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Some Dynamic Market Models
Jan A. Audestad
Norwegian University of Science and Technology, Gjøvik University College
ABSTRACT
In this text, we study the behavior of markets using models expressible as ordinary differential equations. The markets studied are those where each customer buys only one copy of the good, for example, subscription of smartphone service, journals and newspapers, and goods such as books, music and games. One of the underlying models is the Bass model for market evolution. This model contains two types of customers: innovators who buy the good independently of whom else have bought the good, and imitators who buy the good only if other customers have bought the good. Section 2 investigates the dynamics of markets containing innovators only. Section 3 investigates markets with imitators. The main goal is to determine the temporal market evolution for various types of feedback from the market. One particularly important result is to determine the latency time, in this text, defined as the time it takes the market to reach 10% of all potential customers. This is a strategically important parameter since long latency time may result in premature closedown of services that eventually will become very lucrative. Sections 2 and 3 study the evolution of the entire market without taking into consideration the effects of competition. Section 4 is about markets with several suppliers where the purpose is to study the evolution of the market toward stable equilibria for markets with and without churning. For markets with churning it is shown that for some churning functions the final state will consist of only one surviving supplier. Section 5 presents models for interactive games. The effects of no feedback or positive feedback from the market for entering the game or quitting the game are studied. Altogether, six cases are discussed. The model consists of three states representing potential buyers (B), active players (P) and quitters (Q), respectively. The models consist of three coupled first order differential equations, one for each of the three states. One of the models is identical to the SIR (susceptible-infected-recovered) model of epidemiology (Section 5.2.3). The model in Section 5.2.2 allows a closed form analytic solution. The models in Sections 5.2.5 and 5.2.6 requires numerical solution of an integral, while the model in Sections 5.3.4 requires numerical integration of two coupled first order differential equations. The model in Section 5.2.7 requires numerical integration of a single first order differential equation. Section 5.4 studies the effect of complementary games.
1. INTRODUCTION
1.1 Market dynamics
In a dynamic market model, temporal evolution of the market is studied. We are then, for example, concerned with how the number of subscribes to a service or players of a game evolves as a function of time. The dynamic behavior of the market can be described using analytic tools or simulation methods based on, for example, system dynamics (see, for example, [Ste]). System dynamics is a useful tool in cases where the market behavior is so complex that it is impractical or not even meaningful to use simple analytic tools. As with all simulation methods, system dynamics will not provide us with general solutions and thus general insight into the problem. The use of interacting agents is another simulation method that may be used to study the evolution of the market, see [Tes]. However, in this text, we shall only look at some simple market models where analytic solutions exist, and from which, important conclusions can be drawn. We are only concerned with the dynamics of the market and not how factors such as price, availability, and ability to buy the good influence the desire to buy the good or subscribe to a service. The participants in the market are modeled as a uniform group where everyone has the same desire to buy the good or subscribe to the service. This is the same probabilistic method that is used in several fields in science and technology, for example, radioactive decay, epidemiology, teletraffic engineering, and population dynamics, just to mention a few. The markets we are considering are markets where each customer buys at most one copy of each good. Typical examples of such markets are:
subscription for telecommunications services (mobile phone, internet) newspapers, journals, and magazines, books, music, and films, games, subscription for energy, club memberships. Other markets with a similar behavior are insurance and banking. In these cases, the customer may have similar contracts with more than one supplier, or have several contracts with the same supplier. The models are also applicable to the market for certain types of commodities such as refrigerators, furniture, and automobiles. These commodities have long working lives. Let the number of potential users of a service (or game) be 𝑢(𝑡) at time 𝑡. The demand for the service during a short time interval 𝛥𝑡 is then the number of users 𝛥𝑢
buying the service during that interval; that is, the demand per unit time is defined as ∆𝑢∆𝑡 or,
in the continuous limit, the demand per unit time becomes the time derivative 𝑑𝑢𝑑𝑡 , or 𝑢̇ for
short.^1 The expressions for the number of customers having bought the good (𝑢) and the time derivative of this expression (𝑢̇) will be computed for each model. The latter is then the
(^1) Following the normal convention in physics, we are using the terminology 𝑢̇, 𝑢̈, 𝑢⃛ ... (the dot derivative or
Newton’s derivative) for the time derivatives to distinguishing them from derivatives with respect to other variables. This enhances the readability of the formulas.
Normalization means that 𝑢 = 𝑈 𝑁⁄ is the relative number of users, where 𝑈 is the absolute number of users and 𝑁 is the total population of potential users. The unnormalized Bass equation can then be written
𝑈̇ = 𝑎(𝑡)(𝑁 − 𝑈(𝑡)) +
Section 2 is about markets consisting of innovators only. These are markets where market feedback does not exist or is so weak that it may be neglected. The mobile phone market and the internet are examples of such markets. The basic model is
𝑢̇(𝑡) = 𝑎(𝑡)(1 − 𝑢(𝑡)), In Sections 2.4 to 2.6, the model is extended to include customers hesitating to buy the service and customers terminating the service. Section 3 contains several models where there are different types of feedback from the market. The general model is
𝑢̇(𝑡) = 𝑎(𝑡)(1 − 𝑢(𝑡)) + 𝛾(𝑡)(1 − 𝑢(𝑡))𝐹(𝑢(𝑡)),
where 𝐹(𝑢(𝑡)) is a general feedback term. Section 3 describes the full Bass model as well as several models without innovators, that is, models with market equations of the form
𝑢̇(𝑡) = 𝛾(𝑡)(1 − 𝑢(𝑡))𝐹(𝑢(𝑡)),
The effect of various forms of the feedback function 𝐹(𝑢(𝑡)) is studied. This includes weak, medium and strong feedbacks, and the type of feedback that may be expected in markets dominated by trends. Examples of telecommunications markets without innovators are SMS, email, facsimile and Facebook. These services are meaningless to a person if there is no one else with whom he or she may communicate. In other words, the desire to use the service depends on the number of people already using it. Section 4 contains models where several suppliers compete for market shares. The Bass model (with variations) is used as the basic market model. In addition, the effect of churning is investigated, in particular to determine the long term market shares. One important question is to determine the final equilibrium state of the market. Churning may result in rather stable market shared between several suppliers such as the market for mobile communication, or in winner-take-all markets where one supplier will end up as a monopoly. Examples include markets where there are competing standards such as VHS and Betamax offering essentially the same service. Games require more complex models. They are also, to some extent, based on the Bass model but are also related to epidemiological models (see for example [Hop], [Mur]). These models are considered in Section 5.
1.3 Equilibrium
One important definition is that of equilibrium points. The definition is particularly simple if the function 𝑓(𝑢; 𝑡) does not depend explicitly on time, that is, 𝑓(𝑢; 𝑡) = 𝑓(𝑢). If the function depends explicitly on time, the equilibrium points may not be found by the method explained next. Examples 2 and 3 in Section 2.2 show two cases where the method fails.
The system is in equilibrium at the point 𝑢̃ if 𝑓(𝑢̃) = 0. An equilibrium point is also a fixed point.^2 It is well-known from physics that we must have 𝑢̇ = 0 and 𝑢̈ = 0 simultaneously at the equilibrium point; that is, both the velocity and the acceleration vanish at the equilibrium point. The latter condition is equivalent to the requirement that no net external force acts upon the system. Observe that if 𝑓(𝑢) is an everywhere smooth function of 𝑢, 3 then all the time derivatives of 𝑢 vanish at the equilibrium point 𝑢̃. In this case, the fixed points are equilibrium points. For the 𝑘-dimentional case, the equilibrium points satisfy 𝒖̇ = (𝑢̇ 1 , 𝑢̇ 2 , … , 𝑢̇𝑘) = (0,0, … ,0) and 𝒖̈ = (𝑢̈ 1 , 𝑢̈ 2 … , 𝑢̈𝑘) = (0,0, … ,0). The equilibrium points may be stable or unstable. A stable equilibrium point is a point where the system will return to the equilibrium point after a small perturbation away from the point. Stable equilibrium points are called attractors. An unstable equilibrium point is a point where the system is at rest if no external forces act upon it. However, any perturbation (or small force) will cause the system to move away from the equilibrium point. Unstable equilibrium points are called repellers. There are also equilibrium points where some perturbations will cause the system to move back to equilibrium, while other perturbations will cause the system to move away from equilibrium. Such points are called saddle points. The saddle points are unstable. In the 𝑘-dimentional case, all points on a hypersurface of dimension less than 𝑘 may be equilibrium points, for example, if 𝑘 = 2, all points on a line may be equilibrium points. We will encounter examples of such equilibrium points later.
1.4 Churning
If there are several suppliers of a service or good, churning may take place. Churning implies that a customer changes from one supplier to another at a certain time. Churning may then take place back and forth between different suppliers so long as the subscription lasts. The customer will still have just one subscription for the service (for example, mobile services) but obtaining the service from different suppliers at different times. The churning function for supplier 𝑖 can be written in the form 𝐶𝑖 = ∑ 𝑥𝑗𝑖(𝑢𝑗 , 𝑢𝑖) − ∑ 𝑥𝑖𝑗(𝑢𝑖, 𝑢𝑗 ) 𝑗≠𝑖 𝑗≠𝑖
where the first sum is the flow of customers from all suppliers 𝑗 to supplier 𝑖, and the second sum is the flow of customers from supplier 𝑖 to all other suppliers 𝑗. 𝐶𝑖 is then the net change in the number of customers of supplier 𝑖. 𝐶𝑖 may be negative or positive. Note that ∑ 𝐶𝑖 𝑖 = 0 since churning does not alter the total number of customers but only the distribution of customers among the suppliers. A reasonable assumption is that the rate by which a supplier is losing customers due to churning is proportional to the number of customers of that supplier since these are the potential churners. The churning function then takes the form
(^2) A fixed point of a transformation T (for example a motion) is a point 𝑥 where 𝑇(𝑥) = 𝑥. In dynamical systems,
an attractor or a repeller may be regarded as fixed points since no motion takes place at these points; we then regard motion as a transformation from one point in space to another. However, a fixed point may not be an equilibrium point. See also [Str]. (^3) A function is everywhere smooth if the derivatives of any order of the function exist and are continuous at all
points.
This is exactly the evolution of mobile services and internet we have observed in several countries. If 𝑢 0 = 0, the time it takes until 50% of the market has been captured (that is,
𝑢(𝑡) = 12 ) is
ln 2 𝑎
or 𝑎 = ln2/𝑇 50. The formula is useful since it is easy to select a suitable value for 𝑇 50 , say 5 years, and from this value determine 𝑎 (for 𝑇 50 = 5 years, 𝑎 = 0.14 year^1 ). The latency time, defined here to be the time it takes until 10% of the market is captured, is
𝑇 10 =
ln(10 9⁄ ) 𝑎
This is a good measure for how fast the market increases initially, and is thus an important strategic parameter concerning whether services is likely to become lucrative in the long run. If 𝑇 50 = 5 years, then 𝑇 10 = 9 months. This is a little faster than a linear evolution since for which, 𝑇 10 would be one year.
2.2 Time-dependent adaptation rate
If the adaption rate depends on time, that is, 𝑎 = 𝑎(𝑡), then the equation for the number of subscribers becomes
We may expect that the attracting equilibrium point is 𝑢 = 1 where the asymptotic solution ends up; however, as will be evident from Examples 2 and 3, this is not always the case. For 𝑢 0 = 0, the solution is obviously^4
𝑢(𝑡) = 1 − exp [− ∫ 𝑎(𝑥)𝑑𝑥
𝑡
0
].
The initial condition 𝑢(0) = 0 is fulfilled since ∫ 𝑎(𝑥)𝑑𝑥 = 0
0 0 for a well-behaved function 𝑎(𝑢). The demand is
𝐷(𝑡) = 𝑎(𝑡)𝑁 exp [− ∫ 𝑎(𝑥)𝑑𝑥
𝑡
0
].
Example 1
For the linearly increasing adaptation rate 𝑎(𝑡) = 𝑎 0 + 𝑎 1 𝑡, we find:
(^4) For the purpose of readability of formulas with complex expressions in the exponential, we use the notation
𝑒𝑥^ = exp(𝑥).
𝑢(𝑡) = 1 − exp [−𝑎 0 𝑡 − (
) 𝑡^2 ].
We have 𝑢(∞) = 1 so that this is an attracting equilibrium point. The demand is
𝐷(𝑡) = (𝑎 0 + 𝑎 1 𝑡)𝑁 exp [−𝑎 0 𝑡 − (
) 𝑡^2 ].
Example 2
For the exponentially decreasing adaptation rate 𝑎(𝑡) = 𝑎 0 𝑒−𝛽𝑡^ we find:
𝑢(𝑡) = 1 − exp [−
(1 − 𝑒−𝛽𝑡)].
The demand is
𝐷(𝑡) = 𝑎 0 𝑁𝑒−𝛽𝑡^ exp [−
(1 − 𝑒−𝛽𝑡)].
For 𝑡 = ∞, we then have
𝑢(∞) = 1 − 𝑒
−𝑎 𝛽^0 < 1;
that is, not everyone will become subscribers of the service. Hence, the point 𝑢 = 1 is never reached even if it apparently is an equilibrium point. Therefore, we must be careful not drawing premature conclusions concerning stability when the differential equation depends explicitly on time. The solution approaches asymptotically another equilibrium point which cannot be determined from the differential equation directly. However, it is easily seen that both 𝑢̇ = 0 and 𝑢̈ = 0 for 𝑡 = ∞ so that the point 𝑆(∞) < 𝑁 is, in fact, an equilibrium point. On the other hand, 𝐷(𝑡) → 0 for 𝑡 → ∞ as it should.
Example 3
The adaptation rate is constant up to time 𝑇; thereafter, the adaptation rate is zero; that is,
𝑎(𝑡) = {𝑎,0,^ 𝑡 ≤ 𝑇𝑡 > 𝑇
This gives
𝑢(𝑡) = { 1 − 𝑒
This example is also a case where the asymptotic solution does not correspond to the expected equilibrium point. For the demand, we find
𝐷(𝑡) = {𝑎𝑁𝑒
Figure 2.2 Model with hesitant customers The set of differential equations describing the temporal behavior of 𝑝, ℎ, and 𝑠 is:
We see immediately that 𝑝 + 𝑢 + ℎ = 1. Moreover, we may assume that there are no initial customers, that is, 𝑝(0) = 1, 𝑢(0) = 0, ℎ(0) = 0. The solution for 𝑝 is readily found:
𝑝 = 𝑒−(𝑎+𝑏)𝑡. From 𝑝 + 𝑢 + ℎ = 1 we also see that ℎ = 1 − 𝑝 − 𝑢 = 1 − 𝑒−(𝑎+𝑏)𝑡^ − 𝑢, giving 𝑢̇ = 𝑎𝑒−(𝑎+𝑏)𝑡^ + 𝑐[1 − 𝑒−(𝑎+𝑏)𝑡] − 𝑐𝑢. The solution of this inhomogeneous linear equation is readily found using the method of integrating factor [Inc], [Kor].
𝑢(𝑡) = 1 −
[𝑏𝑒−𝑐𝑡^ + (𝑎 − 𝑐)𝑒−(𝑎+𝑏)𝑡].
The demand is then
𝐷(𝑡) =
[𝑏𝑐𝑒−𝑐𝑡^ + (𝑎 − 𝑐)(𝑎 + 𝑏)𝑒−(𝑎+𝑏)𝑡].
2.5 Adaptation by hesitation 2
In this model we assume that a customer entering the hesitation state may go back to the potential subscriber state with intensity c as shown in Figure 2.3. We are using the same normalized variables as in Section 2.4.
Figure 2.3 Second model with hesitating customers The differential equations are now:
First, we observe that the equation for 𝑢 is
𝑢(𝑡) = 𝑎 ∫ 𝑝(𝑥)𝑑𝑥
𝑡
0
and that 𝑝 and ℎ are determined by the first and the last equation. Eliminating 𝑝 and ℎ from these equations will lead to a new second order differential equation (that can be readily solved since it is linear with constant coefficients). However, we will solve the equations using a more elegant method only using first order derivatives. The different steps in the procedure are shown without proof, see [Goe] or [Kor]. The equations for 𝑝 and ℎ can be written in matrix form
(
where 𝑀 = (−𝑎 − 𝑏^ 𝑐 𝑏 −𝑐
). Formally, treating the vector (
ℎ) = 𝑽^ as a single object, we may write the equation in the form 𝑽̇ = 𝑀𝑽 or 𝑑𝑽𝑽 = 𝑀𝑑𝑡 , and seek the solution of this
equation in the form
𝑽 = (
where
𝑒𝑀𝑡^ ≝ 1 +
(𝑀𝑡)^2
(𝑀𝑡)^3
The exponential of a matrix is itself a matrix since the right-hand side of the defining equation consists only of the elementary matrix operations addition and multiplication. The matrix 𝑒𝑀𝑡^ can be evaluated by diagonalization as follows.
𝜆 1 𝑡 + 𝛽 (𝑣^12
that is,
𝑝 = 𝛼(𝑐 + 𝜆 1 )𝑒𝜆^1 𝑡^ + 𝛽(𝑐 + 𝜆 2 )𝑒𝜆^2 𝑡 ,
ℎ = 𝛼𝑏𝑒𝜆^1 𝑡^ + 𝛽𝑏𝑒𝜆^2 𝑡.
The constants of integration 𝛼 and 𝛽 are determined from the initial conditions: 𝑝(0) = 1 ⇒ 1 = 𝛼(𝑐 + 𝜆 1 ) + 𝛽(𝑐 + 𝜆 2 ),
The constants of integration are then
𝛼 = −𝛽 =
This gives
𝑝 =
[(𝑐 + 𝜆 1 )𝑒𝜆^1 𝑡^ − (𝑐 + 𝜆 2 )𝑒𝜆^2 𝑡]
and
ℎ =
[𝑒𝜆^1 𝑡^ − 𝑒𝜆^2 𝑡].
Finally, the equation for the number of subscribers is
𝑡
0
[𝜆 2 (𝑐 + 𝜆 1 )𝑒𝜆^1 𝑡^ − 𝜆 1 (𝑐 + 𝜆 2 )𝑒𝜆^2 𝑡] +
Setting 𝑟 = √(𝑎 + 𝑏 + 𝑐)^2 − 4𝑎𝑐, then 2𝜆 1 = −(𝑎 + 𝑏 + 𝑐) + 𝑟, and 2𝜆 2 = −(𝑎 + 𝑏 + 𝑐) − 𝑟; that is, 𝜆 1 − 𝜆 2 = 𝑟 and 𝜆 1 𝜆 2 = 𝑎𝑐. The formula for 𝑢 can then be written in the form
𝑢(𝑡) = 1 −
𝑒𝜆^1 𝑡[(𝑎 + 𝜆 1 )𝑒−𝑟𝑡^ − 𝑎 − 𝜆 2 ].
The demand is
𝐷(𝑡) =
𝑒𝜆^1 𝑡[𝑎 + 𝜆 1 − (𝑎 + 𝜆 2 )𝑒−𝑟𝑡].
2.6 Model with birth and death rates but without hesitation
The model in Figure 2.4 shows the case where the number of potential subscribers, 𝑃, increases at rate 𝑑 and decreases at rate 𝑓, while the number of subscribers decreases at rate 𝑔; 𝑑, and f and 𝑔 may be called birth and death rates, respectively. Moreover, we assume that 𝑎 + 𝑓 > 𝑑 + 𝑔. The differential equations are:
Figure 2.4 Model with birth and death rates Here, we again determine 𝑝 from the first equation and, from that solution, deduce the equation for 𝑢:
𝑢̇ = 𝑎𝑒−(𝑎+𝑓−𝑑)𝑡^ − 𝑔𝑢. This equation is a non-homogeneous equation which is solved using the method of integrating factor. The solution is (for 𝑢 0 = 0):
𝑢(𝑡) =
𝑒−𝑔𝑡[1 − 𝑒−(𝑎+𝑓−𝑑−𝑔)𝑡].
3. MARKETS WITH FEEDBACK
Feedback from the market implies that the attractiveness of a product depends on the number of users of the product. The feedback is also referred to as a network effect or network externality [Sha]. We start with studying the Bass model. The effect of the strength of the feedback is also investigated in the case when all users are imitators. In Section 3.2, we consider network effects where the attractiveness of the product decreases as a function of customers, illustrating the case where the market stagnates after a fierce initial increase. The general market equation is: 𝑢̇ = 𝑎(1 − 𝑢)𝐹(𝑢),
where 𝐹(𝑢) is the feedback term: 𝐹(𝑢) = 1 means that there is no feedback. Moreover, all coupling parameters are constants and the feedback term does not contain time explicitly. Then the differential equation is separable with general solution
∫
where 𝑐 is a constant of integration determined by the initial value for 𝑢 = 𝑢(0) = 𝑢 0. The solutions for the different models are compared by requiring that the time it takes to capture 50% of the market, 𝑇 50 , is the same in all models. This parameter is a simple strategic decision variable where, for example, a service is implemented provided that it takes no more than five years to capture 50% of the market. Otherwise the service is terminated.
The inflexion point (if it exists for positive 𝑢) corresponds to the case where the sales per unit time is maximum. This gives 𝑢̈ = 0 = (−𝑎 + 𝛾)𝑢̇ − 2𝛾𝑢𝑢̇. Solved for 𝑢 we find 𝑢𝑖𝑛𝑓𝑙 = (𝛾 − 𝑎) 2𝛾⁄^.
The inflexion point exists for positive 𝑢 if 𝛾 > 𝑎. If we set 𝑢 0 = 0, the inflexion point (the maximum sales rate) is reached after time 𝑡𝑖𝑛𝑓𝑙 = (ln 𝛾 − ln 𝑎) (𝛾 + 𝑎)⁄^.
3.1.2 Only imitators: Linear positive feedback
This case corresponds to the Bass model with imitators only. This model applies to services such as SMS, telefax and Facebook where there is no reason to subscribe to the service unless there is at least one other subscriber to communicate with. The differential equation for the market evolution is then 𝑢̇ = 𝛾𝑢(1 − 𝑢).
The feedback term is now 𝐹(𝑢) = 𝑢. The coupling factor 𝛾𝑢(𝑡) represents linear^7 positive feedback from the market; that is, if the number of customers increases, more people are stimulated to buy the product.^8 This type of market feedback is sometimes also called a network externality. The equation is called the logistic differential equation.^9 The differential equation that can be written 𝑑𝑢 𝑢(1 − 𝑢)
The solution is
𝑢(𝑡) =
where 𝑢 0 = 𝑢(0) is the initial value of 𝑢(𝑡). Solved for 𝑡, we find
𝑡 =
ln
The parameter 𝛾 is then estimated from this formula for 𝑢 = 12 and 𝑡 = 𝑇 50. This gives
(^7) Since the feedback function is of the first order, or linear, in 𝑢.
(^8) In a system with positive feedback , a small perturbation in the output from the system will result in a bigger
perturbation in the output. The result is a runaway system where the output from the system increases toward saturation, becomes empty, or oscillates in a regular or irregular manner. Oscillations may occur if the feedback signal is delayed, for example, in markets where the amount produced must be chosen before the prices are known, or in education where the expected future demand for professionals in a certain field is based on the demand when the training starts. In a system with negative feedback , the feedback will counteract any perturbation in the output such that the perturbation is reduced, or, in other words, the negative feedback stabilizes the output from the system. (^9) There is also a logistic difference equation, which for particular choices of parameter 𝑎, gives rise to chaotic
solutions with bifurcations; see for example [Str] and [Sch]. This equation is not important for the markets studied in this text.
ln
The inflexion point 𝑇𝑖𝑛𝑓𝑙𝑙𝑖𝑛^ of the function 𝑢 is the point where 𝑢̈ = 0. Since 𝑢̇ = 𝛾𝑢(1 − 𝑢) it follows that 𝑢̈ = (1 − 2𝑢)𝑢̇; that is, the coordinates for the inflexion point are
𝑢𝑖𝑛𝑓𝑙 = 12 and 𝑇𝑖𝑛𝑓𝑙 = 𝑇 50. Moreover, the gradient at the inflexion point is:
𝑢𝑖𝑛𝑓𝑙𝑙𝑖𝑛^ =
ln
For 𝑢 0 = 0.01, this gives
𝑢𝑖𝑛𝑓𝑙𝑙𝑖𝑛^ =
The latency time is (again taken as the time it takes to capture 10% of the market)
𝑇 10 = 𝑇 50
ln[0.1(1 − 𝑢 0 )] − ln[0.9𝑢 0 ] ln(1 − 𝑢 0 ) − ln 𝑢 0
Table 1 shows 𝑇 𝑇^10 50
for some values of 𝑢 0 and 𝑇 50 = 5 years.
Table 3.1 Latency time for some values of 𝑢 0
𝑢 0 𝑇 10 /𝑇 50 𝑇 10 for^ 𝑇 50 =^5 years 0.001 0.67 3 years and 4 months 0.005 0,58 2 years and 11 months 0.01 0.52 2 years and 7 months 0.02 0.44 2 years and 2 months 0.04 0.31 1 year and 6 months
Observe that if 𝑢 0 = 0 then 𝑢(𝑡) = 0 for all 𝑡; that is, the customers will buy the service only if there already are customers who have bought the service. This is one of the strategic difficulties in markets with positive feedback and no innovators: the supplier must in one way or another establish an initial customer base before the service is launched, for example, offering the service for free to some trial customers. Examples of information services with positive feedback are SMS, Facebook, LinkedIn, telephone service, data communication, telefax, and interactive games. Note that there is no significant positive feedback for mobile telephony since the customers of mobile services can communicate with customers of the fixed network; there is no significant positive feedback for MMS since SMS is a subset of this service with an established customer base. Table 3.1 illustrates a difficult strategic problem, namely the time it takes for the market to increase to an acceptable level (the latency time). If it is expected that the product will be bought by 50 percent of the potential customers after 5 years, the table illustrates that the supplier may face a severe problem since the market share is only 10 percent after almost three years for an initial customer base is 0.5%. If the initial market share is 2%, it
where the constant of integration, satisfying the initial condition 𝑢(0) = 𝑢 0 , is
Solving for 𝑢 gives, after some simple calculations,
𝑢 = [
1 + √𝑢 0 − (1 − √𝑢 0 )𝑒−𝛾1 2⁄^ 𝑡
1 + √𝑢 0 + (1 − √𝑢 0 )𝑒−𝛾1 2⁄^ 𝑡
]
2 .
As we shall see, 𝑢(0) = 𝑢 0 = 0 is not an equilibrium point. The reason for such behavior is that (^) √𝑢 is not regular at the point 𝑢 = 0 (the point is a branch point separating the positive and the negative branch of the square root). The growth curve starting from 𝑢(0) = 𝑢 0 = 0 is then
𝑢 = [
1 − 𝑒−𝛾1 2⁄^ 𝑡
1 + 𝑒−𝛾1 2⁄^ 𝑡
]
2 .
The second derivative at 𝑢(0) = 0 is easily found by derivation of the original equation:
𝑢̈ = 𝛾1 2⁄ [
(1 − 𝑢) − 𝑢̇√𝑢].
Inserting 𝑢̇ √𝑢⁄ = 𝛾1 2⁄ (1 − 𝑢) and setting 𝑢 = 0, we get 𝑢̈(0) = 12 𝛾1 2⁄ 2 > 0. Since
the acceleration at 𝑢(0) = 0 is nonzero, 𝑢(0) = 0 is not an equilibrium point. The inflexion points are given by 𝑢̈ = 0, that is,
𝑢̈ = 𝛾1 2⁄ [
(1 − 𝑢) − 𝑢̇√𝑢] = 0,
with the single solution (𝑢̇ ≠ 0) 𝑢𝑖𝑛𝑓𝑙𝑠𝑞𝑟^ = 13.
The inflexion point is located at a smaller 𝑢 than for linear positive feedback. The gradient at the inflexion point is (using the formula for 𝑇 50 computed below):
𝑢̇𝑖𝑛𝑓𝑙𝑠𝑞𝑟^ =
Solving the equation for 𝑡 gives
ln
The 50% point is found by setting 𝑢 = 12. This gives for 𝑢(0) = 0
ln
The growth parameter 𝛾1/2 is given by
𝛾1 2⁄ =
The 10% latency time is found by setting 𝑢 = 0.1. This gives
ln
or if 𝑢 0 = 0, 𝑇 10 = 0,37𝑇 50 ; that is, 1 year and 10 months if 𝑇 50 = 5 years. For 𝑢 0 = 0. we find 𝑇 10 = 0.29𝑇 50 , or 1 year and 5 months. The demand is
4𝑁(1 − 𝑢 0 )𝛾1/2𝑒−𝛾1/2𝑡[1 + √𝑢 0 − (1 − √𝑢 0 )𝑒−𝛾1/2𝑡]
[1 + √𝑢 0 + (1 − √𝑢 0 )𝑒−𝛾1/2𝑡]
3.1.4 Only imitators: Strong feedback
If the feedback is proportional to the square of the number of subscribers, the market equation is 𝑢̇ = 𝛾 2 𝑢^2 (1 − 𝑢). The feedback term is 𝐹(𝑢) = 𝑢^2. This equation is used to study the effect of a strong positive feedback from the market. The differential equation is:^10
𝑑𝑢 𝑢^2 (1 − 𝑢)
𝑢^2
Integrating term by term gives:
ln 𝑢 −
− ln(1 − 𝑢) = ln
1 𝑢 1 − 𝑢
For an initial value 𝑢(0) = 𝑢 0 , the solution is 𝑢 1 − 𝑢
1 𝑢 (^) =
exp (𝛾 2 𝑡 −
This is a transcendental equation that cannot be solved explicitly for 𝑢; however, the solution is easily plotted using 𝑡 as dependent variable:
𝑡 =
ln [
exp (
)] =
[ln
]
We see that 𝑢 = 0 is a repelling equilibrium point and 𝑢 = 1 is an attracting equilibrium point. This also implies, just as for linear feedback, 𝑢 0 > 0 for a non-zero solution to emerge. The time 𝑇 50 is given by the formula
𝛾 2 𝑇 50 = ln
(^10) Note that the inverse of a polynomial factored into real factors can be written as a sum over the inverse
factors, for example, (^) (𝑥−𝑎)(𝑥−𝑏) (^2) (𝑥−𝑐)(𝑥^12 +𝑑𝑥+𝑒) can be written as (^) 𝑥−𝑎𝛼 + (^) 𝑥−𝑏𝛽 + (^) (𝑥−𝑏)𝛾 2 + (^) 𝑥−𝑐𝛿 + (^) 𝑥 2 𝜀𝑥+𝜁+𝑑𝑥+𝑒 for unique values of 𝛼, 𝛽, etc, only dependent on the constants 𝑎, 𝑏, etc. Note how multiple factors such as (𝑥 − 𝑏)^2 and nonlinear terms are handled.
