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The orchard residual mean square in ANOVA is on the original tree-response scale: it equals 30 times the corresponding residual mean square in ANOVA for orchard means. Thus the estimated variance of one orchard's sample mean is $240/30=8$.

Each pruning marginal sample mean averages four independent orchard means, giving variance $8/4=2$. Two different pruning marginals use disjoint orchards, so the variance of their estimated treatment contrast isEach spray marginal sample mean uses six orchards. Similarly,The corresponding standard errors are $2$ and $\sqrt{8/3}$ in the units of weight per tree. These compare per-tree marginal sample means, averaging equally over the other factor, even when an interaction term is fitted. Comparing orchard totals instead would multiply these variances by $30^2$.

Report the six treatment-combination means and their uncertainty, together with useful marginal treatment contrasts, standard errors and confidence intervals. Means are needed to interpret direction and practical importance, while the fitted interaction term determines whether one factor's effects should be reported separately at levels of the other. For the pairwise marginal comparisons in part (c), ordinary model-based 95% Student t confidence intervals use $t_{6,0.975}$ times the corresponding standard error; a simultaneous set of comparisons needs an appropriate adjustment.

Also describe orchard selection, actual randomization, treatment delivery, numbers of trees observed, missing data and response units. Check orchard-level residuals for unequal variance, outliers or systematic patterns using a residual-versus-fitted plot and an appropriate distributional diagnostic. Explain that twelve orchards provide independent treatment replication; 360 tree measurements provide subsampling precision. No numerical treatment means or numerical confidence intervals can be recovered from the supplied sums of squares in ANOVA alone.

This produces a split-plot design: assign spray to whole orchards, with six sprayed and six unsprayed, then independently randomize ten trees to each pruning method inside every orchard. Orchards remain experimental units for spray; individual trees become experimental units for pruning. Pruning contrasts and the spray-by-pruning interaction term now lie in the within-block ANOVA stratum, while spray is tested between orchards.

The between-orchard ANOVA stratum has eleven statistical degrees of freedom, split into one for spray and ten for error. The within-orchard ANOVA stratum has 348, split into two for pruning, two for the interaction term and 344 for error. This pooling of within-orchard error is appropriate under the stated compound-symmetry covariance model; additional orchard-specific pruning variation would need its own variance component rather than this simplified error model.

The shared orchard effect cancels in a pruning difference within an orchard. Its variance is $2a/10$, so averaging across twelve orchards givesThe spray contrast still compares means of six orchards per group, each based on 30 trees, so its estimated variance remains $8/3$. For a difference of pruning differences between the two spray groups, each group's pruning difference has estimated variance $2\cdot 180/(10\cdot 6)=6$, and the resulting interaction contrast has estimated variance $12$, compared with $16$ in the original allocation.

Splitting pruning within orchards improves pruning and interaction precision without extra trees, and gives spray a less sparse error estimate; it does not reduce the spray contrast's variance. This option requires tree-level pruning to be practical without interference between neighboring trees. The numerical gains, like those in the other options, assume the current variance components remain applicable next year.

Allocate the extra orchards evenly, giving three independent orchards to each of the six combinations. Each orchard still has mean variance $8$ under the fitted model. Each pruning marginal now uses six orchards and each spray marginal nine, givingBoth are two-thirds of their original values. The orchard residual statistical degrees of freedom increase from $12-6=6$ to $18-6=12$.

Extra orchards improve independent replication, give a predicted one-third reduction in contrast variances, and improve error estimation. Both this option and using 45 trees require 540 tree measurements, but extra orchards offer the greater predicted statistical gain under the fitted model. They may cost more to recruit and administer; their suitability and comparability also matter.

Under the fitted compound-symmetry covariance model, an orchard mean based on $k$ trees has varianceWith the current estimates this falls from $v_{30}=2+180/30=8$ to $v_{45}=2+180/45=6$. The numbers of orchards contributing to each marginal remain unchanged, so both marginal treatment-contrast variances fall by a factor $6/8=3/4$: pruning differences have estimated variance $3$ and the spray difference has estimated variance $2$.

Adding trees gives a predicted 25% reduction in these variances, but no extra independent orchard replication. It requires 50% more tree measurements and leaves six orchard residual statistical degrees of freedom. Increasing $k$ reduces the $a/k$ term but cannot remove the shared orchard component $c$, so gains eventually diminish.

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 $A_{jr}\in \{0,1\}$, positive integer capacities and finitely many fixed routes. Let $B_j$ be resource blocking and $q_j=1-B_j$. A route-$r$ call contributes to the offered traffic at resource $j$ after surviving the other resources on its route. Thus the reduced-load approximation iswhere the Erlang B formula isThe estimated route acceptance is ${\prod }_{j\in r}{q}_j$. 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 $m_C(v)=v[1-E(C,v)]$. It increases strictly from zero to $C$ as $v$ increases: differentiating the expected value of its upper-truncated Poisson distribution with respect to $\log v$ gives the strictly positive occupancy variance. Blocking also increases strictly, as is evident on dividing the Erlang denominator by its final term.

Use $p_j=-\log q_j\ge 0$. Let $v_C(p)$ be the unique offered load with $1-E(C,v_C(p))=e^{-p}$, and set $g_C(p)=v_C(p)e^{-p}=m_C(v_C(p))$, with $g_C(0)=0$. This function is strictly increasing and tends to $C$. DefineEach exponential term is a convex function, and each integral is a strictly convex function because its derivative is strictly increasing. Therefore $F$ 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 $C_j$. A unique minimizer exists.

If resource $j$ carries some positive offered route, its inward derivative at $p_j=0$ is negative, so its minimizing coordinate is positive. At such a coordinate the first-order equation isDividing by $q_j$ gives precisely $v_j={\sum }_{r:j\in r}{\rho }_r{\prod }_{k\in r\setminus \{j\}}{q}_k$. A resource with no positive offered route uniquely has $p_j=0$, hence $B_j=0$. 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.

For any $s\ge 0$, on the event $Y\ge 0$ one has $e^{sY}\ge 1$. The Markov inequality therefore gives $P(Y\ge 0)\le \mathbb{E}{e}^{sY}$, including the trivial value one at $s=0$. Taking the infimum proves the Chernoff bound.

For $s>0$ with finite moment-generating functions, independence givesConsequentlyAn effective bandwidth is an exponential-moment measure of demand at a chosen tail parameter $s$: independence makes these quantities additive. The spare capacity $\gamma /s$ pays for the desired exponential tail bound. For a cumulative-demand process over time $t$, the corresponding bandwidth would be $\log \mathbb{E}{e}^{sX(t)}/(st)$; here the time horizon is one. It is generally larger than mean demand because it charges for fluctuations, and mathematical optimization over $s$ selects the useful tradeoff.

For independent normal distributions, putThe Gaussian effective bandwidth is ${\alpha }_j(s)={\lambda }_j+s{\sigma }_j^2/2$. For $\gamma >0$ and $v>0$, the sufficient condition becomes $m+sv/2+\gamma /s\le C$. Its left side is minimized at $s_{\ast }=(2\gamma /v{)}^{1/2}$, givingThis is a sufficient Chernoff safety margin, not the exact normal tail quantile.

Indeed $X\sim N(m,v)$, so, writing $\Phi$ for the standard normal distribution function,The exact Gaussian chance constraint is thereforeThis is necessary and sufficient when $v>0$, even when $\varphi$ is negative. The Chernoff coefficient $\sqrt{2\gamma }$ is more conservative.

There is an important boundary qualification. If $v=0$, then $X=m$ deterministically, and for $\gamma >0$ the exact requirement is $C>m$, not $C\ge m$. For example, $X\equiv 0$, $C=0$, $\gamma =1$ satisfies the printed square-root condition but has $P(X\ge C)=1>e^{-1}$. Thus both Gaussian non-strict displayed forms require positive total variance, which follows if at least one flow is present and its variance is positive. With no positive variance, the deterministic boundary in an upper-tail chance constraint must be treated separately. If $\gamma \le 0$, the target upper bound is at least one and imposes no restriction; the displayed square-root discussion naturally assumes $\gamma >0$.