Carried load of an Erlang loss resource 2026-10-06
For offered load and capacity , the carried load equals the mean occupancy:Differentiating the truncated-Poisson equilibrium weights gives for . The carried load increases from zero to . This monotonicity supplies the strictly increasing term in the convex potential for the Erlang fixed point.
Update one link blocking probability at a time using the reduced load accepted by all other links, then cycle through the links. In logarithmic acceptance coordinates this is exact coordinate descent for the convex potential for the Erlang fixed point. Potential values decrease within a compact sublevel set. Continuity of a whole-sweep update forces every limit point to minimize every coordinate, hence to equal the unique global minimizer. This proves convergence without requiring a simultaneous substitution scheme.
Past exam of the mathematics course of the University of Cambridge 2012 iii Paper 32 2 Solution Created 2026-10-03 Updated 2026-10-07
For a single resource, assume unit-capacity calls arrive as a Poisson process of rate , holding times have independent exponential distributions with mean , and a call is rejected when all units are occupied. Let . Its occupancy is a birth-death process on , with birth rate below capacity and death rate in state . Detailed balance for a birth-death process givesBy Poisson arrivals see time averages, the Erlang loss formula isThis derivation assumes exponential holding times; the standard insensitivity result extends the same formula to independent holding times of the same mean, but is not needed here.
A loss network with fixed routing has finitely many resources , capacities , and call types with a fixed nonempty set of resources. Type has Poisson arrivals of rate , independent holding times of mean , and offered load . It is accepted only if every resource on its route has a free unit, and otherwise is lost. In the displayed equations use the unit-incidence matrix for and zero otherwise. The equations concern this unit-demand model, rather than arbitrary multi-unit requirements.
The Erlang fixed point approximation replaces the correlated resource availabilities by independent ones. A type- call survives blocking at resources other than with approximate probability . Treating this filtering as Poisson thinning gives the reduced offered loadApply the single-resource formula to obtain . Independence and Poisson thinning here are approximations; no such approximation is claimed for the exact network's joint stationary distribution.
Assume first . The right side is a continuous map from into itself, so the Brouwer fixed-point theorem gives existence. To prove uniqueness, let be the carried load of an Erlang loss resource, also the mean of the truncated Poisson occupancy. Differentiation yieldsThus increases continuously from zero to one and increases from zero to .
Write , and define and , with . Each is continuous and strictly increasing. A fixed point satisfiesThese are precisely the stationary equations for the convex potential for the Erlang fixed pointThe first term is convex; the second is strictly convex because each strictly increases. Hence is strictly convex. Its gradient cannot vanish at two distinct points: the difference of gradients dotted with the difference of points is strictly positive, with a positive contribution from every changed coordinate in the integral term. This also covers boundary coordinates , using the right derivative there. Every fixed point has , since its offered load is finite, so the change of variables is valid. There is exactly one fixed point. If a capacity is zero, its blocking probability is one; delete it and all routes using it before applying the argument to the remaining positive capacities. Unused resources have blocking probability zero.
Past exam of the mathematics course of the University of Cambridge 2013 iii Paper 37 1 ii Solution Created 2026-10-03 Updated 2026-10-07
The exact normalizing sum can be expensive on a large loss network. The Erlang fixed point approximation replaces joint resource acceptance by a product of marginal acceptances. Here consider unit requirements , positive integer capacities and finitely many fixed routes. Let be resource blocking and . A route- call contributes to the offered traffic at resource after surviving the other resources on its route. Thus the reduced-load approximation iswhere the Erlang B formula isThe estimated route acceptance is . This is an independence approximation, not an alternative exact factorization of the stationary law in part (i).
Uniqueness follows from a convex potential for the Erlang fixed point. For a single resource, the carried load of an Erlang loss resource is . It increases strictly from zero to as increases: differentiating the expected value of its upper-truncated Poisson distribution with respect to gives the strictly positive occupancy variance. Blocking also increases strictly, as is evident on dividing the Erlang denominator by its final term.
Use . Let be the unique offered load with , and set , with . This function is strictly increasing and tends to . DefineEach exponential term is a convex function, and each integral is a strictly convex function because its derivative is strictly increasing. Therefore is a strictly convex function. It is also a coercive function on the nonnegative orthant: an unbounded coordinate makes its integral grow asymptotically linearly with positive slope . A unique minimizer exists.
If resource carries some positive offered route, its inward derivative at is negative, so its minimizing coordinate is positive. At such a coordinate the first-order equation isDividing by gives precisely . A resource with no positive offered route uniquely has , hence . Thus the minimizer and the fixed point coincide, proving existence and uniqueness of the Erlang fixed point for fixed unit-resource routing. This does not by itself guarantee convergence of every simultaneous substitution algorithm; the uniqueness claim concerns the solution of the equations.
Past exam of the mathematics course of the University of Cambridge 2015 iii Paper 39 2 Solution Created 2026-10-03 Updated 2026-10-06
For the Erlang loss formula, assume a Poisson process of arrivals with rate , independent holding times with exponential distribution of mean , unit capacity usage per accepted call, and rejection without waiting when all units are occupied. Here is a positive integer and the offered load is . The call population on is a birth-death process; detailed balance givesBy Poisson arrivals see time averages, the fraction of arrivals rejected is the time probability of full occupancy. ThereforeThe paper's is the offered load, rather than necessarily an arrival rate. Exponential holding times suffice for this derivation; no unproved insensitivity of loss networks assumption is needed.
A loss network with fixed routing has capacities , call types , and a link-route incidence matrix with exactly when a type- call requires link . An accepted call holds one unit on every link of its route for its holding time; if any required link is full, the call is rejected everywhere. Take independent Poisson processes of calls, independent holding times, and route offered loads . For the finite-variable formulation below, take ; a zero-capacity link and all routes using it may be removed first.
The Erlang fixed point approximation treats blocking at different links as independent and thins the traffic offered to link by acceptance at the other links. Its reduced offered load isThe equality uses : only routes with contribute. Treating this reduced stream as Poisson leads to . Independence and Poisson thinning here are the approximation, not exact properties of a general loss network.
To construct the convex potential for the Erlang fixed point, write and , the carried load of an Erlang loss resource, equivalently the mean occupancy in the single-link stationary distribution. Differentiating its finite sums gives, for ,The variance is positive because the finite occupancy law is nondegenerate for . Furthermore maps increasingly onto , while increases from zero to . SetIn particular , is continuous and strictly increasing, and as .
Forthe first sum is convex, and each integrated strictly increasing function is strictly convex. Thus is a strictly convex function on . Since tends to a positive limit on every coordinate, is a coercive function there; it has a unique minimizer. Its coordinate derivative isThe Erlang fixed point equation is exactly . For an unused link, and the optimum is ; for a used link with positive offered load, the derivative at zero is negative and the optimum is positive. Hence every fixed point gives the unique minimizer, and the minimizer gives a fixed point. The Erlang fixed point is unique under these fixed-routing assumptions.
A convergent cyclic substitution for the Erlang fixed point updates one coordinate at a time, repeatedly cycling through all links:Always use the most recently available values of the other coordinates. Start, for example, with . In coordinates this is exact coordinate descent for : with other coordinates fixed, the displayed update is the unique coordinate minimizer.
Here is a convergence proof. All iterates stay in the initial compact sublevel set of the coercive function , and their potential values decrease to a limit. Each single-coordinate update is continuous, since its reduced load is continuous and finite and . Thus a whole-sweep map is continuous. If a full-sweep subsequence converges to , continuity gives , because both consecutive potential values converge to the same limit. Strict coordinate convexity means equality can occur only when no coordinate changes during that sweep. Thus minimizes every coordinate, satisfies the nonnegative-orthant first-order conditions, and is the unique global minimizer of . Every subsequential limit is therefore the same point, proving convergence of the full-sweep iterates and, by continuity of the updates, all intermediate iterates. This argument applies to sequential substitution; it does not assume simultaneous updates converge.