Digital bundling

Abstract

Increasingly, we see that firms offer many items on information-intensive channels and the Internet. Especially with low-marginal-cost digital goods, bundling may be a beneficial strategy. Different bundles may help customers become more discriminating and maximize profits. However, the marketplace evidence provides mixed observation of bundling rigorously pursued. In this study, we provide a general framework to analyze when and how bundling may be beneficial. We compare and contrast the firm and customer characteristics on bundling strategy. We analyze when individual sales can be beneficial. We find that when costs do not increase relative to the bundle valuation, firms find it beneficial to limit the number of bundles offered in the market. A low (zero) marginal cost firm benefits from offering just one big bundle. Such a firm obtains a higher profit compared to a firm that offers many bundles. With high marginal costs, the number of bundles increases with increasing customer valuation and/or heterogeneity. When a firm offers all the bundles, prices and profit increase as customer heterogeneity and valuation increase. When the number of offered items is high, individual sale (unbundling) may be preferred over bundling and bundling becomes an inferior strategy. Interestingly, a firm may benefit from unbundling the items when customers have higher valuations.

Appendix

Table 2.

Table 2 Parameter definitions for the main model

Proposition 1

Given a maximum number of items in a bundle (B), relative values (k _b ) and low marginal costs (c _b ), if the increase in cost relative to the bundle valuation ck _b $\left( { = \frac{{c_{b} - c_{b - 1} }}{{k_{b} - k_{b - 1} }}} \right)$is in the range (r, c ₁ /k ₁ ) and increasing in b for b = 2 to B, and if the increase in the relative value (k _b ) is decreasing as bundle size increases, the optimal incentive compatible solution calls for the firm to offer all the bundles with prices given as p ^* _b  = (rk _b  + c _b )/2 for b = 1 to B. Otherwise, if there exists a bundle b such that ck _b does not increase, the firm does not offer that bundle as part of its optimal solution.

Proof

First, we solve the profit maximization problem. Then, we show that it provides an incentive compatible solution. The profit function is given as

$$ \Uppi = \frac{{p_{n} - c_{n} }}{r}\left( {r - \frac{{p_{n} - p_{n - 1} }}{{k_{n} - k_{n - 1} }}} \right) + \frac{{p_{n - 1} - c_{n - 1} }}{r}\left( {\frac{{p_{n} - p_{n - 1} }}{{k_{n} - k_{n - 1} }} - \frac{{p_{n - 1} - p_{n - 2} }}{{k_{n - 1} - k_{n - 2} }}} \right) + \cdots + \frac{{p_{1} - c_{1} }}{r}\left( {\frac{{p_{2} - p_{1} }}{{k_{2} - k_{1} }} - \frac{{p_{1} }}{{k_{1} }}} \right). $$

Note that the profit function is concave with respect to individual prices and the second derivative is negative. By taking the derivative of the objective function and equating it to zero, we find that the optimal solutions satisfy:

$$ \begin{aligned} p_{N} & = \frac{{c_{N} }}{2} + \frac{{r\left( {k_{N} - k_{N - 1} } \right) + 2p_{N - 1} - c_{N - 1} }}{2} \\ p_{N - j} & = \frac{{c_{N - j} }}{2} + \frac{{\left( {2p_{N - j + 1} - c_{N - j + 1} } \right)\left( {k_{N - j} - k_{N - j - 1} } \right) + \left( {2p_{N - 1} - c_{N - j - 1} } \right)\left( {k_{N - j + 1} - k_{N - j} } \right)}}{{2\left( {k_{N - j + 1} - k_{N - j - 1} } \right)}}, \quad j = 1{\text{ to }}N - 2 \\ p_{1} & = \frac{{c_{1} }}{2} + \frac{{\left( {2p_{2} - c_{2} } \right)k_{1} }}{{2k_{2} }} \\ \end{aligned} $$

or, the solution of the following:

$$ \left[ \begin{array}{l} p_{N} \hfill \\ p_{N - 1} \hfill \\ p_{N - 2} \hfill \\ . \hfill \\ . \hfill \\ . \hfill \\ p_{1} \hfill \\ \end{array} \right] = \left[ {\begin{array}{lllllll} 1 & { - 1} & 0 & 0 & \ldots & 0 & 0 \\ {\frac{{ - k_{N - 1} + k_{N - 2} }}{{k_{N} - k_{N - 2} }}} & 1 & {\frac{{ - k_{N} + k_{N - 1} }}{{k_{N} - k_{N - 2} }}} & 0 & \ldots & 0 & 0 \\ 0 & {\frac{{ - k_{N - 2} + k_{N - 3} }}{{k_{N - 1} - k_{N - 3} }}} & 1 & {\frac{{ - k_{N - 1} + k_{N - 2} }}{{k_{N - 1} - k_{N - 3} }}} & \ldots & 0 & 0 \\ . & {} & {} & {} & {} & {} & {} \\ . & {} & {} & {} & {} & {} & {} \\ . & {} & {} & {} & {} & {} & {} \\ 0 & 0 & 0 & 0 & \ldots & {\frac{{ - k_{1} }}{{k_{2} }}} &1 \\ \end{array} } \right]^{ - 1} \left[ {\begin{array}{*{20}l} \begin{array}{l} \frac{{c_{N} }}{2} + \frac{{r\left( {k_{N} - k_{N - 1} } \right) - c_{N - 1} }}{2} \hfill\\ \frac{{c_{N - 1} }}{2} - \frac{{c_{N} \left( {k_{N - 1} - k_{N - 2} } \right) + c_{N - 2} \left( {k_{N} - k_{N - 1} } \right)}}{{2\left( {k_{N} - k_{N - 2} } \right)}} \hfill \\ \end{array} \\ {\frac{{c_{N - 2} }}{2} - \frac{{c_{N - 1} \left( {k_{N - 2} - k_{N - 3} } \right) + c_{N - 3} \left( {k_{N - 1} - k_{N - 2} } \right)}}{{2\left( {k_{N - 1} - k_{N - 3} } \right)}}} \\ . \\ . \\ . \\ {\frac{{c_{1} }}{2} - \frac{{c_{2} k_{1} }}{{2k_{2} }}} \\ {} \\ \end{array} } \right] $$

which gives:

$$ \left[ \begin{array}{l} p_{N} \hfill \\ p_{N - 1} \hfill \\ p_{N - 2} \hfill \\ . \hfill \\ . \hfill \\ . \hfill \\ p_{1} \hfill \\ \end{array} \right] = \left[ \begin{array}{l} \frac{{rk_{N} + c_{N} }}{2} \hfill \\ \frac{{rk_{N - 1} + c_{N - 1} }}{2} \hfill \\ \frac{{rk_{N - 2} + c_{N - 2} }}{2} \hfill \\ . \hfill \\ . \hfill \\ . \hfill \\ \frac{{rk_{1} + c_{1} }}{2} \hfill \\ \end{array} \right] $$

or, at the optimal $p_{i} = \frac{{rk_{i} + c_{i} }}{2}$ should hold. Then, the optimal profit is given by:

$$ \left( {\frac{{k_{n} - c_{n} /r}}{4}} \right)\left( {r - \frac{{c_{n} - c_{n - 1} }}{{k_{n} - k_{n - 1} }}} \right) + \left( {\frac{{k_{n - 1} - c_{n - 1} /r}}{4}} \right)\left( {\frac{{c_{n} - c_{n - 1} }}{{k_{n} - k_{n - 1} }} - \frac{{c_{n - 1} - c_{n - 2} }}{{k_{n - 1} - k_{n - 2} }}} \right) + \cdots + \left( {\frac{{k_{1} - c_{1} /r}}{4}} \right)\left( {\frac{{c_{2} - c_{1} }}{{k_{2} - k_{1} }} - \frac{{c_{1} }}{{k_{1} }}} \right) - f_{n}. $$

Note that we assume $h_{i} = \frac{{p_{i + 1} - p_{i} }}{{k_{i + 1} - k_{i} }}$ is increasing in i. At optimum, $h_{i} = \frac{{k_{i + 1} + c_{i + 1} - k_{i} - c_{i} }}{{2\left( {k_{i + 1} - k_{i} } \right)}} = \frac{1}{2}\left( {1 + \frac{{c_{i + 1} - c_{i} }}{{k_{i + 1} - k_{i} }}} \right)$ holds. For h _i to be increasing in i, $\frac{{c_{i + 1} - c_{i} }}{{k_{i + 1} - k_{i} }}$ should be increasing in i.

We also check to see whether the customer surplus constraints are satisfied. Since customers should at least have a zero surplus, k ₁ h − p ₁ ≥ 0 should hold. But, $h \ge \frac{{p_{1} }}{{k_{1} }}$ holds since $h_{0} = \frac{{p_{1} }}{{k_{1} }} = \frac{{k_{1} + c_{1} }}{{2k_{1} }}$ and $h_{i} = \frac{1}{2}\left( {1 + \frac{{c_{i + 1} - c_{i} }}{{k_{i + 1} - k_{i} }}} \right) \ge h_{0} = \frac{{k_{1} + c_{1} }}{{2k_{1} }}$ as long as $\frac{{c_{i + 1} - c_{i} }}{{k_{i + 1} - k_{i} }} \ge \frac{{c_{1} }}{{k_{1} }}$ for i = 1 to n. But since, $\frac{{c_{i + 1} - c_{i} }}{{k_{i + 1} - k_{i} }}$ is increasing in i and providing that it has a minimum value of $\frac{{c_{1} }}{{k_{1} }},$ minimum surplus that the customers receive is greater than or equal to zero.

Next, we show that the solution provides an incentive compatible solution. In specific, a customer buys the bundle that maximizes his surplus. However, given a surplus maximizing bundle, the surplus that the customer obtains from this bundle should be always greater than the sum of the surpluses the customer may obtain by buying any other combination of alternatives offered. Let us assume that the customer buys bundle i. But, there may be many different ways the customer may obtain i items in the bundle. Let S(i) represent the set of all these possibilities, and S(i, t) represent the set of each of these possibilities where t = 1,…, T(i) and t is the specific set’s index number. Let S(i, t, j) represent the jth item in the set S(i, t) for j = 1,…, J(i, t). J(i, t) represents the necessary number of purchases the customer needs to make. As an example take a bundle with three items for a customer with a quality valuation at $\hat{h}.$ Assume that the monopolist offers one product and/or a bundle with two items as other possible alternatives for the customer. In this case S(3) = {{3}, {1, 2}, {1, 1, 1}}. That is, the customer can buy a three item bundle (make one purchase only), or buy a one item plus a two item bundle (make two purchases), or buy three items separately (make three purchases). Hence, J(3) can be 1, 2 or 3 depending on the specific alternative given in S(3). Consider the second alternative, S(3, 2), or {1, 2} where the customer makes two choices and J(3, 2) = 2. When the customer makes the first purchase, the surplus of that purchase is k _S(3,2,1) $\hat{h}$ − p _S(3,2,1) which equals to k ₁ $\hat{h}$ − p _1. The two item bundle purchase results in k ₂ $\hat{h}$ − p ₂ where k ₂ equals k ₃ − k ₁. Although the customer is paying the price for a two-item bundle, the quality valuation of buying the additional two-item bundle on top of the already purchased first item will be the difference of the quality valuations of a three item bundle minus the one already purchased item. Hence k ₂ captures this incremental quality valuation. In general, we expect $k_{i} h_{i} - p_{i} \ge \sum\nolimits_{j = 1}^{{J\left( {i,t} \right)}} {\left( {\Updelta_{{S\left( {i,t,j} \right)}} h_{i} - p_{{S\left( {i,t,j} \right)}} } \right)} ,\quad t = 1, \ldots ,T\left( i \right),\quad i = 1, \ldots ,n ,$ where $\Updelta_{{S\left( {i,t,1} \right)}} = k_{{S\left( {i,t,1} \right)}} , \Updelta_{{S\left( {i,t,j} \right)}} = k_{{\sum\nolimits_{{j^{'} = 1}}^{j} {S\left( {i,t,j^{'} } \right)} }} - k_{{\sum\nolimits_{{j^{'} = 1}}^{j-1} {S\left( {i,t,j^{'} } \right)} }} ,\quad j = 2, \ldots ,J(i,t),\quad t = 1, \ldots ,T\left( i \right),\quad i = 1, \ldots ,n.$The above equations can be simplified further. Since $k_{i} = \sum\nolimits_{j = 1}^{{J\left( {i,t} \right)}} {\Updelta_{{S\left( {i,t,j} \right)}} } ,$ as long as $p_{i} < \sum\nolimits_{j = 1}^{{J\left( {i,t} \right)}} {p_{{S\left( {i,t,j} \right)}} } ,$ we expect the customer to buy only the surplus maximizing offer. Then, we can formulate the problems. First, let us define the problem P as follows:

Problem P

$$ \begin{array}{l} \mathop {\max }\limits_{{\left\{ {p_{i} } \right\}_{i = 1}^{n} }} \quad \Uppi = \sum\limits_{i = 1}^{n} {\left( {p_{i} - c_{i} } \right)d_{i} } - f_{n} \hfill \\ {\text{subject}}\,{\text{to}}\,h_{0} = \frac{{p_{1} }}{{k_{1} }},\quad h_{n} = 1 \hfill \\ d_{i} = h_{i} - h_{i - 1} ,\quad i = 1, \ldots ,n \hfill \\ k_{i} h_{i} - p_{i} \ge 0,\quad i = 1, \ldots ,n \hfill \\ k_{i} h_{i} - p_{i} \ge k_{i + 1} h_{i} - p_{i + 1} ,\quad i = 1, \ldots ,n \hfill \\ \end{array} $$

Problem P′ is the same as Problem P with the additional constraint that ensures the incentive compatibility. That is $p_{i} < \sum\nolimits_{j = 1}^{{J\left( {i,t} \right)}} {p_{{S\left( {i,t,j} \right)}} } ,\quad t = 1, \ldots ,T\left( i \right),\quad i = 1, \ldots ,n.$

Proposition A1

Assume $\frac{{c_{i + 1} - c_{i} }}{{k_{i + 1} - k_{i} }}$ is increasing, has a minimum value of $\frac{{c_{1} }}{{k_{1} }}$ for i for i = 1 to n − 1 and the increase in k _i  + c _i is decreasing (a concave function). Than, the solution of the Problem P also solves Problem P′.

Based on the optimal solution, we check the conditions when $p_{i} < \sum\nolimits_{j = 1}^{{J\left( {i,t} \right)}} {p_{{S\left( {i,t,j} \right)}} }$ constraint holds. Note that, the constraint $p_{i} < \sum\nolimits_{j = 1}^{{J\left( {i,t} \right)}} {p_{{S\left( {i,t,j} \right)}} }$ with the optimal values inserted, is equivalent to $k_{i} + c_{i} < \sum\nolimits_{j = 1}^{{J\left( {i,t} \right)}} {\left( {k_{{S\left( {i,t,j} \right)}} + k_{{S\left( {i,t,j} \right)}} } \right)} .$ This constraint is not binding, if the increase in k _i decreasing given low marginal cost (c _i ).

Next, we show that if there exists some i such that $\frac{{c_{i + 1} - c_{i} }}{{k_{i + 1} - k_{i} }}$ is nonincreasing and i ≤ n − 1, the firm does not offer that specific quantity bundle as part of its optimal solution. Assuming that the firm offers the quantity bundle i while $\frac{{c_{i + 1} - c_{i} }}{{k_{i + 1} - k_{i} }}$ is non increasing, that is, $\frac{{c_{i + 1} - c_{i} }}{{k_{i + 1} - k_{i} }} - \frac{{c_{i} - c_{i - 1} }}{{k_{i} - k_{i - 1} }} \le 0$ holds. For this to be part of the optimal solution, there should be a positive demand for bundle i. However, the incentive compatibility constraints, k _i−1 h _i−1 – p _i−1 ≥ k _i h _i−1 – p _i and k _i h _i  – p _i  ≥ k _i+1 h _i  – p _i+1 should also hold. Then, since the demand for bundle i, or d _i equals h _i − h _i−1, we find $d_{i} = \frac{1}{2}\left( {\frac{{c_{i + 1} - c_{i} }}{{k_{i + 1} - k_{i} }} - \frac{{c_{i} - c_{i - 1} }}{{k_{i} - k_{i - 1} }}} \right) \le 0,$ a contradiction.

Proposition 2 (Significant Bundling)

The firm finds it more profitable to design the bundles such that $k_{b} - k_{b - 1} < k_{b - 1} - k_{b - 2}$ when the items share similar low marginal costs.

Proof

Assume that different bundling schemes can be ordered based on customer valuations such that k _n  > k _n−1 > ⋯ > k _i  > ⋯ > k ₂ > k ₁ holds. In this case, bundle i denotes the significant bundle that corresponds to the ith lowest valued bundle for customers among different significant bundling options. This maximizes the number of bundles at the optimum. The profit is maximized when the firm can better discriminate its customers.

Proposition 3

The firm finds it beneficial to offer the maximum number of bundles/items when

$$ c < \frac{{2rk\left( {1 - \alpha } \right)}}{{2^{{B^{\prime}}} }}, $$

and B* = B′. Otherwise, when $c > \frac{{2rk\left( {1 - \alpha } \right)}}{{2^{{B^{\prime}}} }},$ the firm limits the bundled items and B* < B^′. With such high costs, as the heterogeneity and valuation increase, the optimal number of bundles increases. As the steepness of valuation increase between different bundles decreases (α) or cost (c) increases, the number of bundles decreases.

Proof

Given $k_{i} = \sum\nolimits_{i' = 1}^{i} {\frac{k}{{2^{i - 1} }}} = k\left( {2 - 1/2^{i - 1} } \right)$ marginal costs c, we find that $\frac{{c_{i} - c_{i - 1} }}{{k_{i} - k_{i - 1} }} = \frac{c}{{k\left( {2 - 1/2^{i - 1} } \right) - k\left( {2 - 1/2^{i - 2} } \right)}} = 2^{i - 1} \frac{c}{{k\left( {1 - \alpha } \right)}}$ which is increasing in i. The profit for different bundles is given by $\frac{{r\left( {2k^{\prime}\left( {1 - 2^{ - i + 1} } \right) + \alpha - ic^{\prime}} \right)2^{i - 1} c^{\prime}}}{{4k^{\prime}}},$ i = 2 to B − 1, B > 3. Offering bundle i, i = 2 to B − 1 is profitable when 2 k′(1 − 2⁻ⁱ⁺¹) + α − ic′ > 0. This requires c to be less than $\frac{{r\left( {2k\left( {1 - \alpha } \right)\left( {1 - 2^{ - i + 1} } \right) + \alpha } \right)}}{i}$ for i = 2 to B − 1. Offering first item is profitable when $\frac{{r\left( {k^{\prime} + \alpha - c^{\prime}} \right)\left( {k^{\prime} + 2\alpha - c^{\prime}} \right)c^{\prime}}}{{4k^{\prime}\left( {k^{\prime} + \alpha } \right)}}$ or (k′ + α − c′)(k′ + 2α − c′) > 0 this requires c to be less than r(k(1 − α) + α). Offering the last item is profitable when $\frac{{r\left( {2k^{\prime}\left( {1 - 2^{ - n} } \right) + \alpha - nc^{\prime}} \right)\left( {k^{\prime} - 2^{n - 1} c^{\prime}} \right)}}{{4k^{\prime}}}$ or (2 k′(1 − 2⁻ⁿ) + α − nc′)(k′ − 2ⁿ⁻¹ c′) > 0. This requires c to be less than $\frac{{2rk\left( {1 - \alpha } \right)}}{{2^{B} }}.$ Note that when c is less than $\frac{{2rk\left( {1 - \alpha } \right)}}{{2^{B} }},$ all the above requirements are satisfied.

Solution (bundling). For the last bundle to be offered, we need $r - \frac{{c_{n} - c_{n - 1} }}{{k_{n} - k_{n - 1} }} = r\left( {1 - 2^{n - 1} \frac{c/r}{k}} \right) > 0.$ All the bundles are offered with prices $p_{i}^{*} = \frac{{2\left( {1 - 2^{ - i} } \right)k^{\prime} + \alpha + ic}}{2}.$ The profit equals $\frac{{r\left( {2k^{\prime}\left( {1 - 2^{ - n} } \right) + \alpha - nc^{\prime}} \right)\left( {k^{\prime} - 2^{n - 1} c^{\prime}} \right)}}{{4k^{\prime}}} + \sum\nolimits_{i = 2}^{n - 1} {\frac{{r\left( {2k^{\prime}\left( {1 - 2^{ - i + 1} } \right) + \alpha - ic^{\prime}} \right)2^{i - 1} c^{\prime}}}{{4k^{\prime}}}} + \frac{{r\left( {k^{\prime} + \alpha - c^{\prime}} \right)\left( {k^{\prime} + 2\alpha - c^{\prime}} \right)c^{\prime}}}{{4k^{\prime}\left( {k^{\prime} + \alpha } \right)}}.$ This results in $\frac{r}{{4k^{\prime}}}\left( {2k^{'2} \left( {1 - 2^{{ - B^{*} }} } \right) + 2^{{B^{*} }} c^{'2} - c^{\prime}\left( {k^{\prime}\left( {1 + 2B^{\prime}} \right) + 2\alpha } \right) + \alpha k^{\prime} + \frac{{\left( {k^{\prime} + \alpha - c^{\prime}} \right)\left( {k^{\prime} + 2\alpha } \right)c^{\prime}}}{{\left( {k^{\prime} + \alpha } \right)}}} \right),$ where k′ = k(1 − α) and c′ = c/r.

Lemma 1

When a firm offers all the bundles, prices and profit increase as heterogeneity, valuation increase. As the steepness of valuation increase between different bundles decreases (α) or cost (c) increases, the profit and prices decrease.

Proof

We take the derivative of the profit equation given above with respect to the parameters. The derivative with respect to heterogeneity equals:

$\frac{{k^{\prime}}}{2}\left( {1 - 2^{{ - B^{*} }} } \right) - 2^{{B^{*} }} \frac{{c^{2} }}{{4r^{2} k^{\prime}}} + \frac{\alpha }{4} + \frac{{\left( {k^{\prime} + 2\alpha } \right)c^{2} }}{{\left( {k^{\prime} + \alpha } \right)4r^{2} k^{\prime}}},$ which is always higher than $\frac{{k^{\prime}}}{2}\left( {1 - 2^{{ - B^{*} }} } \right) - \frac{{k^{\prime}}}{{2^{B} }} + \frac{\alpha }{4} + \frac{{\left( {k^{\prime} + 2\alpha } \right)c^{2} }}{{\left( {k^{\prime} + \alpha } \right)4r^{2} k^{\prime}}}$ since maximum marginal cost should be less than $\frac{{2rk\left( {1 - \alpha } \right)}}{{2^{B} }}.$ But, $\frac{{k^{\prime}}}{2}\left( {1 - 2^{{ - B^{*} }} } \right) - \frac{{k^{\prime}}}{{2^{B} }} + \frac{\alpha }{4} + \frac{{\left( {k^{\prime} + 2\alpha } \right)c^{2} }}{{\left( {k^{\prime} + \alpha } \right)4r^{2} k^{\prime}}} > 0.$ Derivative with respect to the cost is proportional to $2^{{B^{\prime} + 1}} c^{\prime} - \left( {k^{\prime}\left( {1 + 2B^{\prime}} \right) + 2\alpha } \right) + \frac{{\left( {k^{\prime} + \alpha - 2c^{\prime}} \right)\left( {k^{\prime} + 2\alpha } \right)}}{{\left( {k^{\prime} + \alpha } \right)}}$ or $2^{{B^{\prime} + 1}} c^{\prime} - 2k^{\prime}B^{\prime} - \frac{{2c^{\prime}\left( {k^{\prime} + 2\alpha } \right)}}{{\left( {k^{\prime} + \alpha } \right)}}.$ Note that when the cost is zero, it is negative. When it is positive, the maximum it can reach is $2k^{\prime}\left( {2 - B^{\prime}} \right) - \frac{{2c^{\prime}\left( {k^{\prime} + 2\alpha } \right)}}{{\left( {k^{\prime} + \alpha } \right)}},$ which is negative. Similarly, we find the respective derivatives.

Proposition 4

Azero marginal cost firm that offers just one big bundle earns a higher profit compared to a slightly positive marginal cost firm that offers all possible bundles.

Proof

It follows from the discussions given above, see Lemma 1 and Proposition 1.

Solution (individual sales). A firm that sells individual items only charges $p = c\left( {W\left( {\frac{{ce^{2} }}{{2k\left( {1 - \alpha } \right)rB^{\prime}\ln 2}}} \right)} \right)^{ - 1} ,$ the fraction of items purchased by customer h equals $\frac{1}{{B^{\prime}\ln 2}}\left( {\ln h/r - W\left( {\frac{{ce^{2} }}{{2k\left( {1 - \alpha } \right)rB^{\prime}\ln 2}}} \right)} \right),$ the firm profits $\left( {c\left( {W\left( {\frac{{ce^{2} }}{{2k\left( {1 - \alpha } \right)rB^{\prime}\ln 2}}} \right)} \right)^{ - 1} - c} \right)\left( {1 - W\left( {\frac{{ce^{2} }}{{2k\left( {1 - \alpha } \right)rB^{\prime}\ln 2}}} \right)} \right)/\ln 2,$

Proof

Customers maximize their surplus and at the optimum $\frac{d}{{db^{\prime}}}\left( {h\left( {\alpha + 2k\left( {1 - \alpha } \right)\left( {1 - 2^{{ - b^{\prime}B^{\prime}}} } \right)} \right)} \right) = p.$ We find $b^{\prime} = \frac{1}{{B^{\prime}}}\left( {1 + \log_{2} \frac{{hk\left( {1 - \alpha } \right)B^{\prime}\ln 2}}{p}} \right).$ The firm maximizes $\int\limits_{0}^{r} {\frac{1}{r}\left( {p - c} \right)b^{\prime}B^{\prime}dh} ,$ which is equivalent to maximizing $\left( {p - c} \right)\left( {1 - \frac{1}{\ln 2} + \log_{2} \frac{{rk\left( {1 - \alpha } \right)B^{\prime}\ln 2}}{p}} \right).$ Note that the second derivative equals $- \frac{p + c}{{p^{2} \ln 2}} < 0.$ By equating the first derivative to zero, we find the optimal price that the firm charges $p = c\left( {W\left( {\frac{{ce^{2} }}{{2k\left( {1 - \alpha } \right)rB^{\prime}\ln 2}}} \right)} \right)^{ - 1} .$ Since $\ln \left( {W\left( x \right)} \right) = \ln x - W\left( x \right),$ we find $b^{\prime} = \frac{1}{{B^{\prime}\ln 2}}\left( {\ln h/r - W\left( {\frac{{ce^{2} }}{{2k\left( {1 - \alpha } \right)rB^{\prime}\ln 2}}} \right)} \right).$ The profit equals $\left( {c\left( {W\left( {\frac{{ce^{2} }}{{2k\left( {1 - \alpha } \right)rB^{\prime}\ln 2}}} \right)} \right)^{ - 1} - c} \right)\left( {1 - W\left( {\frac{{ce^{2} }}{{2k\left( {1 - \alpha } \right)rB^{\prime}\ln 2}}} \right)} \right)/\ln 2.$ When marginal cost is low, based on Taylor series expansion, we can approximate $W\left( {\frac{{ce^{2} }}{{2k\left( {1 - \alpha } \right)rB^{\prime}\ln 2}}} \right) \approx \frac{{ce^{2} }}{{2k\left( {1 - \alpha } \right)rB^{\prime}\ln 2}}$ providing the term $\left( {\frac{{ce^{2} }}{{2k\left( {1 - \alpha } \right)rB^{\prime}\ln 2}}} \right)^{i} ,$ for i ≥ 2 is negligible. Then, profit and price equal to $\frac{{\left( {2k\left( {1 - \alpha } \right)rB^{\prime}e^{ - 1} \ln 2} \right)^{2} + c^{2} e^{2} }}{{2k\left( {1 - \alpha } \right)rB^{\prime}\ln^{2} 2}} - \frac{2c}{\ln 2}$ and $\frac{{2k\left( {1 - \alpha } \right)rB^{\prime}\ln 2}}{{e^{2} }},$ respectively. As c goes to zero, the profit equals $\frac{{2k\left( {1 - \alpha } \right)rB^{\prime}}}{{e^{2} }}.$

Lemma 2

A firm with low marginal costs, increases its price and profit in proportion to the maximum items it has in its product portfolio when engaging in direct selling.

Proof

Given profit of $\frac{{2k\left( {1 - \alpha } \right)rB^{\prime}}}{{e^{2} }}$ and a price of $\frac{{2k\left( {1 - \alpha } \right)rB^{\prime}\ln 2}}{{e^{2} }},$ both terms increase in proportion to B′ suggesting that the firm should increase its price with individual sales and increasing product portfolio.

Proposition 5

When marginal cost is low, individual sale is always preferred over quantity bundling when customers have a high valuation for an individual item while additional items increase the valuation at a steeper rate. Otherwise, with low valuation, bundling provides the highest profits.

Proof

We see this by comparing the respective profits. Note that individual sales profit as marginal cost approaches to zero is $\frac{{2k\left( {1 - \alpha } \right)rB^{\prime}}}{{e^{2} }}.$ We see that $\frac{{2k\left( {1 - \alpha } \right)rB^{\prime}}}{{e^{2} }} > \frac{r}{4}\left( {2k\left( {1 - \alpha } \right) + \alpha } \right) > \frac{r}{4}\left( {2k\left( {1 - \alpha } \right)\left( {1 - 2^{{ - B^{\prime}}} } \right) + \alpha } \right),$ when $B^{\prime} > \frac{{e^{2} }}{8}\left( {2 + \frac{\alpha }{{k\left( {1 - \alpha } \right)}}} \right).$