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CS6998-3: Solutions to Problem Set # 1
Etienne Vouga
Problem 1 (10 points)
(a) We know from the theorem stated (but not proven) in class that it suffices to restrict our attention to two-link, two-node graphs, with edge latency functions l 1 (x) = axi^ and l 2 (x) = 1. In other words, letting Pd(N ) denote the price of anarchy of a graph N with edge latencies in Md, and [f (x); g(x)] a two-link, two-node graph with latency functions f (x) and g(x), we have
Pd(N ) ≤ max a, 0 ≤i≤d
Pd([axi; 1]).
We now compute the right-hand side in five steps.
Lemma 0.1. If both latency functions are constant, Nash and optimal flows have equal cost:
max a Pd([ax^0 ; 1]) = 1.
Proof. If both latency functions are constant, obviously the flow routing all supply through the edge with least latency is both optimal and Nash.
Lemma 0.2. If i 6 = 0 and a ≤ (1 + i)−^1 , Nash and optimal flows have equal cost:
max a≤(1+i)−^1 , 1 ≤i≤d
Pd([axi; 1]) = 1.
Proof. We check that the flow S routing all supply through the first edge is both optimal and Nash. Since S sends no supply through the second edge, S is optimal if the marginal cost of the first edge is at most that of the second:
a ≤ (i + 1)−^1 a(i + 1) ≤ 1 [ a(i + 1)xi
]
d dx
xaxi
d dx
(x) (0)
c′ 1 (1) ≤ c′ 2 (0),
as required. Thus S is optimal. To check S is Nash, we check that the latency of the first edge is at most the latency of the second:
a ≤ 1 axi ≤ 1 l 1 (1) ≤ l 2 (0),
so S is also Nash.
Lemma 0.3. If i 6 = 0 and a > 1 ,
max a> 1 , 1 ≤i≤d
Pd([axi; 1]) =
1 − d(d + 1)−(d+1)/d^
Proof. Since every step of the inequality manipulation in Lemma 0.2 is reversible, and a > 1 (and thus a > (1 + i)−^1 ), we know the flow routing all supply through the first edge is neither Nash nor optimal. We also check briefly that the flow routing all supply through the second edge is neither optimal nor Nash:
The marginal costs for such a flow are
c′ 1 (0) = (i + 1)axi = 0
since i 6 = 0, and c′ 2 (1) = 1,
so c′ 2 (1) 6 ≤ c′ 1 (0) and such flow is not optimal. The latencies are
l 1 (0) = axi = 0
l 2 (1) = 1
so similarly l 2 (1) 6 ≤ l 1 (0) and such flow is not Nash.
We conclude that both the Nash and optimal flow sends a little bit of supply through the first edge, and a little bit through the second. Thus, writing S∗^ for the Nash flow, C(S∗) for the cost of the Nash flow, and y as the amount of supply sent through the first edge, we have equality of latencies:
l 1 (y) = l 2 (1 − y) ayi^ = 1. (1)
The formula for the C(S∗) is C(S∗) = yayi^ + (1 − y). (2)
Plugging (1) into (2) gives C(S∗) = y + (1 − y) = 1.
What about the cost of the optimal solution? Let S be the optimal solution, which sends z supply through the first edge. We have equality of marginal costs:
c′ 1 (z) = c′ 2 (z) (i + 1)azi^ = 1, (3)
and a formula for C(S), C(S) = zazi^ + (1 − z). (4)
Solving for z in (3) yields z = [a(i + 1)]−^1 /i.
Substituting (3) into (4) gives
C(S) =
z i + 1
= 1 − i i + 1
z
i i + 1
[a(i + 1)]−^1 /i.
Unlike in the previous lemma, it is no longer obvious how changing a changes f , so we resort to taking a derivative using the quotient rule:
f ′(a) =
1 − (^) i+1i [a(i + 1)]−^1 /i^ − a[a(i + 1)]−^1 /i−^1 ( 1 − (^) i+1i [a(i + 1)]−^1 /i
1 − (^) i+1i [a(i + 1)]−^1 /i^ − (^) i+1^1 [a(i + 1)]−^1 /i ( 1 − (^) i+1i [a(i + 1)]−^1 /i
1 − [a(i + 1)]−^1 /i ( 1 − (^) i+1i [a(i + 1)]−^1 /i
since a(i + 1) > 1 and i ≥ 1. So we maximize the price of anarchy by maximizing a, which in this case means setting a = 1: max (1+i)−^1 0, and consider a second path p′ i from si to ti. Since the supply travelling through pi,j is acting greedily, the latency of the pi,j must be at most that of p′ i, or some supply would switch to flowing through there instead. Thus ∑
e∈pi,j
le(xe) ≤
e∈p′ i
le(xe).
where le is the latency function of edge e, and xe the total amount of supply (from all sources) flowing through e. Since le(x) = aex + be is linear, ∑
e∈pi,j
aexe + be ≤
e∈p′ i
aexe + be. (5)
Now suppose S is optimal, and again let pi,j be any path with xi,j > 0. Optimality means transferring any amount δ of supply from pi,j to another path p′ i from si to ti cannot improve the total cost of S, so the change in cost of such a switch must be nonnegative. Consider the edges e on pi,j. We can partition such edges into two sets: those edges U also on the path p′ i, and those V 1 that are not. Let V 2 be the set of edges of p′ i not on pi,j , that is, those not in U. Then, writing ce(x) = xle(x), the change in cost of switching δ supply from pi,j to p′ i is ∑
e∈V 1
ce(xe) +
e∈V 2
ce(xe) −
e∈V 1
ce(xe − δ) −
e∈V 2
ce(xe + δ).
Thus (^) ∑
e∈V 1
ce(xe) +
e∈V 2
ce(xe) −
e∈V 1
ce(xe − δ) −
e∈V 2
ce(xe + δ) ≤ 0.
We now proceed exactly as in the lecture notes. Manipulating this equation, and applying the definition of the derivative, yields ∑
e∈V 1
ce(xe) −
e∈V 1
ce(xe − δ) ≤
e∈V 2
ce(xe + δ) −
e∈V 2
ce(xe)
∑
e∈V 1
[ce(xe) − ce(xe − δ)] ≤
e∈V 2
[ce(xe + δ) − ce(xe)]
∑
e∈V 1
[ce(xe) − ce(xe − δ)] +
e∈U
[ce(xe + δ) − ce(xe)] ≤
e∈V 2
[ce(xe + δ) − ce(xe)] +
e∈U
[ce(xe + δ) − ce(xe)]
∑
e∈V 1
ce(xe) − ce(xe − δ) δ
e∈U
ce(xe + δ) − ce(xe) δ
e∈V 2
ce(xe + δ) − ce(xe) δ
e∈U
ce(xe + δ) − ce(xe) δ ∑
e∈V 1
c′e(xe) +
e∈U
c′e(xe) ≤
e∈V 2
c′e(xe) +
e∈U
c′e(xe)
∑
e∈pi,j
c′e(xe) ≤
e∈p′ i
c′e(xe).
Since le is linear, ce(xe) = aex^2 e + bexe, and c′ e(xe) = 2aexe + be, so ∑
e∈pi,j
2 aexe + be ≤
e∈p′ i
2 aexe + be. (6)
(b) Let S be a Nash solution. Then by (5), ∑
e∈pi,j
aexe + be ≤
e∈p′ i
aexe + be.
Now consider the flow S′^ found by halving the amount of supply flowing through each path in S. This flow routes 12 of a unit from each si to each ti, and if xe and x′ e are the total supply passing through an edge for the flow S and S′^ respectively, 2x′ e = xe. Substituting into the above equation gives ∑
e∈pi,j
2 aex′ e + be ≤
e∈p′ i
2 aex′ e + be,
which is exactly (6). Thus S′^ is optimal for routing 12 of a unit of supply from the sources to the sinks.
The least non-zero value of Φ is 1, so if we ever have Φ ≤ 1, we know we will reach Φ = 0, which must be a Nash equilibrium, in the next step. Thus, to bound the worst-case number of steps s needed to reach an equilibrium, we solve
1 =
k
)j C|E|(log k + 1), (7)
and know s ≤ 1 + j.
Taking the logarithm of both sides of (7) gives
0 = j log
k
- log C + log |E| + log(log k + 1)
j =
log C + log |E| + log(log k + 1) log
k k−
s ≤ 1 +
log C + log |E| + log(log k + 1) log k − log(k − )
To show s is polynomial in several variables, it is enough to show that it is polynomial in each individual variable with the other variables treated as constant.
Lemma 0.6. s ∈ O(log |E|) ⊂ O(|E|).
Proof. Obvious.
Lemma 0.7. s ∈ O(k log log k) ⊂ O(k^2 ).
Proof. Since log is analytic on its domain, by Taylor’s Theorem,
log(x − ) = log x +
∑^ ∞
i=
i!
(−)i^
di dxi^
log x
= log x +
∑^ ∞
i=
i!
i(−1)i^ (−1)i−^1 (i − 1)! xi
= log x +
∑^ ∞
i=
−i ixi
≤ log x −
x
so log k − log(k − ) ≥ (^) k and
1 log k − log(k − )
k
s ≤ 1 +
k(log C + log |E| + log(log k + 1))
so s ∈ O(k log log k).
Lemma 0.8. s ∈ O
Proof. Obvious from (8).
Figure 1: Under certain conditions described in part B of problem 4, this network design problem allows player 1 to decrease his cost by an infinite factor from the solution found by the proposed algorithm.
(b) Consider the graph depicted in Figure 1 for a two-player network design problem, where a is arbitrary, b = max
0 , 2 a−a
. Take as an initial guess S that player 1 takes the top link, and player 2 takes the bottom link. The potential Φ(S) is
Φ(S) = aH 1 + bH 1 = (a + b).
Player 2 clearly cannot improve his cost by choosing a different path. Player 1, on the other hand, can switch to the middle path. The resulting candidate solution S′^ then has potential
Φ(S′) = bH 2 = b.
We have
b >
2 a − a b > 2 a − a (a + b) > 2 a 1 2 (a + b) > a
(a + b)
k
> a k
Φ(S) > a,
so the change in potential a is not large. Thus the proposed algorithm terminates at S. However by switching to S′^ player 1 decreases his cost from a to 0, an infinite factor.
