Exercises

By AM-GM: $d_1 d_2 \leq ((d_1+d_2)/2)^2 = D^2/4$ with equality iff $d_1 = d_2$. The maximum of $d_1 d_2$ is $D^2/4$ attained at $d_1 = d_2 = D/2$.

$d_1^n d_2^n = (d_1 d_2)^n$ is monotonically increasing in $d_1 d_2$ for $n > 0$, so the same $d_1 = d_2$ maximizer holds. The RIS incremental SNR gain therefore grows maximally at the midpoint (product path loss is $\propto 1/(d_1^2 d_2^2)$, maximized SNR gain at minimized path loss).

$d_1 = 10$: $d_2 = 90$; $1/d_1^2 d_2^2 = 1/(100 \cdot 8100) = 1.23 \times 10^{-6}$. $d_1 = 25$: $d_2 = 75$; $1/d_1^2 d_2^2 = 1/(625 \cdot 5625) = 2.84 \times 10^{-7}$. $d_1 = 50$: $d_2 = 50$; $1/d_1^2 d_2^2 = 1/(2500 \cdot 2500) = 1.6 \times 10^{-7}$.

$d_1 = 10$ (near-BS) gives the strongest relative signal: $7.7\times$ stronger than the midpoint, and $4.3\times$ stronger than $d_1 = 25$. Near-endpoint placement wins by almost an order of magnitude. The midpoint actually LOSES in this product-path metric.

$p_{\text{cov}} = 0.9 \cdot 0.6 + 0.9 \cdot 0.4 = 0.54 + 0.36 = 0.90$.

Coverage rises from $0.9 \cdot 0.6 = 0.54$ (no RIS) to $0.90$. The missing 10% is the 10% of blocked UEs not covered by the RIS (far from panel, geometry unfavorable).

$\lambda \cdot \pi \cdot 2500 = 3$ → $\lambda = 3/(7854) = 3.82 \times 10^{-4}$/m² $= 382$ panels/km².

$\lambda = 3/(\pi \cdot 10000) = 9.55 \times 10^{-5}$/m² $= 95.5$ panels/km².

Doubling useful range quarters the required density — physics is on the operator's side. Larger $N$ panels provide this via extended useful range.

$N_{\min} = \sqrt{4 \cdot 316 \cdot 0.5} = \sqrt{632} = 25.1$.

Any RIS with $N \geq 32$ (typical smallest commercial size) beats this relay. RIS wins handily at commercial scales.

Capex = $6000. OpEx 10yr =$5000. Total = $11,000.

CapEx = $6000. OpEx 10yr =$7500. Total = $13,500.

CapEx = $5000. OpEx 10yr =$5000. Total = $10,000.

RIS is cheapest at $10k; small cell at$11k; relays at $13.5k. RIS and small cell are within 10% of each other; choice depends on capacity need (small cell) vs. blockage fill-in (RIS). Relays lose on all counts.

Multi-RIS placement reduces to the Maximum Coverage problem: given $n$ sets over a universe of $m$ elements and a budget $k$, pick $k$ sets to cover as many elements as possible. Mapping: each candidate panel's coverage footprint = one set; covered UEs = elements; budget = number of panels allowed.

Maximum Coverage is known NP-hard. It is submodular, so greedy gives $(1-1/e) \approx 63\%$-optimal solutions — the best achievable in polynomial time unless P = NP (Feige 1998).

Marginal gains: $p_1: 3, p_2: 4, p_3: 2$. Pick $p_2$. Covered = $\{2,3,4,5\}$.

Marginal gains: $p_1$ adds $\{1\}$ → 1; $p_3$ adds $\{6\}$ → 1. Tie-break alphabetically: pick $p_1$. Covered = $\{1,2,3,4,5\}$ — 5 UEs.

Optimal: $\{p_2, p_3\}$ covers $\{2,3,4,5,6\}$ — also 5 UEs. Or $\{p_1, p_2\}$ — 5 UEs. Or $\{p_1, p_3\}$ — 5 UEs. Greedy achieves optimal in this instance. In general, greedy achieves $\geq (1-1/e) \cdot \text{OPT}$.

$dp/d\lambda = A e^{-\lambda A}$. $p/\lambda = (1-e^{-\lambda A})/\lambda$. Condition: $A e^{-\lambda A} = (1-e^{-\lambda A})/\lambda$ → $\lambda A e^{-\lambda A} = 1 - e^{-\lambda A}$.

Let $x = \lambda A$: $xe^{-x} = 1 - e^{-x}$ → $x = e^x - 1$, i.e., $x + 1 = e^x$.

Solution: $x \approx 1.146$ (this is the Omega constant form). Therefore $\lambda^\star = 1.146/A$. For $A = 5000$ m²: $\lambda^\star \approx 229$ panels/km². Coverage at optimum: $p^\star = 1 - e^{-1.146} \approx 0.682$ → 68%.

$p_{\text{all-blocked}} = 0.4^5 = 0.01024$ → 1%.

99% of the time, at least one RIS is usable. This is the case for distributed, redundant deployment vs. one large central panel (where blockage of that one panel kills the link).

One panel with $N$ elements. Per-UE gain: $N^2$.

20 panels with $N/20$ elements each. Each UE served by its own RIS, gain per-UE: $(N/20)^2 = N^2/400$.

Centralized beats distributed by $400 \times$ in per-UE gain — the $N^2$ coherent combining wants lots of elements in one panel, not split. This is why "one large panel per cluster" dominates over "many small panels per UE". However, distributed wins on blockage diversity (see Ex 10). Optimal deployment: few-large-panel-per-cluster.

$c(100) = 300 \cdot 100^{-0.3} = 300/3.98 = \$75.3$. $c(1000) = 300 \cdot 1000^{-0.3} = 300/7.94 = \$37.8$.

At 1000 units: $37.8/panel vs. $75.3/panel at 100 units. Per-panel savings: 50%. This is the commercial lever: scale-up deployment halves the per-panel price, making dense RIS deployments financially viable.

$5000n + 200m \leq 100000$ → $25n + m \leq 500$. Maximize: $300n + 10m$.

All small cells: $n = 20, m = 0$: 6000 UEs. All RIS: $n = 0, m = 500$: 5000 UEs. Mix: $n = 15$, $m = 125$: $4500 + 1250 = 5750$ UEs.

Maximum is all-small-cells (6000 UEs). But this ignores that small cells also need backhaul, installation time, and OpEx that RIS avoids. With realistic TCO, the hybrid approach wins.

$0.3 \cdot 500 = 150$ facades available in 5 km² = 30/km².

Even if the operator wants 100 panels/km², the leasing constraint limits them to 30. This leasing bottleneck is the single biggest real-world challenge for commercial RIS. Mitigation: partnerships with building owner aggregators; public right-of-way (lamp posts, bus stops) which is operator-accessible.

$\$1.5\text{M} / \$300 = 5000$ panels over 3 years.

Year 1: 500 panels → 50/km² (hotspots). Year 2: 2000 panels (cumulative 2500) → 250/km² (dense urban). Year 3: 2500 panels (cumulative 5000) → 500/km² (ubiquitous).

Year 1: focus on highest-traffic downtown blocks (ROI proof). Year 2: expand to residential/office zones. Year 3: fill coverage holes. This phasing matches the industry roadmap in the chapter.