Network Analytics - Homework 3 - Msc Business Analytics - Imperial College London

1. Network Analytics
Homework 3
Jonathan Zimmermann
14-12-2015

2. Problem 1
Convert the following problems (Do not give a solution, but a transformation
as a weighted perfect-matching problem).
Question a)
Finding a maximum matching in this graph (note need not be a perfect match-
ing) by converting it to a weighted perfect matching problem
Answer
This is a bipartite graph where nodes on the left side should be matched with
one node of the right side. Assuming that the weight on each edge is the same
(for any arbitrary weight), we have a ”minimum weight maximum matching
problem”. We want to convert it into a ”minimum weight perfect matching
problem”. So first, we redraw the graph by adding weights on edges, as well as
an additional ad hoc node with ad hoc edges of very high weights to every of
the other edges to allow for a perfect matching:
1

3. Concretely, in the figure above, we assigned a weight of 0 to all the pre-
existing edges, then created node h as well as edges h-a, h-b, h-c and h-d with
a weight of 1 for each. Instead of 0 and 1, we could have used any other value
as long as the weight on the newly created edges is higher than the weight of
the original edges.
We now have the following linear optimisation problem to solve:
wi,j = weight between node i and node j,
mi,j = 1 if we match i with j, 0 otherwise (binary variable)
∀i ∈ {a, b, c, d}, ∀j ∈ {e, f, g, h}
minimise: wa,e + wa,f + wa,g + wa,h + wb,e + wb,f + wb,g + wb,h + wc,e + wc,f
+wc,h + wc,h + wd,e + wd,f + wd,g + wd,h
subject to:
ma,e + ma,f + ma,g + ma,h = 1
mb,e + mb,f + mb,g + mb,h = 1
mc,e + mc,f + mc,g + mc,h = 1
md,e + md,f + md,g + md,h = 1
ma,e + mb,e + mc,e + md,e = 1
ma,f + mb,f + mc,f + md,f = 1
ma,g + mb,g + mc,g + md,g = 1
ma,h + mb,h + mc,h + md,h = 1
Question b)
Find the largest subset of edges in this graph such that the number of these
edges incident on the nodes satisfy the conditions given next to the nodes by
converting it to a min-cost flow problem
2

4. Answer
First, let’s redraw this graph by adding a source and a sink node:
We now have the following linear optimisation problem to solve:
fi,j = amount of flow we send through edge i-j
∀ pairs i, j forming an existing edge
f = maximum flow going to sink node
maximise: f
3

5. subject to:
(Outgoing flow = incoming flow for each node, except source and sink:)
fe,t + ff,t + fg,t = f
fs,a + fs,b + fs,c + fs,d = −f
fs,a − fa,e − fa,f − fa,g = 0
fs,b − fb,e − fb,f − fb,g = 0
fs,c − fc,e − fc,f − fc,g = 0
fs,d − fd,e − fd,f − fd,g = 0
fe,t − fa,e − fb,e − fc,e − fd,e = 0
ff,t − fa,f − fb,f − fc,f − fd,f = 0
fe,g − fa,g − fb,g − fc,g − fd,g = 0
(Sink restrictions:)
me,t = 2
mf,t = 1
mg,t ≤ 2
(Source restrictions:)
ms,a ≤ 1
ms,b = 1
ms,c ≤ 1
ms,d ≤ 2
(All other edges have a max capacity of 1:)
0 ≤ ma,e ≤ 1
0 ≤ ma,f ≤ 1
0 ≤ ma,g ≤ 1
0 ≤ mb,e ≤ 1
0 ≤ mb,f ≤ 1
0 ≤ mb,g ≤ 1
0 ≤ mc,e ≤ 1
0 ≤ mc,f ≤ 1
0 ≤ mc,g ≤ 1
0 ≤ md,e ≤ 1
0 ≤ md,f ≤ 1
0 ≤ md,g ≤ 1
4

6. Problem 2
Suppose we have a set of 3 sellers labeled a, b, and c, and a set of 3 buyers
labeled x, y, and z. Each seller is offering a distinct house for sale, and the
valuations of the buyers for the houses are as follows.
Describe what happens if we run the bipartite graph auction procedure from
Chapter 10, by saying what the prices are at the end of each round of the auc-
tion, including what the final market-clearing prices are when the auction comes
to an end.
(Note: In some rounds, you may notice that there are multiple choices for the
constricted set of buyers A. Under the rules of the auction, you can choose any
such constricted set. It’s interesting to consider - though not necessary for this
question - how the eventual set of market-clearing prices depends on how one
chooses among the possible constricted sets.)
Answer
Round 1
The starting prices are 0 for each seller. The preferred-seller graph looks like
that:
5

7. Set of buyers x, y is constricted to neighbour b, so no perfect matching. So
b increases its price by 1. Prices of a and c are still 0, so no need to reduce the
prices.
Round 2
With the new prices, the preferred-seller graph looks like that:
Set of buyers x, y is constricted to neighbour b, so no perfect matching. So
b increases its price by 1. Prices of a and c are still 0, so no need to reduce the
prices.
Round 3
With the new prices, the preferred-seller graph looks like that:
Set of buyers x, y, z is constricted to neighbours b, c so no perfect matching.
So b and c increase their price by 1. Price of a is still 0, so no need to reduce
the prices.
6

8. Round 4
With the new prices, the preferred-seller graph looks like that:
We now have a perfect matching:
• Buyer x is indifferent between all the sellers: he has a payoff of 3 for each
of them (3-0=3 for a, 6-3=3 for b, 4-1=3 for c).
• Buyer y prefers seller b (payoff of 8-3=5, compared to payoff of 2-0=2 for
seller a and payoff of 1-1=0 for seller c)
• Buyer z prefers seller c (payoff of 3-1 = 2, compared to payoff of 1-0 = 0
for seller a and payoff of 2-3=-3 for seller b)
• If we attribute b to y (his preferred seller), c to z (his preferred seller) and
a to x (one of his preferred seller), then everybody receives his preferred
seller and we have a perfect matching at the market clearing prices of 0,
3, 1.
Problem 3
Consider a product that has network effects in the sense of our model from
Chapter 17. Consumers are named using real numbers between 0 and 1; the
reservation price for consumer x when a z fraction of the population uses the
product is given by the formula r(x)f(z), where r(x) = 1-x and f(z) = z.
Question a)
Lets suppose that this good is sold at cost 1/4 to any consumer who wants
to buy a unit. What are the possible equilibrium number of purchasers of the
good?
7

9. Answer
We know that
r(x)f(z) = (1 − x)z
From Chapter 17, we know that equilibriums can only be found in this model
when expectations are rational, i.e. when the expectations of the number of
buyers is equal to the number of buyers, i.e. z = x. We also know that for an
equilibrium to be found, the marginal value for a new individual must be equal
to the marginal price (which is here p = 0.25). Then, we must satisfy:
r(x)f(z) ≡ r(z)f(z)
= (1 − z)z
= z − z2
= 0.25 = p∗
The value of z that satisfies this equation is z = 0.5 = x (i.e. with 50% of the
market purchasing. There is only one solution).
We also have an additional equilibrium at z = x = 0, where no consumer
will buy anything. Since at this point the network effect f(0) = 0, then the value
of the product is 0 ¡ p. This wouldn’t be an equilibrium if a negative number
of consumers could buy the product, but since our model is bounded by this
non-negativity constraint, z = x = 0 is an equilibrium.
This gives us a total of 2 equilibriums.
Question b)
Suppose that the cost falls to 2/9 and that the good is sold at this cost to any
consumer who wants to buy a unit. What are the possible equilibrium number
of purchasers of the good?
Answer
Our equation now becomes:
r(x)f(z) ≡ r(z)f(z)
= (1 − z)z
= z − z2
= 2/9 = p∗
8

10. The values of z that satisfy this equation are z = 1/3 = x and z = 2/3 = x.
As before, we still have the additional equilibrium at additional equilibrium
at z = x = 0. Now, we have a total of 3 equilibriums.
Question c)
Briefly explain why the answers to parts (a) and (b) are qualitatively different.
Answer
The properties of these three equilibriums are not the same, in particular the
properties of the points between each of these equilibriums (i.e. when we don’t
have self-fulfilling predictions of z). When we only have three equilibriums like
in (b), the second equilibrium represents the tipping point - the minimum level
of expectations you need to reach for your product to converge towards a non-
zero equilibrium -, and the third equilibrium represents where your sales will be
in the long term if you have reached the tipping point. In our example of (b),
what this means is that if you want to launch the product, you need to build
expectations around a minimum level of 1/3 of the market to be successful, and
that if you are successful you will eventually sell your product to 2/3 of the
market.
However, if you only have two equilibriums like in (a), then the tipping point
and the long term point are the same. This doesn’t look like it would make a
huge difference except from the fact that the required effort to reach the tipping
point is higher, and that the output of that effort is lower as well, but the
implications are actually much bigger than that: maintaining your product
above the tipping point will require a permanently higher effort. If we live
in a market with different kinds of shocks and variations over time that could
temporarily affect our number of consumers (e.g. economic crisis, press scandal,
seasonality, etc.), then any loss of consumers below the tipping point needs to
be immediately compensated by some marketing efforts to prevent the model
to start naturally convering towards the lower 0 equilibrium. In other words,
we don’t have any margin between the tipping point and the long term stable
equilibrium.
Question d)
Which of the equilibria you found in parts (a) and (b) are stable? Explain your
answer.
Answer
As explained in (c), we have for (a):
• Equilibrium z = 0 = x is stable: any upper deviation from the equilib-
rium is automatically compensated by a downward pressure and we will
9

11. converge again towards the same equilibrium. Lower deviations are not
possible since we are already at the lower bound of the model. This equi-
librium is the one we reach if we are below the tipping point.
• Equilibrium z = 1/2 = x is semi-stable: any upper deviation from the
equilibrium is automatically compensated by a downward pressure and
we will converge again towards the same equilibrium. However, lower
deviations will lead to a downward pressure and to a convergence towards
the first equilibrium z = 0 = x. So this equilibrium is only stable on
one side, making it much more difficult to maintain that a fully stable
equilibrium.
And we have for (b):
• Equilibrium z = 0 = x is stable: any upper deviation from the equilib-
rium is automatically compensated by a downward pressure and we will
converge again towards the same equilibrium. Lower deviations are not
possible since we are already at the lower bound of the model. This equi-
librium is the one we reach if we are below the tipping point.
• Equilibrium z = 1/3 = x is not stable: both lower and upper deviations
will create a downward/upward pressure which will lead to a convergence
towards a different equilibrium. Staying at this equilibrium requires a lot
of stabilization efforts.
• Equilibrium z = 2/3 = x is stable: any lower deviation from the equi-
librium is automatically compensated by a upward pressure and we will
converge again towards the same equilibrium. This equilibrium is the one
we reach if we are above the tipping point.
Problem 4
From the Bass Model description (first, relate it to the model and terminology
we did in class) and the data for the Terminator 3 movie, obtain a rolling
horizon estimate of the parameters (using the optimization model described)
for a forecast after observing the sales till week 4. Compare them with the
estimates in the article.
Answer
Note: I actually use the data of the movie ”The Doctor” instead of ”Termi-
nator 3” to answer this question, as allowed by Kalyan on the Hub, since the
input data is more accurate.
The solution to the non-linear Rolling Horizon optimisation model can be found
in the attached file ”Bass model.xlsx”. As shown in the file, by using only the
data of the first 4 weeks to predict the parameters of the model, we obtain:
10

12. • p = 0.045706097
• q = 0.954293903
• m = 31.7938402
• Sum of squared errors = 2.173489444
where q is the coefficient of imitation, p is the coefficient of innovation, and m
is the number of people who will eventually adopt the product.
Using the estimates of the article instead, we would have had:
• p = 0.074
• q = 0.49
• m = 34.85
• Sum of squared errors = 9.294759418
The reason why those values are so different is that the estimation in the article
was using the data over the first 12 weeks of sales to estimate the model. As
we can see, with a sum of squared errors of 2.173489444, it looks like the Excel
Solver was able to reach a close to global optimal solution. But that the results
change that much by simply adding a few more observations is quite worrisome.
In this specific case, using only the first four weeks of data will always give
strongly inaccurate estimates, independently from the model used, because sales
only start dropping after week 4. Week 5 is indeed the first week where we ob-
serve a drop of revenues, i.e. the right side of the tipping point. The model
needs to know this data point in order to recognize the tipping point. Because
of this, our model predicts an extremely high level of revenues for week 5 but
then extremely low values for the remaining weeks, whereas the actual revenues
are much more spread over time.
11

13. Of course, the estimates of the article were less good at predicting the sales
of the first four weeks (week 4 in particular) since that wasn’t the whole focus of
its objective function, but then much better for the last weeks than our rolling
horizon model (week 7 to 12). In particular, our model predicts levels of sales
close to 0 for weeks 8 to 12, whereas the Bass model using the 12 weeks as input
predicts some fairly high residual sales.
Still, both models seem to fail at predicting the overall shape of the curve:
while their sum of squared residuals have an acceptable value, they both lack
the complexity necessary to explain features such as the angle formed by the
curve at weeks 4-5-6:
But we could use a slightly different variant for the rolling horizon predic-
tions. Instead of comparing a model fully predicted from week 1 to week 12,
we could use a mixed model where week 1 to week 4 use the actual values, and
predicted values for week 5 to week 12, where week 5’s prediction is based on
the actual sales (instead of the predicted sales) of week 4.
The results are shown in the graph below:
12

14. As we can see, our new prediction curve (in purple) looks very similar to our
old prediction curve (in red), but has the advantage of showing correct values
for week 1 to 4. In weeks 5 to 12, the values from the purple curve are slightly
different from those of the red curve, sometimes closer to the actual sales and
sometimes further. It it not clear whether the mixed prediction curve would
give better results that the regular prediction curve since the method we used
stands away from the pure Bass model. The quality of the results might also
depend on the kind of sales we are trying to predict.
Problem 5
Read the Anatomy of a scientific rumor article by De Domenico, Lima, Mougel
and Musolesi and write one paragraph each on the following questions.
Question a)
Summarize the model.
Answer
The model tries to estimate the spreading of information and rumors over time
over a large network, in this case through Twitter. The model was built around
the rumor spreading following the discovery of the Higgs Boson in the CERN in
July 2012, using data acquired directly through the Twitter API. More specif-
ically, the model tries to predict the number of active Twitter users at each
point of time, where ”active” is defined as whether the user published a Tweet
linked to the news in the given timeframe. The model and its intuition can be
summarised with the following equation:
A(t + 1) = (1 − ˜β(t))A(t) + (N − A(t)) ˜λ(t) (t)
Where:
13

15. • A(t + 1): number of active users at time t + 1
• A(t): number of active users on the previous period t
• N: total number of users (counting only those who published at least
one tweet related to the topic during the timeframe considered in the
experiment)
• ˜β(t): probability at time t that an active user becomes non-active
• ˜λ(t)(t): probability at time t that an inactive user becomes active
Then, the intuition behind that equation becomes:
Number of active users at a given time = (the number of active
users of the previous period * a dynamic proportionality factor) +
(the number of inactive users of the previous period * a dynamic
proportionality factor)
What makes this model especially complex is that ˜β(t) and ˜λ(t)(t) are not
constants but functions that depend on multiple parameters (not simply on
time). In addition, the model’s behaviour changes depending on the phase of
the rumour spreading (e.g. before and after the official announcement of the
discovery)
Question b)
How does it differ from the Bass Model?
Answer
This model has many similarities with the Bass model, but is overall a funda-
mentally different method.
Both models try to predict the spread of something in a population: of a product
(or decision to purchase) for the Bass model, and of information (or the decision
to tweet) for this model; even though products are tangibles and information is
intangible, the spread of both can be measured very similarly.
Where this model is very different, however, is in its complexity. Just as the
Bass model, this model expresses the current year’s value as a proportion of last
year’s values; but here, instead of the constants p and q of the Bass model, we
use the complex function ˜β(t) and ˜λ(t)(t) (which I won’t explain in detail).
In addition, this model uses Network Data whereas the Bass model uses simple
numeric sales data.
Of course, it can be argued that the Bass model could also be extended into
something more complex where the proportionality parameters are functions
(even graph related functions) instead of constants. In this case, the previous
14

16. arguments would become less valid, but the proportionality constant/function
multiplies something very different in both models (the (1-F(t)F(t) of the Bass
model is very different from the A(t) or from the N-A(t) of this model), and
this is something even more fundamental that we cannot consider as a mere
extension of the Bass model.
Question c)
How can the model potentially be used in a business application?
Answer
This model could be directly applied in settings where Twitter or similar social
media are used as a communication tool, such as viral marketing or public
relations (notably during a media crisis if the company is negatively impacted
by a scandal). This tool not only has the power to increase the managers’
understanding of what is happening in the ”social media blackbox”, but can
also help them target the right individuals and choose the right time slots to
boost their campaigns/mitigate the impact of bad press. It is especially useful
to know the impact that a certain campaign will have when you need to decide
of the budget to attribute to that campaign, or when you need to assess the
effectiveness post-campaign (e.g. using residual analysis to determine whether
the failure was due to natural variations or bad management).
15