# Lesson 31 Flashcards

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<summary><strong>1. How does Ball define survivability, and what are its two halves?</strong></summary>
<div class="card-answer"><p>Survivability is an aircraft's capability to <em>avoid</em> or <em>withstand</em> a man-made hostile environment. Failing to avoid is <strong>susceptibility</strong>, measured by <span class="math notranslate nohighlight">\(P_H\)</span>; failing to withstand is <strong>vulnerability</strong>, measured by <span class="math notranslate nohighlight">\(P_{K|H}\)</span>. A kill needs both: <span class="math notranslate nohighlight">\(P_K = P_H\,P_{K|H}\)</span> and <span class="math notranslate nohighlight">\(P_S = 1 - P_K\)</span>.</p></div>
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<summary><strong>2. Write one engagement's <span class="math notranslate nohighlight">\(P_K\)</span> along the L1 kill chain.</strong></summary>
<div class="card-answer"><p><span class="math notranslate nohighlight">\(P_K = P_D\,P_{T|D}\,P_{I|T}\,P_{H|I}\,P_{K|H}\)</span> — detect, track given detect, identify given track, hit given identify, kill given hit. The first four make up the susceptibility <span class="math notranslate nohighlight">\(P_H\)</span>; the last is the vulnerability.</p></div>
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<summary><strong>3. Why is it legitimate to multiply the five links?</strong></summary>
<div class="card-answer"><p>Each factor is <strong>conditional</strong> on every link before it succeeding — <span class="math notranslate nohighlight">\(P_{T|D}\)</span> is the chance of holding a track <em>given</em> a detection. Multiplying conditionals is the chain rule of probability, so no independence between the links is assumed.</p></div>
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<summary><strong>4. How does L1's "break any link and the aircraft survives" appear in the product?</strong></summary>
<div class="card-answer"><p>It is the limit case: drive one factor to zero and <span class="math notranslate nohighlight">\(P_K = 0\)</span>. Real levers rarely zero a link — they <strong>shrink</strong> it, and a shrunken factor lowers <span class="math notranslate nohighlight">\(P_K\)</span> by multiplication.</p></div>
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<summary><strong>5. Two levers cut different links. Which one lowers <span class="math notranslate nohighlight">\(P_K\)</span> more?</strong></summary>
<div class="card-answer"><p>The one that removes the larger <strong>fraction</strong> of its link — whichever link it sits on. A lever that multiplies one factor by <span class="math notranslate nohighlight">\(f\)</span> multiplies the whole product <span class="math notranslate nohighlight">\(P_K\)</span> by <span class="math notranslate nohighlight">\(f\)</span>, so taking 0.15 off a link at 0.3 (half of it) beats taking 0.25 off a link at 0.75 (a third). What the algebra leaves out decides the rest: the cost of that cut, and its <strong>reach</strong> — how many threats one lever acts on.</p></div>
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<summary><strong>6. Name Ball's six susceptibility-reduction concepts.</strong></summary>
<div class="card-answer"><p>Threat warning; noise and deception jamming; signature reduction; expendables (chaff, flares, decoys); threat suppression (SEAD); and tactics (route, altitude, timing, terrain masking, maneuver). All of them lower <span class="math notranslate nohighlight">\(P_H\)</span>.</p></div>
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<summary><strong>7. Name Ball's six vulnerability-reduction concepts.</strong></summary>
<div class="card-answer"><p>Component redundancy with separation; component location; passive damage suppression; active damage suppression; component shielding; and component elimination or replacement. All of them lower <span class="math notranslate nohighlight">\(P_{K|H}\)</span>.</p></div>
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<summary><strong>8. What single test classifies a lever as susceptibility or vulnerability reduction?</strong></summary>
<div class="card-answer"><p>Does it change <em>whether they hit you</em> (susceptibility, <span class="math notranslate nohighlight">\(P_H\)</span>) or <em>what a hit does to you</em> (vulnerability, <span class="math notranslate nohighlight">\(P_{K|H}\)</span>)? Where it is installed does not matter — radar-absorbing material is bolted to the airframe but is a susceptibility lever.</p></div>
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<summary><strong>9. Which block bought a lever on which link?</strong></summary>
<div class="card-answer"><p>Detect: B1's range law and threshold, B3's signatures and EMCON, B2's noise jamming. Track: B2's deception and chaff, B1's Doppler notch. Identify: emission discipline (B2, B3). Engage: B3's flares and DIRCM, B2's towed decoys. Kill: airframe design (vulnerability). Block 4 adds no link — it decides how many times the chain runs.</p></div>
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<summary><strong>10. Why is <span class="math notranslate nohighlight">\(P_{I|T}\)</span> often near 1, and what does that make it as a lever?</strong></summary>
<div class="card-answer"><p>In defended airspace with no friendly traffic, anything unknown is treated as hostile, so identification rarely fails. A link already near 1 has little room to shrink, which makes Identify usually the weakest lever.</p></div>
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<summary><strong>11. How do independent encounters combine into a mission <span class="math notranslate nohighlight">\(P_S\)</span>?</strong></summary>
<div class="card-answer"><p>You must survive all of them, so survivals multiply: <span class="math notranslate nohighlight">\(P_S = \prod_i (1 - P_{K,i})\)</span>. Small risks compound — twenty 4% shots leave <span class="math notranslate nohighlight">\(P_S = 0.96^{20} = 0.44\)</span>. Independence is an assumption: a radar that cues every SAM couples their detections.</p></div>
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<summary><strong>12. Define additive risk and its small-probability shortcut.</strong></summary>
<div class="card-answer"><p><span class="math notranslate nohighlight">\(r = -\ln P_S = \sum_i -\ln(1 - P_{K,i})\)</span>, and <span class="math notranslate nohighlight">\(P_S = e^{-r}\)</span>. For small <span class="math notranslate nohighlight">\(P_K\)</span>, <span class="math notranslate nohighlight">\(r_i \approx P_{K,i}\)</span>, but <span class="math notranslate nohighlight">\(-\ln(1-p) = p + p^2/2 + \dots\)</span> is always larger than <span class="math notranslate nohighlight">\(p\)</span>, so the shortcut understates the risk.</p></div>
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<summary><strong>13. Why does a route planner want <span class="math notranslate nohighlight">\(-\ln(1-p)\)</span> as its cost rather than <span class="math notranslate nohighlight">\(p\)</span>?</strong></summary>
<div class="card-answer"><p>A shortest-path search (Dijkstra, L34) minimizes a <strong>sum</strong> of edge costs. The logs sum exactly to <span class="math notranslate nohighlight">\(-\ln P_S\)</span>, so the least-cost route is exactly the most survivable one. Summing the <span class="math notranslate nohighlight">\(p\)</span> values is right only to first order.</p></div>
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<summary><strong>14. What is a site's hazard <span class="math notranslate nohighlight">\(\lambda_i\)</span>?</strong></summary>
<div class="card-answer"><p>Its risk per kilometer of exposure. Inside its lethal zone (detection contour and MEZ) the site shoots every <span class="math notranslate nohighlight">\(\Delta s\)</span> km; each shot kills with the chain after detection, <span class="math notranslate nohighlight">\(p_\text{shot} = P_{T|D}\,P_{I|T}\,P_{H|I}\,P_{K|H}\)</span>. Then <span class="math notranslate nohighlight">\(\lambda_i = -\ln(1 - p_\text{shot})/\Delta s\)</span> and <span class="math notranslate nohighlight">\(P_{S,i} = e^{-\lambda_i \ell_i}\)</span> for <span class="math notranslate nohighlight">\(\ell_i\)</span> km inside.</p></div>
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<summary><strong>15. Why does risk grow in proportion to the distance flown inside a lethal zone?</strong></summary>
<div class="card-answer"><p>Because the site gets one shot per <span class="math notranslate nohighlight">\(\Delta s\)</span> km, so <span class="math notranslate nohighlight">\(\ell\)</span> km buys it <span class="math notranslate nohighlight">\(\ell/\Delta s\)</span> independent shots, and each shot adds the same <span class="math notranslate nohighlight">\(-\ln(1 - p_\text{shot})\)</span> to the risk. Twice the kilometers, twice the risk: <span class="math notranslate nohighlight">\(r_i = \lambda_i \ell_i\)</span>.</p></div>
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<summary><strong>16. What does L31 contribute to the Final Project, and what are its key dates?</strong></summary>
<div class="card-answer"><p>The survival metric (task T2): per-site chain, hazard and route <span class="math notranslate nohighlight">\(P_S\)</span>, ingress and egress separately. The project is issued at L31; a required, feedback-only checkpoint is at L35; the complete draft is due at the start of L39 for peer review; the final paper is due at the start of L41.</p></div>
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