Temporal product bundling with myopic and strategic consumers: Manifestations and relative effectiveness

Abstract

Bundling in this era of eCommerce and high technology is a potent and widespread selling tool. The literature has focused on three static bundling strategies under which the products are sold separately (pure components or PC) or only in a bundled form (pure bundling or PB) or both (mixed bundling or MB). In a generalization, and motivated by real world examples, this paper examines the relative effectiveness of temporal bundling. We consider a firm that sells to a market of myopic and strategic consumers, and a selling season consisting of two stages. We compare four strategies – PC-PC (i.e., pure components in each of two stages), PB-PB, PB-PC and PC-PB – relative to MB. Our results show that PB-PB maximizes profits under low marginal costs; PC-PC prevails under high marginal costs given a large proportion of myopic consumers; and PB-PC is profit maximizing under moderate marginal costs when most consumers are strategic. These temporal strategies dominate MB except when the market is comprised entirely of strategic consumers. Finally, while temporal mixed bundling – MB-MB – is weakly superior to other temporal strategies, the latter are much easier to implement, as shown by real-world uses, and suffice to capture most of the profits. Related interesting pricing implications are discussed. Three extensions to the main model are also proposed.

Appendices

Appendix 1

1.1 Proofs of propositions 1, 2 and 3

Proof of proposition 1

The maximization problem is,

$$ {\mathrm{max}}_{P,\delta}\kern0.5em 2\left(P-c\right)\left(1-P\right)\alpha +2\left(P-\delta -c\right)\left[\alpha \delta +\left(1-\alpha \right)\left(1-P+\delta \right)\right]. $$

The objective function is separable in the two products and it is globally concave. The necessary conditions for maximum yield:

$$ {\displaystyle \begin{array}{c}\hfill P=\left(1+c\right)/2+\delta \left(2-\alpha \right)/2,\hfill \\ {}\hfill \delta =\left(\left(2-\alpha \right)P-\left(1+c\right)+\alpha \right)/2.\hfill \end{array}} $$

Inserting the first into the second, we get δ = (1 − c)/(4 − α). The optimal profit expression can be obtained by substituting these back into the objective function.

The prices in the two stages can be put in a simplified form as ${P}_1=\frac{1+c}{2}+\frac{\left(2-\upalpha \right)\left(1-c\right)}{2\left(4-\upalpha \right)}$ for the regular price and ${P}_2=\frac{1+c}{2}-\frac{\upalpha \left(1-c\right)}{2\left(4-\upalpha \right)}$ after substituting for the discount.

Proof of proposition 2

• Step 1: Demand and profit derivations for the three cases:

  • Case (a): P _B1 > P _B2 ≥ 1. From the α myopic consumers, demand in the first and second stages are α(2 − P _B1)²/2 and α(2 − P _B2)²/2 − α(2 − P _B1)²/2 respectively, at bundle prices of P _B1 and P _B2 respectively (Fig. 2(a) (iii)). From the 1 − α strategic consumers, purchases only occurs in the second stage at the discounted price P _B2 and demand is (1 − α)(2 − P _B2)²/2 (Fig. 2(b) (iii)). Thus, the profit function is,

$$ {\mathrm{max}}_{P_{B1},{P}_{B2}}\kern0.5em \alpha \left({P}_{B1}-2c\right)\frac{{\left(2-{P}_{B1}\right)}^2}{2}+\left({P}_{B2}-2c\right)\left(\left(1-\alpha \right)\frac{{\left(2-{P}_{B2}\right)}^2}{2}+\alpha \left(\frac{{\left(2-{P}_{B2}\right)}^2}{2}-\frac{{\left(2-{P}_{B1}\right)}^2}{2}\right)\right). $$

• Case (b): P _B1 ≥ 1 ≥ P _B2. The demands from the myopic consumers at the bundle prices of P _B1 and P _B2 are α(2 − P _B1)²/2 and $\alpha \left(1-{P}_{B2}^2/2\right)-\alpha {\left(2-{P}_{B1}\right)}^2/2$ in the first and second stages, respectively (Fig. 2(a) (ii)). The strategic segment purchases in the second stage and generates demand $\left(1-\alpha \right)\left(1-{P}_{B2}^2/2\right)$ (Fig. 2(b) (ii)). Thus, the profit function is,

$$ {\mathrm{max}}_{P_{B1},{P}_{B2}}\kern0.5em \alpha \left({P}_{B1}-2c\right)\frac{{\left(2-{P}_{B1}\right)}^2}{2}+\left({P}_{B2}-2c\right)\left(\left(1-\alpha \right)\left(1-\frac{P_{B2}^2}{2}\right)+\alpha \left(1-\frac{P_{B2}^2}{2}-\frac{{\left(2-{P}_{B1}\right)}^2}{2}\right)\right). $$

• Case (c): 1 ≥ P _B1 > P _B2. Myopic consumer demands are $\alpha \left(1-{P}_{B1}^2/2\right)$ at price P _B1 in the first stage and $\alpha \left({P}_{B1}^2-{P}_{B2}^2\right)/2$ at price P _B2 in the second stage (Fig. 2(a) (i)). The strategic consumers have demand $\left(1-\alpha \right)\left(1-{P}_{B2}^2/2\right)$ in the second stage (Fig. 2(b) (i)). Thus, the profit function is,

$$ {\mathrm{max}}_{P_{B1},\kern0.5em {P}_{B2}}\kern0.5em \alpha \left({P}_{B1}-2c\right)\left(1-\frac{P_{B1}^2}{2}\right)+\left({P}_{B2}-2c\right)\left(\left(1-\alpha \right)\left(1-\frac{P_{B2}^2}{2}\right)+\alpha \left(\frac{P_{B1}^2}{2}-\frac{P_{B2}^2}{2}\right)\right). $$

• Step 2: The next step is to examine the derived maximization problems.

  First consider Case (b): P _B1 ≥ 1 ≥ P _B2. Its objective simplifies to:

  $$ {\mathrm{max}}_{P_{B1},\kern0.5em {P}_{B2}}\kern0.5em \alpha \left({P}_{B1}-{P}_{B2}\right)\frac{{\left(2-{P}_{B1}\right)}^2}{2}+\left({P}_{B2}-2c\right)\left(1-\frac{P_{B2}^2}{2}\right). $$

The necessary conditions are:

$$ {\displaystyle \begin{array}{c}\kern1.00em \frac{\partial \uppi}{\partial {P}_{B1}}=\left(2-{P}_{B1}\right)\left(2-3{P}_{B1}+2{P}_{B2}\right)=0\Rightarrow {P}_{B1}=\frac{2}{3}\left(1+{P}_{B2}\right),\kern1.00em \\ {}\kern1.00em \frac{\partial \uppi}{\partial {P}_{B2}}=2-4\alpha -\alpha {P}_{B1}^2+4\alpha {P}_{B1}-3{P}_{B2}^2+4{cP}_{B2}=0.\kern1.00em \end{array}} $$

Inserting P _B1 = 2(1 + P _B2)/3 into the second equation, we get,

$$ \left(27+4\alpha \right){P}_{B2}^2-\left[36c+16\alpha \right]{P}_{B2}-\left(18-16\alpha \right)=0. $$

This has the quadratic solution given in the Proposition,

$$ {P}_{B2}=\frac{18c+8\alpha +\sqrt{{\left(18c+8\alpha \right)}^2+\left(27+4\alpha \right)\left(18-16\alpha \right)}}{27+4\alpha } $$

and only the positive root gives a positive bundle price.

We can find the boundaries between this case and Cases (a) and (c). The former given by P _B2 = 1 is given in the next part of the proof. For the latter, put P _B1 = 1 which implies from P _B1 = 2(1 + P _B2)/3 that P _B2 = 1/2. Insert these into the second necessary condition and it gives 5/4 − α + 2c = 0 which is not possible unless the parameters go out of bounds. Thus Case (c) is never optimal.

Next consider Case (a) P _B1 > P _B2 ≥ 1. The objective can be simplified to:

$$ {\mathrm{max}}_{P_{B1},\kern0.5em {P}_{B2}}\kern0.5em \alpha \left({P}_{B1}-{P}_{B2}\right){\left(2-{P}_{B1}\right)}^2/2+\left({P}_{B2}-2c\right){\left(2-{P}_{B2}\right)}^2/2 $$

The necessary conditions are:

$$ {\displaystyle \begin{array}{c}\kern1.00em \frac{\partial \uppi}{\partial {P}_{B1}}=\alpha \left(2-{P}_{B1}\right)\left(1-\frac{3{P}_{B1}}{2}+{P}_{B2}\right)=0,\kern1.00em \\ {}\kern1.00em \frac{\partial \uppi}{\partial {P}_{B2}}=\alpha \frac{{\left(2-{P}_{B1}\right)}^2}{2}+\left(2-{P}_{B2}\right)\left(1-\frac{3{P}_{B2}}{2}+2c\right)=0.\kern1.00em \end{array}} $$

Thus, P _B1 = 2(1 + P _B2)/3. Inserting it into the second equation, ${P}_{B2}=\frac{18\left(1+2c\right)-8\alpha }{27-4\alpha }$. Since the lowest price possible for this case is given by P _B2 = 1, inserting it into the expression for P _B2 gives the boundary condition, which is c = 0.25 + α/9. Finally, we verify the sufficiency conditions at the solution point.

Proof of proposition 3

• Step 1: Demand and profit derivations for the two cases (Refer to Fig. 4):

The analysis of the strategic consumer segment (Fig. 4(b)) when the product prices are lower than the bundle price is the same as for MB: Demand is (1 − α)(1 − P)(P _B  − P) for each product at a margin of P − c and (1 − α)[(1 + P − P _B )² − (2P − P _B )²/2] for the bundle at a margin of P _B  − 2c. Thus, (1 − α)[2(P − c)(1 − P)(P _B  − P) + (P _B  − 2c)((1 + P − P _B )² − (2P − P _B )²/2)] is the profit from strategic consumers when P _B  ≥ P. If the product price is higher than the bundle price, the derivation is the same as for pure bundling: Demand is $\left(1-\alpha \right)\left(1-{P}_B^2/2\right)$ for the bundle at a margin of P _B  − 2c. Thus, $\left(1-\alpha \right)\left({P}_B-2c\right)\left(1-{P}_B^2/2\right)$ is the profit from strategic consumers when P _B  ≤ P.

For the segment of myopic consumers, the first stage sales mirrors PC and hence the demand is α(1 − P) for each product at a margin of (P − c). The second stage is PB with margin (P _B  − 2c). Consider the case P _B  ≥ P. The demand for the bundle is α(2P − P _B )²/2. Next, consider the case P _B  ≤ P. The demand for the bundle is $\alpha \left[2\left(1-P\right)\left(P-{P}_B\right)+\left({P}^2-{P}_B^2/2\right)\right]$. Thus, the overall profit objectives are:

If P _B  ≥ P,

$$ {\mathrm{max}}_{P_B\in \left[P,2P\right];P\in \left[c,1\right]}\kern0.5em \left(1-\alpha \right)\left[2\left(P-c\right)\left(1-P\right)\left({P}_B-P\right)+\left({P}_B-2c\right)\left({\left(1+P-{P}_B\right)}^2-\frac{{\left(2P-{P}_B\right)}^2\kern0.5em }{2}\right)\right]+2\alpha \left(P-c\right)\left(1-P\right)+\alpha \left({P}_B-2c\right)\frac{{\left(2P-{P}_B\right)}^2\kern0.5em }{2} $$

(1)

If P _B  ≤ P,

$$ {\mathrm{max}}_{P_B\in \left[2c,P\right];P\in \left[c,1\right]}\kern0.5em \left(1-\alpha \right)\left({P}_B-2c\right)\left(1-\frac{P_B^2}{2}\right)+2\alpha \left(P-c\right)\left(1-P\right)+\alpha \left({P}_B-2c\right)\left(2\left(1-P\right)\left(P-{P}_B\right)+{P}^2-\frac{P_B^2}{2}\right) $$

(2)

• Step 2: The boundary between the cases P _B  ≥ P and P _B  ≤ P can be expressed analytically by setting P _B  = P in the necessary conditions. Consider the case P _B  ≤ P. The necessary conditions yield:

$$ {\displaystyle \begin{array}{c}\hfill \frac{\partial \uppi}{\partial P}=2\alpha \left(1+c-2P\right)+2\alpha \left({P}_B-2c\right)\left(1+{P}_B-P\right)=0\hfill \\ {}\hfill \frac{\partial \uppi}{\partial {P}_B}=\left(1-\alpha \right)\left(2{cP}_B+1-\frac{3{P}_B^2}{2}\right)+\alpha \left(2\left(1-P\right)\left(P-{P}_B\right)+{P}^2-\frac{P_B^2}{2}\right)+\alpha \left({P}_B-2c\right)\left(2P-2-{P}_B\right)=0\hfill \end{array}} $$

If we set P _B  = P in the first of these, we get P = 1 − c. If we substitute P _B  = P = 1 − c into the second equation, it yields,

$$ \alpha \left(14{c}^2-8c\right)=7{c}^2-10c+1. $$

We can obtain the condition on the marginal cost for this equation to be satisfied with α in [0, 1] by setting α first to 0 and then to 1 and solving the equation. We find that $c\le \left(-1+2\sqrt{2}\right)/7\approx 0.261$ for α < 1, and $c\ge \left(5-3\sqrt{2}\right)/7\approx 0.108$ for α > 0. Within the range [0.108, 0.261], the boundary between the solution cases P _B  < P and P _B  > P is $\alpha =\frac{7{c}^2-10c+1}{14{c}^2-8c}$.

Appendix 2

1.1 Derivation of demand and profit under MB-MB

In this appendix we outline the derivation of MB–MB demand and profit. We assume and subsequently verify that the second stage component and bundle prices are less than their corresponding first stage prices. Then strategic consumers will postpone purchases to the second stage, and face the standard MB choices. This is also the case in the first stage for myopic consumers.

Thus, in the first stage, the α proportion of myopic consumers contributes,

$$ 2\alpha \left({P}_1-c\right)\left(1-{P}_1\right)\left({P}_1-{\phi}_1\right)+\alpha \left(2{P}_1-2c-{\phi}_1\right)\left[{\left(1-{P}_1+{\phi}_1\right)}^2-{\phi}_1^2/2\right], $$

where P ₁ is the component price and ϕ ₁ the bundle discount. The 1 − α proportion of strategic consumers in the second stage contributes

$$ 2\left(1-\alpha \right)\left({P}_2-c\right)\left(1-{P}_2\right)\left({P}_2-{\phi}_2\right)+\left(1-\alpha \right)\left(2{P}_2-2c-{\phi}_2\right)\left[{\left(1-{P}_2+{\phi}_2\right)}^2-{\phi}_2^2/2\right], $$

where P ₂ is the component price and ϕ ₂ the bundle discount.

Now consider the second stage demand calculation for myopic consumers. Those consumers who bought the bundle in the first stage have no subsequent impact, but consumers who bought one component in the first period might buy the other component if it is cheap enough in the second stage. They do so only if P ₂ < P ₁ − ϕ ₁, i.e., it is cheap enough, and their contribution is:

$$ \max\ \left\{0,2\alpha \left({P}_2-c\right)\left({P}_1-{\phi}_1-{P}_2\right)\left(1-{P}_1\right)\right\}. $$

Finally, we consider the segment of myopic consumers who did not buy in the first stage, but buy in the second stage. Two demand patterns occur depending on whether Δ ≡ (P ₂ − ϕ ₂) − (P ₁ − ϕ ₁) is positive or negative. The case of Δ ≤ 0 is shown schematically in the Figure. The other case is analogous.

If Δ ≤ 0 then $2\alpha \left({P}_2-c\right)\left({P}_1-{P}_2\right)\left({P}_2-{\phi}_2\right)+\alpha \left(2{P}_2-2c-{\phi}_2\right)\left[{\left({P}_1-{P}_2+{\phi}_2\right)}^2-\frac{\phi_1^2}{2}-\frac{\phi_1^2}{2}\right].$

If Δ ≥ 0 then $2\alpha \left({P}_2-c\right)\left[\left({P}_1-{P}_2\right)\left({P}_2-{\phi}_2\right)-\frac{\varDelta^2}{2}\left]+\alpha \left(2{P}_2-2c-{\phi}_2\right)\right[\frac{{\left({P}_1-{P}_2-\varDelta +{\phi}_2\right)}^2}{2}-\frac{\phi_2^2}{2}\right].$

Let P ₂ < P ₁ − ϕ ₁ (i.e., we don’t expect much bundle discount in the first stage –we verify the conditions ex post.) Then Δ ≤ 0. We now proceed by backwards induction. The Stage 2 problem is:

$$ {\displaystyle \begin{array}{c}\hfill \underset{\begin{array}{c}\hfill {P}_2,{\phi}_2;\hfill \\ {}\hfill 1\ge {P}_1-{\phi}_1\ge {P}_2\ge c;\hfill \\ {}\hfill 1\ge {P}_2\ge {\phi}_2\ge 0\hfill \end{array}}{\max \limits }2\left(1-\alpha \right)\left({P}_2-c\right)\left(1-{P}_2\right)\left({P}_2-{\phi}_2\right)+\left(1-\alpha \right)\left(2{P}_2-2c-{\phi}_2\right)\left[{\left(1-{P}_2+{\phi}_2\right)}^2-\frac{\phi_2^2}{2}\right]\hfill \\ {}\hfill +2\alpha \left({P}_2-c\right)\left({P}_1-{\phi}_1-{P}_2\right)\left(1-{P}_1\right)+2\alpha \left({P}_2-c\right)\left({P}_1-{P}_2\right)\left({P}_2-{\phi}_2\right)\hfill \\ {}\hfill +\alpha \left(2{P}_2-2c-{\phi}_2\right)\left[{\left({P}_1-{P}_2+{\phi}_2\right)}^2-\frac{\phi_1^2}{2}-\frac{\phi_2^2}{2}\right]\hfill \end{array}} $$

We obtain necessary conditions from this. Then we maximize the first period problem, which is the sum of the profits from the first and second periods, with respect to P ₁ and ϕ ₁, subject to these necessary conditions. The first stage maximization problem is thus,

$$ {\displaystyle \begin{array}{c}\hfill \underset{\begin{array}{c}\hfill {P}_1,{\phi}_1;\hfill \\ {}\hfill 1\ge {P}_1\ge c;\hfill \\ {}\hfill {P}_1\ge {\phi}_1\ge 0\hfill \end{array}}{\max \limits }2\alpha \left({P}_1-c\right)\left(1-{P}_1\right)\left({P}_1-{\phi}_1\right)+\alpha \left(2{P}_1-2c-{\phi}_1\right)\left[{\left(1-{P}_1+{\phi}_1\right)}^2-\frac{\phi_1^2}{2}\right]\hfill \\ {}\hfill +2\left(1-\alpha \right)\left({P}_2-c\right)\left(1-{P}_2\right)\left({P}_2-{\phi}_2\right)\hfill \\ {}\hfill +\left(1-\alpha \right)\left(2{P}_2-2c-\phi 2\right)\left[{\left(1-{P}_2+{\phi}_2\right)}^2-\frac{\phi_2^2}{2}\right]\hfill \\ {}\hfill +2\alpha \left({P}_2-c\right)\left({P}_1-{\phi}_1-{P}_2\right)\left(1-{P}_1\right)+2\alpha \left({P}_2-c\right)\left({P}_1-{P}_2\right)\left({P}_2-{\phi}_2\right)\hfill \\ {}\hfill +\alpha \left(2{P}_2-2c-{\phi}_2\right)\left[{\left({P}_1-{P}_2+{\phi}_2\right)}^2-\frac{\phi_1^2}{2}-\frac{\phi_2^2}{2}\right]\hfill \end{array}} $$

subject to the first order conditions from the previous stage. The Maximize function in Maple was used to obtain the numerical solution of the problem for different values of the parameters.