ch10

Chapter 10

I Chapter Outline

10.1Management Science Models in the Airline Industry

?Discussion of the following applications of management science in the airline industry:

1.Hub-and-spoke configuration

2.Demand forecast

3.Flight schedule

4.Fleet assignment

5.Crew pairing

6.Overbooking

7.Revenue management

?Further discussion on revenue management

?Shadow prices, simulation modeling, and bid-price control policies

10.2Management Science Models in the Financial Investment Management Industry

?Management science applications in:

1.Analysis of financial data

2.Forecasting future performance of assets

3.Construction and management of efficient portfolios

4.Determining trading strategies that minimize trading cost

5.Evaluation of financial assets such as options and other derivative

securities

10.3 A Year in the Life of a Manufacturing Company

?Analysis of several modeling problems for the Lexington Laser Corporation II Answers to Chapter Cases

SLOAN INVESTORS, PART II

(a)We use the non-linear optimization model for the portfolio optimization

problem discussed in Chapter 8. Below we show the efficient frontiers for

each of the six months in the period under consideration.

(b) The next table summarizes the computations.

(c) The next table summarizes the computations.

(d)Student answers will vary.

REVENUE MANAGEMENT AT ATLANTIC AIR

(a)In this problem, P Q = 250 and P Y = 750. Hence, we want to find L* such that

P(N Y<= L*) = (750 –250)/750 = 0.667. Using the standard normal

distribution tables, we find that P(Z <= 0.43) 0.667. Therefore, we have L* =

0.43 x 5.57 + 27.86 = 30.26 is the optimal protection level. The

corresponding optimal expected revenue is $24,911 per flight leg.

(b)Using the computer model for a plane with a capacity of 75 seats, we obtain

that the optimal expected revenue per flight leg is $31,161. Since we have 4 x

365 = 1,460 flight legs per year, this represents an expected revenue of

$45,495,060 per year, that is, an increment of $45,495,060 - $24,911 x 1,460

= $9,125,000 per year over the revenues from the current 50-seat plane. The

additional operation costs are $2,000 x 365 x 4 = $2,920,000 per year. Hence,

the net cash flow if Atlantic Air accepts the offer will be $9,125,000 -

$2,920,000 - $1,500,000 = $4,705,000. Therefore, Atlantic Air should accept

the offer.

(c)By accepting CCG’ s group, the capacity of the plane will be redu ced to 34

seats. This represents a drop to $20,911 in optimal expected revenues, that is,

we have a reduction of $4,000 in revenues. Therefore, Atlantic Air should

charge to CCG at least $4,000/16 = $250 for a one-way ticket.

A STRATEGIC ALLIANCE FOR THE LEXINGTON LASER CORPORATION

(a)The decision tree is shown on the next page. We denote by B the best case

scenario for Aspen and by W the worst case scenario for Aspen. We denote by

"B forecast" the scenario in which the market study is commissioned and the

report is that the best case scenario for Aspen will occur. Similarly, "W

forecast" represents the scenario in which the worst case scenario for Aspen is

forecast. According to the information given, notice that P(B forecast | W) =

P(W forecast | B) = 0.30. Using the laws of conditional probabilities, as well

as Bayes' law, we obtain that P(B forecast) = P(B forecast | B) x P(B) + P(B

forecast | W) x P(W) = (0.7)(0.7) + (0.3)(0.3) = 0.58. Similarly, P(W forecast) = 0.42. Also, P(B | B forecast) = P(B forecast | B) x P(B)/P(B forecast) =

(0.7)(0.7)/(0.58) = 0.845. Furthermore, P(W | B forecast) = 0.155, P(B | W

forecast) = 0.5, and P(W | B forecast) = 0.5. We assume that if LLC waits for six months, then it will make $5,500,000 x 0.20 = $1,100,000 after tax if the best case scenario occurs.

(b)The decision tree does not capture intermediate scenarios, that is, the

probability distribution of the scenarios. Assuming that there only are two scenarios (best and worst) is too simplistic in practice. Simulation is recommended in this case. In addition, the analysis does not take into consideration Aspen' s competition (which can affect profits) or the possibility of using the new facility to produce custom lasers for other companies (other than Aspen).

(c)The optimal strategy is to start right away building the new facility and not

commission the market study. The EMV of this strategy is $750,000. We would recommend further study of the problem by using simulation of the possible scenarios concerning the potential revenue.

YIELD OF A MULTI-STEP MANUFACTURING PROCESS

(a)We create a simulation model as follows. Let NS be the sub-mount assembly

total yield, then NS = 800X, where X is a random variable with normal

distribution μ = 80% and σ = 5%. Let NM be the module assembly total yield,

then NM is a random variable with binomial distribution p = 95% and n = NS.

Finally, let NF be the final test total yield, and hence, the output of the

manufacturing process, then NF = U x NM, where U is a random variable

with uniform distribution between 90% and 100%. Let D denote the demand,

then the difference between supply and demand is given by NF - D, where D

is a random variable with a normal distribution μ = 500 and σ = 50.

(b)After running the model on the computer, we summarize the distribution of

the output in the frequency chart below.

(c)The distribution of the weekly demand is normal with μ = 500 and σ = 50.

(d)The distribution of the weekly surplus/deficit in supply/demand is shown

below. According to this model, LLC will miss a shipment 6.7% of the time.

(e)The table below shows the proportion of time that supply meets demand

according to the number of starts per week. To meet demand 95% of the time,

LLC will approximately need at least 820 starts per week. To meet demand

99% of the time, LLC will approximately need at least 870 starts per week. To

meet demand 99.5% of the time, LLC will approximately need at least 890

starts per week.

PREDICTION OF YIELDS IN MANUFACTURING

(a)The regression output is shown below. By looking at the t-student factors (or

to the P-values), we determine that the relevant factors are recent exposure to

pigtail and welder maintenance. For each 1% of increase in exposure to pigtail,

yield increases in 0.5%. For each month of increase in the time since the last

welder maintenance, yield decreases in 1.87%.

Regression Statistics

Multiple R 0.847964405

R Square 0.719043633

0.690374615

Adjusted R

Square

Standard

10.90715342

Error

Observations 55

ANOVA

df SS MS F Significance F

Regression 5 14918.84802 2983.769605 25.080861 1.90197E-12

Residual 49 5829.333794 118.9659958

Total 54 20748.18182

t Stat P-value Lower 95% Upper 95% Coefficients Standard

Error

Intercept 87.83143021 30.95137187 2.837723335 0.006592177 25.63235746 150.030503 Experience 6.494858487 3.917200084 1.658035931 0.103699446 -1.377045026 14.366762 Exposure 0.488027009 0.046510282 10.4928843 3.97829E-14 0.394561155 0.581492864 Time 1.295904974 1.856548065 0.698018542 0.488464969 -2.434965782 5.026775729 Temperature -0.866444799 0.451491552 -1.919072007 0.060809297 -1.773750492 0.040860894 Maintenance -1.868019572 0.471033429 -3.965789811 0.00023797 -2.814596112 -0.921443032

(b)Among many other factors, operator distractions, human errors, management

pressure on employees, learning curves, etc.

(c)We would recommend studying further the two factors identified by the

regression model, recent exposure and welder maintenance, to make sure that

they are the reason for the low yields. LLC might increment the training

period for cross-trained operators. LLC might also review maintenance

schedules and perhaps they could implement one more "tune up" in the year. ALLOCATION OF PRODUCTION PERSONNEL

(a)Let XL i, XS i, and XD i represent the amount of employees with pay grade i

assigned to the long-haul, short-haul, and Datacom production, respectively,

for i = 1, …, 6. Using these variables, we introduce the following linear

optimization model.

Min 15(XL1 + XS1 + XD1) + 14.5(XL2 + XS2 + XD2) +

13(XL3 + XS3 + XD3) + 12(XL4 + XS4 + XD4) +

10.5(XL5 + XS5 + XD5) + 9.75(XL6 + XS6 + XD6).

Subject to:

Long-haul demand: 2XL1 + 1.8XL2 + 1.62XL3 + 1.8XL4 +

1.62XL5 + 1.3XL6 >= 48.5,

Short-haul demand: 1.2XS1 + 1.08XS2 + 2.5XS3 + 2.16XS4 +

1.93XS5 + 1.74XS6 >= 25,

Datacom demand: 2XD1 + 1.8XD2 + 1.62XD3 + 1.45XD4 +

1.31XD5 + 1.2XD6 >= 251,

Staffing pay grade 1: XL1 + XS1 + XD1 <= 4,

Staffing pay grade 2: XL2 + XS2 + XD2 <= 9,

Staffing pay grade 3: XL3 + XS3 + XD3 <= 20,

Staffing pay grade 4: XL4 + XS4 + XD4 <= 54,

Staffing pay grade 5: XL5 + XS5 + XD5 <= 102,

Staffing pay grade 6: XL6 + XS6 + XD6 <= 40,

Non-negativity: XL1, XS1, XD1, XL2, XS2, XD2, XL3, XS3, XD3,

XL4, XS4, XD4, XL5, XS5, XD5, XL6, XS6,

XD6 >= 0.

The optimal employee allocation is shown in the table below. The labor costs related to this solution are $2,475 per week. This represents savings of $2,559.5 - $2,474.7 = $84.8 per week.

(b)The optimal allocation in this case is shown in the table below. The cost of

this solution is $2,327 per week, this represents savings of $2,559.5 - $2,326.9 = $232.6 per week.

(c)Intuitively, the difference is because we use the most efficient

(productivity/wage) employees (pay grade 6 workers) to produce the highest demand product (Datacom). The same happens with the demand on long-haul and short-haul, for which we use the most efficient (productivity/wage) employees (pay grade 5 and 3 workers, respectively) to satisfy demands. (d)There might be missing other labor related factors, such as union costs, cost of

training new employees, production bonuses, etc.