dec sci presentation

DECISION SCIENCE DECISION MAKING USING

PROBABILITIESALLAN EMIL C. BALDERAS R.N

Random Variable – variable that takes on different values as a result of a random experiment

Example:

Using The expected Value criterion with continuously distributed random variables

Speakers sold(random variable

No. Of days this quantity is sold( Frequency)

Probability that random variable take on this value(relative frequency)

100 10 0.10

105 20 0.20

110 30 0.30

115 40 0.40

100 1.00

A RANDOM VARIABLE CAN BE

a. Discrete – variable which can only take a countable number of values

Example: If a coin is tossed three times, the number of heads obtained can

be 0, 1, 2 or 3. The number of heads can only take 4 values (0, 1, 2, 3) and so

the variable is discrete

b. Continuous – variable that us allowed to take ANY value within a given range

Example: Daily sales of speakers range from 100 - 120


Expected Value criterion – a criterion which requires the decision maker to calculate the expected value for each decision alternative and then choose that decision in which has the highest expected value among all decision alternatives

Situation 1:Jacque sells cherry tomatoes every spring. She buys them at 9 pesos / crate

and resells them for 16 pesos / crate. If a crate is not sold on the first selling day, it is worth 3 pesos as salvage. Jacque’s examination of past sales record indicate that demand for this particular item is normally distributed, with a mean of 120 crates and standard deviation of 38 crates. What should Jacque’s stock be? Steve can use this equation to justify his stocking of that unit with the date she has available to her.

P = ML / (MP + ML )

Where P = probability that she is sure of selling at least an additional unit ML = marginal loss = 9 pesos / crate - 3 pesos / crate as salvage = 6 MP = marginal profit = 16 pesos / crate - 9 pesos / crate = 7


Where S = standard

deviation N = number of

values in the population

X = values of the random variable in distribution

µ = mean of population

Standard Deviation Formula

P = ML / (MP + ML )

Where P = probability that she is sure of selling at least an additional unit ML = marginal loss = 9 pesos / crate - 3 pesos / crate as salvage = 6 MP = marginal profit = 16 pesos / crate - 9 pesos / crate = 7

P = ML / (MP + ML ) P = 6 / ( 7 + 6 )P = .462

The (P) probability means Jacque must be 0.462 sure of selling at least an additional unit before it will pay him to stock that unit


Continuosly distributed random variable past demand

.538 area.462 area

Z = 0.1

µ = 120Ơ = 38

120

Point Q

- Since the area under the normal curve is 0.462 of the total area under the curve, the open area must be 1.00 - 0.462 or 0.538 of the area under the curve.

- Looking at the Standard Normal Probability Table for 0.538, we see that the value is 0.1

.538 area .462 area

Z = 0.1

µ = 120Ơ = 38

Mean = 120 crates

Point Q


- Looking at the Standard Normal Probability Table for 0.538, we see that the value is 0.1- This means that point Q is 0.1 standard deviations to the right of the mean- Since we know that the standard deviation of the distribution past demand for this item is 38 crate, Point Q is as follows

.538 area .462 area

Z = 0.1

µ = 120Ơ = 38

Mean = 120 crates

Point Q

Point Q = mean + 0.1 (standard deviation)= mean + 0.1 ( 38 crates )= 120 crates + 3.8 crates= 123.8 crates= 124 crates

The Optimal for Jacque to order is 124 crates


Situation 2:Daily sales of bread by the Maalat Baking Company are normally distributed

with a mean of 6000 loaves and a standard deviation of 900 loaves. It costs the bakery 50 cents to produce a loaf of bread which sells for 95 cents. Any Bread unsold at the end of the day is sold to the Botcha lovers compound for 35 cents a loaf. How many loaves should the bakery produce each day to maximize expected profit

P = ML / (MP + ML )

Where P = ML = marginal loss = 50 cents / loaf – 35 cents / loaf salvage = 15 cents MP = marginal profit = 95 cents/ loaf - 50 cents / loaf = 45 cents

P = ML / (MP + ML ) P = 15 / (45 + 15)P = .0.25


Continuosly distributed random variable past demand

.75 area.25area

Z = 0.68

µ = 6000Ơ = 900

6000

Point Q


- Looking at the Standard Normal Probability Table for 0.75, we see that the value is about 0.68

0.75 area 0.25 area

Z = 0.68

µ = 6000Ơ = 900

Mean = 6000 crates

Point Q


- Looking at the Standard Normal Probability Table for 0.75, we see that the value is about 0.68 - This means that point Q is 0.68 standard deviations to the right of the mean- Since we know that the standard deviation of the distribution past demand for this item is 900 crate, Point Q is as follows

0.75 area 0.25 area

Z = 0.68

µ = 6000Ơ = 900

Mean = 6000 crates

Point Q

Point Q = mean + 0.68 (standard deviation)= mean + 0.68 ( 900 loaves )= 9000 loaves + 612 loaves= 9612 loaves

The Optimal for Maalat Baking Company to order is 9612 loaves


This is done where past data are missing or at best incomplete.

SUPPLYING THE NUMBERS

Situation 1: Suppose that you were contemplating the purchase of a machine which replaces manual labour on an operation. The machine will cost $5000 per year to operate and will save $4 for each hour it operates. to break even, it must operate at least $5000/4 = 1250 hours annually. If we are interested in the probability that it will run more than 1250 hours, we must know about the distribution of running times, specifically the mean and standard deviation of the distribution. But where would we find these figures with no past history of machine operation?

Suppose we went to the supervisor of this operation and asked him to guess the mean running time of the machine. With his close knowledge of the process involved, he would probably be able to give a “guessed-at”mean: let us say his best estimate here is 1,400 hours. But how would he react if you then asked him to give you the standard deviation of the distribution? This term is probably not meaningful to him and yet intuitively, he probably has some notion of the DISPERSION of the distribution of running times. Since most people understand betting odds, you will appeal to him on that basis

SUPPLYING THE NUMBERS

Let us count off an equal distance on each side of his mean, say 200 hours, this gives us an interval from 1200 to 1600 hours. Then ask him “what are the odds that the number of hours will lie between 1200 and 1600? If he has had any experience with betting, he should be able to reply. Suppose he says “odds that it will run between 1200 and 1600 hours are 5 to 3. we show his answer on an probability distribution.

The figure indicates exactly what the supervisor replied, that the odds are 5 to 3 that the machine will run between 1200 to 1600 hours rather than outside those limits

35

Mean 1400 hours

Point Q = 1600

35

1200

Now what to do? On the figure. We have called the 1600 hour point Q. WE see that the area under the entire curve as (3+5+5) / ( 3 + 5 + 5 + 3)

Since 13/16 is approximately 0.813. Look for the value of 0.813 in the standard normal probability table which is equal 0.89 standard deviation to the right of the mean

at the distance from the mean to Q is 200 hours, wee see that0.89 s.d.= 200 hours and thus 1 s.d.= 225 (200/89) * 100

3 5

Mean 1400 hours

Point Q = 1600

Z = 0.89

5

1200

3

Now that we can specify the mean and the S.D. Of the distribution of running times, we can calculate the probability of the machine’s running less than its breaking point of 1250 hours

z = (x - µ) / ơ

where z = the number of S.D from the mean x = the value of the random variable concerned, µ = the mean ơ = the standard deviation

3 5

Mean 1400 hours

Point Q = 1600

Z = 0.89

5

1200

3

Now that we can specify the mean and the S.D. Of the distribution of running times, we can calculate the probability of the machine’s running less than its breaking point of 1250 hours

z = (x - µ) / ơ z = (1250 – 1400) / 225z = 0.6667

Then find the value of 0.6667 S.D in the standard normal probability distribution table which is equal to 0.747 of the total area under the curve

?

Mean 1400 hours

Point Q = 1600

Z = 0.89

1200

0.747 of area

Because 0.747 is the probability that the machine will operate more than 1250 hours, the chance that the machine will operate fewer than 1250 hours is

1 – 0.747 or 0.253

0.253 would seem that this is not too risky a situation

?

Mean 1400 hours

Point Q = 1600

Z = 0.6666

1200

0.747 of area

situations in which it becomes necessary to combine both the extensive experience and intuition of operating personnel and statistical evidence

one must not discard judgement simply because its intuitive and similarly not to accept statistical survey results

Both sources of information are subject to estimating errors

The decision maker prefers in most cases to combine the 2 sources of information instead of discarding one or both

COMBINING EXPERIENCE AND NUMBERS

Situation:The sales manager’s probability distribution of monthly sales for the

new product. She puts the mean sales volume per month at 110 units and from the same kind of questions used in the previous example, we are able to calculate the manager’s standard deviation of monthly sales as 10 units.

A recently completed market survey indicates potential monthly sales at 70 units with a standard deviation of 7 units

Now suppose the financial manager, has determined the break-even volume of 80 units a month

At this point management can1. Reject new product on the assumed greater precision of market

survey2. Choose to accept the product if they consider the favorable

sales estimate given by the sales manager

3. Combine the 2 estimates


GIVEN:

µ1 or mean by manager = 110 µ2

or mean by survey =70 Ơ1 or S.D by manager = 10 Ơ2 or S.D by survey = 7 Breakeven point = 80


Combining estimates like this involves weighting each mean by its reliability:

Reliability is the weight of each mean by the reciprocal of its variance(square of the S.D)


Estimate Source

Mean µi Standard deviation ơi

Variance Ơi

2

Reliability1/Ơi

2

Sales manager

µ1 = 110 Ơ1 = 10 Ơ12 = 100 1/100

Market Survey

µ2 = 70 Ơ2 = 7 Ơ2

2 = 49 1/49

Now to combine the estimate mean:

Combined estimate of mean ΰ = µ1 ( 1/ 1/Ơ12 ) + µ2 ( 1/ Ơ2

2 )

( 1/ 1/Ơ12 ) + ( 1/ Ơ2

2 )

= (110 x 1/100) + ( 70 X 1/49)1/100 + 1/49

ΰ = 83.17 units


Then to combine the estimate standard deviation:

Combined estimate of S.D. = √2

√ (1/ Ơ12 ) + (1/ Ơ2

2 )

= √2

√ 1/100 + 1/49

= 8.11 units


Now the we have the Combined mean estimate = 83.17 units = 83 units Combined S.D = 8.11 unitsThe best combined estimate of the new product’s sales results in a mean of

slightly higher than 83 units/month and a S.D. Of distribution around this mean of about 8.11 units. Using these two values, we illustrate the distribution of the proposed product.


83 89868077 9274

Break even point (80 units)

Loss Profit

Because 1 S.D. For the combined distribution is 8.11 units we calculate that the breakeven point is (83 – 80) / 8.11which is

3/8.11 or 0.37 S.D to the left of the mean of the combined distribution. From the standard normal probability table we find the area under the normal curve from one tail to 0.37 S.D past the mean is 0.64431 of the total area


83 89868077 9274


Loss Profit

3/8.11 or 0.37 S.D to the left of the mean of the combined distribution. From the standard normal probability table we find the area under the normal curve from one tail to 0.37 S.D past the mean is 0.64431 of the total area

Thus the portion of the area under the curve representing sales volume less than the breakeven point is calculated as 1.0 – 0.64431 equals to 0.356 area under the curve

On the strength of the combined estimates, it appears that the chances for a loss are less than 0.40, not bad odds at all for a new product


83 89868077 9274


Loss Profit

0.644310.356

Measure of relative satisfaction. Value of a certain outcome or payoff to

someone, the pleasure or displeasure that person would derive from that outcome

UTILITY AS A DECISION CRITERION

SituationYou are a STRUGGLING graduate student

with 2 children and just enough money to get through the semester and a friend offers you

a. 0.9 chance of winning $10 for just $1b. 0.9 chance of winning $1000 for $100c. 0.9 chance of winning $10000 for $100

Now what would you probably do?


Now what would you probably do?

a. 0.9 chance of winning $10 for just $1

you would think of expected value and reason as 0.9 x $10 greater than $1? Because the expected value of a bet is ($9) 9 times greater than the value of the bet($1) you might feel inclined to take the offer. Even if you lose, the loss of $1 would no affect your situation materially

b. 0.9 chance of winning $1000 for $100

expected value of bet is ($900) 9 times greater than the ($100) bet, but you would probably think twice before putting up your money. Why? Even thought the pleasure of winning $1000 is high, the pain of losing $100 might hurt a lot.

c. 0.9 chance of winning $10000 for $100

expected value of bet is ($9000) 9 times greater than bet ($1000). You would probably refuse the offer, not because the expected value of the bet is unattractive, but the thought of losing all your assets is completely unacceptable as an outcome


UTILITY CURVES

1. Matakutin is a cautious and conservative businessman, a move to the right of the zero profit point increases his utility only very slightly, whereas a move to the left of the zero profit point decreases his utility rapidly ( $0 - $50000 profit increases utility value by 1 on the vertical scale, while moving by only $20000 decreases his utility by the same value). He is averse to risk

2. Ramipera. Wee see from her utility curve that a profit increases her utility by more than a loss of the same amount decreases it ( profits from $40000-$50000 raises her utility from -0.5 to +5 on the vertical scale, but lowering her profits $10000(from $0 to -$10000 decreases her utility by only 0.25, from -4 to -4.25. she is risk affine

3. Steady is the kind of businessman who would not suffer from a$50000 loss nor increase his wealth significantly from a $50000 gain. Steady’s curve is linear and can effectively use expected value as his decision criterion. He is said to be risk neutral



-60000 -40000 -20000 0 20000 40000 60000

-6

-4

-2

0

2

4

6

UTILITY CURVES OF MATAKUTINSTEADY, AND RAMIPERA

STEADYMATAKUTINRAMIPERA

CASH PROFIT OR LOSS

UTIL

ITY

Cost-volume-profit analysis - Allows management to determine in advance the effects of

certain contemplated decisions or expected states of nature will have on the revenues, cost and therefore, profits.

- Management have less control over revenue than costs, thus estimation of revenues is a good example of decision making under conditions of risk, where management can usually specify a distribution of revenues

THE NORMAL PROBABILITY DISTRIBUTION AND COST-VOLUME-PROFIT ANALYSIS

Break even point: point at which total costs equals total revenue Situation: we use an a example of a company with theses financial

data concerning a proposed new product:

◦ SELLING PRICE = $7.50◦ VARIALBE COST/UNIT = $4.5◦ FIXED COST/YEAR = $1,500,000

Breakeven point(in units) =

Total fixed cost/ (Price/unit – Variable cost/unit)

- = $1,500,000 / ( $7.50 - $4.50)- = $1,500,000 / $3.00- Breakeven point = 500,000 units anually


Situation: now suppose the sales manager estimates that the mean expected sales volume for the new product for the coming year is 600,000 units. When asked about the variability of this estimate, she indicates chances are to 2 to 1 that sales will be within 300,000 of the mean she has estimated.


12 2

1

600,000 (mean)

900,000

300,000

For the figure, we can see that the area between the left-hand tail of the distribution and the 900,000 point is 5/6 or 0.8333 of the total area under the curve. From the standard normal distribution table we can determine that the 900,000 point is about 0.97 S.D. To the right of the mean of the distribution. Since

0.97 S.D = 300,000 units

(300,000/97) * 100 1.00 S.D = 309,278 units


12 2

1

600,000 (mean)

900,000

300,000

1. what is the probability of at least breaking even? z = (x - µ) / ơ

z = (600000 units– 500000 units) / 309,278z = 0.323 S.D below the mean of distribution sales

Z = 0.323 S.D = 0.626 chance of at least breaking even





2. what is the probability that profits from the new product would be at least $500,000? Since the new product’s contribution per unit(price – variable cost) is $3, for profits to be at least $500,000, sales will have to be at least $500,000/$3 = 166,667 units above the break even point of 500,000 units. Sales of 666,667 units would be (666,667 – 600,000) / 309.728 or 0.216 S.D. Above the mean of the sales distribution. Again from the table of standard normal distribution table, we can see that the chances are only about 0.586 that sales would be less than 666,667 units, thus the chance that sales would be no less than that figure is 1 – 0.586.





2. what is the probability that profits from the new product would be at least $500,000? Since the new product’s contribution per unit(price – variable cost) is $3, for profits to be at least $500,000, sales will have to be at least $500,000/$3 = 166,667 units above the break even point of 500,000 units. Sales of 666,667 units would be (666,667 – 600,000) / 309.728 or 0.216 S.D. Above the mean of the sales distribution. Again from the table of standard normal distribution table, we can see that the chances are only about 0.586 that sales would be less than 666,667 units, thus the chance that sales would be no less than that figure is 1 – 0.586 = 0.414


3. What are the chances that the new product would lose $250,000 or more.

To lose that much money, sales would have to be at least $250,000/$3 = 83,333 units below the 500,00 break even point, that is no higher than 416,667 units. This sales level would be (600,000 – 416,667) / 309,728 = 0.593 S.D. Below the mean of the sales distribution. The Standard Normal Probability table tells us that 0.593 S.D. Is equal to about 0.723 chance of sales being above this point. Therefore, the chance of sales being no more than 416,667 units is 1 – 0.723 = 0.277


In comparing this new product with other possible uses of company resources; management now has these probability estimates:

1. The chance of breaking even is better than 0.625

2. The chance of making at least $500,000 on this product is 0.415

3. The chance of losing $250,000 on this product is 0.277

With the probability estimates, most managements would consider the new product to be an attractive alternative.

The contribution that this analysis has made to managerial decision making is simply that is has allowed the new product decision to be made under marketing conditions involving risk as to future sales volumes

This decision analysis of this type will be of considerable use in choosing among alternatives


The decision maker needs to know not only what his or her probability of loss is, but also what the expected dollar/peso value of that loss is.

USING UNIT LOSS AND EXPECTED LOSS

COMBINING UNIT MONETARY VALUES AND PROBABILITY DISTRIBUTIONS

1,500 hours

Breakeven point

µ = 1,500 hoursƠ = 500 hours

Probability distribution of operating times for a proposed new machine

900 hours

In the figure we assumed the distribution of operating times for a proposal of a new machine

Given:µ = 1500 hoursƠ = 500 hoursBreakeven point = 900 hoursLoss = $6 loss per hour below the breakeven level


USING UNIT LOSS AND EXPECTED LOSS


1,500 hours

Breakeven point



900 hours

Unit loss ($6/hour)

A

B

Using the methods discussed earlier the probability of the machine operating fewer than 900 hours is 0.116. How?

Z = (x - µ) / ơ z = (1500 hours – 900 hours) / 500z = 1.2 S.D

Z = 1.2 S.D = 0.88493 Probability of the machine operating lower than 900 hours = 1 – 0.88493 = 0.11507 = 0.116

THE USE OF UNIT NORMAL LOSS INTERGRAL (UNLI) COMES INTO PLAY - this makes the calculation of expected loss a very straight forward process.

To calculate for expected loss there are 3 steps

1. determine how many S.D. There are between the mean(1500 hours) and the break even point (900 hours)

= (1500-900) / 500 = 1.2 S.D 2.Look up the value corresponding to 1.2 S.D. In the UNLI table


UNIT NORMAL LOSS INTEGRAL

THE USE OF UNIT NORMAL LOSS INTERGRAL (UNLI) COMES INTO PLAY - this makes the calculation of expected loss a very straight forward process.

To calculate for expected loss there are 3 steps


= (1500-900) / 500 = 1.2 S.D

2.Look up the value corresponding to 1.2 S.D. In the UNLI table

1.2 SD = a table value of 0.0561

3. Calculate the total expected loss, multiply together the unit loss, the S.D and the value obtained in step 2:

Expected loss = $6 x 500 x 0.0561 =$168.30


Expected value of additional information in a decision process

Lots of new information is available to decision makers, all which cost money but may reduce one’s uncertainty about the future

Goal is to evaluate the cost of additional information against the expected benefits the information will provide to the decision making process

EXPECTED VALUE OF PERFECT INFORMATION

Situation: We shall assume that we have been offered a detailed analysis of operating time of similar machines that was stated earlier. The asking price of this new information is $50. the information will allow us to reduce our estimate of the S.D of the distribution of operating hours from 500 to 350 hours. Is the new info worth $50?


1,500 hours

Breakeven point



900 hours

Unit loss ($6/hour)

A

B

How to determine if the info was worth $50?


= (1500-900) / 350 = 1.71 S.D

2.Look up the value corresponding to 1.71 S.D. In the UNLI table

1.71 SD = a table value of 0.01785

3. Calculate the total expected loss, multiply together the unit loss, the S.D and the value obtained in step 2:

Expected loss = $6 x 350 x 0.01785 =$37.49

4. we can see from these results that our expected loss ( with new info) is $37.49. This is a reduction in loss from the previous amount and a saving of

$168.30 - $37.49 = $130.81

We should buy the new information as the $50 info will enable us to reduce the expected loss by $130.81, a good return of investment in new information


used to analyze the entire set of outcomes represented by a normal distribution, both those that are favorable (profit) and non-favorable (loss)

SITUATION: Purchase of machineWe estimate the mean of the distribution of operating

hours as 1850 hours and the standard deviation as 600 hours. In this case,our cost information indicates that we will break even at 1600 operating hours. Below 1600 hours we will lose $15/hr and above 1600 hours will earn $11/hr from the operation of the machine line

Determine expected loss, expected profit and net profit

UNIT MONETARY VALUES AND BOTH SIDES OF THE NORMAL DISTRIBUTION ( PROFIT AND LOSSES )

CALCULATING EXPECTED PROFIT, EXPECTED LOSS AND EXPECTED NET PROFIT


Mean = 1850 hours

$11/hour profit

breakeven = 1600 hours

$15/hour loss

A

B


CALCULATING EXPECTED PROFIT, EXPECTED LOSS AND EXPECTED NET PROFIT

NET PROFT = EXPECTED PROFIT – EXPECTED LOSS= $4227.96 - $2012.40= $2215.56


Step 1 (Expected loss) Step 2 ( expected profit )

(1850 – 1600) / 600 = 0.417 S.D (1850 – 1600) / 600 = 0.417 S.D

UNLI table value of 0.417 S.D. = 0.2236 UNLI table value of 0.417 S.D. = 0.2236

Expected loss = $15 x S.D x 0.2236 = $2012.40

Add 0.417 to 0.2236 = 0.6406( on two sided cases such as this when the loss or profit line corsses the mean, the procedure is to add z score to the value in the table)

Expected profit = $11 x 600 x 0.6406 = $4227.96

dec sci presentation

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