Volume, Fibonacci & Elliott Wave

Question 1 (10)

(a) In an uptrend, retracement measures a pullback down from the high by fraction $r$ of the full range $(H-L)$: $P(r) = H - r\,(H - L)$ Here $H-L = 180-120 = 60$. (formula 1, correct setup 1)

(b) With $P(r)=180-60r$:

(1 mark each, 5 total)

(c) The golden ratio $\phi = \frac{1+\sqrt5}{2} \approx 1.6180$. Its reciprocal $1/\phi = \phi - 1 \approx 0.6180$, giving the 0.618 level. The 0.382 level is $1 - 0.618 = 0.382 = 1/\phi^2$. These self-similar ratios appear repeatedly in the Fibonacci sequence limit $F_{n}/F_{n+1}\to 0.618$, which is why markets "respect" them. (concept 1, $1/\phi$ 1, $1-1/\phi$ 1)

Question 2 (8)

(a) Extension target from retracement point C: $T(e) = C + e\,(B - A)$ where $A$=start, $B$=impulse peak, $C$=retrace low. (2)

(b) $A=120$, $B=180$, so $B-A=60$; $C=150$.

• 1.272: $T = 150 + 1.272\times60 = 150 + 76.32 = 226.32$
• 1.618: $T = 150 + 1.618\times60 = 150 + 97.08 = 247.08$

(2 each)

(c) Extensions project beyond the prior high — into unvisited price where there is no historical trading, so no support/resistance exists yet. They serve as forward-looking profit targets where reactions are statistically likely, not as reaction levels formed by past order flow. (2)

Question 3 (12)

(a) $\text{MFM} = \frac{(C-L)-(H-C)}{H-L}, \qquad \text{MFV} = \text{MFM}\times V$ MFM ranges $[-1,+1]$. (MFM 2, MFV 1)

(b)

def ad_line(highs, lows, closes, volumes):
    ad = []
    cum = 0
    for h, l, c, v in zip(highs, lows, closes, volumes):
        if h == l:
            mfm = 0            # avoid div by zero
        else:
            mfm = ((c - l) - (h - c)) / (h - l)
        mfv = mfm * v
        cum += mfv
        ad.append(cum)
    return ad

(loop 2, mfm 2, cumulative sum + zero-guard 2)

(c) $H=110,L=100,C=108,V=10000$:

• MFM $= \frac{(108-100)-(110-108)}{110-100} = \frac{8-2}{10} = 0.6$
• MFV $= 0.6\times10000 = 6000$
• A/D contribution $= +6000$ (first bar → line = 6000).

(MFM 1, MFV 1, A/D 1)

Question 4 (12)

(a) Three inviolable rules: (1 each)

1. Wave 2 never retraces more than 100% of Wave 1 (never below Wave 1 start).
2. Wave 3 is never the shortest of waves 1, 3, 5.
3. Wave 4 never overlaps the price territory of Wave 1 (in a standard impulse).

(b) Endpoints: W1=50, W2=42, W3=68, W4=55, W5=74. Wave 1 start assumed below 42 (say the origin), lengths in price:

• Wave 1 length $=50-\text{start}$; using origin as pre-W1 low, treat W1 travel; measured wave magnitudes: W1: |50−start|. Using the standard test on the given data:
  • Rule 1: Wave 2 bottom (42) must stay above Wave 1 start. 42 > start (origin) ✓
  • Rule 2: Wave 3 length $=68-42=26$; Wave 1 length $=50-\text{start}$; Wave 5 length $=74-55=19$. Wave 3 (26) is not the shortest, since it exceeds Wave 5 (19) and Wave 1 magnitude. ✓
  • Rule 3: Wave 4 bottom = 55; Wave 1 top = 50. Since $55 > 50$, Wave 4 does not overlap Wave 1's territory. ✓

All three rules satisfied → count is VALID. (2 per rule tested with numbers)

(c) The 5 is the impulse (motive) phase moving with the main trend — waves 1,2,3,4,5. The 3 is the corrective phase against the trend — waves A,B,C. Together one full cycle = 8 waves. They nest fractally: each impulse wave subdivides into a smaller 5, each corrective wave into a smaller 3, at every degree. (5 role 1, 3 role 1, nesting 1)

Question 5 (10)

(a)

• No demand bar: an up-bar with a narrow spread, closing on or near the high but on low volume — the rally lacks buying interest, bearish in an uptrend context. (2)
• Stopping volume: a down-bar with ultra-high volume, wide spread, but closing well off the low (mid/high) — heavy buying absorbs selling, halting the decline. (2)

(b) Confluence = multiple independent technical factors pointing to the same price zone (here a prior swing support at 150 and the 0.5 Fib retracement at 150). Independent signals aligning raise the statistical odds of a reaction because more market participants act on that level, so a reversal/bounce there is higher-probability than a single-factor level. (2 concept, 1 higher-probability reasoning)

(c) Any three: (1 each)

1. Highly subjective — analysts count the same chart differently.
2. Retrospective/hindsight-fitting; poor real-time predictive reliability.
3. No fixed timeframe — waves can extend indefinitely (Rule 2 still allows many valid counts).
4. Alternate counts require constant revision (invalidation levels).

Question 6 (8)

(a) Wave 1–3 range $=260-200=60$. Wave 4 at 0.382 retracement (down from top): $260 - 0.382\times60 = 260 - 22.92 = 237.08$ (range 1, formula 1, answer 1)

(b) Heavy down-volume stalling with a close near the high = stopping volume / absorption — sellers are being absorbed. Implies the correction (Wave 4) is ending and Wave 5 upward is likely to begin. (phenomenon 2, implication 1)

(c) Wave 5 is commonly projected as 0.618 or 1.0 (equality) of the Wave 1 length, or the 0.618 extension of Wave 1–3. (2)

[
  {"claim":"Q1 0.618 retracement of 120-180 = 142.92","code":"H,L=180,120; r=0.618; result = round(H - r*(H-L),2)==142.92"},
  {"claim":"Q1 0.5 retracement = 150","code":"result = (180 - 0.5*(180-120))==150"},
  {"claim":"Q2 1.618 extension from C=150 = 247.08","code":"A,B,C=120,180,150; result = round(C+1.618*(B-A),2)==247.08"},
  {"claim":"Q3 MFM for H110 L100 C108 = 0.6 and MFV=6000","code":"H,L,C,V=110,100,108,10000; mfm=((C-L)-(H-C))/(H-L); result = (mfm==Rational(6,10)) and (mfm*V==6000)"},
  {"claim":"Q6 Wave4 0.382 retrace of 200-260 = 237.08","code":"result = round(260-0.382*(260-200),2)==237.08"}
]