Game Theory
Game theory studies decisions whose outcomes depend on what other decision-makers do. We first find best responses and equilibria in simple games. We then apply the same method to contracts between a supplier and a retailer.
1. Games, best responses, and Nash equilibrium
A game is a model with several decision-makers whose payoffs depend on their combined choices. In our example, a retailer and its supplier each decide whether to invest in a shared forecasting IT system. Each firm's profit depends on both firms' choices, not only on its own.
A game is described by its players, their strategies and their payoffs. The players are the actors who participate in the economic setting, here the Retailer and the Supplier. A strategy is the rule for the individual action taken by a specific player, here Invest or Don't invest. A strategy plays the role of the choice variable in Sessions 2 and 3. The combination of both players' choices is called a strategy profile. A player's payoff is its profit for every possible profile. The payoff is the player's objective function, which it maximizes. Importantly, each firm's payoff depends on its own choice and on its partner's choice, unlike the objective functions of Session 2.
We write $a_R$ for the Retailer's strategy, $a_S$ for the Supplier's strategy and $\pi_R(a_R,a_S)$ for the Retailer's payoff. Given the Supplier's strategy $a_S$, a best response of the Retailer is a strategy that maximizes the Retailer's payoff, holding $a_S$ fixed: $$BR_R(a_S)=\operatorname*{arg\,max}_{a_R}\{\pi_R(a_R,a_S)\}.$$ The Supplier's best response $BR_S(a_R)$ is defined in the same way. There may be several best responses if payoffs tie. A Nash equilibrium is a strategy profile in which every player's strategy is a best response to the other player's strategy. In other words, no player can increase its payoff by changing its own strategy alone. Such a change is called a unilateral deviation. If one strategy is a best response whatever the other player does, it is called a dominant strategy.
Checking each strategy profile for a profitable deviation
Payoffs are (Retailer's profit, Supplier's profit), in €000s per season, for each combination of the two firms' decisions on shared forecasting IT. Click a cell to check whether either firm gains by a unilateral deviation. The buttons show each player's best responses.
In this example, (Invest, Invest) gives the highest joint payoff, $60+60=120$, and both firms prefer it to (Not, Not). It is not an equilibrium, however. Each firm can raise its own payoff from 60 to 70 by not paying for the IT system while still using its partner's forecast. When both firms reason in this way, they end at (Not, Not). This situation is called a prisoner's dilemma: in the equilibrium, both players earn less than in another strategy profile.
Why is (Not, Not) the only Nash equilibrium here?
We check every cell against the definition. At (Invest, Invest) the Retailer earns 60. If it switches to Not while the Supplier keeps investing, it earns 70 instead. The Retailer therefore wants to deviate, and (Invest, Invest) is not an equilibrium. At (Invest, Not) the Retailer earns 20, and switching to Not raises its payoff to 40, so this profile is not an equilibrium either. By the same argument for the Supplier, (Not, Invest) is not an equilibrium. Only at (Not, Not) does neither firm have a profitable unilateral deviation, because switching to Invest would lower either firm's payoff from 40 to 20. In fact, "Not" is a dominant strategy for both players here. It is the best response whatever the partner does, so (Not, Not) is the unique equilibrium.
2. Sequential games: a leader and a follower
In the forecasting-IT game, both firms choose at the same time. Many supply-chain relationships are instead sequential. For example, the supplier sets a wholesale price first, and the retailer places an order only after seeing that price. In a Stackelberg game, one player, the leader, chooses first and commits to that choice. The other player, the follower, observes the leader's choice and then chooses its response.
We solve these games by backward induction. We first find the best response of the last mover to each possible choice of the first mover. We then solve the first mover's optimization problem, using that best response in place of the last mover's choice.
A sequential game tree, solved from the last move
Illustrative payoffs (πR, πS), in €000s per season, for a simplified version of the wholesale-price game in §4 with two prices and two order sizes. The supplier chooses the price first, and the retailer then chooses an order size. The retailer's best response after the price not chosen still matters, because the supplier uses it to compare the two prices.
Swipe or scroll horizontally to see the whole tree →
We start at the bottom of the tree. After each price, the retailer, the last mover, chooses the order size with the higher payoff for the retailer. After €14 it orders high, because $18>14$, and after €20 it orders low, because $9>6$. This rules out one branch under each price. The supplier then compares its payoffs given these responses: 11 at €14 and 15 at €20. It therefore chooses €20. The supplier's problem cannot be solved without first solving the retailer's problem.
The same sequential game in normal form
In the tree, each node names the player who moves there, each branch is one of that player's actions, and the payoffs sit at the leaves. The supplier moves once, so its two strategies are its two actions, €14 and €20. The retailer moves after observing the price, so a retailer strategy states its order after each price, including a price that is not chosen. We write HL for High after €14 and Low after €20, with H for High and L for Low. Without randomization, two actions at each of its two nodes give the retailer $2\times2=4$ pure strategies: HH, HL, LH and LL. For each strategy pair, follow its path in the tree and copy the payoffs at the leaf. The resulting normal form lists all strategy combinations:
The sequential game in normal form
Rows: retailer strategies. Columns: supplier strategies. Every cell keeps the tree's payoff order (retailer, supplier), in €000s. Bold payoff components are best responses. Marked cells are pure-strategy Nash equilibria.
| Retailer plan | Supplier: €14 | Supplier: €20 |
|---|---|---|
| HH | (18, 11) | (6, 19) |
| HL | (18, 11) | (9, 15)Nash · backward induction |
| LH | (14, 9) | (6, 19) |
| LL | (14, 9) | (9, 15)Nash |
At price €14, only the first letter is used, so HH and HL share a payoff, as do LH and LL. At price €20, only the second letter is used. Because each retailer strategy specifies a response to both prices, the matrix keeps the information that the retailer observes the price before ordering.
To check a best response, keep the other player's strategy fixed. In the €14 column, the retailer prefers 18 to 14, so HH and HL are best responses. In the €20 column, it prefers 9 to 6, so HL and LL are best responses. In every row, the supplier earns more at €20 than at €14: 19 rather than 11 in row HH, 15 rather than 11 in row HL, 19 rather than 9 in row LH and 15 rather than 9 in row LL. The two pure-strategy Nash equilibria are therefore (HL, €20) and (LL, €20), writing the retailer's strategy first.
Both equilibria lead to price €20 and a low order, with payoffs $(9,15)$. However, LL prescribes a low order even after price €14, where the retailer would prefer a high order because $18>14$. With Nash equilibrium alone, we cannot rule out this suboptimal response at a decision point that is not reached.
A subgame is the part of the game that starts at one decision point, together with everything that follows it. A subgame-perfect equilibrium is a strategy profile that is a Nash equilibrium in every subgame, including subgames that are not reached when the equilibrium is played. In a subgame-perfect equilibrium, we therefore also need optimal play off the path. In this game, the retailer must choose its best order after each price. By backward induction, the retailer's strategy is HL: high after €14 and low after €20. Anticipating this, the supplier chooses €20 because $15>11$. Thus (HL, €20) is the unique subgame-perfect equilibrium of this game. The path of play, price €20 followed by a low order, describes what happens. The strategy profile also describes what the retailer would do after the other price.
From a best response to a best-response function
In the matrix, we found a best response by comparing a few payoffs. When the choice is a quantity, we solve an optimization problem instead. The other firm's choice is held fixed during that calculation. The answer is a function of that choice, because a different quantity chosen by the other firm changes the firm's profit. We call this function the best-response function.
Two manufacturers sell the same product in one market. Firm 1 commits to a quantity $q_1\ge0$ first. Firm 2 observes it, then chooses $q_2\ge0$. Firm 1 cannot revise its commitment. Firm 1's choice variable is $q_1$, and Firm 2's choice variable is $q_2$. Both firms have a constant production cost of €20 per unit. For this example, the market price is $$P(q_1,q_2)=\max\{100-q_1-q_2,0\}\quad\text{euros per unit}.$$ This is an inverse demand function: it gives the price that buyers will pay for the total quantity supplied. Each additional unit lowers the price by €1 while the price is positive. At 100 units or more, the price is zero. Each firm's objective function is its own profit, which it maximizes: $$\pi_i(q_1,q_2)=[P(q_1,q_2)-20]q_i,\qquad i\in\{1,2\}.$$ We often omit the multiplication sign between symbols: $[P(q_1,q_2)-20]q_i$ means $[P(q_1,q_2)-20]\times q_i$, the margin per unit times the number of units. Between numbers we write the sign explicitly, as in $20\times40$. The quantities are treated as continuous. This example has no random demand, so that we can focus on how each firm's decision affects the other firm's profit.
Step 1: solve the follower's problem
After observing $q_1$, Firm 2 solves $$\max_{q_2\ge0}\{\pi_2(q_1,q_2)\}.$$ For $0\le q_1<80$, a positive quantity can be profitable. In the region where the price is positive, we first simplify Firm 2's profit by expanding the bracket:
$$\begin{aligned} \pi_2(q_1,q_2)&=(100-q_1-q_2-20)\,q_2 \\ &=(80-q_1-q_2)\,q_2 \\ &=(80-q_1)\,q_2-q_2^2. \end{aligned}$$The partial derivative $\partial/\partial q_2$ means “differentiate with respect to $q_2$, keeping $q_1$ fixed.” We compute it with the power rule. At the optimum the derivative is zero, which gives the first-order condition:
$$\begin{aligned} \frac{\partial\pi_2}{\partial q_2}&=80-q_1-2q_2, \\ 80-q_1-2q_2&=0, \\ 2q_2&=80-q_1, \\ q_2&=40-\frac{q_1}{2}. \end{aligned}$$The second derivative is $-2$, so this stationary point maximizes profit in the positive-price region. Producing enough to make the price zero gives a loss and cannot improve on it. If $q_1\ge80$, no positive quantity gives Firm 2 a positive margin, so its best choice is zero. Combining these cases gives the best-response function:
$$BR_2(q_1)=\max\left\{0,\;40-\frac{q_1}{2}\right\}.$$For example, $BR_2(20)=40-20/2=30$, $BR_2(40)=20$ and $BR_2(60)=10$. The first-order condition did not produce one fixed answer. It produced an optimal quantity for each possible quantity chosen by Firm 1. As in the two-letter strategies of the tree, Firm 2's strategy specifies its response even after choices that are not made in equilibrium.
Step 2: let the leader anticipate that response
Firm 1 solves $$\max_{q_1\ge0}\{\pi_1(q_1,BR_2(q_1))\}.$$ It must account for how Firm 2 reacts when $q_1$ changes. We therefore substitute the best-response function into Firm 1's profit before differentiating. For $0\le q_1\le80$,
$$\begin{aligned} \pi_1(q_1,BR_2(q_1)) &=\left[100-q_1-BR_2(q_1)-20\right]q_1 \\ &=\left[80-q_1-\left(40-\frac{q_1}{2}\right)\right]q_1 \\ &=\left(40-\frac{q_1}{2}\right)q_1 \\ &=40q_1-\frac{q_1^2}{2}. \end{aligned}$$By the power rule, the derivative with respect to $q_1$ is $40-q_1$. Setting it equal to zero gives the leader's first-order condition and its optimal quantity $q_1^*$:
$$\begin{aligned} \frac{d}{dq_1}\pi_1(q_1,BR_2(q_1))&=40-q_1, \\ 40-q_1^*&=0, \\ q_1^*&=40, \\ q_2^*&=BR_2(40)=40-\frac{40}{2}=20. \end{aligned}$$The second derivative is $-1$, so the leader's profit is strictly concave on this interval. Both endpoints give zero profit, and quantities above 80 give no positive profit, so 40 is the global optimum. Total output is $40+20=60$, and the price is $100-60=\text{€}40$. The profits are
$$\begin{aligned} \pi_1&=(40-20)\times40=\text{€}800, \\ \pi_2&=(40-20)\times20=\text{€}400. \end{aligned}$$The subgame-perfect strategy profile is “Firm 1 chooses 40, and Firm 2 follows $BR_2(q_1)$ after every possible $q_1$.” Its outcome is $(q_1,q_2)=(40,20)$. Firm 2 holds $q_1$ fixed while optimizing. Firm 1 anticipates that $q_2$ changes when it changes $q_1$. This is why their first-order conditions differ.
The follower's best response and the leader's profit
Top: Firm 2's best-response function. Bottom: Firm 1's profit after that response. Both axes stay fixed.
Increasing $q_1$ by two units makes the follower reduce $q_2$ by one unit over this range. Producing more helps the leader only up to $q_1=40$. Beyond that, the lower market price outweighs the extra units sold. At $q_1=40$, each orange dot lies inside the larger green dot.
Simultaneous quantity choices: the Cournot game
In a Cournot game, both firms choose their quantities at the same time, so neither observes the other's quantity. Each firm optimizes while holding the other's quantity fixed. With the same demand and costs, Firm 1's problem has the same form as Firm 2's problem above, so the best-response functions are
$$\begin{aligned} BR_1(q_2)&=\max\{0,40-q_2/2\}, \\ BR_2(q_1)&=\max\{0,40-q_1/2\}. \end{aligned}$$At a Nash equilibrium the quantities are mutual best responses: $q_1=BR_1(q_2)$ and $q_2=BR_2(q_1)$. We substitute the second equation into the first and solve for $q_1$:
$$\begin{aligned} q_1&=40-\frac12\left(40-\frac{q_1}{2}\right) \\ &=40-20+\frac{q_1}{4} \\ &=20+\frac{q_1}{4}, \\ \frac34\,q_1&=20, \\ q_1^C&=\frac{80}{3}\approx26.7. \end{aligned}$$By symmetry, $q_2^C=80/3$ as well. The price and each firm's profit are
$$\begin{aligned} P&=100-\frac{80}{3}-\frac{80}{3}=\frac{140}{3}\approx\text{€}46.67, \\ \pi_1=\pi_2&=\left(\frac{140}{3}-20\right)\times\frac{80}{3} \\ &=\frac{80}{3}\times\frac{80}{3}=\frac{6400}{9}\approx\text{€}711.11. \end{aligned}$$In this example, committing first raises Firm 1's profit from about €711 to €800. The difference comes from the timing and the ability to commit, not from a change in costs or demand.
The quantity example used $q_2=BR_2(q_1)$. In Sections 4–6, the follower is a retailer and its response is an order quantity determined by contract terms, such as $Q^*(w)$. We will solve the retailer's optimization first, then study the supplier's choice of terms.
3. The integrated benchmark: maximizing total supply-chain profit
Amélie runs a retail chain that sells fashion sunglasses. Each season she must place one order before she observes demand. Her supplier, LunetteCo, manufactures the sunglasses at a constant unit cost. We scale the Session 1 weekly sunglasses model to a 12-week season, and we assume that season demand is normally distributed, $D \sim N(\mu, \sigma^2)$. Here $D$ is the random season demand, $\mu=E[D]$ is its mean and $\sigma=\operatorname{SD}(D)$ is its standard deviation. We use $\mu=12\times230.5=2766$ pairs.
For $\sigma$ we do not use the $\sqrt{12}$ scaling for independent weeks, which would give $74.8\times\sqrt{12}\approx259$ pairs. That scaling assumes that the weekly demands are independent. A fashion season, however, has uncertainty about the frame's popularity that affects every week of the season. We instead assume the larger value $\sigma = 898$ pairs, which keeps the coefficient of variation $\sigma/\mu$ at about 0.32, as in the weekly data. This is an illustrative calibration for the teaching example, not an estimate from the data. Amélie sells at $p = 25$€, LunetteCo produces at $c = 10$€, and unsold pairs salvage at $s = 4$€. These five parameters, $\mu$, $\sigma$, $p$, $c$ and $s$, stay fixed for the rest of this session. Only the contract terms change.
Before we study two separate firms, we consider a benchmark. Suppose that one company owns both the factory and the stores. We call this company the integrated chain. Its choice variable is the order quantity $Q$, and its objective function is the expected total profit $E[\pi(Q,D)]$, which it maximizes. It solves the newsvendor problem of Session 3 with unit cost $c$, because no wholesale payment is made inside one company. Its underage cost is $c_u=p-c$ and its overage cost is $c_o=c-s$, so its critical ratio is
$$\begin{aligned} F(Q^0)&=\frac{c_u}{c_u+c_o} \\ &=\frac{p-c}{(p-c)+(c-s)} \\ &=\frac{p-c}{p-s} \\ &=\frac{25-10}{25-4} \\ &=\frac{15}{21}\approx0.714. \end{aligned}$$We denote the integrated optimal order quantity by $Q^0$. The superscript $0$ labels this integrated benchmark. It is not an exponent. Because demand is normal, $F(Q)=\Phi\big((Q-\mu)/\sigma\big)$, where $\Phi$ is the standard normal CDF. Solving $F(Q^0)=15/21$ for $Q^0$ gives
$$\begin{aligned} Q^0&=\mu+\sigma\,\Phi^{-1}\!\left(\frac{p-c}{p-s}\right) \\ &=2766+898\times\Phi^{-1}(0.714) \\ &\approx2766+898\times0.566 \\ &\approx3274\text{ pairs}. \end{aligned}$$Session 3 wrote the realized profit as $\pi(Q,d)=(p-c)d-c_o(Q-d)^+-c_u(d-Q)^+$. Holding $Q$ fixed and taking expectations, as in Session 3, gives the expected profit: $$E[\pi(Q,D)]=\underbrace{(p-c)\mu}_{\text{margin on mean demand}} -\underbrace{c_o\,E[(Q-D)^+]}_{\text{expected leftover cost}} -\underbrace{c_u\,E[(D-Q)^+]}_{\text{expected shortage cost}}.$$ Recall also from Session 3 that, with $z=(Q-\mu)/\sigma$, the expected shortage is $E[(D-Q)^+]=\sigma L(z)$. Here $L(z)=\phi(z)-z[1-\Phi(z)]$ is the standard normal loss function and $\phi$ is the standard normal density. The expected leftovers are $E[(Q-D)^+]=Q-\mu+E[(D-Q)^+]$. For example, here $c_u=15$€ and $c_o=6$€. At $Q=\mu=2766$ we have $z=0$ and $L(0)\approx0.3989$, so the expected shortage and the expected leftovers are both $898\times0.3989\approx358.25$ pairs, and $$E[\pi(2766,D)]\approx15\times2766-6\times358.25-15\times358.25\approx\text{€}33{,}967.$$ The integrated optimal quantity maximizes this function: $$Q^0\in\operatorname*{arg\,max}_{Q\ge0}\{E[\pi(Q,D)]\}.$$ Its maximum value is $E[\pi(Q^0,D)]\approx$ €35,080 per season. We call this value the first-best profit. It is the highest expected total profit in this model.
Expected chain profit as a function of the order quantity
Choose Q with the slider. The curve is the expected chain profit E[π(Q,D)], and its maximum is at Q⁰.
Profit changes slowly near the optimum, so small order errors have a modest cost in this example. At $Q=3200$, expected profit is about €35,058, only €22 below the maximum. Larger departures reduce total profit more. At $Q=2000$, it is about €27,937. The next section shows why a supplier's preferred wholesale price can lead to such a departure.
In practice, LunetteCo and Amélie are two separate firms, and each maximizes its own expected profit. The newsvendor decision then becomes part of a two-player Stackelberg game. LunetteCo, the leader, proposes the terms of a contract, which set the payments between the two firms, such as a wholesale price per unit. Amélie, the follower, observes the terms and chooses her order quantity as a best response. A contract coordinates the chain if the retailer's best response is $Q^0$. Sections 4–6 study three contracts that LunetteCo could offer. For each one, we ask whether the retailer chooses $Q^0$ and how the expected profit is divided between the firms.
4. Wholesale prices and double marginalization
The simplest contract is a wholesale-price contract. Now we have two players: a retailer, Amélie, written with subscript $R$, and a supplier, LunetteCo, written with subscript $S$. LunetteCo sells each pair to Amélie at a fixed wholesale price $w$ per pair, with $c \le w < p$. Amélie pays $w$ for every pair she orders, whether or not it sells. Her realized profit $\pi_R(Q,D)$ is the profit of Session 3, with $w$ in place of $c$. LunetteCo's profit $\pi_S(Q,D)$ is its margin $w-c$ on every pair ordered, which does not depend on demand. Taking expectations as in Session 3, the expected profits are
$$\begin{aligned} E[\pi_R(Q,D)]&=(p-\textcolor{#fc6255}{w})\mu-(\textcolor{#fc6255}{w}-s)\,E[(Q-D)^+]-(p-\textcolor{#fc6255}{w})\,E[(D-Q)^+], \\[6pt] E[\pi_S(Q,D)]&=(\textcolor{#fc6255}{w}-c)\,Q. \end{aligned}$$For example, if $w=15$€ and $c=10$€, the supplier earns $(15-10)Q=5Q$. The wholesale price $w$ is fixed by the supplier, and Amélie chooses $Q$. Amélie pays $wQ$ and LunetteCo receives it, so the payments cancel in the sum: $E[\pi_R(Q,D)]+E[\pi_S(Q,D)]=E[\pi(Q,D)]$. Total expected profit changes only if the order changes.
Recall from Session 3 that we solved the newsvendor's first-order condition and found $F(Q^*)=c_u/(c_u+c_o)$. In Session 3 the unit cost was $c$. Here Amélie buys each pair from LunetteCo at unit cost $w$. Her overage cost, the loss on an unsold pair, is $w-s$. Her underage cost, the margin lost on a missed sale, is $p-w$. These are the coefficients of the expected leftovers and the expected shortage in her expected profit. Her critical ratio therefore uses $w$ instead of $c$:
$$\begin{aligned} c_u&=p-\textcolor{#fc6255}{w},\qquad c_o=\textcolor{#fc6255}{w}-s, \\[6pt] F(Q^*(w))&=\frac{c_u}{c_u+c_o} \\ &=\frac{p-\textcolor{#fc6255}{w}}{(p-\textcolor{#fc6255}{w})+(\textcolor{#fc6255}{w}-s)} \\ &=\frac{p-\textcolor{#fc6255}{w}}{p-s}. \end{aligned}$$For fixed $p$ and $s$, the critical ratio depends on $w$, so the optimal quantity is a function of $w$. Because demand is normal, we solve for $Q^*(w)$ with the standard normal quantile function, as for $Q^0$. If the formula gives a negative quantity, the constraint $Q\ge0$ binds and the constrained optimum is $Q=0$. Hence
$$Q^*(w)=\max\!\left\{0,\;\mu+\sigma\Phi^{-1}\!\left(\frac{p-w}{p-s}\right)\right\}.$$| Wholesale price $w$ | Critical ratio $(25-w)/21$ | Order $Q^*(w)$ |
|---|---|---|
| 10€ | $15/21\approx0.714$ | 3274 pairs |
| 15€ | $10/21\approx0.476$ | 2712 pairs |
| 20€ | $5/21\approx0.238$ | 2126 pairs |
The wholesale price $w$ is a choice of the supplier. A higher $w$ gives a lower critical ratio and a smaller order. As in the quantity game, the follower's best response depends on the leader's choice.
The retailer orders less than the integrated chain
Because $w>c$, the retailer's critical ratio is smaller than the chain's. Since the CDF $F$ is increasing, a smaller critical ratio leads to a smaller order:
$$\begin{aligned} p-w&<p-c, \\ \frac{p-w}{p-s}&<\frac{p-c}{p-s}, \\ F(Q^*(w))&<F(Q^0), \\ Q^*(w)&<Q^0. \end{aligned}$$This holds for every $w$ with $c<w<p$. The supplier's margin $w-c$ raises the retailer's unit cost above the chain's production cost $c$, so the retailer orders less than the integrated chain would. Each firm adds its own margin: the supplier through $w-c$ and the retailer through $p-w$. The resulting reduction in the order quantity and in total expected profit, compared with the integrated chain, is called double marginalization.
The supplier chooses the wholesale price, anticipating the retailer's response
LunetteCo anticipates how Amélie will respond to each wholesale price. Its choice variable is $w$, and its objective function is its own expected profit when Amélie orders $Q^*(w)$, which is $(w-c)Q^*(w)$: $$w^*\in\operatorname*{arg\,max}_{c\le w<p}\{(w-c)Q^*(w)\}.$$ To differentiate $(w-c)Q^*(w)$, we need the derivative of $Q^*(w)$. We differentiate both sides of the first-order condition $F(Q^*(w))=(p-w)/(p-s)$ with respect to $w$. By the chain rule, the left-hand side gives $f(Q^*(w))\,Q^{*\prime}(w)$, so $$\begin{aligned} f(Q^*(w))\,Q^{*\prime}(w)&=-\frac{1}{p-s}, \\ Q^{*\prime}(w)&=-\frac{1}{(p-s)\,f(Q^*(w))}. \end{aligned}$$ By the product rule, the derivative of the supplier's profit is $$\begin{aligned} \frac{d}{dw}\big[(w-c)Q^*(w)\big]&=Q^*(w)+(w-c)\,Q^{*\prime}(w) \\ &=Q^*(w)-\frac{w-c}{(p-s)\,f(Q^*(w))}. \end{aligned}$$ The first term is the gain from a higher margin on every pair ordered. The second term is the loss from the smaller order. At $w=c$, the derivative equals $Q^0>0$, so the supplier gains by raising $w$ above its cost. Setting the derivative equal to zero gives an equation with no closed-form solution, so we solve it numerically. The supplier's best price is $w^*\approx\text{€}21.40$, and Amélie then orders $Q^*(w^*)\approx1914$ pairs, about 1360 pairs fewer than $Q^0$.
The retailer's complete strategy is the best-response function $w\mapsto Q^*(w)$. It plays the role of the two-letter strategy in the small game. Solving the retailer's problem first, then the supplier's problem, is backward induction.
The retailer's order and the firms' profits as functions of the wholesale price
Top: demand density with the integrated order Q⁰ and Amélie's actual order Q*(w). Bottom: each firm's profit and the chain total as w varies, against the Q⁰ benchmark.
At $w=c$, Amélie orders $Q^0$ and the supplier earns zero profit. A higher price initially raises supplier profit but reduces the retailer's order and total chain profit. The supplier's preferred price maximizes its purple profit curve, which is a different objective from maximizing the green total-profit curve.
5. The buyback contract: the supplier shares the cost of leftovers
Under a wholesale markup, Amélie optimizes against a unit cost above the chain's production cost. A buyback contract can offset this incentive by sharing the cost of leftovers. Under a buyback contract, LunetteCo still charges a wholesale price $w$ per unit. In addition, it pays Amélie a rebate of $b$ euros for every unit that is unsold at the end of the season. The rebate is paid on top of the salvage value $s$ that Amélie already recovers. The contract terms are the pair $(w,b)$.
We require $b\ge0$ and $s + b \le w$. In words, what Amélie recovers on an unsold unit, salvage plus rebate, may not exceed what she paid for it. Otherwise, ordering units that she never expects to sell would be profitable in itself. With the rebate, Amélie's effective salvage value rises to $s+b$. For a finite optimum in the normal model, we use the strict inequality $s+b<w$. At equality, an unsold unit costs Amélie nothing, and her best quantity grows without bound.
As in Session 3, Amélie's critical ratio is $c_u/(c_u+c_o)$. With the rebate, she receives $s+b$ for each leftover pair instead of $s$. Her underage cost is unchanged, but her overage cost falls by $b$:
$$\begin{aligned} c_u&=p-w,\qquad c_o=w-(\textcolor{#fc6255}{s+b})=w-s-b, \\[6pt] F(Q^*(w,b))&=\frac{c_u}{c_u+c_o} \\ &=\frac{p-w}{(p-w)+(w-s-b)} \\ &=\frac{p-w}{p-s-\textcolor{#fc6255}{b}}. \end{aligned}$$For example, at $w=18$€, the rebate $b=0$ gives the ratio $7/21\approx0.333$, and $b=5$€ gives $7/16\approx0.438$. When this quantile is positive, her best order is
$$Q^*(w,b)=\mu+\sigma\Phi^{-1}\!\left(\frac{p-w}{p-s-b}\right).$$For a realized demand $D$, the profits of the two firms are $$\begin{aligned} \pi_R(Q,D)&=p\min\{Q,D\}+\underbrace{(\textcolor{#fc6255}{s+b})}_{\text{salvage plus rebate}}(Q-D)^+-wQ, \\ \pi_S(Q,D)&=(w-c)Q-\textcolor{#fc6255}{b}\,(Q-D)^+. \end{aligned}$$ Amélie's profit is the profit of Session 3 with $w$ in place of $c$ and $s+b$ in place of $s$. Taking expectations as in Session 3, the expected profits are $$\begin{aligned} E[\pi_R(Q,D)]&=(p-w)\mu-\big(w-(\textcolor{#fc6255}{s+b})\big)E[(Q-D)^+]-(p-w)\,E[(D-Q)^+], \\ E[\pi_S(Q,D)]&=(w-c)Q-\textcolor{#fc6255}{b}\,E[(Q-D)^+]. \end{aligned}$$ For example, if $w=18$€, $c=10$€ and $b=11.20$€, the supplier earns 8€ on each pair ordered and pays 11.20€ on each leftover pair. When we add the realized profits, the wholesale payments $wQ$ and the rebate payments $b(Q-D)^+$ cancel, because each is paid by one firm and received by the other: $$\begin{aligned} \pi_R(Q,D)+\pi_S(Q,D)&=p\min\{Q,D\}+s(Q-D)^++b(Q-D)^+-wQ \\ &\quad+wQ-cQ-b(Q-D)^+ \\ &=p\min\{Q,D\}+s(Q-D)^+-cQ \\ &=\pi(Q,D). \end{aligned}$$ This holds for every demand, so $E[\pi_R(Q,D)]+E[\pi_S(Q,D)]=E[\pi(Q,D)]$. The contract therefore changes total expected profit only through the order quantity. If we choose $(w,b)$ so that Amélie's critical ratio equals the chain's, $$\frac{p-w}{p-s-b} = \frac{p-c}{p-s},$$ then she orders $Q^0$ voluntarily, as her own best response, and the contract coordinates the chain. The terms $w$ and $b$ then determine how $E[\pi(Q^0,D)]$ is divided between the two firms.
Buyback contracts that coordinate the chain
Top: the $(w,b)$ plane and the line of coordinating contracts. Bottom: each firm's expected profit and their total at the current terms. The total equals $E[\pi(Q^0,D)]$ on the coordinating line and is lower away from it. The upper limit of the b slider changes with w so that $s+b<w$, which keeps the optimal quantity finite. A negative supplier profit means that the rebate is so large that the supplier loses money. Coordination alone does not guarantee that both firms will accept a contract.
A yellow dot marks an uncoordinated contract, and a green dot means that the retailer orders $Q^0$. Use “Coordinate at this w” to set the rebate exactly. Along the coordinating line, total profit stays at its maximum while the firms' shares change.
Where does the coordinating line b(w) come from?
For a given $w$, we want to solve for $b$ such that the retailer's critical ratio equals the integrated critical ratio. The retailer then orders $Q^0$. We multiply both sides by $(p-s-b)\times(p-s)$, divide by $p-c$, isolate $b$ and factor out $p-s$: $$\begin{aligned} \underbrace{\frac{p-w}{p-s-b}}_{\text{retailer's ratio}}\quad&=\quad\underbrace{\frac{p-c}{p-s}}_{\text{integrated ratio}} \\[10pt] (p-w)(p-s)&=(p-c)(p-s-b), \\[6pt] p-s-b&=\frac{(p-s)(p-w)}{p-c}, \\[6pt] b(w)&=(p-s)-\frac{(p-s)(p-w)}{p-c} \\ &=(p-s)\left[1-\frac{p-w}{p-c}\right] \\ &=(p-s)\,\frac{(p-c)-(p-w)}{p-c} \\ &=\frac{(p-s)(w-c)}{p-c}. \end{aligned}$$ With $p=25$, $c=10$ and $s=4$, the coordinating line is $$b(w)=\frac{21\times(w-10)}{15}=1.4\times(w-10).$$ For example, at $w=18$ the coordinating rebate is $b=1.4\times8=11.2$ euros. At $w=c$ the formula gives $b=0$. No rebate is needed there, because the retailer already faces the chain's own unit cost. As $w$ rises, $b$ must rise with it to offset the larger margin that the retailer pays to the supplier.