Transcript OPIM 5894 Advanced project management

1
OPIM 5984
ANALYTICAL CONSULTING IN
FINANCIAL SERVICES
SURESH NAIR, Ph.D.
Financial Services Analytical Consulting
2




There is increasing convergence between operations, marketing and
finance.
Nowhere is this more evident than in the financial services industry –
banking, credit cards, brokerage, insurance, mortgages, etc.
What differentiates financial services from other services

Large number of customers

Repeat nature of interactions over the customer’s lifetime,

Lots of data available for analysis and decision making, and a

Wide variety of tools and techniques are applicable – from deterministic to
stochastic modeling, from analytical methods to simulation.
There is huge potential for analytical consulting in financial services
3
Attributes of a good Management Consultant
Outline
4


Management consulting situations
Attributes of a good consultant – Lessons learnt
Time is of the essence – Quick analysis is very important
 It is far more difficult to start from a clean slate than to
improve an existing process/idea.




85% of the benefit from a good idea, however implemented.
Optimization only improves from there.
Be Rumpelstiltskin – learn to spin straw into gold. Learn to
work on unstructured problems
“Socialize” recommendations – don’t surprise client
Consulting situations
5
Available
No need for
consultants
Creative Modeling
Not Available
Data Availability
Modeling/Solution Techniques
Known
Not Obvious
Creative Data
Gathering
Qualitative
Inductive
Recommendations
Time is of the essence
6




It is more important to be timely than perfect.
Problems are unstructured. No such thing as a perfect solution to a problem
that is hard to define.
Learn the tradeoff between time and performance
If you take too long, the problem changes by then. You have the perfect
solution to the wrong problem.
Breakthrough vs. Incremental ideas
7

It is far more difficult to start from a clean slate than to improve an existing
process/idea.

85% of the benefit from a good idea, however implemented. Optimization only
improves from there.
Be Rumpelstiltskin – spin straw into gold
8

Learn to work on unstructured problems

Quadrant 2: Creative Modeling



Quadrant 3: Creative Data Gathering


Retail Bank Sweeps
Credit Card solicitations
End of life planning for a blockbuster drug going off exclusivity
Quadrant 4: Qualitative Inductive Recommendations

Impact of Comparative Effectiveness Research on drug sales
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Simulating Alternative Recommendations
in Financial Services
An Example (recap for students who took
my OPIM 5641)
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Jack sells insurance. His records on the number of policies sold per week over a
50 week period are:
Number policies sold
Frequency

0
8
1
15
2
17
3
7
4
3
Suppose we wanted to simulate the policies Jack sells over the next 50
weeks.
Example (contd.)
11

Life is random
Give Chance a Chance
iPod Shuffle
It is fairly simple to evaluate different alternative order quantities quickly
using simulation.
Step 1

Compute Probabilities, Cumulative Probabilities and assign Random
Numbers
Number policies sold
Frequency
Probability
Cumulative Probability
Random Numbers
0
8
0.16
0.16
00-15
1
15
0.30
0.46
16-45
2
17
0.34
0.80
46-79
3
7
0.14
0.94
80-93
4
3 Total
0.06
1.00
1.00
94-99
The trick for assigning random numbers is easy. Compute the cumulative
probability, start from 00 to 1 less than the cum frequency. For the next row,
start from the next random number to 1 less than the cum prob., etc.
Step 2

Simulate the next 50 orders
#Policies Simulation
12
#Policies Example (contd.)
13


Suppose 30% of the policies are Life and 70% are
Supplemental, simulate the type of policies for the next 50
weeks.
Suppose 25% of the Life policies are for $100K, 50% for
$250K, and 25% for $500K, simulate the value of the policies
for the next 50 weeks.
Break-out Exercise
14
For the Credit Cards data file on the website, please simulate the following
for the next 24 months for a customer:

Current Balance

Payment

Purchase + Cash advance
What are the assumptions you made?
What else would you have done in modeling future behavior, if you had more
time?
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Simulating Standard Distributions
Simulating Standard Distributions (another
recap for my OPIM 5641 students)
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


In Excel, use \Data\Data Analysis and then select Random Number
Generation. This tool can simulate the following distributions:

Normal

Uniform

Exponential

Poisson

Discrete
The random numbers generated do not change when F9 is pressed (that is,
once generated, they stay fixed).
Excel functions can be used to generate some of these and other
distributions. This can be done on the fly and is very handy, as we will see
next.
Standard Distributions (contd.)
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
Random numbers following certain distributions can
be generated to change with every press of F9.
This can be very useful in practice.
Generating Normally distributed random numbers:

Suppose you wanted to generate Normal random
numbers with a mean of 50 and standard
deviation of 5.

=NORMINV(RAND(),50,5)
Generating Uniformly distributed random numbers:

Suppose you wanted to generate sales per day
that were Uniformly distributed between 6 and 12
(inclusive).

=RANDBETWEEN(6,12)
Standard Distributions (contd.)
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Generating Exponentially distributed random numbers:

Suppose you want to simulate the next breakdown of a
machine that fails exponentially with a mean of 5 hours,
then use

= – 5*LN(RAND())

Exponential distribution can be used for time
between arrivals of events (breakdowns, customers
to a restaurant, customer service calls, cars at a toll
plaza, customer orders, etc.)

For number of arrivals of events, use Poisson
distribution (number of breakdowns/day, #customers
to a restaurant/hour, #customer service calls/hour,
etc.)

Exponential and Poisson distributions are sister
distributions, both require only one number, the Mean
time between arrivals (unlike the Normal distribution
which requires the mean and standard deviation).
Standard Distributions (contd.)
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Generating Binomially distributed random numbers:

Use Binomial when you have a yes/no, response/no response, kind of
binary situation

= CRITBINOM(n,p,rand())

Where n is the number of trials, and p is the probability of success
Standard Distributions (contd.)
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Generating Poisson distributed random numbers:

You need the average for the Poisson distribution.

Use Random Number Generator under
\Data\Data Analysis
Generating Discrete distributed random numbers:

Use Random Number Generator
Real Estate Exercise
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


After working a few years, you have been promoted and reassigned to a
new location by your company. You need to buy a new house sometime in
the next 30 days. Houses come up on the market at the rate of 2 every day
(Poisson). Suppose 50% of these homes meet your requirements. House
prices follow Normal (275,000, 20,000) dollars.
While looking for a house, you must stay in a hotel at a cost of $100 per
day. Your goal is to find a house that satisfies your needs and that
minimizes the expected value of the sum of the costs of staying in the hotel
and the cost of the house. Houses sell quickly at the new location, so if you
do not buy one on a day, it is not available again.
Suppose you decide to buy the house the first time you see a price below
some value H. What should be that value?
Break-out Exercise (Retail Banking)
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For the Retail Banking data file on the website, please simulate the following for the next 60
days for a customer:

Net deposit

Net withdrawal

Daily balance
What are the assumptions you made?
What else would you have done in modeling future behavior, if you had more time?
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Financial Options Valuation
Financial Options Valuation
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
Financial options are popular products in the financial industry

An option give you a right (but not an obligation) to buy or sell a stock.


It establishes a specific price, called the Strike Price, at which the contract
may be Exercised, or acted on. And it has an Expiration date. When an
option expires, it no longer has value and no longer exists.
Suppose today’s (Dec 1) Apple (AAPL) stock price is $388, the option prices
for Dec 17 are
Financial Options Valuation
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
There are two kinds of Options – Calls and Puts

Calls give you the right to buy stock at a price – e.g., at $400 on Dec 17 (recall
the stock is at 388 today, Dec 1), and it will cost us $2.87 to buy an option
today.

If on Dec 17, the price is $410, we would have $10-2.87=$7.13 of profit

If on Dec 17, the price is 395, the option has zero value

Puts are the opposite of Calls. If you buy a Put, it gives you the right to sell at a
particular price.

You could also sell (called write) a Call or a Put
Financial Options Valuation
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
Given past prices of the underlying stock, say AAPL, you can figure out the
value of the Option today using simulation. Suppose the prices are as
shown below (see spreadsheet)
Financial Options Valuation
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

Simulation can give us the value of the Option.
If the value we compute is better than the market price of the option, we
could purchase the option, otherwise, if the price is higher, we could write
an option
Break-out Exercise (Financial Options)
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
Determine the value of the following financial option

AXP160115C00110000 (meaning for American Express, Call option, Jan 2016,
$110 strike price)

What is the current price of the option?

Would you buy it?
Critical Thinking (Financial Options)
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


It is assumed tomorrow’s price depends only on today’s price.
Meaning it does not matter if today’s price was part of an
increasing or decreasing trend over the past few days. It is
path independent, and memoryless.
If in fact there is path dependence, then the methodology can
be modified to accommodate it.
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Acquisition Modeling for Financial
Services
Financial Services Optimization
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


In most business situations, managers have to achieve objectives while
working within several resource constraints. For example, maximizing sales
within an advertising budget, improving production with existing capacity,
reducing costs while maintaining service metrics, etc.
Mathematical modeling can help in such situations. Linear Programming (LP)
is the most important of these techniques.
It is used in a wide array of applications, such as


Determining the credit card acquisitions, risk management, optimal product mix,
advertising and media planning, investment decisions, branch/ATM location
siting, assignment of people to tasks, etc.
We will learn about how LP helps decision making by considering several of
these applications.
LINEAR PROGRAMMING
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Example: (Maximization)

A insurance broker sells 2 kinds of products, Homeowners Insurance (H) and
Life Insurance (L). The profit from H is $300, and the profit from L is $250.
The limitations are



Direct personnel: It takes 2 hours to effort for sale of H, and 1 hour of
effort for every sale of L. There are only 40 hours in a week.
Support staff: It takes 1 hour support work for each H and 3 hours for L.
There are only 45 support staff hours in a week.
Marketing: The broker determines she cannot sell more than 12 units of H
per week.
How many of H&L should she aim to sell each week to maximize profits?
Example: (Minimization)
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A credit card company wishes to have a balance of balance carrying and
monthly usage customers in its portfolio of new accounts. It is required that the
portfolio have a usage rating of at least 300 units, and a monthly balance
carrying level of at least 250 units. These can be produced by two types of
accounts, Revolvers and Transactors.

Both revolvers and transactors provide 1 unit of monthly usage per account.

Only revolvers carry balance, of 3 units per account.

Acquiring revolvers costs $45 and acquiring transactors costs $12/account.
How many revolvers and transactors should the credit card company acquire to
minimize costs while achieving its portfolio profile?
Binary (0-1) Assignment Example
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
A manager Global Financial Corp, a commercial loan firm, wishes to
minimize turn around time for loan processing. He has 5 associates and the
task requires 4 steps. He needs to pick the best 4 associates depending on
their time for each of the tasks. The average times (in minutes) for each of
task was recorded as below:
Task
Eval and Analysis

John
482
Susan
444
David
459
Ben
370
Melissa
429
Interest Rate
295
321
264
347
317
Loan Terms
379
341
384
306
397
Final Issuing
120
120
124
109
115
Who should be assigned to which task to minimize turn-around time for loan
applications?
Credit Card Solicitation Optimization
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A credit card company wishes to optimize it direct mail campaign for profitability and
risk. It divides the mailbase into 90 segments by risk, response and balance scores. Use
data file provided

The company wishes to maximize pre-tax profits

It wishes to pick segments to mail or not mail

Each segment’s marginal risk for charge-off should be below 7.5%

The total risk of charge-offs should be less than 4.5% over all segments being
mailed.
Break-out Exercise
36
Using the Credit Card data file do the following:
Identify the optimal segments to mail for the following scenario
1.
Maximize size of mailing (same constraints as before – Total Net Credit Losses
< 4.5%, Marginal Net Credit Losses < 7.5%)

What is the % increase in mailing from the classroom solution? What is the reduction in
profit?
2.
Do the above with the additional constraint that total $ Charge off is less than
$50MM
3.
Complete the following table
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Conjoint Analysis for New Product
Development in Financial Services
Conjoint Analysis
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


Conjoint Analysis is a widely used statistical Market Research technique to
figure out how consumers make trade-offs in making product/service
preference choices.
The product/service can be thought to be a bundle of attributes, each
having different levels
For example, for a credit card, the attributes and levels may be

Introductory rate(attribute)

0 APR(level)
 3.99 APR
 5.99 APR
3 months
 6 months
 12 months


Go to rate
9.99 APR
 12.99 APR

Duration


Rewards
Cash back
 Airline Miles

Conjoint Analysis (contd.)
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
Prospective customers are shown a set for products and asked to rank or
rate them (say from 1-100 points)


Say 16 credit cards are shown, designed by random combinations of attribute
levels

Card#1: 3.99 APR, 9.99 Goto, 12 month duration, Airline Miles

Card #2: 0 APR, 12.99 Goto, 3 month duration, Cash back

:

Card #16: 3.99 APR, 9.99 Goto, 6 month duration, Cash back
Based on the ranks or ratings, CA tries to tease out the value (part-worths) to
each consumer for each attribute level of the product/service
Example – Credit Cards
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

There are software packages available for CA. However, we will do it
using Solver in Excel, using a methodology called Goal Programming
Suppose 12 products (called profiles) are shown to prospects
Profile
1
2
3
4
5
6
7
8
9
10
11
12
Introductory rate
Go to rate
Duration
Rewards
0 APR
3.99 APR 5.99 APR 9.99 APR 12.99 APR 3 months 6 months 12 months Cash Back Air Miles
0
1
0
0
1
0
0
1
0
1
1
0
0
1
0
0
1
0
0
0
0
1
0
1
0
1
0
0
1
0
1
0
0
1
0
0
0
1
0
0
0
0
1
1
0
0
1
0
0
0
0
1
0
1
0
1
0
0
1
0
0
0
1
0
1
1
0
0
1
0
1
0
0
0
1
0
1
0
0
0
0
0
1
0
1
0
0
1
0
0
0
1
0
0
1
0
1
0
0
1
1
0
0
0
1
1
0
0
0
1
0
0
1
1
0
0
0
1
0
1

Check: for each profile, there is a one 1 in each attribute, others are 0

Check: A random set of designs will have evenly balances column sums for each
attribute
Example – Credit Cards (contd.)
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

There are software packages available for CA. However, we will do it
using Solver in Excel, using a methodology called Goal Programming
Suppose 12 products (called profiles) are shown to prospects
Profile
1
2
3
4
5
6
7
8
9
10
11
12
Introductory rate
Go to rate
Duration
Rewards
0 APR
3.99 APR 5.99 APR 9.99 APR 12.99 APR 3 months 6 months 12 months Cash Back Air Miles
0
1
0
0
1
0
0
1
0
1
1
0
0
1
0
0
1
0
0
0
0
1
0
1
0
1
0
0
1
0
1
0
0
1
0
0
0
1
0
0
0
0
1
1
0
0
1
0
0
0
0
1
0
1
0
1
0
0
1
0
0
0
1
0
1
1
0
0
1
0
1
0
0
0
1
0
1
0
0
0
0
0
1
0
1
0
0
1
0
0
0
1
0
0
1
0
1
0
0
1
1
0
0
0
1
1
0
0
0
1
0
0
1
1
0
0
0
1
0
1

Check: for each profile, there is a one 1 in each attribute, others are 0

Check: A random set of designs will have evenly balances column sums for each
attribute
Example – Credit Cards (contd.)
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


Suppose our first prospect, Jay, is shown the 12 profiles and
asked to rate them on a scale of 1-100 as per his preference
He does this on the right
Based on this, using Solver, we can figure out that his valuation
of part-worths is as follows (details in Excel)
Introductory rate
Go to rate
Duration
Rewards
0 APR
3.99 APR 5.99 APR 9.99 APR 12.99 APR 3 months 6 months 12 monthsCash Back Air Miles
35.0
35.0
0.0
35.0
10.0
0.0
15.0
30.0
0.0
5.0


We can similarly figure out part-worths for all the prospects in
our sample.
Using this data, we can then mix and match attribute levels to
design a credit card that could potentially be preferred by
the most prospective consumers and they would respond to our
offer of a new credit card.
Profiles shown
Rel Ratings
to prospect
1
90
2
85
3
65
4
100
5
70
6
70
7
10
8
60
9
35
10
65
11
50
12
65
Example – Credit Cards (contd.)
43


Suppose the prospects have the following part-worths
Then several products could be designed and the Utility of each product to
each prospect can be evaluated
Example – Credit Cards (contd.)
44

You can see that Jay prefers Product 1 to Product 2

Susan prefers Product 2 to Product 1

We can produce 3*3*2*2=36 different products this way and pick the best

When there are many attributes and levels, the possible products can be in
the thousands.

Thousands of full products would have been difficult to ask prospects to evaluate
and rate

CA allows doing this by showing only a small set of randomly designed products
(say 15-30).

The fundamental assumption is that a product is a bundle of attributes, and the
part-worths are additive to give the utility of the product to a consumer.
Break-out Exercise (Retail Bank)
45

Suppose you have been tasked with designing the concept for a new type
of retail bank so as to beat the competition.

What would be the attributes you would use?

What levels for the attributes would you use?

Create 12 random designs to test.

Show the designs to 5 prospective customers and have them rate the designs.
(Simulate this by using fake ratings).

Determine part-worths for the attribute levels and pick the best design.

Present the two designs and compare them.
Critical Thinking (Conjoint Analysis)
46







There is a lot of creativity involved in determining attributes and levels.
These are not obvious.
Too many attributes and levels complicate the data collection. It will
necessitate more profiles to be shown to consumers.
It is assumed that part-worths are additive.
Part-worths are assumed to be compensatory, meaning a small value in one
will be compensated by a large value in another.
Some are obvious “more is better” levels – use price as an attribute and
eliminate infeasible combinations as explained below.
Conflicting attributes should be combined. For example, two attributes –
Engine size (Big, Small) and Price (High, Low) should be combined into one
attribute with 2 levels; Engine-Price(BigHigh, SmallLow)
Using Goal Programming, you can add constraints (such as individual S+ or
S- should be less than 30), which you cannot if you used a software
package for Conjoint Analysis.
47
Markov Analysis
Markov Analysis
48
Markov Analysis is a probabilistic model useful in analyzing
complex systems. It describes and predicts the movement of a
system over time.
Example: (Marketing)
A firm has Brand A in the market. A competitor introduces Brand C,
which is promoted heavily and attracts buyers away from A. The
firm decides to introduce Brand B. It fears that B may
cannibalize customers from A, in addition to getting market share
back from C. The customers switch brands as shown below:
C u rre n t M o n th
F o llo w in g M o n th
0 .2
B ra n d A
0 .2
B ra n d A
0 .6
B ra n d B
B ra n d B
0 .5
B ra n d C
0 .2
0 .3
B ra n d C
49
Thus, if a customer uses brand A this month, s/he has a 20% chance of buying A again
next month, a 20% chance of buying B next month, and a 60% chance of buying C
next month. Similarly, if the customer purchases B this month, she has a 10% chance
of switching to A next month, a 50% chance of staying with B, and a 40% chance
of switching to C next month, and so on.
Note that the probability of making a particular purchase next month depends only on
the purchase made this month, and not on the purchases made in the past. This is an
assumption made to simplify analysis, and is called the memory-less or Markov
property. It is found to apply in many real applications. If two or three periods of
past data influences the present decision, the situation can be adapted to Markov
analysis by suitably redefining the state.
The state of a system is all we need to know to describe the system at any instant.
Transition Matrix
50
In our example, the transition matrix can be written as:
F ro m
To
1
2
3
1
0 .2
0 .1
0 .5
2
0 .2
0 .5
0 .2
3
0 .6
0 .4
0 .3
Note that all rows add to 1.
Suppose we want to calculate the probability that a customer will be in each of the
three states next month. This is called state probabilities, qt(i), where i is the state
and t is the period .
To do this, we will first need to know what the customer is buying this month. Let the
probabilities of her buying A,B,C be [0.275, 0.375, 0.350], or
q0(1)=0.275, q0(2)=0.375, q0(3)=0.350
To compute q1(1) note that you could get to state 1 in the next period from state 1
(with probability 0.2), or state 2 (with probability 0.1), or state 3 (with probility
0.5). Since you know the probability you are in these three states in period 0 from
q0(i), we can condition on this.
51
Thus,
q1(1)=0.275(0.2)+0.375(0.1)+0.350(0.5)=0.2675
Similarly,
q1(2)=0.3125, q1(3)=0.420
The easier method of doing this would be in matrix form
[q1(1),q1(2),q1(3)]=[q0(1),q0(2),q0(3)] x P
or,
Q1=Q0P
Similarly, we can say that
Q2=Q1P
which in turn is
Q2=Q1P=Q0P2
and in general,
Qt=Q0Pk
Market Share
52
If you were to use this formula over and over, you could see how the market
shares of the three brands evolve.
M a rke t S h a re
P e rio d s
A
B
C
0
0.275
0.375
0.35
1
0.2675
0.3125
0.42
2
0.29475
0.29375
0.4115
4
0.296528
0.286438
0.417035
8
0.296702
0.28572
0.417578
16
0.296703
0.285714
0.417582
Note that after some periods, the market shares stabilize. These are called
steady state probabilities. These steady state probabilities are the same
irrespective of what the initial probabilities were at time 0.
It is easy to compute steady state probabilities. We know for steady state
Qt=Qt-1
Steady State Probabilities
53
which by definition is
Qt=QtP
Thus, in our example
 .2 .2 .6 
[ q 1 , q 2 , q 3 ]  [ q 1 , q 2 , q 3 ] .1 .5 .4 


 .5 .2 .3
q1  q 2  q 3  1
which is a set of 2 independent equations (1 is dependent) and 3 unknowns. To
solve, we add the equation
q 1  0 .2 q 1  0 .1q 2  0 .5q 3
to get
q 2  0 .2 q 1  0 .5q 2  0 .2 q 3
q1  q 2  q 3  1
Assumptions in Markov Models
54
which gives us q  0 .297 , q  0 .286 , q  0 .417
1
2
3
as we had expected from the market share table.
These limiting (or steady state) probabilities will occur only if the following
main assumptions hold:


There are a finite number of states, none of which are absorbing (if you get
there, you cannot come out - a row in the transition matrix with all zeros and
a 1).
The transition probabilities are constant over time (called stationary or
homogeneous)

Markovian assumption holds (next state depends only this period’s state).

There is only one transition per period.
Markov Decision Processes
55



Other useful information that can be obtained from Markov Analysis:
Mean first passage times: Mean time before a person switches from a
state to another specified state.
Expected recurrence times: The average number of periods before you
return to the same state.
Other Related Topics


Semi Markov Processes: Before transition from a state to another, there is
a holding time, e.g., car rentals and returns to other towns after a particular
length of rental.
Markov Decision Processes: In each state i, at time t, you can take an
action At=a, which will take you to state j, with a probability Ptj(a). These
probabilities are the transition matrix as before. In addition, in state i, you
collect a reward R(i,a). The problem is to identify a policy that will
maximize revenue (or min costs) over a time horizon, e.g., in equipment
replacement. Dynamic Programming can be used to solve MDPs.
Example (Banking channel)
56

In order to plan channel capacity (# branches, ATMs, etc.), consider the
following channel switching model for banking that you have developed to
get a handle on demand. If a customer uses channel Branch in period 1,
then in period 2 there is 50% chance that she will use channel Branch
again, a 36% chance that she will switch to ATM, and a 14% chance that
she will switch to channel Phone/Online. Similar explanation for switching
from the channels in the other two rows.
Channel used,
Period i
Channel Used, Period i+1
Branch
ATM
Phone
Branch
0.50
0.22
0.27
ATM
0.26
0.30
0.17
Phone
0.14
0.30
0.24
Online
0.05
0.25
0.25
Online
0.10
0.18
0.32
0.45

What would be the steady state channel share for each channel?

Can you simulate this and confirm the numbers?
Retail Banking Example
57

We can use the Retail Banking file to create a Transition Matrix for Net
transactions.

Create two Long columns, From and To, offset by one row.

Then use Pivot Tables to create the Transition Matrix.

We could then use the Transition matrix to find the Stationary Probabilities
of Net transactions every day.
Break-out Exercise (Credit Cards)
58

Using the Credit Card data file, create a transition matrix for Purchases.

Find the steady state probabilities for Purchases.
59
Modeling using dynamic programming
DYNAMIC PROGRAMMING
60

Dynamic Programming is a technique used for sequential decision making.
Typically, a large problem is broken into smaller parts and solved.
Example: (Sample Path Problem)

Suppose you wanted to go from A to B. What would be the shortest path?
Sample Path Problem
61




To get from A to B, it is only necessary to know

the best way to go from C to B,

the best way to go from D to B, and

the cost of going from A to C and D.
Further, to know the best way to go from C to B, we need only know

the best way to go from E to B,

the best way to go from F to B, and

the cost of going from C to E and F.
and so on, until we get to the trivial case of finding the best way to go from
O to B and P to B, which is 2 and 1 respectively.
These last values are called the Boundary Conditions.
Principle of Optimality
63

The principle of optimality (Bellman) states that (for this example):
The best path from A to B has the property that, whatever the initial decision at A,
the remaining path to B, starting from the next point after A, must be the best path
from that point to B.


Now that we know the minimum “cost” of going from A to B, we can go
back and figure out the choices of paths at each intersection.
This is what Dynamic Programming is all about. There are no further key
ideas in DP. However, there is an art in formulating DPs. This has to do with
deciding what to use as a state and stage, and what to use as the value
function. More on these later.
In co-ordinate axes
64
S ( x , y )  th e v a lu e o f th e m in im u m e ffo rt p a th
c o n n e c tin g ( x , y ) to ( 6 , 0 )
a u ( x , y )  e ffo rt in g o in g u p fro m ( x , y ) to ( x + 1, y + 1)
a d ( x , y )  e ffo rt in g o in g d o w n fro m ( x , y ) to ( x + 1, y  1)
 a u ( x , y )  S ( x  1, y  1)
S ( x , y )  m in 
 a d ( x , y )  S ( x  1, y  1)
a n d th e B o u n d a ry c o n d itio n is S ( 6 , 0 )  0
Terminology
65


The above equation is called the functional equation. It is a recursive
relationship (it feeds on itself).
Stage: The problem is solved at different points in time, or places. These
are called stages. Typically in the argument of the value function, the stage
is incremented or decremented by 1 on the RHS compared to the LHS of the
functional equation, (e.g., x above).

State: All other variables in the argument constitute the state, e.g., y above.

Immediate Reward: The profit (or cost) that is collected in the current state.

Action: In each state you have to take one of several available actions

Policy: A predetermined plan of selecting a course of action for every
circumstance. In DP we want to identify the optimal policy.
Example: Equipment Replacement
66


Consider the case of a high speed solicitations mailing machine that
deteriorates with age. The issue is when should one replace it, if the cost of
a new machine, cost of operating an old machine, and salvage value from
selling an old machine is known.
Suppose we want to solve this problem to minimize costs over an N period
horizon.



ci = cost of operating for one year an i year old machine
p = purchase price of a new machine
si = salvage value for an i year old machine
Data
67
Suppose
N 5
p  50
c 0  10 , c1  13 , c 2  20 , c 3  40 , c 4  70 , c 5  100 , c 6  100
s1  25 , s 2  17 , s 3  8 , s 4  0 , s 5  0 , s 6  0
Let Ft(i) be the minimum cost of owning a machine from year t when the machine is of
age i, through N.
 Buy : p  s i  c 0  Ft  1 (1)
Ft ( i )  min 
 Keep : c i  Ft  1 ( i  1)
The terminal condition (reward) is
FN (i )   s (i )
Solution
68

In the above example, t is the stage, and i is the state. The formulation is
complete when the recursion and the boundary condition is given. This can
then be solved using a simple computer program.
Period
|
|
V
0
1
2
3
4
5
Age ---->
0
1
91
108
62
81
35
52
6
25
-15
-4
0
-25
0
Keep
Keep
Keep
Keep
Keep
Salvage
2
124
95
68
39
12
-17
3
133
104
77
48
27
-8
4
141
112
85
56
35
0
5
100
100
85
56
35
0
2
3
4
5
1
Keep
Keep
Keep
Keep
Keep
Salvage
Keep
Buy
Keep
Buy
Keep
Salvage
Buy
Buy
Buy
Buy
Buy
Salvage
Buy
Buy
Buy
Buy
Buy
Salvage
Keep
Keep
Buy
Buy
Buy
Salvage
Break-out Exercise (Sales Territory Management)
69
An insurance company has 5 sales representatives available for assignment
to 3 sales districts. The sales in each district during the current year depend
on the number of sales reps assigned.
Reps assigned
0
1
2
3
North
1
2
3
4
Central
2
4
6
8
South
3
5
6
9
Use dynamic programming to determine an assignment of sales reps to
districts that maximizes the expected sales.
70
Elementary VBA Coding
Instructions to see and edit VBA code
71

Click on Developer tab and then Visual Basic

If you don’t see the Developer tab, go to File, Options, Customize Ribbon… and
then put a check against Developer on the right panel.
Instructions to see and edit VBA code
72

Now click on sheet1 on the left panel

You will now see the code on the right panel.
Simple “Hello, world” program in VBA with
Randbetween function
73

Worksheet and Code

Output
Break-out Exercise (Hello, world and
MSBAPM rocks!!)
74


Modify the “Hello, world” code to write “Hello, world” randomly ten times
in cells (1,1) to (10,10) and another message, say “MSBAPM rocks”
randomly ten times in cells (11,11) to (20,20)
Modify the above program to write “MSBAPM rocks” randomly 10 times in
all cells in (1,1) to (20,20) other than (1,1) to (10,10).
Simple Factorial calculation using Recursion
75

Worksheet and Code

Output
Simple “MSBAPM Rocks” program in VBA with Recursive
function for Fibonacci number
76

Worksheet and Code

Output
Instructions to debug VBA code
77

Go to Debug
Instructions to debug VBA code
78

Suppose you want to go to a particular place in the program and see the current
value of a particular variable. This is extremely important to check if your program
is working correctly.

Click on the program where you want to run up to, and choose Debug, Run to Cursor (or
Ctrl F8)

Suppose you want to see the value of a particular variable at this point in the run, choose
Debug, Add Watch, and then enter the variable name (say N, in the example below)
Instructions to debug VBA code
79

Following the logic of the program – Step Into

It is extremely important to follow the logic of your program to see if it is going where you
want it to go, and doing what you want it to do. For this use Step Into, or F8. It will trace
the logic of your program. Very important to debug code.

Keep pressing F8 to see how the logic works, and fix any bugs as necessary.
Simple Equipment Replacement VBA Code
80

Worksheet and Code

Output
Break-out Exercise (Fortune Splitting)
81



Reliance Group’s founder just passed away and in his will has left his assets to his
two sons, Mike and Abe. The only direction in the will is to ”divide the assets as
evenly as possible.” As the executor of his will, you have priced each of the assets
to obtain an accurate monetary value. You are to decide how to divide up the
assets into two groups to minimize the difference in the sum of values of each group.
For example assume you have the following n = 8 assets:
Asset
C1
C2
C3
C4
C5
C6
C7
C8
Value
2
1
3
1
5
2
3
4
After a lot of work, you figure out that you could divide the assets in the following
way:
Mike
C1
C3
C5
Total
10
Abe
C2
C4
C6
C7
C8
Total

Can you formulate this problem as a DP?

Can you write a DP to do this for any number of assets?
11
Why not just use LP (Solver), instead of DP?
82

Suppose you have $8000 to spend on advertising a new Exchange Traded
Fund product. You want to spend this in increments of one thousand, and
collect the following information on effectiveness.
Media

Eyeballs per Thousands of
dollars spent
Daily newspaper
24
Sunday newspaper
15
Radio
20
Television
20
This can be solved using LP Solver. Do that now.
Breakout Exercise (Why not just use LP
Solver, instead of DP?)
83

Now suppose the data was as below.
Media
Daily newspaper
Sunday newspaper
Radio
Television
1
24
15
20
20
2
37
55
30
40
Eyeballs for Thousands of dollars spent
3
4
5
6
46
59
72
80
70
75
90
95
45
55
60
62
55
65
70
70
7
82
95
63
70
8
82
95
63
70

Here is the way the effectiveness increases with $ spent (clearly not linear)

You then HAVE to use DP. There is no other recourse.

Formulate the model now, no need to code it (yet).
Protecting Market Share
84

Suppose you have $4 million to spend on a new Mileage Rewards product.
You could do this over three years. The initial year expense gets you some
market share, and the second and third year expenditure protects your
market share from competition as follows.
Millions of $
expended
0
1
2
3
4



Year 1 Market Share
0
20%
30%
40%
50%
Year 2
Year 2 Multiplier Multiplier of
of Market Share Market Share
20%
30%
40%
50%
50%
60%
60%
70%
For example, an expenditure of 4,0,0 in the three years will result in an end
of year 2 market share of 50%X20%X30%=1.5%
You want to maximize end of Year 2 market share. Fractions of1 million of
investment need not be considered.
Formulate the model now, no need to code it (yet).
85
Stochastic modeling using dynamic
programming (Markov Decision Processes)
Stochastic Model (Equipment)
86
Suppose, instead of age we talk about “status” of the machine. The status can improve
or deteriorate in the next period as follows

With probability 20% the status improves in the next period

With probability 50% the status stays the same in the next period

With probability 30% the status deteriorates in the next period
Let Ft(i) be the minimum cost of owning a machine from year t when the machine is of age i,
through N.
The functional equation then is
 Buy : p  s i  c 0  0 . 7 Ft  1 (1)  0 . 3 Ft  1 (1)
Ft ( i )  min 
 Keep : c i  0 . 2 Ft  1 ( i  1)  0 . 5 Ft  1 ( i )  0 . 3 Ft  1 ( i  1)
The terminal condition (reward) is
FN (i )   s (i )
Mergers and Acquisitions in packs
87



M&A underwriters usually work in packs. Suppose GS has a normal burn
rate of $6M per week, and has a cash capacity of a max of $30M.
Meaning it can go without an M&A for 5 weeks. Target firms generate an
average fee revenue of $164M. The process of targeting a firm takes $1M
in addition to the normal burn rate.
Targeting firms in packs increases the chance of success.
# M&A Firms
1
2
3
4
5
Probability of success
15%
33%
37%
40%
42%
>=6
43%
Formulate this as a Markov Decision Process (MDP, aka stochastic DP)
model, where the state is the remaining cash, the action is whether to target
on a particular week and if so what should be the size of the underwriting
group. Assume one targeting per week at most, and the objective is to
maximize the probability of survival of GS over 30 weeks. Assume that if
the M&A is successful, the fees are shared equally by the firms
participating.
Parking hot money offshore
88

You have some hot money to move offshore and would like to avoid detection by
the tax authorities. These are your options:

Cayman Islands – the chance of detection is 10%



Switzerland – the chance of detection is 2%

If detected, there is a 30% chance of a $15 surcharge to be paid

There is no cost if you are not detected
US – the chance of detection is 0%, it is not hidden




There is no cost if you are not detected
But there is a $5 cost per week to park money here
Once detected, you can use legal maneuvers at a cost of $9 per week until you pay
a penalty of $50 for hot money. Once you pay the penalty you need to again find
a place to park the money in one of the three locations, going back to the original
situation.
You can move your money around every week, without cost and no questions asked.
Model this as an MDP for 52 weeks, with the objective of minimizing total costs.
Decide where to park your money each week and what to do if detected (pay the
penalty, or pay the surcharge).
Break-out Exercise (Real Estate)
89

After working a few years, you have been promoted and reassigned to a new
location by your company. You need to buy a new house sometime in the next 30
days. You have been told by your company that one new house satisfying your needs
comes on the market every day and the housing prices P1,P2,…P30 on days 1,…,30
are independent positive integer-valued random variables (measured in thousands of
dollars) lying between $250,000 and $300,000 dollars and have known
distributions. While looking for a house, you must pay your current mortgage and
stay in a hotel at a total cost of $1000 per day. Your goal is to find a house that
satisfies your needs and that minimizes the expected value of the sum of the costs of
staying in the hotel and the cost of the house. Houses sell quickly at the new location,
so if you do not buy one on a day, it is not available again. Let Ct(s) be the minimum
expected cost incurred on days t,…,30 when the sale price of the house offered on
day t is s.

Develop an MDP model that would allow you to calculate the Ct(s) recursively and to
determine whether it is optimal for you to buy a house offered at the price s on day t or to
wait another day.
Break-out Exercise (Real Estate, contd.)
90

Suppose the distribution of sale prices is as follows:
Price
Probability
250,000
17%
260,000
17%
270,000
17%
280,000
17%
290,000
17%
300,000
17%

Write DP code to solve this problem for Ct(s)

Show that there are maximum acceptable prices m1,m2,…m30 on days 1,…,30
such that it is optimal to buy on day t if the price s on that day does not exceed
mt and to wait if the price exceeds that level. (Keep in mind that if you have not
bought a house before day 30, you must do so then).
Credit Cards Individualized Mailing
91



The current method optimizes for each mailing (every 6 weeks)
independently of previous mailings.
We did this using Solver previously in class
How would you use the history of mailings to a particular prospect to create
an individualized mailing strategy?
CANARY Model
Control Algorithm for Navigating Revenue and Yield
92

In each period we evaluate 2 value functions ValP and ValA (for prospects and
accounts respectively). One of two kinds of offers (Basic, Rewards) may be
mailed.
 MailB : c  b[(1  rn b )ValP t  1 ( n b  1, n r )  rn b ValA t  1 ( b ,1 )]

ValP t ( n b , n r )  max  MailR : c  b [( 1  rn )ValP t  1 ( n b , n r  1 )  rn ValA t  1 ( r ,1 )]
r
r
actions
 NoMail : b ValP ( n , n )
t 1
b
r

ValA t ( b , m )  NCF
b ,m
 b [( 1  a b ,m )ValA ( b , m  1 )  a b ,m 0 ]
ValA t ( r , m )  NCF
r ,m
 R  b [( 1  a r ,m )ValA ( r , m  1 )  a r ,m 0 ]
ValA T ( , )  ValP T ( , )  0
Where nb , nr are the number of prior basic and rewards,
c is the mailing cost, r.. is the response rate,
a.. is the attrition rate,
m is the months on book, R is the rewards redemption cost,
b is the discount rate (1/(1+i)), t is the time period, and T the horizon of the problem
As an aside, coming up with backronyms like CANARY are a good tool in consulting
projects, remember “canary in a coal mine” gives advance warning to miners..
CANARY Model
Control Algorithm for Navigating Revenue and Yield
93
Legend
b,r
Prospect
b = #prior BT offers
r = #prior rewards offers
Regular Account
m = MOB
7,2
m
m
No
Response
Regular
Account
Attrite
1
Response
NCF$
Attrite
Attrite
2
NCF$
Rewards Account
m = MOB
T
3
NCF$
Traditional BT
Offer
NCF$
No
Mail
Prospects
No
Mail
Optimal CANARY Sequence
Attrite
Rewards
Account
Attrite
Attrite
Response
6,2
1
2
3
NCF$
NCF$
NCF$
Rewards Offer
T
NCF$
- Rewards
Redemption
Cost $
No
Response
6,3
CANARY Model
94
Segment 4: Risk = L; Attrition
HI LO RW CA 1
2
3
0 0 0 0 R
R
R
1 2 5 1 H
H
C
3 4 2 6 R
R
R
5 0 6 6 L
L
L
= H;
4
R
C
R
L
Response
5
6
R
H
H
H
H
H
NM NM
=H
7
8
9 10 11 12
H
H
H
C
C
C
L
C
L
H
C NM
H NM NM NM NM NM
NM NM NM NM NM NM
Segment 6: Risk = H; Attrition
HI LO RW CA 1
2
3
0 0 0 0 C
C
C
1 0 0 6 H
H
H
2 2 4 5 R
C
H
6 5 5 5 C
R
H
= L;
4
C
H
H
C
Response
5
6
C
C
R
R
H
H
NM NM
=H
7
8
9 10 11 12
R
R
R
R
R
R
R
R
R
H
R
H
R
H
C
L
L
L
NM NM NM NM NM NM
Prior offers to date
Keys
R: Rewards offer
C: Cash rebate offer
H: High value basic offer
L: Low value basic offer
NM: No Mail
CANARY Model
Control Algorithm for Navigating Revenue and Yield
95



Results:

On the basis of prospects, the Canary cohort received 6.5% more responses
over BAU in the same offer mix used in the test

Initial observation shows a 25% cumulative improvements on profitability in the
accounts booked in the first 12 months

By month 13, the lift in Canary profitability completely offsets the mailing cost
differentials in BAU
This model will require a COMPLETE revamp of their data gathering and
archiving policies. It will also require lots more data processing than the
original segment level optimization.
But, it is a neat idea, even if I say so myself.
Critical Thinking (DP)
96

The state should contain ALL information to make a decision in
that period.




For example, if sales in two periods affects the sales in the next
period, the state variables should include sales for two periods.
Though DP is memoryless, it can carry memory in this manner.
DPs are best solved using a simple computer program. Since
DPs come in all shapes and sizes, there are no good Solvers
for it. It is best to code by yourself, using VBA or any other
language with the capability to make recursive calls.
Beware of state space explosion – DPs formulations are
indeed an art.
Critical Thinking (DP)
97

The state should contain ALL information to make a decision in
that period.




For example, if sales in two periods affects the sales in the next
period, the state variables should include sales for two periods.
Though DP is memoryless, it can carry memory in this manner.
DPs are best solved using a simple computer program. Since
DPs come in all shapes and sizes, there are no good Solvers
for it. It is best to code by yourself, using VBA or any other
language with the capability to make recursive calls.
Beware of state space explosion – DPs formulations are
indeed an art.
98
Real Options in Financial Services
Real Options in Financial Services
99


Real Options, are like financial options, only for real items or property.
Usage, risk, profitability, cash flows, etc., may be uncertain (stochastic) over
a time horizon. Several possible actions may be available at any time (give
credit line increase, do not give credit line increase; reprice the product, do
not reprice; sweep funds, do not sweep funds, etc.). Given that these actions
are taken, customer behavior may change (increased use of line, customer
attrition, etc.)

In a sense this is a richer, more complex environment than financial options

Dynamic Programming is a good tool to use for this
Breakout Exercise: Bond Refunding
100


100
Most corporate bonds contain a provision which allows the issuer to Call the
bonds before maturity upon repayment of principal plus a premium. This
provision raises a series of problems for the financial manager, much like
refinancing a home mortgage. First given that the firm has outstanding
callable bonds should it call them today and if so with bonds of what
maturity should it replace them. Second, what is the cost of following an
optimum refunding (calling) policy over time?
While there are a multitude of factors affecting the optimum timing and
value of a call (or refinancing a home), the factor which has been singled
out for special attention is the interest savings that can accrue to the firm
through the execution of a call. When a firm refunds a bond, it incurs a
fixed charge equal to the call premium on the old bond plus the cost of
floating a new bond. The refunding itself results in a change in future
interest payments equal to the difference in interest payments between the
old and the new bonds.
Breakout Exercise: Bond Refunding (contd.)
101

Typical MBAs may try to solve this as a standard capital budgeting
decision. One form of this solution is to refund in any period if the present
value of the interest savings, over the life of the outstanding bond exceeds
the cost of calling the old bond and floating the new bond.
Suppose



101
Management has a 5 year time horizon. It will only issue callable bonds in
$100 denominations and with a maturity of 3 years. Floatation expenses on
new debt involve a fixed charge of $2.
The cost of calling the old debt is $2 if it is one year old and $1 if it is two
years old. This applies at each decision point as well as at the horizon.
The firm currently has a 3 year bond outstanding with a coupon rate of 5%
and one year remaining to maturity. It always needs to have a constant
level of debt, so if it calls a bond, it needs to float another bond of like
value.
Breakout Exercise: Bond Refunding (contd.)
102

Management is willing to base its decision on the point estimates of future interest
Time Interest Rate
rates shown below.
-2
-1
0
1
2
3
4

102
5
5
4
5
5
7
7
At each of 5 periods management has to make a decision as to whether to keep or
call (refund) its outstanding bonds. For example,

an initial decision to Keep the outstanding bond involves a one period cost of $5, the
interest payment on that bond.

If instead the firm decided to Call, the cost would be $1 penalty for calling a 2 year old
bond, plus $2 for floating the new 3-year bond, plus the $4 interest payment on the new
bond. One period later the firm again faces a decision. Assuming that it had initially refunded them the cost of the two alternatives would be:

Keep; $4 interest payment

Call; $2 call penalty for calling a 1-year old bond, $2 floatation expense and $5 interest
payment.
Breakout Exercise: Bond Refunding (contd.)
103

What would be the state of the system?

Write down the functional equation and the boundary condition.

103
How would the model change if the future interest rate was uncertain?
Please show the new state definition and functional equations.
Case: Retail Bank Sweep Programs
104


104
Monetary Control Act (1980) authorized Fed Reserve to require banks to
hold 10% of transaction deposits as reserves.

No reserve requirement for time deposits (savings, money market)

Reserves earn no interest for bank
Banks had an incentive to keep deposits as Savings rather than Checking
deposits
Background of Sweep Programs
105

In 1994 one bank came up with a neat idea.

The bank would maintain two accounts for every customer [a
Bank Transaction Account (BTA) and a Money Market
Deposit Account (MMDA)]

By sweeping funds frequently from BTA into MMDA, banks
can keep checking deposits to a minimum


Win-win for both banks and customers

Banks reserve requirements reduce

Customers get higher interest in MMDA accounts
These sweep accounts are transparent to customer
Background of Sweep Programs
106

There are limitations to sweep programs
 Debits
only serviced from transaction accounts –
need some BTA balance to cover check writing, etc.
 Regulation D limits the number of withdrawals from
savings accounts to 6 per month.
 6th transfer requires full dump to BTA
Existing Sweep Method - Cushion
D ay
1
2
3
:
6
7
:
10
11
:
15
16
:
18
19
20
21
:
30
BTA
1 ,0 0 0
1 ,0 0 0
4 8 ,0 0 0
4 0 ,0 0 0
1 ,0 0 0
2 ,0 0 0
7 ,0 0 0
-
6 ,0 0 0
2 ,0 0 0
8 ,0 0 0
T ra n s fe r # 2
4 0 ,0 0 0
3 6 ,0 0 0
2 ,0 0 0
3 ,0 0 0
3 ,0 0 0
-
1 ,0 0 0
3 ,0 0 0
4 ,0 0 0
T ra n s fe r # 3
3 6 ,0 0 0
3 1 ,0 0 0
3 ,0 0 0
4 ,0 0 0
4 ,0 0 0
-
1 ,0 0 0
4 ,0 0 0
5 ,0 0 0
T ra n s fe r # 4
3 1 ,0 0 0
2 4 ,0 0 0
2 4 ,0 0 0
-
4 ,0 0 0
7 ,0 0 0
7 ,0 0 0
2 3 ,0 0 0
6 ,0 0 0
8 ,0 0 0
2 ,0 0 0
5 ,0 0 0
7 ,0 0 0
T ra n s fe r # 5
1 ,0 0 0
Dum p
2 4 ,0 0 0
2 3 ,0 0 0
107
C u s h io n
1 ,0 0 0
A c tu a l T ra n s fe r
M M D A to B T A C o m m e n ts
2 ,0 0 0 T ra n s fe r # 1
M M DA
5 0 ,0 0 0
4 8 ,0 0 0
4 8 ,0 0 0
-
W ith d ra w a l
1 ,0 0 0
-
M in T ra n s fe r
M M D A to B T A
1 ,0 0 0
M M D A D um ped
Motivation for Model
108
Game Show: Look who is counting
9
0
8
Opponent 1
5
1
7
Opponent 2
2
6
3
5
Whoever gets the
larger 5 digit number
wins!!
108
4
5
Motivation for Model
Game Show: Look who is counting
9
0
8
Opponent 1
5
1
7
6
3
5
Whoever gets the
larger 5 digit number
wins!!

5
4
P la c e m e n t in u n o c c u p ie d c e ll
N um ber
on w heel
1
0
5
1
5
2
5
3
4
4
3
5
3
6
2
7
1
8
1
9
1
Optimal Policy

Opponent 2
2
Source: Puterman, MDP, 1994
This model can be solved using
stochastic dynamic programming
109
2
4
4
4
4
3
2
2
1
1
1
S p in N u m b e r
3
3
3
3
2
2
1
1
1
1
1
4
2
2
2
2
1
1
1
1
1
1
5
1
1
1
1
1
1
1
1
1
1
Modeling Customer Behavior
110



Divide the population into various segments
Divide withdrawals and deposits into “transaction intervals.” For example,
Large withdrawal (<-$1500), Small withdrawal (-$1 to -$1499), no
transaction (0), Small deposit ($1-$750) and Large deposit (>$751).
For each segment create a transition matrix showing the chance of
withdrawal and amount of withdrawal every day.
C u rren t d ay
1
2
3
4
5
1
N ext d ay
3
2
11%
5%
3%
4%
15%
44%
49%
28%
47%
50%
32%
35%
54%
37%
30%
110
4
5
8%
8%
8%
8%
2%
A vg A m t
5%
3%
6%
4%
3%
-2200
-150
0
325
4100
Modeling Customer Behavior
111



Divide the population into various segments
Divide withdrawals and deposits into “transaction intervals.” For example,
Large withdrawal (<-$1500), Small withdrawal (-$1 to -$1499), no
transaction (0), Small deposit ($1-$750) and Large deposit (>$751).
For each segment create a transition matrix showing the chance of
withdrawal and amount of withdrawal every day.
C u rren t d ay
1
2
3
4
5
1
N ext d ay
3
2
11%
5%
3%
4%
15%
44%
49%
28%
47%
50%
32%
35%
54%
37%
30%
111
4
5
8%
8%
8%
8%
2%
A vg A m t
5%
3%
6%
4%
3%
-2200
-150
0
325
4100
Stochastic DP Model


The state of the system may be defined as (m,b,i,x), where m is
the balance in MMDA, b is the balance in BTA, i is the
transaction interval, and x is the transfer count (x<6).
Suppose rmb is the reward from having a balance of m in
MMDA and b in BTA, then
r mb  r ( m  [1   ]b )

112
Where d is the % reserve requirement, and r is the return for
the bank on funds invested.
Stochastic DP Model - Cushion
113

The functional equation of our model will be
ft ( m , b, i , x )  r m b
T


C u sh io n  c z :



 m a x C u sh io n  c :
c  0 ,..., c z
1

C u sh io n  0 :


b [  p ij f t  1 ( m  b  s i  c z , c z , j , x  1)]
T
j
b [  p ij f t  1 ( m  b  s i  c 1 , c 1 , j , x  1)]
T
j
b [  p ij f t  1 ( m  b  s i ,0 , j , x  1)]
T
j
Where b is the one period discount factor and p
the transition matrix.
The functional equation changes a bit for
specific conditions (e.g., x=6).
Sample Results - Cushion
114
D ay
o f M o n th
114
1
2
3
4
5
6
7
8
9
10
11
12
13
14
15
16
17
18
19
20
21
22
23
24
25
M M D A + B T A : 2 5 ,0 0 0
T ra n s fe r # , x
1
2
3
4
500
500
500
250
250
250
250
250
250
250
250
250
0
0
0
0
0
0
0
0
0
0
0
0
0
500
500
500
500
500
500
500
500
250
250
250
250
250
250
250
0
0
0
0
0
0
0
0
0
1000
1000
750
750
750
750
750
500
500
500
500
500
250
250
250
250
250
0
0
0
0
0
0
1500
1500
1250
1250
1250
1000
1000
1000
1000
750
750
750
500
500
500
500
250
250
250
0
0
0
5
3250
3250
3250
3250
3250
3000
3000
3000
2750
2750
2500
2500
2250
2000
1750
1750
1250
1000
500
250
0
M M D A + B T A : 5 0 ,0 0 0
T ra n s fe r # , x
1
2
3
4
500
500
500
500
500
250
250
250
250
250
250
250
250
0
0
0
0
0
0
0
0
0
0
0
0
750
750
500
500
500
500
500
500
500
250
250
250
250
250
250
0
0
0
0
0
0
0
0
0
1000
1000
1000
1000
750
750
750
750
750
500
500
500
500
500
250
250
250
250
250
250
0
0
0
2250
2250
2000
2000
2000
2000
2000
2000
2000
2000
2000
2000
2000
2000
2000
2000
2000
2000
2000
2000
1750
0
5
3750
3750
3750
3750
3750
3750
3750
3500
3500
3250
3250
3250
3000
3000
2750
2750
2500
2500
2000
1750
0
Impact of Model

The model is scheduled to be implemented in a mid size bank.

Savings are expected to be about $3 million per year.

115
In simulations, reduced BTA balances from 13% to 26% over existing
methodology