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At an interior local extremum of a differentiable function, its gradient is zero: restricting the function to every line through the point shows that all directional derivatives vanish. This is only a necessary condition in general. For a strictly concave function, an interior stationary point is the unique global maximizer on its convex domain. Constrained or boundary optima instead need conditions adapted to their feasible directions.

On positive-definite matrices, the log-determinant has differential $d\log \det A=\mathrm{Tr}(A^{-1}dA)$ and is a strictly concave function.

The Lyapunov moment inequality says that higher positive absolute moments give no smaller absolute moment sizes on a probability space. For a real random variable $Z$ and $0<r<s$,allowing an infinite value. For $r\ge 1$ the left expression is an Lp norm; below one it is a quasi-norm, so the usual triangle inequality terminology needs that distinction. The comparison uses a probability measure of total mass one; arbitrary unnormalized measure spaces need a factor depending on their total mass.

For a nonnegative count, the first moment size is its expected value, while the second is the square root of its second moment. Both ten-trial counts in second moment method have first moment size one. The independent count has second moment size $\sqrt{1.9}\approx 1.37840$, whereas the shared-outcome count $N=10B$ has second moment size $\sqrt{10}\approx 3.16228$. The second moment distinguishes models with the same average because its square increases the influence of rare large counts.

More generally, for $N=10B$ with a Bernoulli random variable $B$ of parameter $1/10$,It rises from one at $r=1$ toward ten as $r\to \infty$. Replacing $1/10$ by a parameter $0<q\le 1$ gives $10q^{1/r}$, which decreases toward zero for a fixed positive exponent when $q\downarrow 0$, making the rare-event and large-exponent limits different questions. For a scaled Rademacher random variable taking values $-2$ and two with equal probabilities, every absolute moment size is exactly two: increasing the exponent does not distinguish signs when the absolute value is constant.

For finite $\mathbb{E}|Z{|}^s$, apply the reversed Jensen's inequality to the strictly concave function $t^{r/s}$ and the nonnegative random variable $|Z{|}^s$:Taking the positive $r$th root proves the claim. Equality in this finite case holds exactly when $|Z|$ is constant almost surely. If the higher moment is infinite, the upper bound is infinite and remains valid. Exponent zero is excluded from the statement; logarithmic or geometric limiting summaries require separately stated conventions, especially when $Z$ can vanish.

For an essential Nash bargaining problem, maximizing the Nash product gives a unique vector: on positive gains its logarithm is a strictly concave function. The rule satisfies the four bargaining axioms, and the supporting triangle for Nash bargaining proves their converse characterization. The nonessential case requires additional conventions, since a zero product may have many maximizers.

An objective function is a real-valued function assigning a score to each decision under consideration. An optimization problem seeks a feasible solution maximizing or minimizing this score. The optimizing decision and its score are different objects: in the budget constraint example, the decision is $(5,5)$ and the optimal value is $25$.

Choosing an objective function expresses what matters. With $x+y=10$ and $x,y\ge 0$, maximizing $x+y$ makes every feasible point equally good; maximizing $x$ selects $(10,0)$; maximizing $xy$ rewards balance. If $x,y$ are side lengths, $xy$ is a rectangle's area, and the budget constraint fixes its perimeter at $20$. The same mathematics then says that a square encloses the largest area among these rectangles.

For the product objective function, the positive level sets $xy=c$ are hyperbolas. Each collects decisions with identical scores. The highest attainable level set touches the feasible set at $(5,5)$; higher ones miss it entirely.

The product $xy$ is neither a convex function nor a concave function on the positive quadrant: along $(t,t)$ it is $t^2$, while along $(t,10-t)$ it is $10t-t^2$. Yet its restriction to this feasible set is a strictly concave function. Thus the geometry of the allowed decisions can make an otherwise nonconcave maximization problem tractable. Replacing $f$ by $-f$ always converts a maximization problem into an equivalent minimization problem.

// Target: mathematical-optimization.bigb

Interpret $x_r$ as route $r$'s sending rate, $w_r>0$ as its demand weight, and ${\kappa }_r>0$ as its response speed. Resource $j$ sees aggregate load $y_j={\sum }_{s:j\in s}{x}_s$ and returns the resource congestion price ${\mu }_j=p_j(y_j)$. Route $r$ sums these feedback signals to its route congestion price $P_r={\sum }_{j\in r}{\mu }_j$. Its rate increases when $x_r{P}_r<w_r$ and decreases when $x_r{P}_r>w_r$. At equilibrium, each user balances weighted logarithmic utility against the summed resource prices: $x_r{P}_r=w_r$.

Take finite route and resource sets, nonempty routes, positive $w_r,{\kappa }_r$, and initially positive rates. The standard interpretation of increasing prices supplies a price positive somewhere on each route; strictly increasing nonnegative prices automatically do so. More generally, nondecreasing prices suffice provided every route contains at least one resource with a nonzero price function. Define the primal congestion potentialBecause $p_j$ is continuous and nondecreasing, its integral is a convex function. The positive-weight logarithmic sum is strictly concave; subtracting the convex load costs makes $U$ a strictly concave function on the positive orthant. Its gradient satisfiesThus this is gradient flow with diagonal mobility, ascending the utility-minus-cost potential.

We must also prove boundedness and existence of an optimizer. For each route choose a resource on it whose price is positive somewhere. Monotonicity supplies constants $c_r>0,a_r<\infty$ such that this resource's price is at least $c_r$ for loads at least $a_r$. Its integrated cost is therefore at least $c_r{x}_r-d_r$. A resource may be chosen by several routes, but at most $|R|$ times. Since all costs are nonnegative,for a finite constant $D$. Each scalar expression on the right is bounded above and tends to minus infinity as $x_r\to \infty$. Consequently a superlevel set $U(x)\ge U(x(0))$ is bounded. On a bounded set, if any $x_r\downarrow 0$, its logarithmic term tends to minus infinity while the remaining terms are bounded above. The superlevel set therefore stays away from every zero coordinate and is compact inside the positive orthant. The same argument gives an attained maximum $x^{\ast }$; strict concavity makes it unique, and its interior first-order equations giveAlong every trajectory,with equality exactly at $x^{\ast }$. Thus the trajectory remains in its compact superlevel set, exists for all positive time, and $U(x(t))$ increases to a finite limit. The nonnegative dissipation on the right has a finite time integral. It is also uniformly continuous: it is a continuous function on the trapped compact set, and the continuous vector field is bounded there, making $x(t)$ uniformly Lipschitz in time. The result uniformly continuous integrable functions vanish at infinity gives $dU/dt\to 0$. Every subsequential limit is therefore a critical point, hence $x^{\ast }$. Compactness then proves the full convergenceContinuity, rather than differentiability, of the prices is enough. For completeness, square-root coordinates for primal congestion control also establish uniqueness of the underlying trajectories: with $x_r={\kappa }_r{z}_r^2/4$, the system becomes $\dot{z}=\nabla V(z)$ for $V(z)=U(x(z))$. Each resource cost is a convex nondecreasing function composed with a convex quadratic load, while the logarithmic terms remain strictly concave. Hence $V$ is concave, and for two trajectories $\frac{1}{2}\frac{d}{dt}\Vert z-\tilde{z}{\Vert }^2\le 0$ by monotonicity of its gradient. Identical initial data therefore give identical solutions without assuming Lipschitz price functions. If some initial rates are zero, ${\dot{x}}_r={\kappa }_r{w}_r>0$ at that boundary, so the nonnegative physical trajectory enters the positive orthant and the same proof applies from any positive time.

There is a genuine qualification if the word increasing is meant merely nondecreasing without a nontrivial-price assumption. With one route, $w=\kappa =1$ and $p(y)\equiv 0$, the displayed model has $\dot{x}=1$, so $x(t)=x(0)+t$ and no equilibrium. Positive weights, nonempty priced routes, and positive adjustment rates are necessary model assumptions, not consequences of an arbitrary nonnegative continuous price function. Under the standard positive-data, nontrivially priced interpretation, every physical trajectory converges to the unique point proved above.

A Nash bargaining problem consists of a compact convex set $S$ of feasible two-player payoff vectors and a disagreement point $d\in S$. If bargaining fails, the players receive $d$; agreement selects a feasible vector. The convex set allows jointly agreed correlated payoff lotteries over outcomes. Assume essentiality: some feasible vector strictly improves both coordinates of $d$. The admissible answers satisfy bargaining individual rationality, $u_i\ge d_i$.

The four characterizing axioms are Pareto efficiency (no feasible vector weakly improves both chosen payoffs and strictly improves one), bargaining symmetry (interchanging identical player roles leaves the answer unchanged), positive affine invariance in bargaining (independently replacing utilities by $a_i{u}_i+b_i$, $a_i>0$, transforms the answer by the same maps), and bargaining independence of irrelevant alternatives (restricting to an admissible smaller feasible set that still contains the chosen vector leaves the choice unchanged). The last axiom keeps the disagreement point fixed.

The arbitration procedure maximizes the Nash product over individually rational vectors:A maximizer exists by compactness and has positive gains by essentiality. It is unique, since the logarithm of the product is a strictly concave function of the two gains on their positive domain. The product is strictly increasing in either positive gain, giving Pareto efficiency. A positive affine utility change multiplies the product by $a_1{a}_2$, giving positive affine invariance in bargaining. Uniqueness gives bargaining symmetry and bargaining independence of irrelevant alternatives.

For completeness, these axioms also force the procedure. Normalize the product-maximizing vector to $(1,1)$ and $d$ to zero using positive affine invariance in bargaining. For any normalized feasible vector $z$, the directional derivative of the product along the feasible segment from $(1,1)$ to $z$ is $z_1+z_2-2$. The initial part of that segment has positive gains, so optimality implies $z_1+z_2\le 2$. Choose $M$ sufficiently large that the transformed feasible set is contained in the symmetric triangleOn this triangle, bargaining symmetry requires equal coordinates and Pareto efficiency then forces $(1,1)$. By bargaining independence of irrelevant alternatives, restricting from $T_M$ to the original normalized set still selects $(1,1)$. Undoing the normalization proves the Nash bargaining solution characterization. This is the supporting triangle for Nash bargaining argument.

For the numerical game, the maximin disagreement point is the vector of the players' separate security level payoffs. If player I uses its first action with probability $p$, its worst expected payoff isThe two lines meet at $p=1/4$; before that point the increasing second line is smaller and afterwards the decreasing first line is smaller. Hence $d_1=7/2$. If player II uses its first action with probability $q$, its worst payoff ismaximized at $q=1$, giving $d_2=4$. Thus $d=(7/2,4)$. These are separate worst-case guarantees; they need not equal the actual payoffs when both security strategies are used together.

The cooperative feasible set is the convex hull of the four outcome vectors. The outcomes $(2,4)$ and $(2,3)$ are dominated by $C=(4,5)$. Its Pareto frontier is the segment joining $C$ to $B=(8,2)$, with equation $y=8-3x/4$. Any maximizer of the positive Nash product is on this frontier. Bargaining individual rationality restricts its relevant portion to $4\le x\le 16/3$. There we maximizeThe stationary point lies strictly inside that interval and is the unique maximum. Therefore the Nash bargaining solution isThe gains are $11/12$ and $11/16$, with Nash product $121/192$. The arbitration can implement the vector by the agreed correlated payoff lottery assigning probability $5/48$ to outcome $B$ and $43/48$ to outcome $C$. Indeed $4+4(5/48)=53/12$ and $5-3(5/48)=75/16$. This is a correlated payoff lottery; independent mixed strategies are not required to implement the cooperative agreement.

For maximization of $f(x)$ subject to $g_j(x)\le 0$ and $h_k(x)=0$, the Lagrangian sufficiency theorem says: if a feasible $x^{\ast }$ globally maximizes $L(x)=f(x)-{\sum }_j{\lambda }_j{g}_j(x)+{\sum }_k{\nu }_k{h}_k(x)$, where ${\lambda }_j\ge 0$, and ${\lambda }_j{g}_j(x^{\ast })=0$, then $x^{\ast }$ globally maximizes $f$ over the feasible set. Indeed, at every feasible $x$,This requires a global maximum of the optimization Lagrangian; a stationary point alone is not sufficient without additional concavity or another justification.

Put $q=p/(p-1)$ and $C=({\sum }_i|a_i{|}^q{)}^{1/q}$. If every $a_i=0$, the maximum is zero and every feasible vector is optimal. Otherwise setSince $p(q-1)=q$, ${\sum }_i|x_i^{\ast }{|}^p=1$. For $g(x)={\sum }_i|x_i{|}^p-1$, the optimization Lagrangian is $L(x)={\sum }_i{a}_i{x}_i-\lambda g(x)$. Its gradient vanishes at $x^{\ast }$ becauseThe function ${\sum }_i|x_i{|}^p$ is a strictly convex function for $p>1$, so $L$ is a strictly concave function. Its supporting-hyperplane inequality proves that this stationary point is its global maximum. Complementary slackness and feasibility hold, so the sufficiency theorem applies. Substitution givesThe maximizing vector is unique when $a\ne 0$. It is the norming vector for Holder inequality, and the value is the corresponding dual norm.

Only routes with $n_r>0$ need a per-flow rate: absent flows contribute no utility and have no completion events. Interpret the proportional-change sum and the logarithmic objective on these active routes. With positive capacities and finitely many nonempty routes, a strictly positive feasible active-rate vector exists. Define its weighted logarithmic utility by $U(x)={\sum }_{r:n_r>0}{n}_r\log x_r$, with value $-\infty$ if an active rate is zero.

For positive feasible $x,y$, the elementary concavity inequality $\log u\le u-1$ givesThe last sum is nonpositive by proportional fairness. Feasible competitors with a zero active rate have utility $-\infty$, so they cannot improve it either. Thus a proportionally fair allocation maximizes weighted logarithmic utility. Conversely, if $x$ maximizes this utility, the feasible line segment $x+\theta (y-x)$ has nonpositive right derivative at $\theta =0$, giving precisely ${\sum }_r{n}_r(y_r-x_r)/x_r\le 0$. This also covers a competitor with zero coordinates, since $x+\theta (y-x)$ stays positive for sufficiently small $\theta$. The strictly concave function $U$ makes the active-rate optimizer unique.

For the linear flow network, put $M={\sum }_{i=1}^I{n}_i$ and $N=n_0+M$. If $n_i>0$ and resource $i$ is not saturated, its local per-flow rate could be increased without affecting any other resource. This would strictly increase weighted logarithmic utility. HenceIf $n_0>0$ and $M>0$, substitute $x_i=(1-n_0{x}_0)/n_i$ on all active local routes into the utility. The terms depending on $x_0$ areIts derivative vanishes when $1/x_0=M/(1-n_0{x}_0)$, so $x_0=1/N$. Its second derivative is negative, giving the unique maximizer. If $M=0<n_0$, utility increases up to the capacity endpoint $x_0=1/n_0=1/N$. If $n_0=0$, each active local route uses its entire local resource. Altogether the proportionally fair allocation on a linear flow network isAbsent-route rates may be set to zero, and departures are zero at the empty state $N=0$. Notice that these are per-flow rates: the aggregate through service is $n_0/N$, while each active local route gets aggregate service $M/N$.

In the document-transfer flow-level network model, let $e_r$ be the unit vector for route $r$. Independent Poisson processes of flow arrivals give transitions $n\to n+e_r$ at rate ${\nu }_r$. An active document on route $r$ receives service at rate $x_r(n)$ and has an exponential distribution of size with parameter ${\mu }_r$. By the memoryless property, its instantaneous completion rate is ${\mu }_r{x}_r(n)$. Summing over the active documents gives the continuous-time Markov chain intensitiesThus through-route departures have rate ${\mu }_0{n}_0/N$, and active local departures have rate ${\mu }_{i}M/N$. All other off-diagonal intensities vanish. The diagonal intensity is the negative of the sum of the outgoing intensities. Each route's aggregate service is at most one, so the total jump rate is bounded by ${\sum }_r({\nu }_r+{\mu }_r)$; the process is a nonexplosive continuous-time Markov chain.

Set ${\rho }_r={\nu }_r/{\mu }_r$, assume ${\mu }_r>0$, and define the balance function of a flow-level network byThe adjacent-state ratios of this binomial coefficient areThey equal the aggregate service rates just obtained. Therefore, with $w(n)=H(n){\prod }_r{\rho }_r^{n_r}$,These are the detailed balance for a continuous-time Markov chain equations on every adjacent pair. Summing incoming and outgoing pairs proves stationarity whenever the weights can be normalized. Hence the stationary law of a linear flow network isThe coefficient here is a binomial coefficient, not merely the total number of flows. Both it and the denominator in $x_0=1/N$ are visible in the original PDF but missing from the converted TeX.

We can also determine exactly when the normalizing constant is finite. For $0\le {\rho }_0<1$, summing first over $n_0$ with local counts fixed and applying the negative binomial series givesThe remaining sums are independent geometric series. Thusprovided ${\rho }_0+{\rho }_i<1$ for every $i$. These inequalities are also necessary: if ${\rho }_0\ge 1$, the states with every local count zero already give a divergent sum; otherwise any local ratio at least one makes its geometric series diverge. Thus the stationary formula requires strict offered load below capacity at every resource. With positive arrival rates, the process is irreducible, and these normalizable detailed balance weights give its unique stationary distribution. Outside these inequalities, the displayed weights are not a probability distribution. Zero arrival rates can be handled on the reachable class by setting the corresponding population to zero.

The Lagrangian sufficiency theorem for an equality constraint states that if a feasible $x_{\ast }$ and a multiplier ${\lambda }_{\ast }$ exist such that $x_{\ast }$ globally maximizes $L(x,{\lambda }_{\ast })=f(x)-{\lambda }_{\ast }\cdot (g(x)-b)$ over the underlying domain, then $x_{\ast }$ globally maximizes $f$ over $g(x)=b$. Indeed $L=f$ at every feasible point, so global maximality of $L$ directly compares the feasible objective values. The underlying domain can already impose $x_i\ge 0$.

For $0<p<1$, choose $\lambda =p{d}^{1-p}$. Each function $t^p-\lambda t$ on $t\ge 0$ is a strictly concave function and attains its unique maximum at $t=1/d$, since $p{t}^{p-1}=\lambda$ there. The sum Lagrangian is therefore globally maximized when every $x_i=1/d$, and this point is feasible. For $p=1$ the objective is identically one on the feasible set. For $p>1$, each feasible $x_i$ lies in $[0,1]$, so $x_i^p\le x_i$, with equality only at $0$ or $1$; equality in the total is attained exactly at a simplex vertex. ConsequentlyFor $0<p<1$ the unique maximizer is $(1/d,\dots ,1/d)$; for $p=1$ every feasible point is a maximizer; for $p>1$ the maximizers are exactly the $d$ vertices with one coordinate one and all others zero. When $d=1$ these descriptions coincide.

Each user maintains a transmission rate $x_r$. Resource $j$ measures its aggregate load $y_j={\sum }_{r:j\in r}{x}_r$ and returns the resource congestion price ${\mu }_j=p_j(y_j)$. The user adds these feedback signals to form its route congestion price $c_r={\sum }_{j\in r}{\mu }_j$. Its increase term is the constant ${\kappa }_r{w}_r$, while its decrease term ${\kappa }_r{x}_r{c}_r$ is proportional to both its current rate and the congestion signal. This is a distributed congestion control algorithm: resources need only local loads, and users need only prices along their own routes.

There is a needed distinction in the word increasing. If it permits nondecreasing price functions, the all-zero prices are allowed. Then ${\dot{x}}_r={\kappa }_r{w}_r>0$, so there is no equilibrium. The intended convergence statement holds when each nonempty route meets at least one resource whose price is positive at some load. This is automatic if every price is strictly increasing and nonnegative. We prove it under the more general nondecreasing assumption with this explicit condition, for finite sets of nonempty routes and nonnegative initial rates.

Write $P_j(y)={\int }_0^y{p}_j(u)du$. Each $P_j$ is convex and nondecreasing, and the primal congestion potential isOn positive rates this is strictly concave: the sum of positive-weight logarithms is strictly concave, while the resource term is convex. Its derivatives areConsequently the gradient flow with diagonal mobility satisfieswith equality exactly at an equilibrium.

We must also prove trapping and existence of the equilibrium. For each route choose a resource on it and constants $M_r,c_r^0>0$ such that its price is at least $c_r^0$ beyond load $M_r$. Finiteness gives common constants $M,c>0$. If $X={\max }_r{x}_r$ is large, a resource on a maximizing route has load at least $X$, so the resource penalty is at least $c(X-M)$. With $W={\sum }_r{w}_r$, therefore,Once all rates are bounded above, a rate tending to zero also makes $U\to -\infty$ through its positive-weight logarithm. Thus every nonempty superlevel set of $U$ is compact inside the positive orthant. A maximum exists by restricting to such a superlevel set, and strict concavity makes it unique, say $x^{\ast }$. It is interior and satisfies ${\partial }_{r}U(x^{\ast })=0$, equivalentlyConversely every positive equilibrium is a stationary point of the strictly concave function $U$ and hence is this maximum.

Only continuity of the price functions is given, so it is useful to justify trajectories without silently assuming differentiability. Put $z_r=2\sqrt{x_r/{\kappa }_r}$ and $V(z)=U(({\kappa }_r{z}_r^2/4{)}_r)$. The dynamics become ordinary gradient flow ascent, $\dot{z}=\nabla V(z)$. The function $V$ is strictly concave on the positive orthant: its user terms are $2w_r\log z_r$ plus constants, and each penalty is a convex nondecreasing function $P_j$ composed with the convex load ${\sum }_{r:j\in r}{\kappa }_r{z}_r^2/4$. Hence the supporting-tangent inequalities giveThe squared distance between two solutions is nonincreasing, proving uniqueness even though the vector field need only be continuous.

Nonnegative initial rates enter the positive orthant immediately: along a solution, variation of constants in ${\dot{x}}_r+{\kappa }_r{c}_r(t)x_r={\kappa }_r{w}_r$ expresses $x_r(t)$ as a positive exponential times $x_r(0)$ plus a strictly positive integral. Also ${\dot{x}}_r\le {\kappa }_r{w}_r$, so rates cannot become infinite in finite time. The distance argument applies after any positive time and, by taking that time down to zero, also proves uniqueness for initial zero rates.

For a positive starting time, the increasing $U$ traps the trajectory in a compact superlevel set. Its derivative has finite integral since $U$ is bounded above there. It is uniformly continuous in time, being a continuous function of the state on a compact set with bounded velocity. By uniformly continuous integrable functions vanish at infinity, $\dot{U}(t)\to 0$. Every limit point consequently has ${\partial }_{r}U=0$ for every $r$, since all rates are bounded away from zero. There is only the limit point $x^{\ast }$, soThis establishes global convergence of primal congestion control with the needed nonzero-price condition, and supplies the zero-price counterexample when increasing is read in the weak sense.

For the alternative increasing-supply resource-price dynamics, consider the dual congestion potentialIts domain is ${\mu }_j\ge 0$ with every $c_r>0$. Each integrated supply is strictly convex, because $q_j$ is strictly increasing, and each negative logarithm of a positive linear function is convex. Thus $H$ is strictly convex. Its sublevel sets are bounded: for each $j$, $q_j(1)>0$ gives a linear lower bound on the supply primitive at large ${\mu }_j$, whereas the negative logarithmic terms grow only logarithmically in ${\max }_j{\mu }_j$. A route-price sum tending to zero makes $H\to +\infty$ on a bounded set. A minimum therefore exists in this domain.

The derivatives areIf a used resource had ${\mu }_j=0$ at the minimum, then $q_j(0)=0$ but $y_j>0$, so increasing that coordinate would strictly lower $H$. Therefore the minimizer has strictly positive prices at every used resource. An unused resource has its unique minimizing price zero. At used resources its equations areThe multiplicative dynamics are ${\dot{\mu }}_j=-{\kappa }_j{\mu }_j{\partial }_{j}H$. On strictly positive used-resource prices, equilibrium is equivalent to these equations, and strict convexity proves existence and uniqueness of the positive equilibrium. No full-row-rank assumption on $A$ is needed here: the strictly increasing individual supplies already give strict convexity.

If zero prices at used resources are allowed, the unqualified uniqueness assertion is false because of boundary equilibria of multiplicative resource prices. For one unit-weight route using two resources and $q_1(u)=q_2(u)=u$, all three price vectorsare equilibria. At $(1,0)$ the user rate is one; the first resource has demand equal to supply, while the second has zero derivative because its price is zero. The last two vectors do not minimize $H$. Thus the positive-price domain, or a boundary rule preventing a used resource from remaining at zero price, is essential.

Finally, the primal and dual equilibrium equations agree when $q_j$ and $p_j$ are inverse functions on the relevant ranges. For example, continuous strictly increasing prices with $p_j(0)=0$ and $p_j(y)\to \infty$ have inverses satisfying the stated supply assumptions. Then ${\mu }_j=p_j(y_j)$ is equivalent to $y_j=q_j({\mu }_j)$, while both systems have $x_r=w_r/c_r$. With inverse price and supply functions, the two equilibria describe the same rates and prices. Without this relation they need not agree: for one route and one resource with $w=1$, $p(y)=y$ and $q(\mu )=2\mu$, the primal equilibrium is $(x,\mu )=(1,1)$ while the dual equilibrium is $(\sqrt{2},1/\sqrt{2})$.

The objective approximates the logarithm of the product-form stationary distribution of a loss network. To make the approximation precise, multiply arrival rates and capacities by $N$ and write $x=n/N$. The Stirling formula gives, uniformly on the feasible compact set,with $0\log 0=0$. The feasible set is $K=\{x\ge 0:Ax\le C\}$. Positive capacities and a nonempty route for each class make $K$ compact with a strictly positive feasible point. Since $H_{\nu }$ is a strictly concave function, its maximizer $x^{\ast }$ is unique; its infinite inward derivative at $x_r=0$ makes every component positive.

The Karush-Kuhn-Tucker conditions give nonnegative link Lagrange multipliers $z_j$ satisfyingFor any neighborhood of $x^{\ast }$, continuity and uniqueness give a strictly positive objective gap outside it. There are only polynomially many feasible lattice states, while their weights differ exponentially in $N$. The feasible point $\lfloor N{x}^{\ast }\rfloor$ has objective tending to $H_{\nu }(x^{\ast })$. It follows that the stationary law concentrates at $x^{\ast }$ on the scale $n/N$; in particular,Thus NETWORK predicts the dominant occupancy and mean occupancy in a large loss network. For unscaled inputs it gives a continuous approximation to the modal integer state.

It also gives an exact useful change of variables. Substituting ${\nu }_r=x_r^{\ast }{e}^{(A^{T}z{)}_r}$ into the equilibrium weights yieldsThis is Poisson exponential tilting for a loss network: independent Poisson random variables centered at the predicted occupancies, restricted to the feasible region and penalized for slack in positively priced resources. It explains why an unconstrained Gaussian approximation should not automatically be applied at a saturated capacity boundary.

Let $x^{\ast }$ maximize SYSTEM. Under the usual positive-capacity, nonempty-route assumptions, the feasible set is compact, contains a positive point, and has a Slater condition point. Strict concavity makes the optimum unique. The condition $U_r^{\prime }(0)=\infty$ forces $x_r^{\ast }>0$: moving from any optimum with a zero component slightly toward a positive feasible point gives an infinite positive directional derivative in that component and only finite losses in the positive components.

By the Karush-Kuhn-Tucker conditions, there are link Lagrange multipliers $p_j^{\ast }\ge 0$ withSet the route congestion price $y_r^{\ast }=(A^T{p}^{\ast }{)}_r=U_r^{\prime }(x_r^{\ast })$ and chooseThen $x_r^{\ast }={\nu }_r^{\ast }{e}^{-y_r^{\ast }}$, as required. For USER with fixed $y_r^{\ast }$, the substitution $x={\nu }_r{e}^{-y_r^{\ast }}$ is a bijection of the nonnegative half-line, and its objective becomes $U_r(x)-y_r^{\ast }x$. This is a strictly concave function whose derivative vanishes at $x_r^{\ast }$, so its unique maximizer is $x_r^{\ast }$ and the corresponding USER optimizer is ${\nu }_r^{\ast }$.

For NETWORK at ${\nu }^{\ast }$, its objective derivative at $x^{\ast }$ isThus $x^{\ast },p^{\ast }$ obey its Karush-Kuhn-Tucker conditions, with the same feasibility and complementary slackness. The NETWORK objective is a strictly concave function, so $x^{\ast }$ is also its unique maximizer. The triple $(x^{\ast },y^{\ast },{\nu }^{\ast })$ solves all three problems simultaneously. This is a user-network decomposition of concave utility maximization.

Use the usual essential two-person Nash bargaining problem: $F\subset {\mathbb{R}}^2$ is a compact convex set, $d\in F$ is the disagreement point, and some $u\in F$ satisfies $u_i>d_i$ for both players. The Nash bargaining solution isThe positive maximum exists by compactness and essentiality. On positive gains, maximizing this Nash product is equivalent to maximizing $\log (u_1-d_1)+\log (u_2-d_2)$, a strictly concave function. Convexity then gives a unique maximizer. The essentiality and compactness hypotheses matter: for example, with $F=[0,1]\times \{0\}$ and $d=0$, the product is zero everywhere and its argmax alone is not a single-valued definition.

The rule satisfies all four axioms. Pareto efficiency: a feasible vector dominating the chosen vector with at least one strict improvement would increase its positive Nash product. Bargaining symmetry: if the problem is unchanged by swapping players, uniqueness makes the answer unchanged, so the two payoffs agree. Positive affine invariance in bargaining: for $v_i=a_i{u}_i+b_i$ with $a_i>0$, gains transform to $a_i(u_i-d_i)$ and the product is multiplied by the positive constant $a_1{a}_2$, preserving its maximizer. Bargaining independence of irrelevant alternatives: if $G\subseteq F$ is another admissible feasible set containing the chosen vector and the same disagreement point, that vector remains the unique product maximizer over $G$.

To prove characterization, let $f$ be any feasible single-valued rule satisfying these axioms, and let $u^{\ast }=N(F,d)$. Normalize payoffs by the positive affine transformationThe transformed set $H$ has disagreement point zero and product maximizer $e=(1,1)$. For any $z\in H$, convexity puts $e+t(z-e)$ in $H$; for sufficiently small $t>0$ both gains remain positive. The one-sided derivative of the product at its maximum is therefore nonpositive:Thus $H\subseteq \{z:z_1+z_2\le 2\}$. By compactness choose $M\ge 0$ so every coordinate of every $z\in H$ is at least $-M$. The supporting triangle for Nash bargaining isIt contains $H$, is compact, convex, symmetric and essential, and contains disagreement zero. Symmetry forces $f(T_M,0)$ onto the diagonal; Pareto efficiency then forces it to be $e=(1,1)$. Since $e\in H\subseteq T_M$, bargaining independence of irrelevant alternatives gives $f(H,0)=e$. Undoing the normalization by positive affine invariance in bargaining gives $f(F,d)=u^{\ast }$. Hence the four axioms uniquely characterize the Nash bargaining solution on this domain.

Write $p_r=(A^T\mu {)}_r$ for a route congestion price and $y_j=(Ax{)}_j$ for resource demand. A route with weighted logarithmic utility $w_r\log x_r$ chooses $x_r=w_r/p_r$, since its marginal utility equals its route congestion price. Each resource raises its price when $y_j>C_j$ and lowers it when $y_j<C_j$; ${\kappa }_j$ controls the speed. This is a multiplicative resource-price dynamics model of supply-demand feedback.

For a well-defined finite network, assume positive capacities, nonempty routes, and strictly positive initial resource congestion prices. The relevant rank condition is full row rank, $\mathrm{rank}A=|J|$. Zero initial prices are invariant under the printed differential equation, so the unrestricted phrase “all trajectories” needs a qualification; a concrete counterexample is given below.

Define $g(\mu )=Ax(\mu )-C$. On the domain $\mu \ge 0$ with $p_r>0$, direct differentiation givesThus $V$ is a concave function, and it is a strictly concave function when $A$ has full row rank. It has a maximum: as $\Vert \mu {\Vert }_1\to \infty$, its negative linear capacity term dominates its logarithmic terms, while on bounded sets $V\to -\infty$ if any route congestion price approaches zero. A maximizing price vector ${\mu }^{\ast }$ therefore exists with every $p_r^{\ast }>0$, though some individual resource congestion prices may be zero. The Karush-Kuhn-Tucker conditions areThey express feasibility and complementary slackness: positively priced resources are fully used, and an underused resource has zero price. Full row rank gives uniqueness of this maximizing price vector.

To prove convergence, rather than merely monotonicity of $V$, use the relative-entropy Lyapunov function for resource priceswhere a summand with ${\mu }_j^{\ast }=0$ means ${\mu }_j/{\kappa }_j$. Every summand is nonnegative and vanishes only at its reference price. Along a positive-price trajectory,The last equality follows by substituting $x_r=w_r/p_r$, and complementary slackness removes ${\mu }^{\ast T}g({\mu }^{\ast })$.

A sublevel set of $L$ bounds every price from above and bounds every coordinate with ${\mu }_j^{\ast }>0$ away from zero. Each route contains at least one such coordinate, because $p_r^{\ast }>0$. All route congestion prices consequently remain bounded away from zero, so the vector field is regular on the resulting compact set. Positive coordinates cannot hit zero in finite time, since their equations have the form ${\dot{\mu }}_j={\mu }_j$ times a bounded function. These bounds also give global existence.

The LaSalle invariance principle now applies. Equality in $\dot{L}\le 0$ requires $A^T\mu =A^T{\mu }^{\ast }$ and ${\mu }_j{g}_j({\mu }^{\ast })=0$. Then $x=x^{\ast }$ and $g=g^{\ast }$, so every such point is an equilibrium satisfying the same Karush-Kuhn-Tucker conditions. If $A$ has full row rank, $A^T$ is injective, making this set the singleton ${\mu }^{\ast }$. Thusfor every strictly positive initial price vector. This proves full-row-rank convergence of multiplicative resource prices.