(k\) distinct pages
\sigma_1
k~\text{distinct}
k+1\text{-th}
\sigma_2
k~\text{distinct}
\sigma_3
k~\text{distinct}
\dotsm
\sigma_b
k~\text{distinct}
Claim 1:
LRU suffers at most \(k\) cache misses in each block.
Claim 2:
FIF suffers at least \(b\) faults.
\displaystyle \frac{\text{cost}(A, \sigma)}{\text{cost}(\text{FIF}, \sigma)}
\displaystyle \leq \frac{k b}{b}
\displaystyle =k
LRU is online optimal
\sigma
\sigma_1
k~\text{distinct}
k+1\text{-th}
\sigma_2
k~\text{distinct}
\sigma_3
k~\text{distinct}
\dotsm
\sigma_b
k~\text{distinct}
Claim 1:
LRU suffers at most \(k\) cache misses in each block.
Claim 2:
FIF suffers at least \(b\) faults.
After a page \(p\) has been placed in cache, \(k\) distinct requests need to happen for \(p\) to be evicted.
Each block has \(\leq k\) distinct pages
\(\implies\) A page is placed in cache in a block is not removed during the same block.
LRU is online optimal
\sigma
\sigma_1
k~\text{distinct}
k+1\text{-th}
\sigma_2
k~\text{distinct}
\sigma_3
k~\text{distinct}
\dotsm
\sigma_b
k~\text{distinct}
Claim 1:
LRU suffers at most \(k\) cache misses in each block.
Claim 2:
FIF suffers at least \(b\) faults.
At the end of a block \(\sigma_i\), the cache is full with the \(k\) pages of \(\sigma_i\)
The first page of \(\sigma_{i+1}\) ( Â Â ) is not in cache.
\(\implies\) FIF faults at least once per block
Competitive Ratio as a Measure of Optimality
Competitive ratio shows that the famous LRU has optimal competitive ratio.
The ratio may be good to compare algorithms
Not quite. Another "optimal" algorithm: Flush-When-Full (FWF)
Clears entire cache in the case of a single page fault.
FWF still suffers at most \(k\) loss in a block from the previous proof
Comp. ratio \(k\), but this algorithm is HORRIBLE
Competitive Analysis for Online Paging
Explains performance of algorithms in practice?
Useful to design new algorithms?
No. \(k\) times far from OPT is not what happens in practice.
Useful to compare algorithms?
Kind of. Shows the LRU is optimal, but other stupid algorithms are also optimal.
Not really. FWF shows that competitive ratio ignores somethings which are really important.
Comp. analysis is a incredibly useful technique. But in this case it is not working well out-of-the-box.
Resource Augmentation
Resource Augmentation
OPT is too powerful, it is unfair to compare it with online algorithms
Idea: Compare an algorithm with a handicapped OPT
\text{cost}(A, k, \sigma) =
k
# Faults of \(A\) with cache \(k\) on the seq. \(\sigma\)
Compare \(A\) with cache \(k\) against OPT with cache \(h \leq k\)
\displaystyle \sup_{\sigma } \frac{\mathrm{cost}(A, k, \sigma)}{\mathrm{cost}(\text{OPT}, h, \sigma)}
Still is a worst-case analysis
LRU with Resource Augmentation
Theorem 4.3 LRU has competitive ratio \(k\)
Claim 1:
LRU suffers at most \(k\) cache misses in each block.
Claim 2:
FIF suffers at least \(b\) faults.
What if FIF has only \(h \leq k\) of cache?
\sigma
\sigma_i
k~\text{distinct}
\sigma_{i+1}
k~\text{distinct}
p
p
\(h-1\) of space
\(k\) distinct pages \(\neq p\)
\implies
k - (h - 1) faults per block
\((k - h + 1)b\) faults
Cache
LRU with Resource Augmentation
Claim 1:
LRU suffers at most \(k\) cache misses in each block.
Claim 2:
FIF suffers at least \(b\) faults.
\((k - h + 1)b\) faults
\displaystyle \sup_{\sigma } \frac{\mathrm{cost}(\text{LRU}, k, \sigma)}{\mathrm{cost}(\text{OPT}, h, \sigma)}
\leq \frac{k}{k - h + 1}
\displaystyle
\mathrm{cost}(\text{LRU}, 2k, \sigma) \leq 2 \mathrm{cost}(\text{OPT}, k, \sigma)
\(h = k/2\)
LRU with Resource Augmentation
\displaystyle
\mathrm{cost}(\text{LRU}, 2k, \sigma) \leq 2 \mathrm{cost}(\text{OPT}, k, \sigma)
\(h = k/2\)
How to use this "in practice"
See how much cache you need for your need based on the performance of OPT
Double the amount of cache, so LRU performs almost as well as OPT
Any theoretical use for resource augmentation?
Loosely Competitive
Theorem 6.1
Loosely Competitive
For all \(\varepsilon, \delta > 0\), all \(n\), all request seq. \(\sigma\),
there is \(S \subseteq \{1, \dotsc, n\}\) with \(|S| \geq (1 - \varepsilon)n\) such that, for all \(k \in S\),
\text{cost}(\text{LRU}, k, \sigma)
\leq O(\frac{1}{\varepsilon} \log \frac{1}{\delta}) \text{cost}(\text{OPT}, k, \sigma)
either
\text{cost}(\text{LRU}, k, \sigma)
\leq \delta |\sigma|
or
Good relative performance
Good absolute performance
Caveat: Order of quantifiers
Proof of Theorem 6.1
Fix \(\varepsilon, \delta, n, \sigma\). Let \(k, g \in \mathbb{N}\).
By the resource aug. theorem,
\displaystyle
\mathrm{cost}(\text{LRU}, k + g, \sigma) \leq \frac{k + g}{g + 1} \mathrm{cost}(\text{OPT}, k, \sigma)
When removing \(g\) pages from LRU, one of these happens:
\displaystyle
\mathrm{cost}(\text{LRU}, k, \sigma) \leq 2 \mathrm{cost}(\text{LRU}, k + g, \sigma)
\displaystyle
\mathrm{cost}(\text{LRU}, k, \sigma) > 2 \mathrm{cost}(\text{LRU}, k + g, \sigma)
Good \(k\)
Bad \(k\)
If # bad sizes \(\leq \varepsilon n\), we are done.
Suppose \(> \varepsilon n\) bad k's. Idea:
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t
t - g
\(\sim \varepsilon n\) bad k's
All absol. good!
\(g\) bads
\(\leq \varepsilon n\)
Proof of Theorem 6.1
\displaystyle
\mathrm{cost}(\text{LRU}, k + 1, \sigma) \leq \mathrm{cost}(\text{LRU}, k, \sigma)
I will use without proof:
\displaystyle
\mathrm{cost}(\text{LRU}, t, \sigma) < \frac{1}{2} \mathrm{cost}(\text{LRU}, k_1, \sigma)
Pick \(t \in \{1, \dotsc, n\}\) such that we have \(\sim \varepsilon n\) bad k's between \(1\) and \(t - g\)
1
t - g
k_1
\geq g
k_2
\geq g
k_3
\geq g
\dotsm
k_4
\sim \frac{\varepsilon n}{b}~\text{bad k's}
\displaystyle
< \frac{1}{2^2} \mathrm{cost}(\text{LRU}, k_2, \sigma)
\displaystyle
< \frac{1}{2^3} \mathrm{cost}(\text{LRU}, k_3, \sigma)
\displaystyle
\dotsm
\displaystyle
< 2^{-\varepsilon n/g} \mathrm{cost}(\text{LRU}, 1, \sigma)
\displaystyle
= 2^{-\varepsilon n/g} |\sigma|
\delta
= 2^{-\varepsilon n/g} |\sigma|
\displaystyle g = \frac{\varepsilon n}{\log(1/\delta)} \leq \varepsilon n
\iff
t \geq k_1 + g
k_1 \geq k_2 + g
k_2 \geq k_3 + g
Proof of Theorem 6.1
1
n
t
t - g
\(\sim \varepsilon n\) bad k's
All absol. good!
\(g\) bads
\(\leq \varepsilon n\)
\displaystyle
\mathrm{cost}(\text{LRU}, k, \sigma) \leq 2 \frac{k + g}{g + 1} \mathrm{cost}(\text{OPT}, k, \sigma)
Good \(k\)
Total of \(2 \varepsilon n\) bad k's. Use \(\varepsilon/2\) instead.
\displaystyle
\leq 2 \frac{n + g}{g} \mathrm{cost}(\text{OPT}, k, \sigma)
\displaystyle
\leq O\left(\frac{1}{\varepsilon} \log \frac{1}{\delta}\right) \mathrm{cost}(\text{OPT}, k, \sigma)
Final Remarks
Final Remarks
Resource augmentation is a way to make more fine-grained analysis when to problem is parameterized by some kind of resource
This technique yields non-interpretable claims, but was very useful to derive good results (loosely competivive)
This is still a worst-case analysis. We ignore the form of the input.
LRU works well in practice because of the form of real-world data.
Resource Augmentation
By Victor Sanches Portella
