# Collar: Max Profit

In CAFI Derivatives vignettes Q8.

Isn’t max profit =Maximum profit = X2 – S0

This is stated on p.305 in CAFI

why do we use this equation instead Max profit per collar = ST + max(0, X1 − ST) − max(0, ST − X2) − S0 − (p0 − c0), i.e why did we deduct the  − (p0 − c0)?

HMM Foundation owns 30,000 shares of NASDAQ 100 Index Tracking Stock (QQQQ), which has a current price of \$30 per share. Osborne believes there is substantial risk of downside price movement in the index over the next six months. She recommends HMM use a six-month collar for the entire position of 30,000 shares as protection against the QQQQ price falling below \$27. HMM would maintain the collar strategy until expiration of the put and call options. Exhibit 1 provides data on current QQQQ puts and calls expiring in six months.

EXHIBIT 1

QQQQ PUTS AND CALLS EXPIRING IN SIX MONTHS

Option Type
Exercise Price (\$)

Call
35
0.80

Put
27
0.95

Q. If HMM enters into the collar recommended by Osborne, the maximum profit of the collar at option expiration would be closest to:

1. \$154,500.
2. \$150,000.
3. \$145,500.

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Answer is C. So let’s assume stock price hits \$60 at expiration. So profit on your stock is \$30, payoff on put option is 0 since it became deep out of the money and since call is in the money, payoff = \$25 on call. Value of this strategy = \$5.

Now for profit = (Value - premium paid for put + premium received on call)* 30,000 = (5-.95+.8)*30,000 = \$145,500.

Just did this earlier tonight. C is the answer.

I’d go crazy if I had to memorize these ugly ass formulas for option payouts.

Just think about what a collar does. It puts a collar around the share price, and protects you from extreme swings. It’s a long underlying, long put, the cost of which is offset by the premium from a short call.

To find the max loss, follow this strategy. You only need to plug in the extremes that will cause a big loss (i.e. big swings)

Plug in price =0. What’s the profit? You’ll get the premium from the short call, the payout on the put (strike price minus 0), and lose everything on the underlying.

Now plug in price = 1000 (a high number). The profit is again the premium from the short call, the loss on the short call, the put expires worthless but you’re still paying the premium, and the gain on the underlying.

Whatever you get as the max loss is the answer.

As for max profit, first plug in price = strike price of the call, and then price = strike price of the put. The logic here is that as long as the price hovers around with current underlying price (and the strike prices), you’ll make a profit. You just need to find the max.

Now for this example. For the max profit, you plug in 35, then 27.

If p= 35, then the payout is \$5 on the underlying (\$35-\$30), \$.8 on the call premium, -\$.95 on the put premium. That’s it. The call doesn’t lose anything since the price is equal to strike, and the put is out of the money. Total profit is \$4.85.

If you plug in P=27, then the payout is -\$3 on the underlying, \$.8 on the call premium, -\$.95 on the put premium. The put is at the money, so no payout, and the call out of the money. Total payout is less than when p=35.

So the max profit is \$4.85. For 30,000 shares, the max profit is \$4.85 X 30,000 = \$145,500

I realize if you’ve been reliant on formulas then this may be too much at the last min. But this is a pretty intuitive way to look at options payouts and I’ve never missed a question on this.