Level 3 — ProductionOptions Strategies

Options Strategies

45 minutes60 marksprintable — key stays hidden on paper

Chapter: 5.4 Options Strategies Level: 3 — Production (from-scratch derivations, code-from-memory, explain-out-loud) Time limit: 45 minutes Total marks: 60

Notation: STS_T = underlying price at expiry, KK = strike, premiums are per share. Assume no transaction costs/dividends unless stated. Show every derivation step.


Question 1 — Bull Call Spread from scratch (10 marks)

A trader buys a 100100-strike call for 6.206.20 and sells a 110110-strike call for 2.402.40 on the same expiry.

(a) Derive the net debit and the general expiry payoff function P(ST)P(S_T) across all three regions (ST<100S_T<100, 100ST110100\le S_T\le110, ST>110S_T>110). (4)

(b) Derive the maximum profit, maximum loss, and breakeven price. (3)

(c) Explain out loud (2–3 sentences) the view + IV rationale: why choose a bull call spread instead of just buying the 100100 call outright? (3)


Question 2 — Protective Put vs Collar (10 marks)

You hold 100100 shares bought at 5050. Spot is now 5555. You buy a 5252 put for 1.501.50.

(a) Derive the protected position's payoff and breakeven, and state the floor value of the position. (4)

(b) You now also sell a 6060 call for 1.201.20 to finance the put (forming a collar). Derive the net cost of protection, the capped upside value, and the floor value. (4)

(c) Explain out loud why a collar is attractive when implied volatility is elevated. (2)


Question 3 — Iron Condor derivation & code (12 marks)

Construct an iron condor on an underlying at 200200:

  • Sell 190190 put @ 2.102.10, buy 185185 put @ 1.101.10
  • Sell 210210 call @ 2.302.30, buy 215215 call @ 1.301.30

(a) Derive net credit, max profit, max loss, and both breakevens. (6)

(b) Write a Python function (from memory, numpy allowed) iron_condor_payoff(S) returning total P/L per share at expiry for a scalar or array S. (4)

(c) Explain out loud when (view + IV) you deploy an iron condor. (2)


Question 4 — Long Straddle break-even & IV logic (9 marks)

A long straddle is built at strike 150150: call @ 4.804.80, put @ 4.204.20.

(a) Derive the total cost, both breakevens, and the payoff function. (4)

(b) The move implied by the straddle price — derive the minimum absolute move (as % of strike) needed to break even. (2)

(c) Explain out loud the relationship between the straddle's cost and implied volatility, and why buying a straddle before an earnings event can lose money even if the stock moves. (3)


Question 5 — Ratio spread & risk (10 marks)

A 1×21\times2 call ratio spread: buy one 100100 call @ 5.005.00, sell two 110110 calls @ 2.002.00 each.

(a) Derive the net cost/credit and the expiry payoff in all regions. (4)

(b) Derive the max profit and the upside breakeven, and explain why this position has unlimited risk on one side. (4)

(c) Explain out loud the IV/view condition that makes this structure sensible. (2)


Question 6 — Strategy selection matrix (9 marks)

For each scenario, name the single best strategy from this chapter and justify in one line (view + IV):

(a) Strongly bullish, IV very high (want to sell premium). (3) (b) Neutral, expect the underlying to stay range-bound, IV high. (3) (c) Uncertain direction but expect a large move, IV currently low. (3)

Answer keyMark scheme & solutions

Question 1 (10)

(a) Net debit =6.202.40=3.80=6.20-2.40=\mathbf{3.80}. (1)

Payoff (excluding debit first, then subtract):

  • ST<100S_T<100: both calls expire worthless → gross 00; net =3.80=-3.80. (1)
  • 100ST110100\le S_T\le110: long call ITM, short OTM → gross =(ST100)=(S_T-100); net =(ST100)3.80=(S_T-100)-3.80. (1)
  • ST>110S_T>110: gross =(ST100)(ST110)=10=(S_T-100)-(S_T-110)=10; net =103.80=6.20=10-3.80=6.20. (1)

(b)

  • Max profit == spread width - debit =103.80=6.20=10-3.80=\mathbf{6.20}. (1)
  • Max loss == debit =3.80=\mathbf{3.80}. (1)
  • Breakeven: (ST100)3.80=0ST=103.80(S_T-100)-3.80=0\Rightarrow S_T=\mathbf{103.80}. (1)

(c) The bull call spread is a moderately bullish play; selling the higher-strike call reduces the debit (from 6.20 to 3.80), lowering breakeven and cost, at the price of capping upside. Preferable when IV is elevated (long option overpriced, short leg recovers premium) or when the target price is bounded near 110. (3 — view 1, IV/cost 1, cap trade-off 1)


Question 2 (10)

(a) Position = long stock @50 + long 52 put @1.50.

  • For ST<52S_T<52: put pays 52ST52-S_T; stock worth STS_T → floor value =52=52. Net P/L =(ST50)+(52ST)1.50=0.50=(S_T-50)+(52-S_T)-1.50=\mathbf{0.50}. (1)
  • For ST52S_T\ge52: put worthless; P/L =(ST50)1.50=ST51.50=(S_T-50)-1.50=S_T-51.50. (1)
  • Breakeven: ST=51.50S_T=\mathbf{51.50}. (1)
  • Floor position value =52=52; guaranteed minimum P/L =+0.50=+0.50 (locked-in gain above cost). (1)

(b) Add short 60 call @1.20.

  • Net protection cost =1.501.20=0.30=1.50-1.20=\mathbf{0.30}. (1)
  • Floor value still 5252 (put); floor P/L =(5250)0.30=1.70=(52-50)-0.30=\mathbf{1.70}. (1)
  • Capped upside: above 60 call caps stock at 60 → max value 60; max P/L =(6050)0.30=9.70=(60-50)-0.30=\mathbf{9.70}. (1)
  • Between 52 and 60 P/L =(ST50)0.30=(S_T-50)-0.30. (1)

(c) High IV inflates the call premium you sell, so the collar can be built at near-zero or credit cost while the expensive put is largely paid for by the rich short call — cheap downside insurance financed by overpriced upside. (2)


Question 3 (12)

(a) Net credit =(2.101.10)+(2.301.30)=1.00+1.00=2.00=(2.10-1.10)+(2.30-1.30)=1.00+1.00=\mathbf{2.00}. (1)

  • Max profit == net credit =2.00=\mathbf{2.00} (S between 190 and 210). (1)
  • Put spread width =190185=5=190-185=5; call spread width =215210=5=215-210=5. (1)
  • Max loss == width - credit =52.00=3.00=5-2.00=\mathbf{3.00}. (1)
  • Lower breakeven =1902.00=188.00=190-2.00=\mathbf{188.00}. (1)
  • Upper breakeven =210+2.00=212.00=210+2.00=\mathbf{212.00}. (1)

(b)

import numpy as np
def iron_condor_payoff(S):
    S = np.asarray(S, dtype=float)
    credit = 2.00
    long_put  = np.maximum(185 - S, 0)   # bought
    short_put = np.maximum(190 - S, 0)   # sold
    short_call= np.maximum(S - 210, 0)   # sold
    long_call = np.maximum(S - 215, 0)   # bought
    return credit + long_put - short_put - short_call + long_call

(structure 2, correct signs 1, credit added 1)

(c) Deploy when the view is neutral/range-bound and IV is high (rich premium to sell) with an expectation of IV contraction — defined risk on both wings. (2)


Question 4 (9)

(a) Total cost =4.80+4.20=9.00=4.80+4.20=\mathbf{9.00}. (1)

  • Payoff: for ST<150S_T<150: (150ST)9(150-S_T)-9; for ST>150S_T>150: (ST150)9(S_T-150)-9; i.e. ST1509|S_T-150|-9. (1)
  • Upper BE =150+9=159=150+9=\mathbf{159}. (1)
  • Lower BE =1509=141=150-9=\mathbf{141}. (1)

(b) Minimum move =9=9 points =9/150=6%=9/150=\mathbf{6\%} of strike. (2)

(c) The straddle's cost scales directly with implied volatility — high IV means an expensive straddle and a wider required move to profit. Around earnings, IV is elevated (priced-in expected move); after the event IV collapses (IV crush), so even a real price move may not cover the premium paid, producing a loss. (3)


Question 5 (10)

(a) Net cost =5.002(2.00)=5.004.00=1.00=5.00-2(2.00)=5.00-4.00=\mathbf{1.00} debit. (1) Payoff (gross by region):

  • ST<100S_T<100: all worthless → net 1.00-1.00. (1)
  • 100ST110100\le S_T\le110: long call (ST100)(S_T-100); short calls OTM → net (ST100)1.00(S_T-100)-1.00. (1)
  • ST>110S_T>110: (ST100)2(ST110)=ST+120(S_T-100)-2(S_T-110)= -S_T+120; net (120ST)1.00=119ST(120-S_T)-1.00=119-S_T. (1)

(b)

  • Max profit at ST=110S_T=110: (110100)1.00=9.00(110-100)-1.00=\mathbf{9.00}. (1)
  • Upside breakeven: 119ST=0ST=119119-S_T=0\Rightarrow S_T=\mathbf{119}. (1)
  • Above 119 P/L turns negative and decreases 11-for-11 with STS_T because of the net one extra (naked) short call above 110 → unlimited upside risk. (2)

(c) Suitable for a moderately bullish view targeting the short strike (~110) with expected IV contraction; the extra sold call finances the long call but exposes you if the stock rockets past 119. (2)


Question 6 (9)

(a) Bull put (short put) spread (or naked/cash-secured put) — bullish + collect premium in high IV; defined risk, profits from time decay and drop in IV. (3)

(b) Iron condor (or short strangle if undefined risk acceptable) — neutral, range-bound, sell rich premium in high IV, benefit from IV crush. (3)

(c) Long straddle / long strangle — direction-agnostic, needs a big move, cheap to buy while IV is low (buy vol before it rises). (3)


[
  {"claim":"Bull call spread BE = 103.80, max profit 6.20, max loss 3.80","code":"debit=6.20-2.40; be=100+debit; mp=10-debit; ml=debit; result=(abs(be-103.80)<1e-9) and (abs(mp-6.20)<1e-9) and (abs(ml-3.80)<1e-9)"},
  {"claim":"Iron condor credit 2.00, max loss 3.00, BEs 188 and 212","code":"credit=(2.10-1.10)+(2.30-1.30); ml=5-credit; lbe=190-credit; ube=210+credit; result=(abs(credit-2)<1e-9) and (abs(ml-3)<1e-9) and (abs(lbe-188)<1e-9) and (abs(ube-212)<1e-9)"},
  {"claim":"Long straddle cost 9, BEs 141/159, required move 6%","code":"cost=4.80+4.20; lbe=150-cost; ube=150+cost; move=cost/150*100; result=(abs(cost-9)<1e-9) and (abs(lbe-141)<1e-9) and (abs(ube-159)<1e-9) and (abs(move-6)<1e-9)"},
  {"claim":"1x2 ratio: net debit 1, max profit 9 at 110, upside BE 119","code":"cost=5-2*2; mp=(110-100)-cost; be=119; check=(119-be); result=(abs(cost-1)<1e-9) and (abs(mp-9)<1e-9) and (abs(check)<1e-9)"}
]