Finance-Theory · Unit 10 · Video 7 · Interactive Practice

Why Futures Settle Every Day: Collateral Fails, Markups Fail, Daily Rewrites Work

IKey Formulas

FormulaNameWhat it says
F−ST>dF - S_T > dWalk-away conditionA buyer who posted deposit dd per unit defaults once performing loses more than the deposit
Vlong=St−F1+r,Vshort=−VlongV_{\text{long}} = \dfrac{S_t - F}{1+r}, \quad V_{\text{short}} = -V_{\text{long}}Forward value before settlementSpot StS_t stands in for the value at delivery; rr is the interest rate over the time left
Ft−Ft−11+r\dfrac{F_t - F_{t-1}}{1+r}Daily settlement to the longCancel yesterday's contract, pay the difference, rewrite at today's price FtF_t
∑t=1T(Ft−Ft−1)=FT−F0\displaystyle\sum_{t=1}^{T} (F_t - F_{t-1}) = F_T - F_0Daily payments add upIgnoring interest, the daily cash totals the single forward's gain or loss

Key Insight: A futures contract is a sequence of forward contracts, cancelled and rewritten every day, so the amount at stake is one day's price move instead of three months'.

IIHow Big Must the Collateral Be?

A buyer holding deposit dd walks away whenever the price falls more than dd below the forward price.

💡 Challenge: find the smallest deposit that keeps a fall to $60 a ton out of the walk-away zone, and read off how much cash it ties up.

IIIA Markup Over the Market Locks in Nothing

Paying market price plus 10% at delivery leaves the buyer's cost as uncertain as the spot price — more so.

IVMarking to Market: One Day at Stake

An unmarked forward lets the buyer's loss pile up for three months; daily settlement pays each day's move as it happens.

💡 In the video's version each daily payment is also discounted by 1+r1+r over the remaining term; with the interest handled that way, the daily cash still adds up to what the single forward would have settled at maturity.

VQuiz Questions

Problem 1 · The Walk-Away Price

Given: A miller agreed to buy corn forward at $240 a ton and posted a deposit of $12 a ton, which he forfeits if he fails to take delivery. Find the spot price at delivery below which walking away costs him less than performing.

✅ Correct! Performing loses 240−ST240 - S_T per ton, walking away loses 12, so he walks once 240−ST>12240 - S_T > 12, i.e. once spot is below $228.
❌ Check the direction. The buyer is hurt by a fall in price; a spot price above $240 makes the contract valuable to him, so he performs.
❌ Not quite. $12 is what he loses by walking away, not the price at which he starts to prefer it. Compare the loss from performing, 240−ST240 - S_T, with 12.
❌ Close, but the deposit matters. Just below $240 performing loses only a few dollars, less than the $12 he would forfeit; he performs until the loss exceeds the deposit.
❌ Not quite. Set the loss from performing, 240−ST240 - S_T, equal to the deposit.
Show solution

Per ton, performing costs the forward price minus what the corn is worth: 240−ST240 - S_T. Walking away costs the forfeited deposit, 12. Walking away is cheaper when

240−ST>12  ⟺  ST<240−12=228.240 - S_T > 12 \iff S_T < 240 - 12 = 228.

So below $228 a ton the miller prefers to default. A $12 deposit protects the seller only against price falls of up to $12 a ton.

Problem 2 · Valuing the Farmer's Side

Given: A farmer sold 2,000 tons of wheat forward at $210 a ton. Partway to delivery the spot price is $222 a ton, and the interest rate over the time remaining is r=2%r = 2\%. Using today's spot as the value of the wheat at delivery, find the value of the contract to the farmer.

✅ Correct! The farmer must deliver wheat worth $222 for $210, a shortfall of $12 a ton paid at delivery, so it is discounted one period.
❌ Check the sign. The farmer is short: a rise in spot above the forward price hurts the seller. Vshort=(F−St)/(1+r)V_{\text{short}} = (F - S_t)/(1+r) is negative here.
❌ Close, but discount it. The $24,000 changes hands at delivery, not today; divide by 1+r=1.021 + r = 1.02.
❌ Not quite. Multiplying by 1.021.02 moves the amount forward in time; a value today needs dividing by 1.021.02.
❌ Not quite. Compute (F−St)(F - S_t) per ton for the seller, scale by 2,000 tons, then discount.
Show solution

Per ton, the seller's value is

Vshort=F−St1+r=210−2221.02=−121.02.V_{\text{short}} = \frac{F - S_t}{1 + r} = \frac{210 - 222}{1.02} = \frac{-12}{1.02}.

On 2,000 tons:

−12×20001.02=−240001.02≈−23529.41\frac{-12 \times 2000}{1.02} = \frac{-24000}{1.02} \approx -23529.41

The contract is worth about −$23,529 to the farmer — and about +$23,529 to the buyer. If the farmer wanted out, she would have to pay roughly that much for someone to take her place.

Problem 3 · Four Days of Daily Settlement

Given: A long futures position covers 500 tons. The settlement prices, per ton, are $172 on day 0, then $166, $175, $171 and $180 on days 1–4. Each day the position is settled at the change in price; ignore interest.

What is the long's net cash over the four days?

What is the largest amount the long had to pay on any single day?

✅ Correct! The daily changes telescope to 180−172=8180 - 172 = 8 per ton, the same as one forward held four days, while the most the long ever had to pay in one day was $6 a ton.
❌ Not quite. That adds the size of every move, gains and losses alike. Days 1 and 3 are payments by the long, so they subtract.
❌ Not quite. That counts only the two up-days. The long also paid on days 1 and 3.
❌ Check the sign. The price ended higher than it started, and a long gains when the price rises.
❌ Check the net. Add the daily changes Ft−Ft−1F_t - F_{t-1} with their signs, then multiply by 500.
❌ Not quite. A $9 move happened twice, but both were rises — the long received $4,500 on those days.
❌ Not quite. $4,000 is the net over four days, and the long received it. The question asks for the worst single day.
❌ Check the daily payments. The long pays on the days the price falls; find the largest fall and multiply by 500.
Show solution

Daily changes per ton, Ft−Ft−1F_t - F_{t-1}:

166−172=−6,175−166=9,171−175=−4,180−171=9.166 - 172 = -6,\quad 175 - 166 = 9,\quad 171 - 175 = -4,\quad 180 - 171 = 9.

Net: −6+9−4+9=8=180−172-6 + 9 - 4 + 9 = 8 = 180 - 172 per ton, so on 500 tons the long nets 8×500=40008 \times 500 = 4000, i.e. +$4,000 — exactly what one forward at $172 would have paid at the day-4 price.

The long pays on days 1 and 3: 6×500=30006 \times 500 = 3000 and 4×500=20004 \times 500 = 2000. The largest single payment is $3,000. That, not the whole drift, is what the other side is ever exposed to if the long disappears.

Problem 4 · A Different Markup

Given: Instead of a fixed price, a refiner agrees to buy at 8% over the market price prevailing at delivery. The spot price at delivery could be anywhere from $180 to $260 a ton. Find the width of the range of prices per ton the refiner might pay.

✅ Correct! The cost is 1.08 ST1.08\,S_T, so its range is 1.08 times the spot range: wider than not hedging at all.
❌ Not quite. The markup is fixed, but it is a percentage of a price nobody knows yet. Nothing about the level is locked in.
❌ Close, but the markup scales too. 8% of a higher price is a larger markup, so the cost moves 1.08 times as much as spot.
❌ Not quite. $6.40 is only how much the markup itself varies (0.08×800.08 \times 80); the refiner pays the spot price plus the markup.
❌ Not quite. Compute the price paid at both ends of the spot range.
Show solution

The price paid is 1.08 ST1.08\,S_T. At the two ends of the range:

1.08×180=194.40,1.08×260=280.80.1.08 \times 180 = 194.40, \qquad 1.08 \times 260 = 280.80.

The width is 280.80−194.40=86.40=1.08×80280.80 - 194.40 = 86.40 = 1.08 \times 80, i.e. $86.40 a ton, larger than the spot range of $80. The contract may secure supply, but price uncertainty — the very risk to be hedged — is untouched, and any seller would gladly accept it.

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