QA Suarez (Updated Apr-14)

8/10/2019 QA Suarez (Updated Apr-14)

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K102: Quantitative Analysis

Contents (lecture-wise):

Lecture-01: Course Outline

Lecture-02: Inventory Model (Risk, Certainty, Expected Value, Payoff Matrix, EVR)

Lecture-03: Opportunity Loss method, EVLO, EVPI, EVC, Incremental Analysis

Lecture-04: Lab Class (Payoff table, Opportunity Loss table in Excel)

Lecture-05: Introduction to Probability, Using tree diagrams for mutually exclusive events

Lecture-06: Problems solved with tree diagrams

Lecture-07: Bayes Theorem

Lecture-08: Decision Analysis (Using Bayes theorem), Decision tree diagram

Lecture-09: Introduction to Linear Programming (LP) Modelling

Lecture-10: LP modelling exercises

Lecture-11: Lab Class (Data input in Excel and using Solver)

Lecture-12: Transportation Model and Assignment Model (LP modelling)

Formulae

Contents (topic-wise):

Payoff Table, Opportunity Loss Table, EVR, EVC, EVLO etc: Lectures 02-04

Probability, Tree Diagram, Bayes theorem: Lectures 05 to 08

Decision Analysis, Decision tree: Lecture-08

LP modelling: Lectures 09 to 12

Transportation and Assignment Model: Lecture-12


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Lecture-1

February 18, 2013Course Outline

1. Concepts of Probability

2. Decision Analysis

3. Linear Programming (LP) Model (Use of excel)

4. Forecasting Techniques

5. Critical Path Method (CPM)

6. Waiting Line Model

7. Markov Process Model

8. Others

Solving of problems is not expected, as they will be solved by simulation. The goal is to

build models based on the variables and conditions.

Reference Material

Either Quantitative Methods for Business or An Introduction to Management Science

(both by) Anderson Sweeney and Williams

Mark Distribution

Quiz - 15%

MT - 25%

FT - 35%

Assignments - 15%

Class - 10%

Grade Boundary

A: 90+

B+: 85-90

B: 80-85

C+: 75-80C: 70-75

D+: 65-70

D: 60-65


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Lecture-2

February 20, 2013Decision Making

Uncertain Risk Certain

Risk situation: Where some of the components are certain and some are uncertain.

The idea of probability is related with risk.

=> The more information we collect, the more we move towards certainty.

For example, if a weatherman forecasts that there is a 90% chance of raining tomorrow, it

means that out of 100 days it will rain on 90 days.

The certain component here is that it will rain on 90 days. The uncertain component is thatwe do not know if a given day or group of days fall within the 10 days or the 90 days.

Expected Value:A predicted value of a variable, calculated as the sum of all possible

values each multiplied by the probability of its occurrence.

For XYZ Bread Manufacturing Company (the significance of bread here is that it is a

perishable good i.e. if 42 units are produced when there is a demand of 40, then 2 units will

be wasted as they cannot be stored for sales next day).

Demand = 40-44 [25 different outcomes e.g demand is 40, production is 41 is 1 outcome]

Selling Price = Tk.38; Variables Cost = Tk.25; Fixed Cost = Tk.200

Table 1.1 Payoff Matrix / Profit Matrix

Decision Alternatives (Productions)

Probability Demand 40 41 42 43 44

10% 40 320 295 270 245 220

20% 41 320 333 308 283 258

40% 42 320 333 346 321 296

20% 43 320 333 346 359 334

10% 44 320 333 346 359 372

100% Expected

Value

320 329.2 330.8 317.2 296


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5 expected values (EVs) for each of the production levels (in units) have been found. The

optimum expected value is known as Expected Value under Risk (EVR).

Thus, here EVR = Tk. 330.8 (at 42 units of production).

Note, for the first cell (Demand=40, Production=40) the formula R=38*40-40*25-200 has

been used. However from then on, the values of cells adjacent to each other have been

found using marginal cost, marginal and marginal profit principles.

In the short-term, the models may produce undesired outcome. However, in the long-term,

these models help us make more correct decisions and we would have made without them.


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Lecture-3

February 27, 2013

OL (Opportunity Loss): Loss incurred by not taking the best decision.

CM (Contribution Margin): Marginal profit per unit of sale (Selling price - Variable cost)

OLU(Opportunity Loss of Understocking) = CM (Contribution Margin)

OLO(Opportunity Loss of Overstocking) = VC (Variable Cost)

For XYZ Bread Manufacturing CompanyDemand = 40-44; Selling Price = Tk.38; Variables Cost = Tk.25; Fixed Cost = Tk.200

Table 1.2 Opportunity Loss table

Production -->


10% 40 0 25 50 75 100

20% 41 13 0 25 50 75

40% 42 26 13 0 25 50

20% 43 39 26 13 0 25

10% 44 52 39 26 13 0

ELO 26 16.8 15.2 28.8 50

+ + + + + +

EV 320 329.2 330.8 317.2 296

= = = = = =

EVC 346 346 346 346 346

*Only the ELO are calculated from this table. EV had been calculated in the payoff matrix.

Here, the ELOR (Expected Opportunity Loss under Risk) is 15.2 (for 42 units). It is also

the EVPI (discussed below). Therefore, the best decision is to produce 42 units.


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ELO (Expected Opportunity Loss)

EV (Expected Value):A predicted value of a variable, calculated as the sum of all possible


EVR (Expected Value under Risk)

EVC (Expected Value under Certainty)

EVPI (Expected Value of Perfect Information)

For calculation of EVC, the situation has no uncertain components, i.e. we know exactly

which days the demand will be x. Thus, there will never be any case of overstocking or

understocking. Consequently, there will be no opportunity loss.

Thus, we can say that Opportunity Loss occurs due to lack of information.

Table 1.1 Payoff Matrix / Profit Matrix

Decision Alternatives (Productions)


10% 40 320 295 270 245 220

20% 41 320 333 308 283 258

40% 42 320 333 346 321 296

20% 43 320 333 346 359 334

10% 44 320 333 346 359 372

100% Expected

Value

320 329.2 330.8 317.2 296

Here, EVC = (320 * 0.1) + (333 * 0.2) + (346 * 0.4) + (359 * 0.2) + (372 * 0.1)

= 346

EVC = EVR + EVPI , or EVPI = EVC - EVR

Thus, EVPI = 346 - 330.8 = 15.2 = ELOR (found above)

EVPI = ELOR , i.e. the value of perfect information is the opportunity loss under risk

And for each level of production, EV + ELO = EVC (refer to Table 1.2)


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Incremental Analysis

Probability of selling the additional unit: (refer to Table 1.1 or 1.2 for probability values)

40th- 100%

41st- 90%

42nd- 70%

43rd- 30%

44th- 10%

For 41stunit, probable profit (? term) = (13 * 0.9) - (25 * 0.1) = 9.2

Since the value is greater than zero, it is profitable to produce the 41stunit.

OLup - OLo(1-p) = 0, where p = probability of selling; (1-p) = probability of not selling

The above equation helps us find the minimum value of p for which loss is not made.

Thus, p = OLO/ (OLO+ OLU), where OLO= VC; OLU= CM = SP - VC

This value of p is a minimum value. If, for example, p = 0.68 (or 68%) then up to the 42nd

unit should be produced. The 43rd unit should not be produced because 68% > 30%. Even if

p is 30.5%, then only 42 units should be produced. 43 units should only be produced when p

is below 30%, for example 25%.

Single-Period Inventory Models with Probabilistic Demand

Method Advantages Disadvantages

Payoff Table /

Payoff Matrix

can be used to directly find the

EVC, EVR, EVPI

ELO can be found indirectly by

working backwards from EVC

= EV + ELO

calculations are complex

and time-consuming

ELO cannot be found

directly

Opportunity

Loss

can be used to direclty find the

ELO, EVPI

simpler calculations

does not reveal all

information e.g EVC,

EVR

Incremental

Analysis

very simple and more viable

for larger numbers of

production/demand values

only SP, VC, p values needed

does not reveal any

information about EVC,

EVR, ELO, EVPI

When only the decision is needed, Opportunity Loss method and Incremental Analysis

Method offer quick anwers. However, the Payoff Table is needed when more detailed

information e.g. about EVC and EVRis needed.


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Single-Period Inventory Model with Probabilistic Demand

[Excerpt from An Introduction to Management Science 13E, page-474]

In situations in which the demand rate is not deterministic, other models treat demand as

probabilistic and best described by a probability distribution. In this section we consider a

single-period inventory model with probabilistic demand.

The single-period inventory model refers to inventory situations in which one order is placed

for the product; at the end of the period, the product has either sold out or a surplus of

unsold items will be sold for a salvage value. The single-period inventory model is

applicable in situations involving seasonal or perishable items that cannot be carried in

inventory and sold in future periods. Seasonal clothing (such as bathing suits and winter

coats) is typically handled in a single-period manner. In these situations, a buyer places

one pre-season order for each item and then experiences a stockout or holds a clearance

sale on the surplus stock at the end of the season. No items are carried in inventory and

sold the following year. Newspapers are another example of a product that is ordered one

time and is either sold or not sold during the single period. Although newspapers are

ordered daily, they cannot be carried in inventory and sold in later periods. Thus,

newspaper orders may be treated as a sequence of single-period models; that is, each day

or period is separate, and a single-period inventory decision must be made each period

(day). Because we order only once for the period, the only inventory decision we must

make is how much of the product to order at the start of the period.


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Lecture-4

March 6, 2013[LAB CLASS]

p = quantity produced

d = quantity demanded

SP = selling price

VC = variable cost

FC = fixed cost

CM = contribution margin = SP - VC

Use IF function to formulate:

1. Profit = p x SP - p x VC - FC (when production demand)

[Using IF, for payoff matrix]

2. Opportunity Loss = (d - p) x CM (when production demand)

[Using IF, for opportunity loss method]

3. Units to produce = Index (Production array, 1,

(Match (EVR/ELOR, EV/ELO array, 0)

)

e.g. =Index($C$8:$G$8,1,Match(H14,C14:G14,0))

or =Index($C$8:$G$8,1,Match(Max(C14:G14),C14:G14,0))

Note that H14 = Max (C14:G14) i.e. EVR = Max (EV array)

[Using INDEX and MATCH, for payoff matrix method or OL method]

Excel functions used: SUM, IF, MIN, MAX,

New Excel functions: SUMPRODUCT, INDEX, MATCH


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Lecture-5

March 10, 2013

Probability:A number expressing the likelihood that a specific event will occur, expressed

as the ratio of the number of actual occurrences to the number of possible occurrences

Experimente.g. toss a coin

Event / Outcomee.g. getting heads or tails

Marginal Probability: Probability with only one condition/criteria

e.g. P(A) , P (B)

Joint Probability: Probability with more than one condition/criteriae.g. P(AB) , P(AUB); P(AUB) = P(A) + P(B) - P(AB);

P(AUBUC) = P(A) + P(B) + P(C) - P(AB) - P(AC) - P(BC) + P(ABC)

Conditional Probability: Probability when one or more outcome has already occurred or is

given. The sample size or sample space becomes smaller when using this condition.

e.g. P(A|B); P(A|B) = P(AB) / P(B)

Mutually Exclusive Events: Two events are said to be mutual if one cannot occur

simultaneously with the other.

e.g. heads and tails, king and ace

For mutually exclusive events A and B: P(AB) = 0; P(AUB) = P(A) + P (B)

Independent Events: Two events are said to be independent of each other if the

occurrence of one event does not affect the occurrence of the other.

e.g. picking cards with replacement; a coin landing on heads and die landing on six.

Two events, A and B, are independent only when: P(A) = P(A|B), i.e the marginal probability

of one of an event is the same as its conditional probability (given the other event).

For independent events A and B: P(AB) = P(A).P(B); P(AUB) = P(A) + P(B) - P(A).P(B)

Note: Two events cannot simultaneously be mutually exclusive and independent.

But can two events can simultaneously be mutually inclusive and dependent.


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Picking an integer from 1 to 10. Sample space is {1,2,3,4,5,6,7,8,9,10).

P(T) = probability of multiple of 3

P(E) = probability of even number

P(T) = 0.3; P(E) = 0.5; P(TE) = 0.1; P(TUE) = 0.7; P(T).P(E) = 0.15

Since P(TE) 0, the events are not mutually exclusive

Since, P(TE) P(T).P(E), the events are not independent

Thus the events are neither mutually exclusive nor independent. That is, they are

dependent mutually inclusive events.

Formulae

For all events:

1a. P(AUB) = P(A) + P(B) - P(AB)

1b. P(AUBUC) = P(A) + P(B) + P(C) - P(AB) - P(AC) - P(BC) + P(ABC)

2a. P(A|B) = P(AB) / P(B) or,

2b. P(AB) = P(A|B) / P(B) or,

2c. P(B) = P(AB) / P(A|B)

For mutually exclusive events:

3. P(AB) = 0

4. P(AUB) = P(A) + P (B)

For independent events:

5. P(AB) = P(A).P(B)

6. P(AUB) = P(A) + P(B) - P(A).P(B)

The branches of tree diagrams represent mutually exclusive events


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Problem-1:

There are: 10 cups with milk and sugar

8 cups with milk and no sugar

6 cups with no milk and sugar

4 cups with no milk and no sugar

1. If we randomly pick one cup what is the probability that:

a) it will contain sugar

b) it will contain no milk

c) given that it contains sugar, what is the probability that it will contain milk

d) it will contain both sugar and milk

2. Are the events sugar and milk independent?

Solution:

[Use table, NOT venn diagram]

Milk No Milk Total

Sugar 10 6 16

No Sugar 8 4 12

Total 18 10 28

1a) P(S) = 16/28

1b) P(M) = 10/28

1c) P(M|S) = 10/16

1d) 10/28

2. P(M) = 18/28; P (M|S) = 10/16; P(M) P(M|S)

Therefore, the cup containing milk and the cop containing sugar are not independent events.

Note: Another case of events that are neither independent nor mutually exclusive.


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Problem-2:

Draw 2 cards from a deck and calculate the probability of getting one ace.

Solution:

[Use tree diagram or direct method. Tree diagram is preferred in complex problems, e.g.

those with more than 4 final outcomes.

Probability of getting one ace = P(AA) + P(AA)


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Lecture-6

March 13, 2013Problem-1:

H, C and F represent claims for health, car and flat respectively.

P(H) = 0.4; P(C) = 0.3; P (F) = 0.2

What is the probability that a person will file claim for:

a) car and flat

b) flat or car but not both

c) exactly 2 claims

d) car and health but not house

e) any one claim

Solution:

a) For car and flat, P = P(HCF) + P (HC) = 0.024 + 0.096 = 0.12

Note: Its no mention whether flat or no flat, so both combinations of car and house have

been taken.


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b) For flat or car but not both, P = P(F) + P(C) + P(HF) + P(HC)

= 0.084 + 0.144 + 0.056 + 0.096

= 0.38

(Again, House is not important here. All combinations that have either F or C but not FC

have been taken).

c) For exactly to claims, P = P(FC) + P(HC) + P(HF)

= 0.036 + 0.096 + 0.056

= 0.188

d) P(HC) = 0.096

e) For any one claim, P = P(H) + P(F) + P(C)

= 0.224 + 0.084 + 0.144

= 0.452

Problem-2:

The probability that a customer will buy a pen is 30% and a book is 40%. What is the

probability that:

a) he will buy only a pen

b) he will buy both the pen and the book

c) either the pen or the book (but not both)

d) 2 consecutive customers dont buy anything

Solution-A (with Tree diagram):

a) P(PB) = 0.18; b) P(PB) = 0.12;

c) P(PB) + P(PB) = 0.18 + 0.28 = 0.46; d) [P(PB)]2= 0.422= 0.1764


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Solution-B (without Tree diagram):

a) P(PB) = 30% x 60% = 0.18

b) P(PB) = 30% 40% = 0.12

c) P(Only pen) + P(Only book) = P(PB) + P(PB) = (30% x 60+ (70% x 40%) = 0.46

d) P(two customers buy nothing) = [P(Nothing)]2= (70% x 60%)2= 0.422= 0.1764

Problem-3:

You have 2 red and 2 black marbles. If you randomly pick 2 marbles, what is the probability

that both will be back (without replacement).

Note: Unless replacement is mentioned explicitly, it will always be without replacement.


P(BB) = 1/6

Solution-B (without tree diagram):

P(BB) = 2/4 x 1/3 = 1/6


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Problem-4:

From a deck of cards, you pick 2 cards. What is the probability that you will get an Ace and

a King (in any order)?


P(Ace and King) = P(AK) + P(KA) = 4/663 + 4/663 = 8/663


P(Ace and King) = P(AK) + P(KA)

= (4/52 x 4/51) + (4/52 x 4/51)

= 4/663 + 4/663

= 8/663


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Problem-5:

From a deck of cards, you pick 2 cards. What is the probability that you will get an Ace and

a King of the same suit (in any order)?


P(AK of same suit) = 2/663


P(AK of hearts in this order) = 1/52 x 1/51 = 1/2652

P (AK of hearts in any order) = 2 x 1/2652 = 1/1326

P( AK of any suit in any order) = 4 x 1326 = 2/663


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Lecture-7

March 20, 2013Bayes Theorem

Problem-1:

P(X) = 30%; P(E|X) = 10% where E = error

P(Y) = 20%; P(E|Y) = 20%

P(Z) = 50%; P(E|Z) = 5%

To find: P(X|E)

Sample Space:

Steps:

1.Find joint probabilities [i.e. P(KE) ]

2. Find marginal p of E [ P(E) = sum of all P(KE) ]

3. Find all the P(K|E) using P(K|E) = P (KE) / P(E)


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Solution:

Events Marginal

Probability

P(K)

Conditional

Probability

P(E|K)

Joint

Probability

P(KE)

Conditional P

(given Error)

P(K|E)

X 0.30 0.10 0.03 0.316

Y 0.20 0.20 0.04 0.421

Z 0.50 0.05 0.025 0.263

Total P(E) = 0.095 1

Problem-2:

Three typists are working. The probability of work being doen by each typist and the

corresponding chances of error are given:

P(A) = 25%; P(E|A) = 20%

P(B) = 45%; P(E|B) = 25%

P(C) = 30%; P(E|C) = 10%

Given that there is error, what is the probability that

(1) A typed it (2) B typed it (3) C typed it

Solution:

Events Marginal

Probability

P(K)

Conditional

Probability

P(E|K)

Joint

Probability

P(KE)

Conditional P

(given Error)

P(K|E)

A 0.25 0.20 0.05 0.260

B 0.45 0.25 0.1125 0.584

C 0.30 0.10 0.03 0.156

Total P(E) = 0.1925 1

Probability that (1) A typed it is 0.260, (2) B typed it is 0.584, (3) C typed it is 0.156

Problem-3:

There will be a football match tomorrow. It rained 5 days each year. Weatherman

forecasted rain for tomorrows match. When it rains, he forecasts 90% of the time. When it

doesnt rain, he forecasts rain 20% of the time. So given that he forecasted rain tomorrow,

what is the probability that it will rain?


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Solution:

Let the events Raining be R, not raining be N, and weatherman forecasts rain be F.

Thus,

P(R) = 5/365; P(N) = 360/365; P(F|R) = 90%; P(F|N) = 20%

Events Marginal

Probability

P(K)

Conditional

Probability

P(F|K)

Joint

Probability

P(KF)

Conditional P

(given Error)

P(K|F)

R 5/365 0.90 0.0123 0.0587

N 360/365 0.20 0.1973 0.9413

Total P(F) = 0.2096 1

Thus, the probability of rain tomorrow given that weatherman forecasted rain is 0.0587

*Note: If P(F|N) decreases, e.g. 10%, then P(R|F) will increase since it means accuracy

is increasing. [When P(F|N) = 10%, P(R|F) = 0.11 > 0.0587 ]

Discrete Probability Distribution: (Most probably not in syllabus)

1. Binomial Distribution

2. Poisson Distribution

Conditions of binomial distribution:1. There are 2 events/outcome

2. Outcomes are independent of previous outcome

Binomial distribution:

(p+q)n ; where p is the probability of success, q = (1-p) is the probability of failure and n is

the number of experiments

(p+q)10 = p10+ (AB)/C p9q + ..

where, A = power of p in previous term

B = coefficient of previous term (AB/C of previous term)

C = position no. of previous term (in this case, 1)

Probability of a single term (the rthterm:

P of (r+1)thterm = ncrpn-rqr


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In binomial distribution, n is finite. In poisson distribution n is infinite.

We can only measure success in Poisson distribution, not failure. This is because the

variable is distributed over space and time. For example, we can measure the road

accidents in Dhaka-Chittagong highway from 5pm to 6pm on a day, but we cannot the

measure the number of road accidents averted.


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Lecture-8

March 30, 2013Decision Analysis

Problem-1:

Given: Payoff table (the values express revenue/profit)

Decision High Demand (h) Low Demand (l)

Large Plant 200,000 -180,000

Small Plant 100,000 -20,000

Do nothing 0 0

The company can buy a report which may show the market has favorable conditions (F) or

unfavorable conditions (U). Given that P(F|h) = 70%, P(F|l) = 20%, P(h) = 50%, P(l) = 50%,

find the expected value of the report and the expected value of sample information.

Solution:

When report is Favorable (F):

Variable P(k) P(F|k) P(kF) P(k|F)

h 0.50 0.70 0.35 0.778 [0.35/0.45]

l 0.50 0.20 0.1 0.222 [0.1/0.45]

Sum 0.45 = P(F) 1

EV of Large Plant = (200,000 x 0.778) + (-180,000 x 0.222)

= 115,555.56

EV of Small Plant = (100,000 x 0.778) + (-20,000 x 0.222)

= 73,333.33

EV of Do Nothing = (0 x 0.778) + (0 x 0.222)

= 0

*EV of Do Nothing is always zero as we actually are not producing anything

EVR when favorable = EVRF= 115,555.56 [largest value among the three]


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When report is Unfavorable (U):

Variable P(k) P(U|k) P(kU) P(k|U)

h 0.50 0.30 [1- 0.70] 0.15 0.273 [0.15/0.55]

l 050 0.80 [1 - 0.20] 0.4 0.727 [0.4/0.55]

Sum 0.55 = P(U) = 1 - P(F) 1

EV of Large Plant = (200,000 x 0.273) + (-180,000 x 0.727)

= -76,363.63 [negative value means loss is made]

EV of Small Plant = (100,000 x 0.273) + (-20,000 x 0.727)

= 12,727.27

EV of Do Nothing = 0

EVR when unfavorable = EVRU= 12,727.27

EV of Report = EVRFx P(F) + EVRUx P(U)

= (115,555.56 x 0.45) + (12,727.27 x 0.55)

= 59,000

Without Report:

EV of Large Plant = 200,000 x P(h) + (-180,000) x P(l)

= (200,000 x 0.5) + (-180,000 x 0.5)

= 10,000

EV of Small Plant = 100,000 x P(h) + (-20,000) x P(l)

= (100,000 x 0.5) + (-20,000 x 0.5)

= 40,000

EV of Do Nothing = 0

EV of without report = EVR = 40,000

To find Expected Value of Sample Information (EVSI)

EV of Report = 59,000; EV without report = 40,000

Thus, EVSI = 59,000 - 40,000

= 19,000

So, the company should buy the report if price of the report is less than 19,000.


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Insertbeautiful

decision

treediagram

here.


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Problem-2:

Given two routes, one of which uses the expressway, and the time taken to travel:

Expressway Open (s1) Expressway Jammed (s2)

Queen City Avenue

(Route 1)

30 minutes 30 minutes

Expressway

(Route 2)

25 minutes 45 minutes

Based on their experience with traffic problems, Rona and Jerry agreed on a 0.15

probability that the expressway will be jammed.

In addition, they agreed that weather seemed to affect the traffic conditions on the

expressway. Let: C = clear; O = overcast; R = raining

The following conditional probabilities apply:P(C|s1) = 0.8; P(O|s1) = 0.2; P(R|s1) = 0.0 [Note that 0.8+0.2+0.0 = 1]

P(C|s2) = 0.1; P(O|s2) = 0.3; P(R|s2) = 0.6 [Note that 0.1+0.3+0.6 = 1]

Which route should be selected if the weather is: (a) clear, (b) overcast, (c) raining

(d) What is the expected travel time?

Solution:

For Clear weather:

Variable P(k) P(C|k) P(kC) P(k|C)

s1 0.85 [1-0.15] 0.80 0.68 0.978 [0.68/0.695]

s2 0.15 [given] 0.10 0.015 0.022 [0.015/0.695]

Sum 0.695 = P(C) 1

EV of Route 1 = 30

[Route 1 does not use the expressway, so probability of s1 and s2 has no effect on it]

EV of Route 2 = 25 x P(s1|C) + 45 x P(s2|C)

= (25 x 0.978) + (45 x 0.022)

= 25.44

EVR of Clear = EVRC= 25.44 [Smallest value is EVR as it is regarding time taken]

(a) Route 2 should be selected if weather is clear as it will take less time to travel.


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For Overcast weather:

Variable P(k) P(O|k) P(kO) P(k|O)

s1 0.85 0.20 0.17 0.791 [0.17/0.215]

s2 0.15 0.30 0.045 0.209 [0.045/0.215]

Sum 0.215 = P(O) 1

EV of Route 1 = 30

EV of Route 2 = 25 x P(s1|O) + 45 x P(s2|O)

= (25 x 0.791) + (45 x 0.201)

= 29.18

EVR of Overcast = EVRO= 29.18(b) Route 2 should be selected if weather is overcast as it will take less time to travel.

For Rainy weather:

Variable P(k) P(R|k) P(kR) P(k|R)

s1 0.85 0.00 0.00 0 [0/0.09

s2 0.15 0.60 0.09 1 [0.09/0.09]

Sum 0.09 = P(R) 1

EV of Route 1 = 30

EV of Route 2 = 25 x P(s1|R) + 45 x P(s2|R)

= (25 x 0) + (45 x 1)

= 45

EVR of Rainy = EVRR= 30

(c) Route 1 should be selected if weather is rainy as it will take less time to travel.

(d) Expected time to travel = [P(C) x EVRC]+ [P(O) x EVRO] + [P(R) x EVRR]

= (0.695 x 25.44) + (0.215 x 29.18) + (0.09 x 30)

= 26.65

___________________________________The End_________________________________

Note that for the marginalpvalues of weather found in this problem, P(C) + P(O) + P(R) =1


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Insertbeautiful

decision

treediagram

here.


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Lecture-9

March 31, 2013Linear Programming (LP)

Things to do:

1. Develop the LP model

2. Input the data into MS Excel and use Solver to find the answer

Note: We DONT have to know how to solve the LP problems manually.

LP Requirements:

Maximize or minimize an objective function (MAX or MIN) Constraints acting as a limiting factor, introduced by S.T. (Subject to)

There must be alternatives available

Mathematical relationship is linear (power of decision variables is 1, i.e. no x2, x3...)

Basic Assumptions of LP:

Certainty (e.g. raw materials per unit / profit per unit is fixed/constant)

Proportionality (e.g. profit for 10 units = profit for 1 unit x 10)

Additivity (e.g. Profit from A and B = Profit from A + Profit from B)

Divisibility (e.g. non-integer solutions such as no. of chairs produced = 2.5)

Steps to develop LP model:

1. Understand the problem

2. Identify the decision variables

3. Determine the objective function in terms of decision variables

4. Determine constraints in terms of decision variables

5. State the non-negative constraints

Special Conditions in LP Models:

Alternate optimal solutions (when objective line is parallel to a constraint line) Redundant constraints

Unbounded solutions

Infeasibility


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Problem-1:

Blue Ridge Hot tubs produces two types of tubs, X and Y. Their details are given below:

Type X Type Y

Pumps 1 1

Labor 9 hrs 6 hrs

Tubing 12 feet 16 feet

Unit profit $350 $300

There are 200 pumps, 1566 hours of labor and 2880 feet of tubing available.

Solution:

x1

= number of type X to produce;

x2= number of type Y to produce

MAX: 350x1+ 300x2S.T. : x1+ x2 200

9x1+ 6x21566

12x1+ 16x2 2880

x1 0 ; x2 0

Note: The objective here is to maximize the profit function. The first constraint is about

pump capacity, the second constraint relates to labor capacity, the third constraint is about

tubing capacity, and the final constraints are for non-negativity.

Problem-2:

Two types of paint jobs can be done. A and B are raw materials required:

Exterior Paint Interior Paint

No. of units of A required 1 2

No. of units of B required 2 1

Revenue 3000 2000

Only 6 units of A and 8 units of B can be used. No more than two interior paint jobs can be

done. At least one more exterior paint job than interior paint job needs to be done.

Develop an LP to maximize the revenue.


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Solution:

x1= no. of exterior paint jobs

x2= no. of interior paint jobs

MAX: 3000x1+ 2000x2

S.T. : x1+ 2x2 6

2x1+ x2 8

x1- x2 1

x2 2

x1 0

x2 0

Note: The objective here is to maximize the revenue function.The first constraint is about maximum capacity of resource A (which is 6) and the second

constraint is about capacity of resource B (which is 8).

The third constraint expresses that there is AT LEAST one more exterior paintjob than

interior paintjob.

The fourth constraint is about maximum number of interior paintjobs that can be done.

Last 2 constraints are for non-negativity.


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Lecture-10

April 1, 2013Problem-1

ICT has $5 million available for immediate investment and wishes to do two things:

(1) maximize the interest earned on the investments made over the next one year and

(2) satisfy the diversification requirements as set by the board of directors. The specifics of

the investment possibilities are as follows:

Investment Interest earned (%) Maximum investment($ millions)

Trade credit 7 1.0

Corporate bonds 11 2.5

Gold stocks 19 1.5

Construction loans 15 1.8

In addition, the board specifies that at least 55% of the funds invested must be in gold

stocks and construction loans, and that no less than 15% be invested in trade credit.

Develop the LP model.

Solution:x1= investment in trade credit ($ millions)

x2= investment in corporate bonds ($ millions)

x3= investment in gold stocks ($ millions)

x4= investment in construction loans ($ millions)

MAX: 0.07x1+ 0.11x2+ 0.19x3+ 0.15x4

S.T. : x1+ x2+ x3+ x4 5

x1 1

x2 2.5

x3 1.5

x4 1.8

x3+ x4 0.55 (x1+ x2+ x3+ x4)

x1 0.15 (x1+ x2+ x3+ x4)

x1, x2, x3, x4 0


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Note:

1. Objective is to maximize interest earned, so objective function is sumproduct of array of

interests and the types of investment. The decision variables are the particular types of

investments.

2. First constraint expresses that total investment cannot exceed allotted sum of $5 million.

3. Constraints 2-5 express the maximum investment capacity for each type of investment.

4. Constraint 6 expresses at least 55% of total investment must be in gold + construction.

5. Constraint 7 expresses that at least 15% of total investment must be in trade credit.

6. Non-negative constraints

Problem-2

MSA, a surveying agency, must conduct a survey that satisfied the following requirements:

1. Survey at least 2,300 US households in total.

2. Survey at least 1,000 households whose heads are 30 years of age or younger.

3. Survey at least 600 households whose heads are between 31 and 50 years of age,

4. Ensure at least 15% of those surveyed live in a state that borders Mexico.

5. Ensure that no more than 20% of those surveyed who are 51 years of age or over

live in a state that borders Mexico.

MSA decides that all surveys should be conducted in person. It estimates that the costs of

reaching people in each age and region category are as follows:

Region Age 30 Age 31-50 Age 51

State Bordering Mexico 7.5 6.8 5.5

State not bordering Mexico 6.9 7.25 6.1

MSAs goal is to meet the above sampling requirements at the least possible cost.

Note: With three age groups and two geographic groups, there are 6 different combinations

of population. Moreover, there are specific constraints in the problem such as the group of

people in Age 51 who are living in a state that borders Mexico. Thus there must be 6

decision variables, for each of the combinations. In LP problems like these, it helps to

visualize the decision variables in a table/array format (shown below), as it is more

convenient to input these decision variables when using excel to solve.(This table is to help visualize, and is NOT needed when developing the LP model):

[x1...x6represents no. of people] Age 30 Age 31-50 Age 51

State Bordering Mexico x1 x2 x3

State not bordering Mexico x4 x5 x6


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Solution:

x1= No. of people in age 30 living in a state bordering Mexico

x2= No. of people in age 31-50 living in a state bordering Mexico

x3= No. of people in age 51 living in a state bordering Mexico

x4= No. of people in age 30 living in a state NOT bordering Mexico

x5= No. of people in age 31-50 living in a state NOT bordering Mexico

x6= No. of people in age 51 living in a state NOT bordering Mexico

MIN: 7.5x1+ 6.8x2+ 5.5x3+ 6.9x4+ 7.25x5+ 6.1x6

S.T. : x1+ x2+ x3+ x4+ x5+ x6 2300

x1+ x4 1000

x2+ x5 600

x1+ x2+ x3 0.15 (x1+ x2+ x3+ x4+ x5+ x6)

x3 0.20 (x3+ x6)x1, x2, x3, x4, x5, x6 0

Note: The first 5 constraints correspond to the constraints numbered 1-5 in the problem,

and the last constraint is for non-negativity. The 5th constraint does not take the entire

population of people surveyed as it is mentioned in the problems constraint no.5 of those

surveyed who are 51 years of age or over so only 20% of (x3+ x6) is taken.

Problem-3

You need to buy some filing cabinets. You know that Cabinet X costs $10 per unit, requires

six square feet of floor space, and holds eight cubic feet of files. Cabinet Y costs $20 per

unit, requires eight square feet of floor space, and holds twelve cubic feet of files. You have

been given $140 for this purchase, though you don't have to spend that much. The office

has room for no more than 72 square feet of cabinets. How many of which model should

you buy, in order to maximize storage volume?

Solution

Drawing the table makes the problem very easy to model.

Cabinet X Cabinet Y Maximum

Cost $10 $20 $140

Floor Space 6 sq. ft. 8 sq. ft. 72 sq. ft.

Storage Volume 8 cu. ft. 12 cu. ft. [Objective]

Thus, objective is to maximize storage volume.


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x1= no. of Cabinet X

x2= no. of Cabinet Y

MAX: 8x1+ 12x2

S.T : 10x1+ 20x2 140

6x1+ 8x2 72

x1, x2 0

Problem-4

XYZ bank has 12 full time tellers. The number of teller required per hour is given below:

Time No. of tellers required

9-10 10

10-11 12

11-12 14

12-1 16

1-2 18

2-3 17

3-4 15

4-5 10

The bank is considering hiring some part time tellers. If it is economically viable, they will

reduce the number of full time employees.

Full Time Tellers:

Work from 9 AM 5 PM

Take a 1 hour lunch break, half at 11, the other half at noon

Cost $90 per day (salary & benefits)

Part Time Tellers:

Work 4 consecutive hours (no lunch break)

Are paid $7 per hour ($28 per day)

Part time teller hours cannot exceed 50% of the days minimum requirement

Develop the LP Model


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Solution:

p1= No. of part-time employees starting work at 9




p5= No. of part-time employees starting work at 1 x = No. of full-time employees

MIN: 90x + 28(p1+ p2+ p3+ p4+ p5)

S.T. : x + p1 10 [Note: for 9-10 timeslot]

x + p1+ p2 12 [Note: for 10-11 timeslot]

x/2 + p1+ p2 + p3 14 [Note: for 11-12 timeslot]

x/2 + p1+ p2 + p3+ p4 16 [Note: for 12-1 timeslot]

x + p2 + p3+ p4 + p5 18 [Note: for 1-2 timeslot]

x + p3+ p4 + p5 17 [Note: for 2-3 timeslot]

x + p4 + p5 15 [Note: for 3-4 timeslot]

x + p5 10 [Note: for 4-5 timeslot]

4(p1+ p2+ p3+ p4+ p5) 56

x 12

p1, p2, p3, p4, p5, x 0

Note: The objective here is to minimize daily cost (salary). The daily salary of full-timers is

$90. Part-timers earn $7 an hour and work for 4 consecutive hours (which means minimum

and maximum 4 hours) so their daily salary is $28. The decision variablesare the number

of part-timers that have started to work at each timeslot beginning and the number offull-timers. Note that there are only 5 decision variables of part-timers as part-timers cannot

work 4 hours if recruited after 1-2 timeslot (the work-day ends at 5).

The first 8 constraints are about the minimum number of full-time + part-time workers

required for each of the 8 time-slots. At each slot till the 1-2 slot, a new set of part-timers

join in (from p1to p5). Each set of part-timers leave after 4 hours, for example the ones that

joined in 9-10 timeslot (p1) are found in 9-10, 10-11, 11-12, 12-1: these four timeslots.

Finally, note that in the 11-12 and 12-1 timeslots, the number of full-time workers is halved

as half of the full-time workers leave for lunch.

The 9th constraint: 4(p1+ p2+ p3+ p4+ p5) 56 relates to Part time teller hours cannot

exceed 50% of the days minimum requirement. The days minimum requirement (112) is

found by adding all of the 8 entries for no. of tellers required in the problems table. Then it

is halved (56), and the total part-time hours are found by adding all the part-timers and

multiplying with 4 as each part-timer works for 4 hours.

The 10th constraint expresses that number of full time employees is at most 12.


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Lecture-11

April 3, 2013[LP Lab Class]

Read this if you feel technologically challenged, at least as far as MS Excels Solver is

concerned. Warning: reading the next few lines may slightly lower your self-esteem.

[These steps have been taken (and modified) from Computer Shikkha by Arman Pavel. ]

0. Input all your data.

1. Open Solver. You can find it in the Data pane or Data menu.

2. Click Objective Function / Set Objective box and select the cell that has the function

that you need to maximize or minimize.

3. Select either the Max or Min radio button below the Objective box depending on what you

need to do.

4. Click the By changing variable cells box and select ALL the cells that contain the

decision variables. You should input the data in array format so that you can just select

the array here. For example $A$1:$C$3 is an array that includes the 9 cells A1, A2, A3, B1,

B2, B3, C1, C2 and C3.

5. Click Add beside the Subject to the Constraints box. Then select the appropriate cells

for the cell reference boxes and make sure that the sign is appropriate ( , = , )

6. After adding all the constraints, make sure that Make unconstrained variables

nonnegative box is checked (by default, it IS checked) and change Select a Solving

Method to Simplex LP (usually it is set at GRG Nonlinear by default).

7. Click Solve, then whatever dialog box appears click OK.

8. [Depending on luck] EITHER watch the magix happen OR freak out as it doesn't work

Note: Refer to the excel files in the group in case you have any problems whatsoever.


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Lecture-12

April 7, 2013Transportation Model (LP):

Decision: How much to ship from each origin to each to destination?

Objective: Minimize shipping cost

i) A balanced transportation model is one where total demand is equal to total supply:

Since all supplies will be used supply constraints have =

Since all demands are satisfied demand constraints have =

ii) If demand exceeds supply (unbalanced transportation model):

Since all supplies will be used supply constraints have =Since not all demands can be satisfied demand constraints have

iii) If supply exceeds demand (unbalanced transportation model):

Since not all supplies will be used supply constraints have

Since all demands can be satisfied demand constraints have =

Problem-1

Cost of transporting from each origin to each destination:

Albuquerque Boston Cleveland

Des Moines 5 4 3

Evansfield 8 4 3

Fordville 9 7 5

Capacity of units:

Des Moines 100 Albequerque 300

Evansfield 300 Boston 200

Fordville 300 Cleveland 200

Total 700 Total 700

*Note that total capacity of units demanded (Destination) and supply (Origin) sides are

equal (700 units). Thus, this is a balanced model.


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Solution

Let no. of units to be shipped from Des Moines to Albuquerque = x 11

Let no. of units to be shipped from Des Moines to Boston = x12

Let no. of units to be shipped from Des Moines to Cleveland = x13

Let no. of units to be shipped from Evansfield to Albuquerque = x 21

Let no. of units to be shipped from Evansfield to Boston = x22

Let no. of units to be shipped from Evansfield to Cleveland = x 23

Let no. of units to be shipped from Fordville to Albuquerque = x31

Let no. of units to be shipped from Fordville to Boston = x32

Let no. of units to be shipped from Fordville to Cleveland = x 33

MIN: 5x11+ 4x12+ 3x13+ 8x21 + 4x22+ 3x23+ 9x31+ 7x32+ 5x33

S.T: x11+ x12+ x13= 100

x21+ x22+ x23= 300

x31

+ x32

+ x33

= 300

x11+ x21+ x31= 300

x12+ x22+ x32= 300

x13

+ x23

+ x33

= 200

x11, x12, x13, x21, x22, x23, x31, x32, x33 0

Note:1. The objective is to minimize the cost function, thus objective function is the

sumproduct of the array of no. of units (shipped from each origin to each destination) and

the unit cost of transportation. The decision variables here are the no. of unitsshipped

from each origin to each destination, and as there are six combinations, there are 9

decision variables from x11to x33

2. Constraints 1-3 are for origin/demand-side. The 1st constraint is for Des Moines, 2nd

constraint for Evansfield and so on. = is used as this is a balanced transportation model.

3. Constraints 4-6 are for destination/supply-side. The 4th constraint is for Albuquerque, 5thconstraint is for d-2 and so on. = is used as this is a balanced transportation model.

4. The final (7th) constraint is for non-negativity.


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Assignment Model (LP):

Assignment models are like transportation models, except you decide whether or to assign

a source to a destination (or employee to a task).

Decision: Which assignee (person) to assign to which assignment (task)?

Objective: Minimize cost

1-to-1 relationship between assignments and assignees, i.e. each assignment must

get AT MOST 1 assignee, and each assignee must get AT MOST 1 assignment

Decision variables are binary, i.e. they can only be either 0 or 1. For example,

John-1 = 0 means that John will not perform assignment-1 and John-1 = 1 means

that John will perform assignment-1.

Problem-2

Cost of given programmers performing each task:

Task-1 Task-2 Task-3

Adams 11 14 6

Brown 8 10 11

Cooper 9 12 7

Solution

Adams doing Task-1 = x11 Adams doing Task-2 = x12; Adams doing Task-3 = x13

Brown doing Task-1 = x21 Brown doing Task-2 = x22; Brown doing Task-3 = x23

Cooper doing Task-1 = x31 Cooper doing Task-2 = x32; Cooper doing Task-3 = x33

MIN: 11x11+ 14x12+ 6x13+ 8x21 + 10x22+ 11x23+ 9x31+ 12x32+ 7x33

S.T: x11+ x12+ x13= 1

x21+ x22+ x23= 1

x31+ x32+ x33= 1

x11+ x21+ x31= 1x12+ x22+ x32= 1

x13+ x23+ x33= 1

x11, x12, x13, x21, x22, x23, x31, x32, x33 0

x11, x12, x13, x21, x22, x23, x31, x32, x33 1


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Note:

1. The objective is to minimize the cost function, thus objective function is the

sumproduct of the array of each assignee doing each assignmentand the unit cost of

doing the work. The decision variables here are assignees doing assignments.

2. The last two constraints (constraints 7 and 8) are to express that the value of decision

variables can only be either 1 (for assignee doing the assignment) and 0 (for assignee not

doing assignment).

3. Constraints 1-6 are for 1-to-1 relationship of assignees/people and assignments/tasks.

The first three constraints are to express that each assignee/person can perform AT LEAST

and AT MOST one task only. Since total value is 1 and decision variables can only take

values of either 1 or 0, only one of the three decision variables in the constraint will be

equal to 1 and the other two will equal zero. For example, x11

+ x12

+ x13

= 1 is the

constraint for Adams. If x12= 1, this means x11=x13=0, which means that Adams will be

performing task-2 and not performing task-1 and task-3.

Similarly, the last three constraints are to express that each assignment/task can be

performed by AT LEAST and AT MOST one assignee/person only.


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Single-Period Inventory Models with Probabilistic Demand

p (Quantity Produced)

d (Quantity Demanded)

SP (Selling Price)

VC (Variable Cost)

FC (Fixed Cost)

CM (Contribution Margin): Marginal profit per unit of sale

EV (Expected Value):A predicted value of a variable, calculated as the sum of all possible


EVR (Expected Value under Risk)

EVC (Expected Value under Certainty)

OL (Opportunity Loss): Loss incurred by not taking the best decision.

OLU(Opportunity Loss of Understocking)

OLO(Opportunity Loss of Overstocking)

ELO (Expected Opportunity Loss)

ELOR (Expected Opportunity Loss under Risk)

EVPI (Expected Value of Perfect Information)

CM = SP - VC

EVR = maximum EV value in a risk situation

ELOR = minimum ELO ina risk situation

EVC = EV + ELO (in other words, OL is the gap between EV of certainty and EV without it)

EVC = EVR + ELOR

EVC = EVR + EVPI (difference in EV of certain and risk situation is caused by perfect info)

EVPI = EVC - EVR

EVPI = ELOR (opportunity loss under risk comes from lack of perfect information)ELOC = 0 (in a certain situation, there is no lack of perfect information, thus no OL)

OLO= VC

OLU= CM = SP - VC

p = OLO/ (OLO+ OLU) (this is the minimum value of p used in incremental analysis)


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Excel:

1. Profit = p x SP - p x VC - FC (when production demand)

[Using IF, for payoff matrix]

2. Opportunity Loss = (d - p) x CM (when production demand)

[Using IF, for opportunity loss method]

3. Units to produce = Index (Production array, 1,

(Match (EVR/ELOR, EV/ELO array, 0)

)

e.g. =Index($C$8:$G$8,1,Match(H14,C14:G14,0))

or =Index($C$8:$G$8,1,Match(Max(C14:G14),C14:G14,0))

Note that H14 = Max (C14:G14) i.e. EVR = Max (EV array)

[Using INDEX and MATCH, for payoff matrix method or OL method]

4. probability = p0= VC / (VC + CM)

Units to produce = whichever level of demand has highest probability of selling

additional unit that is less than or equal to p0=if(E32


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Bayes Theorem:

1.Find all the joint probabilities [i.e. P(KE) ]

2. Find marginal probability of E [ P(E) = sum of all P(KE) ]

3. Find all the P(K|E), using P(K|E) = P (KE) / P(E)

Steps to develop LP model:

1. Understand the problem

2. Identify the decision variables

3. Determine the objective function in terms of decision variables

4. Determine constraints in terms of decision variables

5. State the non-negative constraints

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