The Design of Experiential Serviceswith Acclimation and Memory Decay:
Optimal Sequence and Duration
Aparupa Das Gupta, Uday S. Karmarkar, Guillaume RoelsUCLA Anderson School of Management, 110 Westwood Plaza, Los Angeles, CA 90095, USA
For many consumer-intensive (B2C) services, delivering memorable customer experiences to maximize cus-
tomer satisfaction can be a source of competitive advantage. Yet, there is no clear guideline available for
service encounter design that accounts for customer behavior. In this paper, we show how experiential services
should be sequenced and timed to maximize satisfaction of customers who are subject to memory decay and
acclimation. We find that memory decay favors positioning the highest service level near the end, whereas
acclimation favors maximizing the gradient of service level. Although memory decay and acclimation lead to
the same design individually, they can act as opposing forces when considered jointly. Specifically, together,
they maximize the gradient of service level near the end, which may result in (i) sequencing activities in
a U-shaped fashion and (ii) lengthening the duration of activities with the lowest service levels. Based on
the analytical characterization of the optimal sequence and timing of activities, we propose heuristics to
design the service encounter. Overall, this paper shows that service experiences can be engineered to enhance
customer satisfaction.
Key words : service design, experience, scheduling, social psychology, behavioral operations, heuristics
1. Introduction
Managing customer experience during service encounters is a powerful way of improving customer
satisfaction and differentiating from competitors (Pine and Gilmore 1998, Voss and Zomerdijk
2007, DeVine and Gilson 2010). For instance, the Walt Disney Company’s entertainment parks and
Benihana’s restaurants are two well known examples of companies offering unique service expe-
riences to their customers (Heskett et al. 1997). Maximizing customer satisfaction often entails
understanding customer behavior. DeVine et al. (2012) indeed report that companies focusing on
the human side of service delivery have been able to improve customer satisfaction scores by up to
30% while reducing costs.
In practice, many service firms ensure that their customers start or end their service experience
on a high note. For instance, Royal Caribbean cruise line ensures that the on-board entertainment
program builds towards a peak on the second-to-last night of the cruise. On the last night, they
present a Farewell Show during which they build an emotional connection with their customers by
recapitulating key events from the cruise (Voss and Zomerdijk 2007). In contrast, office furniture
1
Author: The Design of Experiential Services2 Article submitted to ; manuscript no. (Please, provide the mansucript number!)
manufacturer Herman Miller enhances the start of the visit of their potential customers by design-
ing special ‘decompression’ rooms (Voss and Zomerdijk 2007).
Though many firms pay attention to design a good first or last impression of their service
encounter, no clear guidelines seem available for designing the operations of a service encounter so
as to maximize customer satisfaction. Cook et al. (2002) however argue that service encounters can
be designed with the same depth and rigor as manufacturing processes, by drawing concepts from
psychology, sociology, and their various subspecialties such as behavioral decision theory. Chase
and Dasu (2008, p.1) also emphasize the significance of “critical psychological variables that lie
at the subconscious level of the customer, and which, if understood could be managed to enhance
customer satisfaction from a service encounter.”
In this paper, we show how to sequence and allocate time to activities in a service encounter so
as to maximize customer satisfaction at the end of the experience. We consider a service encounter
that is composed of a sequence of homogeneous activities with different unidimensional service
levels. For instance, a one-day professional development workshop consists of multiple sessions on
different topics, and these sessions may be associated with different levels of popularity, measured
by, e.g., ratings in end-of-session evaluations. Moreover, we assume that customers do not join in
the middle of or balk from the service encounter.
We refer to the combination of sequence and duration of activities as a service design, and assume
that it is under the control of the service provider, if it can be changed. Depending on the nature of
the service offered, the service provider can alter or not the sequence and duration of service activ-
ities. In particular, in some service encounters (e.g., a music concert with multiple performances)
it is feasible to change the order of activities but not their durations. Other service encounters
(e.g., treatments in a spa) offer a fixed sequence of activities, but their durations can vary. Finally,
some encounters (e.g., fitness classes) offer the flexibility to alter both the sequence and duration of
activities during the service encounter. Based on the nature of services, we consider three stylized
models of a service encounter, 1) variable sequence and fixed duration (VSFD), 2) fixed sequence
and variable duration (FSVD), and 3) variable sequence and variable duration (VSVD).
We define ex-post customer satisfaction as the total remembered utility of a customer from the
service experience, which is influenced by customer behavior. We focus on two recognized behav-
ioral phenomena, namely, memory decay and acclimation. On the one hand, memory of human
beings fade with the passage of time. Hence, individuals tend to pay more importance to events
closer to the present than to the ones farther away from it. As Kahneman (2010) puts it, “we don’t
choose between experiences, we choose between memories of experiences.” On the other hand,
human beings adapt, or acclimate, to a given environment over a period of time. As Ariely (1998)
suggests, because physiological adaptation causes less nerve activation for longer experiences, the
Author: The Design of Experiential ServicesArticle submitted to ; manuscript no. (Please, provide the mansucript number!) 3
intensity of experience becomes less sensitive to a constant service level. While there exist other
behavioral phenomena (e.g., satiation, see Baucells and Sarin 2012), we choose to focus on memory
decay and acclimation given their large empirical evidence in psychology and marketing.
While memory decay favors a high service level near the end, acclimation favors a steep rise
in service levels. Though memory decay and acclimation are distinct phenomena, we show that
they are mathematically symmetric in our setting. Consequently, we find that a service design
that is optimal for memory decay is also optimal for acclimation. In particular without precedence
constraints, it is optimal to sequence activities in increasing order of service levels so as to finish
at a peak and, with variable durations, to lengthen the duration of activities with the highest
service level. Taken together, however, acclimation and memory decay favor a maximum gradient
of service level closer to the end of the encounter. Specifically, when both phenomena are present,
it may be optimal to sequence activities in a U-shape fashion, beginning and finishing with high
service levels, and to lengthen the duration of activities with the lowest service level. Hence, even
if memory decay and acclimation act in the same way when considered individually, they act in
opposite directions when considered jointly. Because the resulting design problems are hard to solve
optimally, we propose heuristics based on the analytical characterization of the optimal solution.
The remainder of this paper is organized as follows. In §2, we review the related literature. In
§3, we introduce the model for customer satisfaction. In §4, we characterize the optimal solution
for the sequencing and duration allocation problem and propose heuristics based on our analytical
characterization. We present our concluding remarks in §5. The proofs, the mathematical program-
ming formulation of the problem, the heuristic algorithms and the derivation of an upper bound
appear in appendix.
2. Literature Review
This paper is related to three bodies of research: the study of human biases in behavioral sciences,
experienced utility theory in decision theory, and the operational aspects of service design. We
next review these three streams of literature and position our paper accordingly.
2.1. Behavioral Aspects
Research in psychology and marketing has identified many factors that influence the retrospective
evaluation of experiences, including the trend of an experience (Loewenstein and Prelec 1993), its
rate of change (Hsee and Abelson 1991), and its maximum and final intensities (Frederickson and
Kahneman 1993). Underlying these factors are psychological phenomena such as adaptation, loss
aversion, or memory decay. We contribute to this literature by adopting a design perspective on
two such phenomena, namely, memory decay and acclimation.
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2.1.1. Memory Decay. In their seminal work, Varey and Kahneman (1992), Frederickson
and Kahneman (1993), and Kahneman et al. (1993) show that the maximum and final intensities of
an experience play pivotal roles in influencing the remembered utility from the overall experience.
Although their work suggests that the overall evaluation of an experience can be summarized by
a weighted average of the peak and final intensities (i.e., the peak-end rule), regardless of the
duration of the experience, subsequent research has also shown the importance of duration and
gradient of intensities. In particular, Price and Tursky (1975) and Ariely (1998) show that people
are sensitive to duration if subject to changing levels of painful stimuli. Ariely (1998) moreover
shows that retrospective evaluation of painful experiences is influenced not only by the final pain
intensity, but also by the intensity trend during the latter half of the experience. Studying memory
of ads, Baumgartner et al. (1997) demonstrate that, even if people prefer advertisements with
peaks and high ends, retrospective evaluation of advertisements is not always independent of their
durations, especially if duration affects the intensity of the peak experience and the final moment.
In this paper, we adopt an agnostic approach on the peak-end rule, i.e., we do not cast the
remembered utility as a function of the peak and the end experience. Rather, we represent memory
decay with a simple model of reverse discounted utility based on the exponential forgetting curve
introduced by Ebbinghaus (1913), and we show that, under certain conditions, the optimal service
design mimics the peak-end rule.
2.1.2. Acclimation. There is considerable evidence available regarding adaptation, or accli-
mation as it is referred for short experiences, in life experiences (Helson 1964, Brickman and
Campbell 1971, Brickman et al. 1978). Ariely (1998) observes that adaptation can explain the
impact of duration on retrospective evaluations of patterned stimuli and its lack thereof for constant
stimuli. Nelson and Meyvis (2008) experimentally verify that, contrary to conventional wisdom,
interruptions may make retrospective evaluation of pleasant experiences more enjoyable and that
of unpleasant experiences more irritating, and they attribute this result to the adaptation effect.
In this paper we model acclimation as a dynamic reference effect consistent with decision theory
(Baucells and Sarin 2012): At each point in time, a customer’s instantaneous utility is a function of
the difference between the current service level and the current reference level, and that reference
level dynamically adapts to the service level.
2.2. Decision Theory
Kahneman et al. (1997) propose that there exist two kinds of experienced utility, the one reported
in real time (instantaneous) and the other based on retrospective evaluation (remembered). Using
this idea, our model represents the remembered utility as a function of the instantaneous utility.
Baucells and Sarin (2012) propose six laws that govern satisfaction including habit formation,
Author: The Design of Experiential ServicesArticle submitted to ; manuscript no. (Please, provide the mansucript number!) 5
satiation, and social comparison. Consistent with the adaptation or habit formation models in Con-
stantinides (1990), Wathieu (1997), and Baucells and Sarin (2012), our acclimation model adapts
the reference level to the current service level. In contrast to habit formation, which results from
adaptation over a long period of time, our acclimation model deals with adaptive processes of short
duration.
Proposing an analytical utility model for adaptation before, during, and after an event, Bau-
cells and Belleza (2012) show that the utility function during anticipation is unimodal (decreas-
ing, increasing, or U-shaped) because the magnitude of effective consumption, i.e., the difference
between the conceptual consumption of experience during anticipation and the current reference
level, decreases over time, whereas the discount function due to adaptation, which is inversely
proportional to the effective consumption, increases over time. We focus here on the event itself,
considering an initial reference level, which captures the customer anticipation at the start of the
event, and a moment of recall at the end of the encounter, when the remembered utility is evalu-
ated. Moreover, the discount factor due to acclimation remains constant, i.e., is independent of the
magnitude of the instantaneous utility. With this utility model, we find that the optimal sequence
of activities under certain conditions is U-shaped in service levels. Consistent with our result, Bau-
cells et al. (2013) empirically find that the sequences of songs in music concerts and individual
playlists exhibit a U-shape, based on their popularity or individual ratings respectively.
We contribute to this literature by introducing an analytical model for remembered utility that
is derived from the total experienced instantaneous utility. Unlike these decision theory papers,
our goal is to demonstrate the implications of memory decay and acclimation on service experience
design.
2.3. Service Operations
Management of service operations typically consists of managing capacity (e.g., staffing), schedul-
ing resources (Pinedo 2005), and pricing (Talluri and Van Ryzin 2004). In the context of service
encounter design, Bellos and Kavadias (2011) study how to allocate tasks between the service
provider and the customer and how to price the service. In contrast to their model, which considers
customer experience as random, we adopt here a behavioral model of customer experience. More-
over, our decision variables are not pricing and task allocation, but task sequencing and timing.
Hence, our approach can be viewed as complementary to theirs.
Lately, some papers have studied how operational decisions are affected by customer behavior.
The literature in this area is diverse in terms of modeling paradigms and behavioral regularities.
In particular, Nasiry and Popescu (2011) study pricing and Dixon and Verma (2013), Dixon and
Thompson (2013a,b) study event scheduling decisions in the light of peak-end rule. Aflaki and
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Popescu (2013), Popescu and Wu (2007), and Nasiry and Popescu (2011) study how loss aversion
affects pricing and service level policies. Caro and Martınez-de-Albeniz (2012) show how satiation
affects assortment decisions. Adelman and Mersereau (2013) study how customer memory effects
impact a supplier’s profits when the supplier dynamically allocates limited capacity among a port-
folio of customers. Focusing on queueing experiences, Plambeck and Wang (2012, 2013) consider
the effect of hyperbolic discounting on unpleasant services, Carmon et al. (1995) show how cus-
tomer dissatisfaction can be reduced by spreading the service across the waiting period in queues,
and Huang et al. (2013) show that estimating waiting times is impeded by customer bounded
rationality.
Our work is most closely related to Dixon and Thompson (2013a,b), who study an event schedul-
ing problem by creating bundles of events in the presence of peak-end, trend, and other outcomes
of psychological phenomena. Unlike these two papers, we directly model the psychological phe-
nomena, and not their outcomes. Moreover we adopt an analytical approach, whereas they focus
on the development of metaheuristics.
3. Model
We consider a service provider who seeks to maximize ex-post customer satisfaction from a service
encounter. The service encounter consists of a sequence of n activities. Each activity i ∈ {1, .., n}
is designed to offer a fixed service level xi, for a duration ti with∑n
i=1 ti = T . We assume that
the service level is unidimensional and that the total duration is sufficiently small to maintain a
certain degree of homogeneity among activities. Our intent is thus not to model long, heterogenous
encounters (e.g., a week-long medical treatment), but rather short, homogeneous encounters. For
instance, each day of a week-long executive education course may consist of different sessions
(e.g., operations, finance), each with a specific duration (e.g., one or two hours) and a specific
service level (e.g., based on instructor’s ratings). Accordingly, a service encounter is defined by two
design variables: 1) the sequence of activities (i1, .., in) where ik ∈ {1, .., n} is the kth activity in the
sequence and 2) the duration for each activity ti,∀i. For any sequence (i1, .., in), let Tk =∑k
j=1 tij
be the time passed by the end of the kth activity and T k =∑n
j=k tij be the time remaining from the
beginning of the kth activity to the end of the encounter. We next build the customer satisfaction
utility model and then consider the service provider’s design problem.
3.1. Customer Satisfaction Utility
We consider a customer who is subject to memory decay and acclimation. We assume the customer’s
instantaneous experienced utility at activity ik is captured by a unidimensional utility function
uik(t) where Tk−1 ≤ t≤ Tk. Acclimation determines the intensity of experience uik(t), and memory
decay determines the importance of the contribution of uik(t) to the overall remembered utility.
Author: The Design of Experiential ServicesArticle submitted to ; manuscript no. (Please, provide the mansucript number!) 7
3.1.1. Customer Memory Process. We model the memory decay process based on the
exponential nature of forgetting proposed by Ebbinghaus (1913). According to this model, memory
loss occurs rapidly initially, but more gradually subsequently. The memory decay process thus
weighs experiences at the end of the encounter more heavily than the ones at the beginning of
the encounter. Accordingly, memory decay operates as a backward discounting process. When the
customer discounts past experiences exponentially with rate w ∈ [0,∞), her remembered utility or
satisfaction from the service experience is given by
S((i1, .., in), t) =n∑k=1
∫ Tk
Tk−1
uik(t)e−w(T−t)dt. (1)
With no acclimation, if uik(t) = xik ,∀k, then satisfaction from memory decay is given by a dis-
counted sum of the difference of service levels across consecutive activities. Formally,
S((i1, .., in), t) =n∑k=1
(xik −xik−1)(1− e−wTk)
w.
3.1.2. Customer Acclimation Process. The acclimation process affects the instantaneous
experienced utility. Let b(t) be the customer’s reference level at time t. The instantaneous experi-
enced utility is moderated by the acclimation level b(t) such that, at any given point in time,
uik(t) = xik − b(t), Tk−1 ≤ t≤ Tk. (2)
As defined by Kahneman and Tversky (2000), the instantaneous experienced utility, uik(t), captures
both the valence (good or bad) and the intensity (mild to extreme) of the instantaneous experience.
In particular, acclimation may yield negative utility even if the service level is positive, in contrast
to satiation (Baucells and Sarin 2012).
Similar to the habit formation model in Baucells and Sarin (2012), the instantaneous acclimation
level b(t) for activity ik is assumed to evolve according to the function
b(t) = xik − (xik − b(Tk−1))e−α(t−Tk−1), Tk−1 ≤ t≤ Tk, (3)
where α> 0 is the rate of acclimation and the initial reference level b(0) captures the history of past
experiences. The value of α can be estimated empirically (e.g., see Dawes and Watanabe 1987).
A high value of α implies that the customer is less influenced by past service levels. Similar to
the adaptation models used in life sciences (Overbosch 1986), this model can be generated when
the reference point is exponentially adapted to the current level of service, i.e., dbdt
= α(x(t)− b(t))
although we do not restrict α to be smaller than 1.
Two implicit assumptions of this model are that (i) α is common across all activities of the
encounter and that (ii) the reference point at the end of an activity carries over to the beginning
Author: The Design of Experiential Services8 Article submitted to ; manuscript no. (Please, provide the mansucript number!)
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Figure 1 Acclimation process leads to a decay in utility over time for a fixed service level and yields negative
utility after a drop in service level.
of the following activity, i.e., limt↑Tk
b(t) = limt↓Tk
b(t).
Figure 1 shows the acclimation process experienced by a customer during a service encounter
with three activities x1, x2, x3 such that x2 > x3 > x1. The acclimation level b(t) always trails the
service level. Therefore, for a constant service level x(t), the instantaneous utility decreases over
time and eventually becomes zero (Ariely 1998). Whereas, an upward or downward jump in service
level across consecutive service levels leads to a rapid increase or decrease in the instantaneous
utility level respectively.
Expanding (3), we obtain
b(t) = xik −
((xi1 − b(0)) +
k∑j=2
(xij −xij−1)eαTj−1
)e−αt, Tk−1 ≤ t≤ Tk.
Therefore, the instantaneous utility (2) is given by
uik(t) = (xi1 − b(0))e−αt +k∑j=2
(xij −xij−1)e−α(t−Tj−1), Tk−1 ≤ t≤ Tk. (4)
The instantaneous experienced utility, uik(t), can thus be expressed as a discounted sum of the
past changes in service levels. In this case, discounting operates forward in time, in contrast to
memory decay, which applies backward discounting.
3.1.3. Customer Satisfaction Model. Combining (1) and (4) leads to the following model
of cumulative remembered utility in the presence of acclimation and memory decay:
S((i1, .., in), t) =n∑k=1
∫ Tk
Tk−1
uik(t)e−w(T−t)dt=n∑k=1
(xik −xik−1)(e−αTk − e−wTk)
w−α. (5)
Equation (5) reveals that, memory decay and acclimation mathematically play a symmetric role
on ex-post satisfaction even though they are different phenomena. A fundamental difference is
Author: The Design of Experiential ServicesArticle submitted to ; manuscript no. (Please, provide the mansucript number!) 9
that memory decay is a macroscopic phenomenon, affecting the weighing of instantaneous utilities
in the remembered utility; whereas, acclimation is a microscopic phenomenon, affecting directly
instantaneous utility. Although both phenomena are backward looking, memory decay applies
backward discounting to past utilities, whereas acclimation applies forward discounting to past
changes in service levels. Given their differences, it may come as a surprise that they turn out to
be mathematically symmetric in our setting.
Due to the overlap of backward discounting with memory decay and forward discounting with
acclimation, the customer effectively weighs a change in service intensity from the (k−1)th to the
kth activity by
Φ(α,w,T k).=e−αTk − e−wTk
w−α=
∫ Tk
0
e−αte−w(Tk−t)dt.
0 0.5 1 1.5 2 2.5 3 3.5 4 4.5 50
0.2
0.4
0.6
0.8
1
1.2
1.4
1.6
1.8
2
time
!(",w,t)
w=0.3,"=0.1
w=0.8,"=0.1
w=3,"=0.1
Student Version of MATLAB
Figure 2 The weight assigned to change in service levels between activity k− 1 and activity k is pseudoconcave
in the remaining time until the end of the encounter.
Given the importance of this function in our characterization of the optimal service design, we
briefly outline its mathematical properties here. First, this function is symmetric in α and w, i.e.,
Φ(α,w,T k) = Φ(w,α,T k). In particular, a model with pure acclimation α and no memory decay
would generate the same satisfaction as a model with pure memory decay w= α and no acclimation.
Second, Φ(α,w,T k) is pseudoconcave in T k for 0< α,w <∞ (see Lemma A-1, Figure 2), with a
stationary point at T k = lnw−lnαw−α , i.e., the inverse of the logarithmic mean of α and w (Carlson
1972). For simplicity of notation we denote this inverse logarithmic mean as
m(α,w).=
lnw− lnα
w−α.
Author: The Design of Experiential Services10 Article submitted to ; manuscript no. (Please, provide the mansucript number!)
The function m(α,w) is decreasing in both α and w (see Figure 3) such that, for any α > 0,
limw→0
m(α,w) =∞ and limw→∞
m(α,w) = 0 and for any w> 0, limα→0
m(α,w) =∞ and limα→∞
m(α,w) = 0.
For ease of notation, we will use Φ(t) and Φ(α,w, t) interchangeably henceforth in the paper.
00.5
11.5
2 00.5
11.5
2
0
1
2
3
4
5
6
7
8
9
10
w!
m(!,w)
Student Version of MATLAB
Figure 3 Plot of m(α,w).
3.2. Service Provider’s Design Problem: Sequence and Duration
We consider a service provider who seeks to optimize the design of a service encounter to maximize
ex-post customer satisfaction. We assume that the population of customers is homogeneous, i.e.,
they have the same values of the parameters α and w.
Every service encounter is composed of activities that offer pleasant experiences (e.g., a perfor-
mance during a concert) and unpleasant experiences (e.g., a painful medical procedure). Ideally
one would want to allocate more time to pleasant experiences and less time to unpleasant ones.
However, doing so may not always be feasible due to resource limitations. As a result, activity
durations are typically constrained to be within certain bounds. For instance, a call center may
choose to serve a call within a certain time window: Above the upper threshold, customer satisfac-
tion may rapidly deteriorate and below the lower threshold, staffing costs would be too important
(Braff and DeVine 2008). Accordingly, we assume that activity durations must lie within certain
bounds, i.e., τ i ≤ ti ≤ τ i,∀i such that τ i > 0, τ i > 0,∀i.
Moreover, we consider a situation in which the total duration of the service encounter is fixed
to T , i.e.,∑
i ti = T , such as a two-hour long concert. Relaxing this constraint can be easily acco-
modated, but it would diminish the time allocation trade-off, making pleasant experiences longer
and unpleasant experiences shorter.
Author: The Design of Experiential ServicesArticle submitted to ; manuscript no. (Please, provide the mansucript number!) 11
Table 1 Three stylized models of a service encounter
Sequence
Fixed Variable
DurationFixed Variable Sequence-Fixed Duration (VSFD)
Variable Fixed Sequence-Variable Duration (FSVD) Variable Sequence-Variable Duration (VSVD)
In addition to having some control over the duration of activities, a service provider may also
have control over the sequence of activities. The admissible sequence of activities can be restricted
by precedence constraints; for instance, taking X-rays must be performed before a medical proce-
dure. Let P be the set of all possible feasible sequences of activities.
Then, using (5), the service provider’s design problem (SPDP) is given by
max(i1,..,in),t
S((i1, ., in), t) =n∑k=1
(xik −xik−1)Φ(α,w,T k) (6)
subject to
τ i ≤ ti ≤ τ i,∀i (7)n∑i=1
ti = T (8)
(i1, .., in)∈P. (9)
In the sequel, we will denote the optimal sequence by (i?1, .., i?n) and the optimal duration by t?.
In Appendix B, we propose a mathematical programming formulation of SPDP. In the following
section, we examine the optimal design strategy to maximize ex-post satisfaction.
4. Analysis
In this section, we characterize the optimal sequence and duration allocation decisions, sequentially
considering variable sequence but fixed duration (VSFD), variable duration but fixed sequence
(FSVD), and variable sequence and variable duration (VSVD); see Table 1. Since the resulting
problems are in general hard to solve, in each case, we first analytically characterize the structure
of the optimal solution and then propose heuristics that leverage our analytical characterizations.
Underlying the optimal design in each scenario are the following two properties of memory decay
and acclimation:
• Memory decay favors high service level at the end of the encounter.
• Acclimation favors large positive gradients of service levels.
We find that, together, they maximize the gradient of the service level near the end of the service
encounter.
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4.1. Variable Sequence and Fixed Duration (VSFD)
VSFD comprises all service encounters that allow the service provider to change the sequence of
activities keeping their respective durations fixed. For instance, in a concert with multiple per-
formances (e.g., a rock festival), the duration of each act is typically fixed, but the organizer can
change the order in which they are performed. The VSFD problem can be obtained from the SPDP
formulation by setting τ i = τ i = ti,∀i such that∑
i ti = T .
VSFD can be interpreted as a single machine job scheduling problem with the objective of
maximizing the total weighted profit∑k
xikf(Tik), where Tik =∑k
j=1 tij is the completion time of
the kth activity in the sequence and f(Tik) = Φ(T − Tik + tik)−Φ(T − Tik) is the corresponding
profit. Because the profit function f(Tik) turns out to be pseudoconvex in Tik (see Lemma A-2 in
Appendix A), the computational complexity of VSFD without precedence constraints is an open
problem (Hohn and Jacobs 2012). With arbitrary precedence constraints, the problem becomes
even more complex. In the following, we consider service encounters with no precedence constraints
to generate analytical insights; specifically we set P as the set of all permutations of activities.
4.1.1. Structural Results. We find that, individually, both memory decay and acclimation
favor designs with increasing service levels. However, when they act together, the optimal sequence
is in general U-shaped, as shown in Proposition 1.
Proposition 1. When durations are fixed, there exists an optimal sequence (i?1, .., i?n) such that
for any two consecutive activities xi?k
and xi?k+1
the following applies:
(i) If both activities start and finish within [0, T − 2m(α,w)], then xi?k>xi?
k+1.
(ii) If both activities start and finish within [T − 2m(α,w), T ] or if k= n− 1, then xi?k<xi?
k+1.
Proposition 1 is illustrated in Figure 4 for different values of α and w. For large m(α,w) the
optimal sequence is increasing over time; otherwise, the optimal sequence is U-shaped with a steep
rise at the end. A decreasing sequence of service levels is however never optimal because of the
recency effect: Memory decay always favors a positive ending of utilities, and to ensure positive
instantaneous utility, the last two service levels should be in increasing order. U-shape sequences
of activities appear in many experiences, such as concerts (Baucells et al. 2013). Although such
a sequence could be justified on the grounds of primacy and recency biases, or on the grounds of
satiation (Baucells et al. 2013), it emerges here only because of acclimation and memory decay.
When either α or w is small (i.e., when T ≤ 2m(α,w)), it is optimal to have activities sequenced
in increasing order of service levels. When w is small, the customer gives equal weight to past and
recent utilities. Because of acclimation, any drop in service level may create a negative instantaneous
utility. Hence with small w, it is optimal to have a monotonically increasing sequence of service
Author: The Design of Experiential ServicesArticle submitted to ; manuscript no. (Please, provide the mansucript number!) 13
Figure 4 The optimal sequence of activities is either increasing or U-shaped, depending on the magnitude of
T − 2m(α,w). The last two activities are always in increasing order.
levels. By contrast when α is small, the customer acclimates to the service level of an activity very
slowly. Therefore, the higher the service level, the larger the magnitude of instantaneous utility.
Because of memory decay, these high utilities are more likely to be remembered if they happen at
the end of the encounter. Hence, with small α, it is also optimal to have a monotonically increasing
sequence of service levels. To summarize, when either α or w is small, the service provider should
sequence activities in increasing order of service levels, as shown in Corollary 1.
Corollary 1. In VSFD, for α= 0 or w= 0, the optimal sequence of activities (i?1, .., i?n) is such
that xi?1 < ... < xi?n, i.e., is in increasing order of service levels.
By contrast when 2m(α,w) < T , Proposition 1 shows that the optimal sequence of activities
is U-shaped: The service level should fall gradually in the beginning and rise steeply at the end.
This U-shaped encounter maximizes the gradient of service level towards the end of the encounter
and is typically desirable when m(α,w) is small, i.e., when both α and w are high (see Figure
3). Intuitively, high memory decay w makes the customer heavily discount past experiences. On
the other hand, high acclimation α makes the current utility depend less on past service levels.
Therefore, the service provider can gradually decline the service level in the beginning, which will
be forgotten due to high memory decay, so as to ensure a steep positive gradient towards the end.
Although one might presume that having the highest service level at the end maximizes sat-
isfaction consistent with the peak-end rule (Kahneman et al. 1993), a peak in service level does
not always lead to a peak in utilities. When the activity with the highest service level lasts for
a long duration, the customer becomes acclimated to that high level of service and her util-
ity diminishes over time. In that case, it may then be optimal to move that activity at the
beginning of the encounter so as to ensure a steep gradient near the end. For instance, when
(x1, x2, x3, x4) = (2,5,7,10) and (t1, t2, t3, t4) = (5,4,3,8), the optimal sequence when w = 1 and
α= 0.7 is (i1, i2, i3, i4) = (4,2,1,3), i.e., the activity with the highest service level is placed at the
beginning of the encounter.
Author: The Design of Experiential Services14 Article submitted to ; manuscript no. (Please, provide the mansucript number!)
Table 2 Optimality Gap of VSFD Heuristicxi ∼Uniform[1,100] xi ∼Gamma(1,2) xi ∼Gamma(9,0.5)
n Mean Median Std Dev Minimum Maximum Mean Median Std Dev Minimum Maximum Mean Median Std Dev Minimum Maximum6 0.77% 0.44% 0.95% 0.00% 4.60% 0.53% 0.15% 0.84% 0.00% 5.97% 0.45% 0.18% 0.61% 0.00% 3.68%7 1.17% 0.86% 1.03% 0.00% 4.29% 1.15% 0.61% 1.34% 0.00% 5.92% 0.75% 0.49% 0.82% 0.00% 3.57%8 1.43% 0.93% 1.36% 0.00% 5.68% 1.23% 0.89% 1.34% 0.00% 7.47% 0.99% 0.68% 1.03% 0.00% 5.59%9 1.88% 1.44% 1.69% 0.00% 7.59% 1.52% 1.19% 1.39% 0.00% 6.14% 1.26% 1.16% 1.18% 0.00% 5.71%10 1.92% 1.39% 1.75% 0.00% 7.68% 1.61% 1.22% 1.45% 0.00% 6.89% 1.36% 0.87% 1.43% 0.00% 6.02%
4.1.2. Computational Study. VSFD can be formulated as a mixed integer non-linear pro-
gramming problem (see Appendix B). Because commercial solvers failed to return a feasible solution
within the set time limit for n≥ 9,1 we propose a heuristic based on Proposition 1.
The heuristic assigns activities sequentially, going backwards in time. It starts allocating activ-
ities in decreasing order of service levels, starting from the activity with the highest service level,
until time T −2m(α,w) is reached. At that point, it allocates the remaining activities in increasing
order of service level, starting from the activity with the lowest service level. Doing so results in a
U-shaped sequence, or in an only increasing sequence if T − 2m(α,w)< 0. The formal algorithm
appears in Appendix C.
We tested the performance of the heuristic on small problem instances that can be solved opti-
mally by exhaustive enumeration. We generated the problem instances with α and w uniformly
drawn between 0 and 1 such that the condition T > 2m(α,w) is satisfied since an increasing
sequence is known to be optimal otherwise. The values of ti’s were uniformly drawn from the
interval [1,100], and the service levels (xi’s) were drawn from a Uniform [1,100] or from a Gamma
distribution with shape (k) and scale (θ) parameters, with either k = 1 and θ = 2 or k = 9 and
θ= 0.5, respectively generating an exponential distribution and a bell-shaped distribution. Table 2
reports the mean, median, standard deviation, minimum, and maximum suboptimality gaps of our
heuristic relative to the optimal solution found through complete enumeration, over 100 randomly
generated instances for each value of n.
Overall, Table 2 shows that the average suboptimality gap for this heuristic is within 2%, and
the maximum gap is within 8%. The suboptimality gaps naturally increase with n, but the dete-
rioration in performance remains moderate relative to the exponential increase in run time of the
enumeration method. Moreover, the gaps appear to be robust to the distribution of service levels.
4.2. Fixed Sequence and Variable Duration (FSVD)
FSVD comprises all service encounters that have a fixed sequence of activities but offer the flex-
ibility to alter their duration. For instance, a spa or dental clinic may have a fixed protocol of
activities, but their individual durations can be varied. The FSVD formulation can be obtained
from the SPDP formulation when P is a singleton set.
1 We solved VSFD with BARON and LINDOGlobal available on NEOS with a maximum run time of 500,000 secondsfor n= 6, . . . ,10.
Author: The Design of Experiential ServicesArticle submitted to ; manuscript no. (Please, provide the mansucript number!) 15
In general, FSVD is neither a convex or concave optimization problem: When the sequence of
activities is increasing in service levels, i.e., xi1 < · · · < xin , the objective function of FSVD is
pseudoconcave in ti’s, whereas when it is decreasing, i.e., xi1 > · · ·> xin , the objective function is
pseudoconvex in ti’s (see Lemma A-4 in Appendix A) with potentially multiple local optima.
4.2.1. Structural Results. For a fixed sequence of activities, memory decay allocates dura-
tion to the activity with the highest service level, whereas acclimation aims at allocating duration
so as to induce a steep rise or gradual decline in service levels. When both act together, the design
pattern for pure acclimation tends to shift closer to the end of the encounter due to memory decay.
Since a service encounter may be composed of increasing and decreasing subsequences of activities,
the following proposition characterizes the optimal duration allocation across a subsequence of
activities.
Proposition 2. For a fixed subsequence of activities (xl, . . . , xr) where l≥ 1, the optimal dura-
tion allocation is such that:
1. Suppose that T < m(α,w), and that there exists an activity m, l < m < r, such that t?m > τm.
Then if xl < · · ·<xr, then t?j = τ j, j =m+ 1, . . . , r; and if xl > · · ·>xr, t?j = τ j, j = l, . . . ,m− 1.
2. Suppose that∑n
i=r τ i <m(α,w)< T . Then if xl < · · ·< xr such that t?l = τ l, t?r = τ r, then t?i =
τ i,∀i, l < i < r; and if xl > · · ·>xr such that t?l > τ l, t?r > τ r, then t?i > τ i,∀i, l < i < r.
3. Suppose that m(α,w)<∑n
i=r τ i and that there exists an activity m, l <m< r, such that t?m > τm.
Then if xl < · · ·<xr, then t?j = τ j, j = l, . . . ,m− 1; and if xl > · · ·>xr, t?j = τ j, j =m+ 1, . . . , r.
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Figure 5 Optimal duration allocation for increasing subsequence. Shaded activities are allocated duration above
their lower bound.
We illustrate Proposition 2 by considering two service encounters: (i) with all increasing service
levels (see Figure 5) and (ii) with all decreasing service levels (see Figure 6). In the figures, the
shaded activities are allocated duration above their lower bound, i.e., t?i > τ i.
Author: The Design of Experiential Services16 Article submitted to ; manuscript no. (Please, provide the mansucript number!)
!"#$%
&'(%
)%
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+%
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Figure 6 Optimal duration allocation for decreasing subsequence. Shaded activities are allocated duration above
their lower bound.
Proposition 2 shows that when either α or w is low (i.e., T <m(α,w)), then duration should be
allocated to the activities in decreasing order of service levels (see (i) in Figures 5 and 6). When
the customer gives equal weight to all instantaneous utilities (w ≈ 0), allocating greater duration
to activities with higher service levels ensures a steep rise or a gradual fall in service level. When
there is no acclimation (i.e., α ≈ 0), this design strategy brings closer to the end of the service
encounter the activities associated with a high service level, so that the customer does not heavily
discount them.
By contrast when both α and w are very high (i.e., when τn >m(α,w)), then duration should be
allocated in increasing order of service levels so as to push the steep rise and gradual fall in service
level towards the end of the encounter (see (iii) in Figures 5 and 6). Large memory decay implies
that the customer forgets the initial experience of the encounter. Therefore with an increasing
sequence, it is indeed optimal to allocate duration to the activities in the beginning so as to bring
closer to the end the steep rise in service levels. Conversely in a decreasing sequence, it is optimal
to allocate duration at the end so as to incur steep drops in service levels initially and make the
gradient of service level near the end tend to zero.
In the intermediate cases when τ in <m(α,w)<T , duration can be allocated in any of the three
possible ways shown in Figures 5 and 6, respectively for an increasing and a decreasing sequence.
However, duration should never be allocated only to the activities in the middle of an increasing
sequence or only to the activities at the beginning and end of a decreasing sequence since this
allocation strategy would not maximize the gradient of service levels near the end.
4.2.2. Computational Study. We solved FSVD with LINDOGlobal. The commercial solver
returned the optimal or a near optimal solution (with a 1% optimality gap) for small problem
instances, but it failed to solve larger problem instances (e.g., n≥ 24) with decreasing subsequences
of service levels within one week. Since the objective function of FSVD with a decreasing sequence
Author: The Design of Experiential ServicesArticle submitted to ; manuscript no. (Please, provide the mansucript number!) 17
Table 3 Heuristic Solution vs LINDOGlobal Solution for FSVD(S(t?solver)−S(t
?heuristic))×100
|S(t?solver
)| Mean Run Time No. of Aborted
n Mean Median Std Dev Minimum Maximum Heuristic Solver Instances by Solver
Incr
easi
ng 8 0.00% 0.00% 0.00% 0.00% 0.00% 0.07 sec 0.10 sec 0 cases
16 0.00% 0.00% 0.00% 0.00% 0.00% 0.08 sec 0.29 sec 0 cases
24 0.00% 0.00% 0.00% 0.00% 0.00% 0.09 sec 1.53 sec 0 cases
32 0.00% 0.00% 0.00% 0.00% 0.00% 0.11 sec 3.15 sec 0 casesD
ecre
asin
g 8 0.00% 0.00% 0.00% 0.00% 0.01% 0.08 sec 0.49 sec 0 cases
16 0.01% 0.00% 0.29% 0.00% 1.25% 0.10 sec 1092.55 sec 0 cases
24 0.02% 0.00% 0.10% 0.00% 1.01% 0.11 sec 1575.66 sec 13 cases
32 0.11% 0.00% 0.30% 0.00% 1.04% 0.12 sec 1888.51 sec 19 cases
U-s
hap
ed
8 0.00% 0.00% 0.00% 0.00% 0.02% 0.08 sec 1.32 sec 0 cases
16 0.00% 0.00% 0.00% 0.00% 0.04% 0.11 sec 1.51 sec 0 cases
24 0.00% 0.00% 0.00% 0.00% 0.07% 0.14 sec 100.01 sec 5 cases
32 0.00% 0.00% 0.00% 0.00% 0.07% 0.15 sec 210.03 sec 13 cases
Λ-s
hap
ed
8 0.00% 0.00% 0.00% 0.00% 0.00% 0.09 sec 2.10 sec 0 cases
16 0.00% 0.00% 0.00% 0.00% 0.00% 0.10 sec 2.31 sec 0 cases
24 0.00% 0.00% 0.00% 0.00% 0.04% 0.11 sec 139.11 sec 6 cases
32 0.00% 0.00% 0.00% 0.00% 0.06% 0.14 sec 256.02 sec 12 cases
is pseudoconvex (Lemma A-4), the maximization problem has potentially multiple local optima
when τ in <m(α,w)<T , and there is no guarantee that a solution can be obtained in a reasonable
amount of time.
We therefore propose a modified coordinate ascent heuristic to solve FSVD that takes less com-
putational time on average than LINDOGlobal. The heuristic starts with t0 = τ and iteratively
allocates time increments ∆t to the activity with the highest marginal return, i.e., arg maxi
∂S(t)
∂tiuntil the total duration is equal to T . The algorithm is detailed in Appendix C.
To compare the heuristic solution against the solution obtained from the solver, we considered
four patterns of sequences of service levels, namely, increasing, decreasing, U-shaped, and Λ-shaped.
For each pattern of sequence, we considered sequences of different lengths, i.e., n= 8,16,24,32. We
randomly generated 100 instances for each n. All instances were generated with α and w uniformly
drawn from (0,1) and with xi, τ i, τ i,∀i uniformly drawn from [1,100]. The total duration of the
service encounter was fixed to T =∑i
τ i +∑i
(τ i − τ i)/n. For each n, the U-shape sequence was
created by forming a decreasing subsequence with n/2 randomly selected activities and an increas-
ing subsequence with the remaining n/2 activities. The Λ-shape was constructed similarly. The
time increment for the heuristic was fixed as ∆t= 0.01.
Table 3 benchmarks the solution as well as the running times from the heuristic to the best solu-
tion found by LINDOGlobal, within a 1% optimality gap. All codes were implemented in MATLAB
7.12.0. and GAMS. LINDOGlobal solver was executed using the NEOS optimization server. The
computational tests on MATLAB were run on a workstation with a 2.26 GHz Intel Core 2 Duo
processor, 8 GB of RAM, and Mac OS X 10.6.8 as the operating system.
Apart from the sequences of activities with increasing service levels, which yield a concave max-
imization problem (Lemma A-4), all other sequences had some cases aborted by NEOS when the
Author: The Design of Experiential Services18 Article submitted to ; manuscript no. (Please, provide the mansucript number!)
solver failed to find an optimal or a near-optimal solution with a 1% optimality gap in 500,000
seconds (reported in the last column in Table 3). Only the cases for which the solver returned an
optimal or near-optimal solution were included when computing the performance gaps.
Overall, the FSVD heuristic gives solutions that closely match the solver solutions (within 1.5
%) in a significantly smaller run time (i.e., in less than 1 second). Moreover, the computation time
of the coordinate ascent heuristic appears less sensitive to the size or specific shape of the sequence.
4.3. Variable Sequence and Variable Duration (VSVD)
Some service encounters, like personal fitness classes, allow the provider to change both the sequence
and the duration of activities. We next consider the VSVD problem without precedence constraints,
i.e., we let P in constraint (9) in SPDP contain all permutations of service activities.
4.3.1. Structural Results. Combining Propositions 1 and 2 leads to the following charac-
terization of the optimal sequence and duration allocation when both are decision variables.
Corollary 2. In VSVD,
1) If T < 2m(α,w), a sequence with increasing service levels is optimal. Moreover,
1.1) If T <m(α,w) the optimal duration allocation is characterized by an index k such that
tij = τ ij ,1≤ j ≤ k− 1, tik = τ, τ ik < τ < τ ik , and tij = τ ij , k < j ≤ n.
1.2) If τ in <m(α,w)≤ T , the optimal duration allocation is such that, if t?il = τ il and t?ir = τ ir
then t?ij = τ ij ,∀j, l < j < r.
1.3) If m(α,w)< τ in, then the optimal duration allocation is characterized by an index k such
that t?ij = τ ij , j = 1, · · · , k− 1, t?ik = τ, τ ik < τ < τ ik , and t?ij = τ ij , j = k+ 1, · · · , n .
2) If 2m(α,w)<T , a U-shaped sequence bottoming out at activity k is optimal such that:
for xik < · · ·<xin, if ∃l, r such that t?il = τ il and t?ir = τ ir then t?ij = τ ij ,∀j, k≤ l < j < r≤ n
for xi1 > · · ·>xik , if ∃l, r such that t?il > τ il and t?ir > τ ir then t?ij > τ ij ,∀j,1≤ l < j < r≤ k.
Overall, Corollary 2 shows that the optimal solution for VSVD is obtained by constructing the
steepest positive gradient near the end.
4.3.2. Computational Study. VSVD is at least as complex as VSFD since an instance of
VSFD can be constructed from VSVD by setting τ i = τ i,∀i. When T < 2m(α,w), the sequence
of service levels is increasing, and therefore the duration allocation problem is pseudoconcave (see
Lemma A-4 in Appendix A). When T > 2m(α,w), Corollary 2 shows that a U-shaped sequence is
optimal and the duration allocation problem may therefore have multiple local optima. In general,
commercial solvers like BARON and LINDOGlobal on NEOS failed to return a feasible solution
in one week for many instances with n ≥ 10. We therefore propose a heuristic for SPDP when
T > 2m(α,w) and we present in Appendix D an upper bound on SPDP against which we benchmark
Author: The Design of Experiential ServicesArticle submitted to ; manuscript no. (Please, provide the mansucript number!) 19
Table 4 Optimality Gaps for the VSVD heuristic with T > 2m(α,w)
Solver Upper Boundn Mean Median Std Dev Minimum Maximum Mean Median Std Dev Minimum Maximum
5 0.72% 0.75% 0.35% 0.00% 1.35% 3.11% 3.92% 2.51% 0.11% 7.13%6 1.09% 0.71% 0.81% 0.00% 2.46% 3.77% 3.01% 2.95% 0.25% 8.31%7 1.42% 1.30% 1.33% 0.08% 4.89% 3.94% 3.31% 3.10% 0.22% 8.91%8 2.19% 2.01% 1.47% 0.22% 5.56% 4.02% 3.80% 3.68% 0.30% 10.01%9 2.67% 2.32% 1.41% 0.52% 5.76% 5.11% 4.10% 3.18% 0.73% 10.19%
the performance of our heuristic.
The heuristic first finds a U-shaped sequence and then allocates duration for the corresponding
FSVD problem so as to maximize the gradient of service levels at the end of the encounter. We
next detail the two steps of the heuristic.
Step 1: Sequencing. The heuristic constructs a U-shaped sequence. Without loss of generality, we
assume that x1 ≥ · · · ≥ xn. The following optimization problem freely allocates duration among all
activities, so as to maximize customer satisfaction by generating a steep gradient in service level.
maxtS(t) =
n−1∑i=1
(xi−xi+1)Φ(i∑
j=1
tj)
subject to 0≤ ti ≤ τ i,∀in∑i=1
ti ≤ T.
Compared to SPDP, the lower bounds on activity duration (7) are set to zero and the constraint on
the total duration of the service encounter (8) is relaxed to an inequality. Moreover, the sequence
is fixed. If an activity is allocated zero duration, it suggests that it does not contribute to creating
a steep gradient and should therefore be moved at the beginning of the encounter. Accordingly, let
L be the set of activities that are allocated zero duration, i.e., if t? denotes the optimal solution
then L = {j|t?j = 0}. Therefore, for all j ∈ L, we order activities in decreasing order of xj’s and
for all j /∈ L, we order activities in increasing order of xj’s. The combination of the decreasing
subsequence followed by the increasing subsequence forms a U-shaped sequence.
Step 2: Duration Allocation. For the U-shaped sequence obtained in Step 1, we apply the coordinate
ascent method described in §4.2.2 and formally presented in Appendix C to allocate duration.
To test the performance of the heuristic against the optimal solution, when computable, and an
upper bound (Appendix D) otherwise, we randomly generated problem instances with α and w
uniformly drawn from [0,1], and xi, τ i, τ i,∀i uniformly drawn from [1,100]. The total duration of
the service encounter was set as T =∑i
τ i +∑i
(τ i − τ i)/n and the condition T > 2m(α,w) was
verified.
Table 4 compares the solution obtained from the heuristic with the optimal solution and the
upper bound when n≤ 9. For every instance of size n, the gaps were evaluated over 100 instances,
Author: The Design of Experiential Services20 Article submitted to ; manuscript no. (Please, provide the mansucript number!)
Table 5 Optimality Gaps for the VSVD heuristic with T > 2m(α,w)
Upper Boundn Mean Median Std Dev Minimum Maximum
10 5.99% 4.91% 3.18% 0.75% 10.53%15 6.91% 5.95% 2.97% 0.83% 11.79%20 7.09% 6.81% 3.01% 1.03% 13.86%25 8.01% 7.77% 3.86% 1.99% 14.01%30 9.81% 8.97% 3.55% 2.08% 14.19%
and the corresponding sample was used for evaluating the mean, median, standard deviation,
minimum, and maximum optimality gap. We observe that the gaps against the optimal solution
are within 3% on an average, whereas against the upper bound the average gaps are within 6%.
Table 5 compares the performance of the heuristic against the upper bound for larger instances,
over a sample of 100 instances for each n. The gap of the heuristic is on average less than 10%
of the upper bound, although it increases with n. Given that the upper bound appears to be less
tight for larger values of n from Table 4, the performance of the heuristic could be within a few
percent of the optimal solution, even for large problem instances.
5. Conclusion
In this paper, we develop an analytical model and heuristics to sequence and time activities in
a service encounter to maximize ex-post customer satisfaction in the presence of memory decay
and acclimation. Memory decay favors high service levels at the end. Whereas, acclimation favors
maximum gradient of service level and thereby induces steep rise or gradual falls.
We show that, individually, the two phenomena give the same monotonic service design. However,
when considered jointly, they can act as opposing forces. In particular, if either memory decay or
acclimation is low, it is optimal to sequence activities in increasing order of service levels and to
lengthen the duration of activities with the highest service levels. On the other hand, when both
memory decay and acclimation are high, it is optimal to sequence activities in a U-shape fashion
and to lengthen the duration of activities with the lowest service levels, located in the middle of
the sequence.
This paper is the first step toward designing service encounters using an analytical framework,
by capturing behavioral phenomena that influence the evaluation of past experiences by customers.
We have thereby developed parsimonious models to obtain insights from this framework. Ample
opportunities exist for research to study the effect on service design of customer heterogeneity,
endogenous choices of service levels and encounter duration, multidimensional service intensities,
other behavioral phenomena (e.g., satiation, loss aversion), and the design implications on the
service provider’s profit. We hope that this paper will induce further interest in engineering service
experiences to maximize customer satisfaction.
Author: The Design of Experiential ServicesArticle submitted to ; manuscript no. (Please, provide the mansucript number!) 21
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Appendix A: Proofs
Lemma A-1. The function Φ(t) is strictly pseudoconcave in t with a stationary point at m(α,w).
Proof. Suppose α<w. We have:
Φ′(t) =e−αt(−α)− e−wt(−w)
w−α=e−wt(w−αet(w−α))
w−α> 0⇔w>αet(w−α),∀w>α⇔ lnw− lnα
w−α> t.
Since Φ′(t)> 0 when t <m(α,w), Φ′(t) = 0 when t=m(α,w), and Φ′(t)< 0 when t >m(α,w), Φ(t) is strictly
pseudoconcave. Note that the result is symmetric for w<α. �
Lemma A-2. The function f(Tik) = Φ(T −Tik + tik)−Φ(T −Tik) is strictly pseudoconvex in Tik .
Proof. Suppose α<w. We have:
f ′(Tik) =αe−α(T−Tik+tik )−we−w(T−Tik+tik )
w−α− αe−α(T−Tik )−we−w(T−Tik )
w−α
Consider t0 = T − 1(w−α) ln w(1−e−wtik )
α(1−e−αtik ). Since f ′(Tik) < 0 when Tik < t0, f ′(Tik) = 0 when Tik = t0, and
f ′(Tik)> 0 when Tik > t0, f(Tik) is strictly pseudoconvex. Note that the result is symmetric for w<α. �
Author: The Design of Experiential Services24 Article submitted to ; manuscript no. (Please, provide the mansucript number!)
Lemma A-3. Φ(t) is concave ∀t∈ [0,2m(α,w)] and convex ∀t∈ [2m(α,w), T ].
Proof. Since, Φ′′(t) = α2e−αt−w2e−wt
w−α , therefore, Φ′′(t) ≤ 0,∀t ∈ [0,2m(α,w)] and Φ′′(t) ≥ 0,∀t ∈
[2m(α,w), T ]. Hence, Φ(t) is concave ∀t∈ [0,2m(α,w)] and convex ∀t∈ [2m(α,w), T ]. �
Proof of Proposition 1. The proof uses Lemma A-3. Let S? be the satisfaction obtained from the optimal
sequence (i?1, . . . , i?n). For any j, let S?j be the satisfaction obtained by interchanging i?j−1 and i?j in the optimal
sequence. Therefore,
S?−S?j = (xi?j−1−xi?
j)(
(−Φ(T j) + Φ(T j+1)) + (Φ(T j−1)−Φ(ti?j−1
+T j+1))).
We next consider three scenarios: (i) when the two activities start and end within [0, T − 2m(α,w)], (ii)
when the two activities start and end within [T − 2m(α,w), T ], and (iii) when j = n.
(i) If i?j−1 and i?j start and finish within [0, T − 2m(α,w)], then by Lemma A-3, the corresponding Φ′(t) is
increasing ∀t∈ [2m(α,w), T ]. Hence,
Φ(T j)−Φ(T j+1)<Φ(ti?j−1
+T j)−Φ(ti?j−1
+T j+1).
Therefore, by optimality of (i?1, . . . , i?n), we have xi?
j<xi?
j−1.
(ii) If i?j−1 and i?j start and finish within [T − 2m(α,w), T ], then by Lemma A-3, the corresponding Φ′(t) is
decreasing ∀t∈ [0,2m(α,w)]. Hence,
Φ(T j)−Φ(T j+1)>Φ(ti?j−1
+T j)−Φ(ti?j−1
+T j+1).
Therefore, by optimality of (i?1, . . . , i?n), we have xi?
j>xi?
j−1.
(iii) By interchanging the last two activities i?n−1, i?n, we obtain a suboptimal sequence with corresponding
satisfaction S?n. Therefore,
S?−S?n = (xi?n −xi?n−1)(
Φ(ti?n) + Φ(ti?n−1
)−Φ(ti?n−1
+ ti?n))
=xi?n −xi?n−1
w−α
{(1− e−wti?n−1 )(1− e−wti?n )− (1− e−αti?n−1 )(1− e−αti?n )
}.
Hence, by optimality of (i?1, . . . , i?n), we have xi?n >xi?n−1
. �
Lemma A-4. Suppose that xi is the service level of the activity at the ith position in the sequence. When
x1 < · · ·<xn, FSVD is strictly pseudoconcave, whereas, when x1 > · · ·>xn, FSVD is strictly pseudoconvex.
Proof. We show that the corresponding Hessian (H) for S(t) is negative definite for pseudoconcave
and positive definite for pseudoconvex, at a stationary point of S(t). By equation (6), S(t) =∑n
i=1(xi −
xi−1)Φ(T i). Taking the partial derivatives of S(t) with respect to ti,∀i we obtain
∂S(t)
∂ti=
i∑j=1
(xj −xj−1)Φ′(T j),∀i.
So the stationary point has to satisfy the following set of equations:
n∑j=i
tj =m(α,w),∀i (A-1)
Author: The Design of Experiential ServicesArticle submitted to ; manuscript no. (Please, provide the mansucript number!) 25
Therefore, t? = (0, . . . ,0,m(α,w)) is the unique stationary point satisfying equations (A-1). We evaluate the
Hessian at t? to obtain for any y ∈<n
yTHy = α(αw
) αw−α
(−(x2−x1)(
n∑i=2
yi)2− (x3−x2)(
n∑i=3
yi)2 · · · − (xn−xn−1)y2n
).
Clearly, for x1 < · · ·<xn, yTHy< 0, and for x1 > · · ·>xn, yTHy> 0 ∀y ∈<n. �
Lemma A-5. In FSVD the necessary conditions for optimality are given by
∂S(t)
∂ti= λi +µ,∀i∈ IU
∂S(t)
∂ti= µ,∀i∈ IM
∂S(t)
∂ti= µ−λi,∀i∈ IL,
(A-2)
and constraints (7) and (8) where µ ∈ <, λi ≥ 0 and λi ≥ 0. IU , IM , IL are three disjoint sets such that
I = IU ∪ IM ∪ IL and ti = τ i,∀i∈ IU , ti = τ, τ i < τ < τ i,∀i∈ IM and ti = τ i,∀i∈ IL.
Proof. The necessary conditions for optimality of t? are given by the following Karush-Kuhn-Tucker
conditions (Boyd and Vandenberghe 2004):
1. The stationarity condition gives ∂S(t)
∂ti−µ+λi−λi = 0,∀i at t?.
2. Complementary slackness gives µ(T −∑n
i=1 t?i ) = 0, λi(τ i− t?i ) = 0,∀i and λi(t
?i − τ i) = 0,∀i.
3. Primal feasibility implies t? satisfies the constraints (7) and (8).
4. Dual feasibility implies µ∈< and λi, λi ≥ 0,∀i.From the complementary slackness condition, λi = 0,∀i∈ IU and λi = 0,∀i∈ IL and both λi = 0, λi = 0,∀i∈IM . �
Proof of Proposition 2. We use Lemmas A-1 and A-5 for this proof. We have
∂S(t?)
∂ti=
i∑j=1
(xj −xj−1)Φ′(T j),∀i,
where x0 = b(0).
1. Suppose that xl < · · ·<xr. Since T <m(α,w), Φ′(t)> 0,∀t∈ [0, T ) by Lemma A-1, and therefore ∂S(t)
∂tl<
· · · < ∂S(t)
∂tr,∀t ≥ (τ1, . . . , τn). Hence, from condition (A-2), if tm > τm, ∂S(t)
∂tm= µ or ∂S(t)
∂tm= µ + λm at t?.
Therefore, ∂S(t)
∂tj= µ+ λj , j =m+ 1, . . . , r at t?. From Lemma A-5, t?j = τ j , j =m+ 1, . . . , r. The proof for a
decreasing subsequence is similar.
2. We show the result by contradiction. Suppose that xl < · · ·< xr and that tl = τ l and tr = τ r. Therefore,
condition (A-2) implies that ∂S(t?)
∂tl= µ− λl and ∂S(t?)
∂tr= µ− λr. If for any i, l < i < r, we have t?i > τ i, then
∂S(t?)
∂ti= µ or ∂S(t?)
∂ti= µ+ λi, by condition (A-2). If
∑n
h=i t?h ≤m(α,w), then Φ′(t)> 0 for t≤
∑n
h=i+1 t?h <
m(α,w) from Lemma A-1 and therefore ∂S(t?)
∂tr> ∂S(t?)
∂ti, thereby a contradiction. If
∑n
h=i t?h >m(α,w), then
Φ′(t)< 0 for t≥∑n
h=i t?h from Lemma A-1 and therefore ∂S(t?)
∂tl> ∂S(t?)
∂ti, thereby a contradiction. Therefore,
from Lemma A-5, t?i = τ i,∀i, l < i < r. The proof for a decreasing subsequence is similar.
3. Suppose that xl < · · · < xr. When∑n
i=r τ i > m(α,w), we have Φ′(t) < 0,∀t ∈ (∑n
i=r τ i, T ] by Lemma
A-1, and therefore ∂S(t?)
∂tl> · · · > ∂S(t?)
∂tr,∀t ≥ (τ1, . . . , τn). From condition (A-2), since tm > τm, ∂S(t?)
∂tm= µ
or ∂S(t?)
∂tm= µ + λm at t?. Therefore, ∂S(t?)
∂tj= µ + λj , j = l, . . . ,m − 1 at t?. Therefore, from Lemma A-5,
t?j = τ j , j = l, . . . ,m− 1. The proof for a decreasing subsequence is similar. �
Author: The Design of Experiential Services26 Article submitted to ; manuscript no. (Please, provide the mansucript number!)
Appendix B: Mathematical Programming Formulation of SPDP
Define zij as a binary variable indicating if activity i directly or indirectly precedes j and define ci as the
completion time of activity i. We keep constraints (7)-(8) and express constraint (9) in SPDP in terms of
zij ’s and ci’s. Constraints (B-1)-(B-2) ensure that for any pair of activities, one has to precede the other.
Let Q be the set of precedence constraints between pairs of activities. Then constraint (B-3) ensures that
activity j precedes activity k for all such pairs of precedence constraints in Q. Constraint (B-4) ensures that
the completion time of an activity must be greater than its duration and smaller than the total duration of
the service encounter. Finally, constraint (B-5) ensures that for any pair of activities, the completion time
of the preceding activity is less than that of the succeeding activity.
maxz,t,c
S(z, t,c) =
n∑i=1
xi (Φ(T − ci + ti)−Φ(T − ci))
subject to constraints (7)-(8)
zij + zji = 1,∀i, j, i 6= j (B-1)
zii = 0,∀i (B-2)
zjk = 1∀(j, k)∈Q (B-3)
ti ≤ ci ≤ T,∀i (B-4)
ci ≤ cj − tj + (1− zij)T,∀i, j, i 6= j (B-5)
zij ∈ {0,1},∀i, j.
Appendix C: Heuristics for VSFD and FSVD
Algorithm 1 Heuristic for VSFD1: Sort activities such that x1 > .. > xn.2: Comment: opt is the optimal sequence.3: Initialize: time= 0, i= n, j = 1, k= n.4: while time≤ 2m(α,w) do5: opt[i]=x[j]6: i=i-17: time=time+t[j]8: j=j+19: end while
10: while i > 0 do11: opt[i]=x[k]12: i=i-1; k=k-1;13: end while
Author: The Design of Experiential ServicesArticle submitted to ; manuscript no. (Please, provide the mansucript number!) 27
Algorithm 2 Heuristic for FSVD1: Comment: ∆t is the step size. I is the set of activities such that ti = τ i,∀i∈ I.2: Initialize: t0 = τ ; m= 0; I = ∅3: while T −
∑ni=1
tmi > 0 do
4: i= arg maxk,k/∈I
∂S(tm)
∂tk5: ∆ti = min{∆t, T −
∑nj=1
tmj , τ i− tmi }6: if ∆ti = 0 then7: I = I ∪{i}8: end if9: tm+1
i = tmi + ∆ti10: m=m+111: end while
Appendix D: Upper Bound Formulation
We introduce an alternate formulation of SPDP that allows the sequence to be either U-shaped or mono-
tonically increasing in service levels. Without loss of generality, we assume that x1 ≥ · · · ≥ xn. A U-shaped
sequence consists of a decreasing subsequence followed by an increasing one. Let ysi ∈ {0,1},∀i,∀s be a vari-
able indicating if activity i belongs to subsequence s, where s = l for the decreasing subsequence on the
left and s = r for the increasing subsequence on the right of the U-shape. The variables tsi ,∀i,∀s denote
the duration allocated to activity i when it is in subsequence s. With these decision variables, SPDP-U is
formulated as:
maxyl,yr,tl,tr
S(yl,yr, tl, tr) =
n∑i=1
(xi−xi−1)Φ(T −i−1∑j=1
tljylj) +
n−1∑i=1
(xi−xi+1)Φ(
i∑j=1
trjyrj )
subject to yli + yri = 1,∀i (D-1)
1≤n−1∑i=1
yri (D-2)
ysi τ i ≤ tsi ≤ ysi τ i, s= l, r,∀i (D-3)n∑i=1
(ylit
li + yri t
ri
)= T (D-4)
yli, yri ∈ {0,1},∀i. (D-5)
where x0 = b(0). The objective function consists of a decreasing and an increasing subsequences, such that
each subsequence is composed of all the activities. When tli = 0, the coefficient of xi in the decreasing
subsequence is equal to zero, i.e., no weight is associated with activity i in the decreasing subsequence.
Similarly if tri = 0, no weight is associated with activity i in the increasing subsequence. Constraints (D-1)
allow an activity to be either a part of the decreasing or a part of the increasing subsequence. Constraint
(D-2) eliminates the possibility of having only a decreasing sequence of activities. Therefore, constraints
(D-1)-(D-2) ensure a U-shaped or a monotonically increasing sequence. The remaining constraints (D-3)-
(D-4) correspond to the duration constraints. In particular, constraints (D-3) ensure that if ysi = 0, then the
corresponding duration tsi = 0. This rules out the possibility of allocating duration to an activity when it
does not belong to the corresponding subsequence.
Lemma D-1. When T > 2m(α,w), SPDP is equivalent to SPDP-U.
Author: The Design of Experiential Services28 Article submitted to ; manuscript no. (Please, provide the mansucript number!)
Proof. We establish the equivalence result in two steps by showing that (i) any optimal solution for SPDP
is feasible in SPDP-U and has the same objective value, and (ii) any optimal solution for SPDP-U is feasible
in SPDP and has the same objective value.
(i) From Proposition 1 we know that any optimal solution of SPDP for T > 2m(α,w) will either be
monotonically increasing or U-shaped in service levels. Let ((i?1, . . . , i?n), t?) be an optimal solution of SPDP
such that xi?1 ≥ · · · ≥ xi?k ≥ xi?k+1≤ xi?
k+2≤ · · · ≤ xi?n , for 0≤ k≤ n− 2. We construct a corresponding feasible
solution for SPDP-U by assigning yrij = 0, trij = 0, ylij = 1, tlij = t?ij ,∀j = 1, . . . , k and ylij = 0, tlij = 0, yrij =
1, trij = t?ij ,∀j = k+ 1, . . . , n.
By construction, (D-1), (D-2), and (D-5) are satisfied. Since∑n
j=1 t?ij
= T and τ ij ≤ t?ij≤ τ ij ,∀j, (D-3)-
(D-4) are also satisfied. Hence, such a solution is feasible in SPDP-U.
We next show that the objective values are equal. The objective value for SPDP at the optimal point is
given by SSPDP ((i?1, . . . , i?n), t?) =
∑n
j=1(xi?j−xi?
j−1)Φ(t?ij + · · ·+ t?in)
=∑k+1
j=1 (xi?j−xi?
j−1)Φ(∑n
p=j t?ip
) +∑n
j=k+2(xi?j−xi?
j−1)Φ(∑n
p=j t?ip
)
=
k+1∑j=1
(xi?j−xi?
j−1)Φ(T −
j−1∑p=1
t?ip) +
n∑j=k+2
(xi?j−xi?
j−1)Φ(
n∑p=j
t?ip)
(D-6)
=
k+1∑j=1
(xi?j−xi?
j−1)Φ(T −
j−1∑p=1
tlip) +
n∑j=k+2
(xi?j−xi?
j−1)Φ(
n∑p=j
trip)
(D-7)
=
k+1∑j=1
(xi?j−xi?
j−1)Φ(T −
ij−1∑p=1
tlp) +
n∑j=k+2
(xi?j−xi?
j−1)Φ(
ij∑p=1
trp)
(D-8)
=
n∑j=1
(xj −xj−1)Φ(T −j−1∑p=1
tlp) +
n−1∑j=1
(xj −xj+1)Φ(
j∑p=1
trp)
(D-9)
=
n∑j=1
(xj −xj−1)Φ(T −j−1∑p=1
tlpylp) +
n−1∑j=1
(xj −xj+1)Φ(
j∑p=1
trpyrp)
(D-10)
= SSPDP−U(yl,yr, tl, tr).
Equality (D-6) is because∑n
j=1 t?ij
= T and (D-7) is because tlij = t?ij ,∀j ≤ k, trij
= t?ij ,∀j ≥ k+ 1. Equalities
(D-8)-(D-9) are due to tlij = 0,∀j ≥ k+ 1, trij = 0,∀j ≤ k, and x1 ≥ · · · ≥ xn by assumption. Finally, equality
(D-10) is due to tsp > 0 if and only if ysp = 1,∀p, and for s= l, r by construction.
(ii) Let (yl?,yr?, tl?, tr?) be an optimal solution of SPDP-U. We construct a sequence of activities (i1, . . . , in)
from (yl?,yr?) and the corresponding duration (ti1 , . . . , tin) as follows.
We define i0 = 0. Let k =∑n−1
j=1 yl?j . For j = 1, . . . , k, let ij = min{p > ij−1 such that yl?p = 1} and set tij =
tl?p . We then assign ik+1 = n and set tik+1= yl?n t
l?n + yr?n t
r?n . Finally, for j = k + 2, . . . , n, let ij = max{p <
ij−1 such that yr?p = 1} and set tij = tr?p .
By equation (D-1), each activity is such that either yl?j = 1 or yr?j = 1. Hence, by construction, all activities
are allocated and no activity is allocated twice, i.e., (i1, . . . , in) represents a sequence of activities. Since,
x1 ≥ · · · ≥ xn, we have xi1 ≥ · · · ≥ xik+1≤ xik+2
≤ · · · ≤ xin . Because of constraints (D-1) and (D-3), τ ij ≤
Author: The Design of Experiential ServicesArticle submitted to ; manuscript no. (Please, provide the mansucript number!) 29
tij ≤ τ ij ,∀j. Moreover by (D-4),∑n
j=1 tij = T . Hence, this constructed solution is feasible for SPDP.
We next show that the objective value of SPDP-U is the same as SPDP at (yl?,yr?, tl?, tr?). The objective
value for SPDP-U at optimality is given by
SSPDP−U(yl?,yr?, tl?, tr?)
=
n∑j=1
(xj−xj−1)Φ(T−j−1∑p=1
tl?p yl?p )+
n−1∑j=1
(xj−xj+1)Φ(
j∑p=1
tr?p yr?p )
=
n∑j=1
(xj−xj−1)Φ(
n∑p=j
tl?p yl?p +
n∑p=1
tr?p yr?p )+
n−1∑j=1
(xj−xj+1)Φ(
j∑p=1
tr?p yr?p ), since
n∑j=1
yl?j tl?j +yr?j t
r?j = T
=
k+1∑j=1
(xij−xij−1)Φ(
n∑p=j
tip)+
n∑j=k+2
(xij−xij−1)Φ(
n∑p=j
tip), since ts?i > 0 if and only if ys?i = 1, s= l, r, by (D-3)
and by construction xi1 ≥ · · · ≥ xik+1≤ · · · ≤ xin
=
n∑j=1
(xij −xij−1)Φ(
n∑p=j
tip)
= SSPDP ((i1, . . . , in), t). �
Upper bound on SPDP-U. The upper bound is obtained by dropping constraints (D-1)-(D-2), relaxing
constraint (D-3) by replacing the lower bounds τ i by zero ∀i, and relaxing (D-4) by converting the equality
sign to an inequality sign, i.e.,∑n
i=1 (ylitli + yri t
ri )≤ T . We furthermore relax constraint (D-5) by allowing yi to
be a continuous variable between 0 and 1, and then introduce a new set of variables, namely, ts
i = tsiysi , s= l, r.
After dualizing the relaxed constraint (D-4), we obtain the upper bound UB = minλ≥0
UL(λ) +UR(λ) +λT , in
which
UL(λ) = maxtUL(λ, t) =
n∑i=1
(xi−xi−1)Φ(T −i−1∑j=1
tj)−λn∑i=1
ti
subject to 0≤ ti ≤ τ i,∀i (D-11)
UR(λ) = maxtUR(λ, t) =
n−1∑i=1
(xi−xi+1)Φ(
i∑j=1
tj)−λn∑i=1
ti
subject to 0≤ ti ≤ τ i,∀i, (D-12)
in which we renamed the duration variables tl
i = ti,∀i in UL(λ) and tr
i = ti,∀i in UR(λ), for simplicity of
exposition. The Lagrangian relaxation UB = minλ≥0
UL(λ) + UR(λ) + λT is then solved by the subgradient
method (Fisher 2004).
Next we show that UR(λ) and UL(λ) can be solved in polynomial time. Without loss of generality, we
assume that the service levels of all activities are unique, i.e., x1 > · · ·>xn. If we have xi = xi−1, for some i,
we can delete activity i− 1 and increase the upper bound on ti to τ i + τ i−1 since
UL(λ, t) =
i−1∑k=1
(xk−xk−1)Φ(T −k−1∑j=1
tj) + (xi−xi−1)Φ(T −i−1∑j=1
tj) +
n∑k=i+1
(xk−xk−1)Φ(T −k−1∑j=1
tj)−λn∑i=1
ti
=
i−1∑k=1
(xk−xk−1)Φ(T −k−1∑j=1
tj) + 0 +
n∑k=i+1
(xk−xk−1)Φ(T −k−1∑j=1
j 6=i,i−1
tj − (ti−1 + ti))−λn∑i=1
ti
=UL(λ, (t1, . . . , ti−2,0, ti−1 + ti, ti+1, . . . , tn)).
and similarly, for UR(λ, t).
Author: The Design of Experiential Services30 Article submitted to ; manuscript no. (Please, provide the mansucript number!)
Lemma D-2. Without loss of generality, suppose that x1 > · · ·>xn and T > 2m(α,w). For any λ≥ 0 only
O(n2) points need to be considered to find an optimal solution for UL(λ).
Proof. Let t? be the optimal solution for UL(λ). Hence, t? should satisfy the first order optimality
conditions:
1. The stationarity condition gives ∂UL(λ,t)
∂ti+µ
i−µi = 0,∀i.
2. Complementary slackness gives µi(τ i− t?i ) = 0,∀i and µi(t?i ) = 0,∀i.
3. Primal feasibility implies t? satisfies the constraints (D-11).
4. Dual feasibility implies µi, µi ≥ 0,∀i.
These conditions reduce to the following:
∂UL(λ, t?)
∂ti= µi ≥ 0,∀i such that t?i = τ i
∂UL(λ, t?)
∂ti= 0,∀i such that 0< t?i < τ i
∂UL(λ, t?)
∂ti=−µ
i≤ 0,∀i such that t?i = 0.
(D-13)
We observe that for any two consecutive activities the following holds
∂UL(λ, t?)
∂ti−1− ∂UL(λ, t?)
∂ti=−(xi−xi−1)Φ′(T −
i−1∑p=1
tp),∀i= 2, . . . , n. (D-14)
Since ∂UL(λ,t?)
∂tn=−λ≤ 0, we may take t?n = 0.
Suppose that T −∑n
p=1 t?p >m(α,w). Then we have ∂UL(λ,t?)
∂t1< ∂UL(λ,t?)
∂t2< · · ·< ∂UL(λ,t?)
∂tn−1< 0 by Lemma
A-1 and condition (D-14). By (D-13), t? = (0, . . . ,0) is an optimal solution.
Suppose next that there exists a k= min{i : i∈ {2, . . . , n+1} such that T −∑i−1
p=1 t?p ≤m(α,w)}. Therefore
by (D-14),∂UL(λ, t?)
∂ti−1<∂UL(λ, t?)
∂ti,∀i < k
∂UL(λ, t?)
∂ti−1≥ ∂UL(λ, t?)
∂ti,∀i≥ k,
(D-15)
since x1 > · · ·>xn and Φ(t) is strictly pseudoconcave and maximized at t=m(α,w) (Lemma A-1), and the
latter inequalities are strict unless there exists an l ≥ k such that T −∑i−1
p=1 t?p =m(α,w) for i= {k, . . . , l}.
Therefore, in case of strict inequalities there exists at most one i < k − 1 and at most one i > l such that
∂UL(λ,t?)
∂ti= 0. By (D-13) the optimal solution is of the form t? = (0, . . . ,0, t?l , τ l+1, . . . , τ r−1, t
?r ,0, . . . ,0).
Suppose next that there exists an index j ≥ k such that T −∑i−1
p=1 t?p = m(α,w) for i ∈ {k, . . . , j} and
T −∑j
p=1 t?p <m(α,w). Then, t?p = 0, p= k, . . . , j− 1. We consider three cases.
(i) Suppose that ∂UL(λ,t?)
∂tk−1> 0. Therefore, ∂UL(λ,t?)
∂ti> 0 for i = k − 1, . . . , j by (D-14). By (D-13), t?p =
τp, p= k− 1, . . . , j, a contradiction since t?p = 0, p= k, . . . , j− 1.
(ii) Suppose that ∂UL(λ,t?)
∂tk−1= 0. Therefore, ∂UL(λ,t
?)
∂ti= 0 for i= k−1, . . . , j by (D-14). By assumption t?p = 0
for p= k, . . . , j− 1. Therefore by (D-15),
∂UL
∂ti−1< 0 for i < k− 1;
∂UL
∂ti< 0 for i > j.
Author: The Design of Experiential ServicesArticle submitted to ; manuscript no. (Please, provide the mansucript number!) 31
By (D-13), the optimal solution is therefore of the form t? = (0, . . . ,0, tk−1,0, . . . ,0, tj ,0, . . . ,0).
(iii) Suppose that ∂UL(λ,t?)
∂tk−1< 0. Therefore, ∂UL(λ,t?)
∂ti< 0 for i = k − 1, . . . , j by (D-14). This implies by
(D-15),∂UL
∂ti−1< 0 for i≤ k− 1;
∂UL
∂ti< 0 for i > k− 1.
By (D-13), the optimal solution is therefore of the form t? = (0, . . . ,0).
In summary, the optimal solution of U(λ, t) can be characterized by two indices (l, r) such that
t? = (0, . . . ,0, t?l , τ l+1, . . . , τ r−1, t?r ,0, . . . ,0), or t? = (0, . . . ,0, t?l ,0, . . . ,0, t
?r ,0, . . . ,0), (D-16)
where t?l and t?r , are such that 0≤ t?l ≤ τ l, 0≤ t?r ≤ τ r and T −∑l
p=1 t?p ≤m(α,w) with r≥ l. We next claim
that, for any such l and r if there exists a solution such that 0< t?l < τ l and 0< t?r < τ r, then there exists
only one such solution. In particular, t?l and t?r have to satisfy (D-13), which reduces to:
f1(tl) = (xl+1−xl)Φ′(T − tl) + · · ·+ (xr −xr−1)Φ′(T − tl− τ l+1− · · ·− τ r−1) = 0
f2(tl, tr) = (xn−xr)Φ′(T − tl− τ l+1− · · ·− τ r−1− tr)−λ= 0.
Solving for f(tl) = 0 gives the closed form solution
tl = T − 1
w−αln
[(xl+1−xl) + · · ·+ (xr −xr−1)ew(
∑r−1p=l+1
τp)
(xl+1−xl) + · · ·+ (xr −xr−1)eα(∑r−1p=l+1
τp)
].
Hence, there exists at most one tl,0 < tl < τ l solving f1(tl) = 0. Moreover, because r ≥ l, T −∑r
p=1 t?p ≤
T −∑l
p=1 t?p ≤ m(α,w), and therefore Φ′′(T −
∑r
p=1 t?p) < 0 by Lemma A-3. Hence, f2(t?l , tr) is a strictly
increasing function of tr given that xn < xr and there can be therefore at most one tr,0 < tr < τ r solving
f2(t?l , tr) = 0.
Hence, for any l, r, there exists only a constant number of potential optimal solutions of type (D-16) such
that T −∑l
p=1 t?p ≤m(α,w) and r ≥ l. Since there are O(n2) different ways of selecting l, r, the number of
candidate solutions is therefore of the order O(n2). �
Lemma D-3. Without loss of generality, suppose that x1 > · · · > xn and T > 2m(α,w). For any λ ≥ 0,
only O(n) points need to be considered to find the optimal solution for UR(λ).
Proof. Let t? be the optimal solution for UR(λ). Hence, t? should satisfy the first order optimality
conditions:
1. The stationarity condition gives ∂UR(λ,t)
∂ti+µ
i−µi = 0,∀i.
2. Complementary slackness gives µi(τ i− t?i ) = 0,∀i and µi(t?i ) = 0,∀i.
3. Primal feasibility implies t? satisfies the constraints (D-12).
4. Dual feasibility implies µi, µi ≥ 0,∀i. These conditions reduce to the following:
∂UR(λ, t?)
∂ti= µi ≥ 0,∀i such that t?i = τ i
∂UR(λ, t?)
∂ti= 0,∀i such that 0< t?i < τ i
∂UR(λ, t?)
∂ti=−µ
i≤ 0,∀i such that t?i = 0.
(D-17)
Author: The Design of Experiential Services32 Article submitted to ; manuscript no. (Please, provide the mansucript number!)
Observe that for any two consecutive activities the following holds
∂UR(λ, t?)
∂ti−1− ∂UR(λ, t?)
∂ti= (xi−1−xi)Φ′(
i−1∑p=1
tp),∀i= 2, . . . , n. (D-18)
Since ∂UL(λ,t?)
∂tn=−λ≤ 0, we may set t?n = 0.
Suppose next that there exists a k= max{i : i∈ {2, . . . , n+ 1} such that∑i−1
p=1 t?p ≤m(α,w)}. Therefore
∂UR(λ, t?)
∂ti−1≥ ∂UR(λ, t?)
∂ti,∀i < k
∂UR(λ, t?)
∂ti−1<∂UR(λ, t?)
∂ti,∀i≥ k,
(D-19)
and the former inequalities are strict unless there exists an l ≤ k such that∑i−1
p=1 t?p = m(α,w) for i =
l+1, . . . , k since x1 > · · ·>xn and by Lemma A-1. If no such k exists, i.e., if t1 >m(α,w), then Φ′(∑r
p=1)t?p <
0,∀r= 1, . . . , n. Therefore, ∂UR(λ,t?)
∂t1< · · ·< ∂UR(λ,t?)
∂tn−1< 0 by (D-17) and since ∂UR(λ,t?)
∂tn≤ 0. Hence, the first-
order conditions (D-17) imply that the optimal solution is t? = (0, . . . ,0).
Suppose that there exists an l≤ k such that∑i−1
p=1 tp =m(α,w) for i= l+ 1, . . . , k and∑l−1
p=1 tp <m(α,w).
By (D-18),∂UR(λ, t?)
∂ti−1>∂UR(λ, t?)
∂ti,∀i= 1, . . . , l
and∂UR(λ, t?)
∂ti−1<∂UR(λ, t?)
∂ti,∀i= k+ 1, . . . , n.
In particular, ∂UR(λ,t?)
∂tk< · · · < ∂UR(λ,t?)
∂tn≤ 0 and therefore, ∂UR(λ,t?)
∂tl= · · · = ∂UR(λ,t?)
∂tk< 0. By (D-17) this
implies that t?l = · · · = t?n = 0. Since ∂UR(λ,t?)
∂ti−1> ∂UR(λ,t?)
∂ti,∀i = 1, . . . , l, there is at most one i such that
∂UR(λ,t?)
∂ti= 0.
Therefore, by (D-17), when there exists an index k such that∑k−1
p=1 t?p ≤m(α,w), the optimal solution has
the form t? = (τ1, . . . , τ r−1, t?r ,0, . . . ,0), with at most one activity t?r assigned a duration 0< t?r < τ r, in which
r≤ k, i.e.,∑r
i=1 t?i ≤m(α,w). We next claim that for any r such that
∑r
i=1 t?i ≤m(α,w), there exists at most
one interior solution tr, 0 < tr < τ r, such that ∂UR(λ,t?)
∂tr= 0. In particular, t?r has to satisfy the first-order
optimality conditions (D-17):
f(tr) = (xr −xn)Φ′(τ i1 + · · ·+ τ r−1 + tr)−λ= 0. (D-20)
Because∑r
p=1 t?p ≤
∑k−1p=1 t
?p ≤m(α,w), Φ′′(
∑r
p=1 t?p)< 0 by Lemma A-3. Moreover, xr > xn by assumption.
Hence, f(tr) is a strictly decreasing function of tr. Therefore, there can be at most one solution for f(tr) = 0.
Since there are O(n) different ways of selecting r, the number of candidate solutions is therefore of the
order O(n). �