Recent trends in globalization have forced businesses around the world to more keenly focus on profitability. This trend is also present in Japan, where historical links between banks and businesses have traditionally blurred the goals of firms. For example, the Japanese business engineering firm, Mitsui & Co. Ltd., recently launched “Challenge 21”. A plan directed at helping the company emerge as Japan’s leading business engineering group. According to a spokesperson for the company.” [ This plan permits us to] create new value and maximize profitability by taking steps such as renewing our management framework and prioritizing the allocation of our resources into strategic areas. We are committed to maximizing shareholder value through business conduct that balances the pursuit of earnings with socially responsible behavior.” Ultimately, the goal of any continuing company must be to maximize the value of the firm. This goal is often achieved by trying to hit intermediate targets, such as minimizing costs or increasing market share. If you as a manager- do not maximize your firm’s value over time, you will be in danger of either going out of business, being taken over by other owners (as in a leveraged buyout), or having stockholders elect to replace you and other managers. Source: “Mitsui & Co., Ltd. UK Regulatory Announcement: Final Results.” Business Wire, May 13, 2004.
Questions
1. What is (Challenge 21 ) plan of Mitsui & Co., Ltd.
2. What were the objectives of the management of Mitsui & Co., Ltd .
3. What are the threats of not maximizing a firm’s value.
In: Operations Management
Concert Nation] Concert Nation, INC. is a nationwide promoter of
rock concerts. The president of
the company wants to develop a model to estimate the revenue of a
major concert event at large venues
(such as Ford Field, Madison Square Gardens) for planning marketing
strategies. The company has
collected revenue data of 32 recent large concert events. For each
concert, they have also recorded the
attendance, the number of concession stands in the venue, and the
Billboard chart of the artist in the
week of each event. This data is available in “Tickets”. They have
two potential models that could
explain the revenue. The two competing models are:
Model A: ??????? = ?? + ???????????? + ???????????? + ??????????? + ?0123?
Model B: ??????? = ?? + ???????????? + ??????????? + ?012?
Run regression on both models. Use only the regression outputs
of the two models and the original data
to answer questions 1 to 7 below.
1. [1 pt] Let’s consider the model A first. What does the result of
F-test indicate?
(a) The p-value of F-test is 100.83. Thus, the model does not
significantly explain the revenue.
(b) The p-value of F-test is close to zero. Thus, all independent
variables in the regression model are
statistically significant.
(c) The p-value of F-test is close to zero. This indicates that at
least some independent variables in the
regression model significantly explain the revenue.
(d) This indicates weak evidence of a linear relationship, because
the p-value is very low.
2
2. [1 pt] If we use model A for prediction, what is the point
estimate for the revenue of a concert that has
attendance of 50,000 people, 5 concession stands, and the song
ranked in no. 15 in the Billboard ranking?
(a) $3.145 M
(b) $2.851 M
(c) $3.252 M
(d) $340K
3. [1 pt] What is an approximate 95% prediction interval for the
concert listed in the previous question?
(a) [$2.757M, $3.533M]
(b) [$2.463M, $3.239M]
(c) [$2.368M, $3.922M]
(d) [$2.074M, $3.628M]
4. [1 pt] Which of the following statement is correct?
(a) The estimated slope for the attendance is only $59.2. This
means that, when keeping everything
else the same, the revenue does not depend much on the
attendance.
(b) The t-statistic associated with the slope for the attendance
variable is 16.9. This means that there is
too much noise to determine if the slope is definitely
positive.
(c) The p-value for the concession variable is 0.933. This means
that the number of concession stands
is not a statistically significant variable to determine the
revenue.
(d) The p-value for the concession variable is 0.933. This means
that the number of concession stands
is a statistically significant variable to determine the
revenue.
5. [1 pt] Is it appropriate to use model A as a final model to
estimate the revenue of a concert?
(a) Yes. All independent variables are statistically
significant.
(b) Yes, because the analysis indicates a linear relationship
between revenue and attendance.
(c) No, because not all independent variables are statistically
important. Thus, revision is necessary.
(d) No, because some of the slopes were negative. Thus, revision is
necessary.
3
6. [1 pt] Now, consider model B. According to model B, what is a
point estimate for a concert that has
attendance of 50000 people, 5 concession stands, and the song
ranked in no. 15 in the Billboard ranking?
(a) $3.147M
(b) $2.839M
(c) $7.139M
(d) $13.637M
7. [1 pt] Based on the regression outputs, which model would you
consider more suitable for predicting the
revenue between the two models– Model A and Model B?
(a) Model A is more suitable, because it has a higher ?2, lower
standard error of the estimates
(??), and lower F-test p-value.
(b) Model A is more suitable because the fraction of SST accounted
for by the residuals is higher than
for model B.
(c) Model B is more suitable, because, while both models have
similar ?2 and F-test p-value, model B
has lower standard error of the estimates (??) and all independent
variables are statistically
significant.
(d) Model B is more suitable, because the slope coefficient is
larger in magnitude.
| Attendance | # of concessions | Billboard Charts | Concert Revenue |
| 30650 | 8 | 56 | 1531762 |
| 80997 | 1 | 87 | 4047180 |
| 93686 | 8 | 24 | 5805972 |
| 44405 | 4 | 99 | 2516538 |
| 77767 | 4 | 39 | 4197208 |
| 95780 | 7 | 35 | 6226065 |
| 82701 | 7 | 86 | 4123048 |
| 50165 | 8 | 29 | 3465110 |
| 50619 | 5 | 93 | 2843474 |
| 36259 | 7 | 86 | 1866318 |
| 52013 | 5 | 35 | 2670798 |
| 97447 | 7 | 71 | 5756817 |
| 69982 | 7 | 97 | 3681670 |
| 31789 | 10 | 72 | 2072149 |
| 39787 | 6 | 89 | 1964361 |
| 63596 | 5 | 65 | 3150802 |
| 73159 | 5 | 41 | 5064323 |
| 51172 | 8 | 1 | 2901564 |
| 54187 | 9 | 17 | 3170058 |
| 56681 | 7 | 1 | 3316764 |
| 78466 | 7 | 86 | 3825369 |
| 65132 | 8 | 86 | 2983563 |
| 52866 | 4 | 8 | 3091641 |
| 39536 | 2 | 20 | 3068049 |
| 32541 | 1 | 53 | 1796727 |
| 36441 | 1 | 60 | 2011990 |
| 74987 | 6 | 58 | 4389931 |
| 33791 | 8 | 81 | 1545359 |
| 64961 | 6 | 94 | 3792136 |
| 61429 | 3 | 86 | 2695672 |
| 68178 | 4 | 50 | 4147528 |
| 85701 | 5 | 52 | 5335423 |
In: Statistics and Probability
The director of admissions of a small college selected 120 students at random from the new freshman class in a study to determine whether a student’s grade point average (GPA) at the end of the freshman year (y) can be predicted from the ACT test score (x1).
GPA ACT ITS RP
3.897 21 122 99
3.885 14 132 71
3.778 28 119 95
2.540 22 99 75
3.028 21 131 46
3.865 31 139 77
2.962 32 113 85
3.961 27 136 99
0.500 29 75 13
3.178 26 106 97
3.310 24 125 69
3.538 30 142 99
3.083 24 120 97
3.013 24 107 55
3.245 33 125 93
2.963 27 121 80
3.522 25 119 63
3.013 31 128 78
2.947 25 106 93
2.118 20 123 22
2.563 24 111 84
3.357 21 113 87
3.731 28 134 98
3.925 27 128 95
3.556 28 126 63
3.101 26 121 79
2.420 28 104 86
2.579 22 113 90
3.871 26 133 97
3.060 21 125 39
3.927 25 128 97
2.375 16 112 57
2.929 28 107 67
3.375 26 115 81
2.857 22 119 75
3.072 24 113 63
3.381 21 115 15
3.290 30 110 95
3.549 27 122 93
3.646 26 118 99
2.978 26 114 90
2.654 30 112 99
2.540 24 106 85
2.250 26 95 84
2.069 29 102 58
2.617 24 114 86
2.183 31 116 82
2.000 15 93 34
2.952 19 120 34
3.806 18 117 23
2.871 27 119 95
3.352 16 115 41
3.305 27 113 28
2.952 26 108 68
3.547 24 116 54
3.691 30 135 77
3.160 21 108 58
2.194 20 110 73
3.323 30 124 94
3.936 29 130 98
2.922 25 118 99
2.716 23 110 91
3.370 25 117 95
3.606 23 123 72
2.642 30 116 65
2.452 21 109 53
2.655 24 110 81
3.714 32 126 41
1.806 18 99 84
3.516 23 121 84
3.039 20 115 35
2.966 23 127 70
2.482 18 99 15
2.700 18 108 47
3.920 29 129 98
2.834 20 103 77
3.222 23 122 72
3.084 26 118 29
4.000 28 135 80
3.511 34 139 88
3.323 20 128 80
3.072 20 120 46
2.079 26 114 89
3.875 32 133 91
3.208 25 123 95
2.920 27 111 83
3.345 27 122 92
3.956 29 136 99
3.808 19 140 41
2.506 21 109 68
3.886 24 133 98
2.183 27 98 59
3.429 25 134 89
3.024 18 124 89
3.750 29 128 92
3.833 24 149 97
3.113 27 121 43
2.875 21 117 52
2.747 19 110 82
2.311 18 104 61
1.841 25 95 72
1.583 18 96 33
2.879 20 117 97
3.591 32 130 97
2.914 24 121 92
3.716 35 125 99
2.800 25 112 61
3.621 28 136 72
3.792 28 129 99
2.867 25 106 76
3.419 22 108 66
3.600 30 138 70
2.394 20 106 44
2.286 20 111 33
1.486 31 101 77
3.885 20 113 57
3.800 29 131 96
3.914 28 140 97
1.860 16 111 65
2.948 28 110 85
1.) Plot the residuals ei against the fitted values ˆyi (in R). What departures from the regression model assumptions can be studied from this plot? What are your findings? (Note:If you are not sure about the validity of any of the assumptions, perform a formal test to verify your answer.)
2.) Prepare a normal probability plot (QQ plot) of the residuals. What assumption can be tested from this plot and what do you conclude? (Note:You can also use the formal test to reinforce your conclusion).
3.) Information is given for each student on two variables not included in the model, namely,intelligence test score (ITS-x2) and high school class rank percentile (RP-x3).Plot the residuals you obtained in part (b) against x2 and x3 on separate graphs to as certain whether the model can be improved by including either of these variables. What do you conclude? (Hint:The residuals represent any variability that was not able to be explained by x1. Therefore, if you see any pattern between the residuals and any other predictor omitted from the model, there is an indication that the predictor will be useful to be added in the model.)
In: Statistics and Probability
a.) A neutron can cause 235U to fission, producing two daughter nuclei (tin and molybdenum) and three more neutrons. These three neutrons can then, in turn, cause three more 235U nuclei to split. One typical reaction is summarized as follows:
n + 235U ® 131Sn + 102Mo + 3 1n
Find the energy released in this process (in MeV), given that M(235U)=235.0439 u; M(131Sn)=130.9169 u; M(102Mo)=101.9103 u.
b.)Lise Meitner’s work in 1938 predicted that the fission of 235U would produce 200 MeV of energy per fission event. How did she do? Find a %-error.
c.)131Sn is radioactive, and typical undergoes three successive b– decays. What is the final product?
d.)How many nuclei are in 1 kg of 235U? This is, incidentally, how many fission events occur if 1 kg of 235U were to fission.
e.)In each “generation” of a chain reaction of 235U, the number of events is three times the number in the previous generation. Hence events go as 1® 3® 9® 27® 81® 243 etc. If n tells you which generation, find the limit of the ratio of (events in nth generation)/(total # of events) as n ® large. (You don’t need to use calculus, just use your calculator and notice the pattern. The series starts 1, 3/4, 9/13, etc.)
f.)Given these results, estimate the number n of chain reaction “generations” that would occur if 1 kg of 235U were to fission.
g.)If each “generation” lasts on the order of 10–7 seconds, estimate the time for the entire fission chain reaction to occur.
In: Physics
To study the effect of temperature on yield in a chemical process, five batches were produced at each of three temperature levels. The results follow.
| Temperature | ||
|---|---|---|
| 50°C | 60°C | 70°C |
| 35 | 30 | 23 |
| 24 | 31 | 29 |
| 36 | 33 | 27 |
| 40 | 22 | 31 |
| 25 | 29 | 35 |
Construct an analysis of variance table. (Round your values for MSE and F to two decimal places, and your p-value to four decimal places.)
| Source of Variation |
Sum of Squares |
Degrees of Freedom |
Mean Square |
F | p-value |
|---|---|---|---|---|---|
| Treatments | |||||
| Error | |||||
| Total |
Use a 0.05 level of significance to test whether the temperature level has an effect on the mean yield of the process.
State the null and alternative hypotheses.
H0: μ50°C ≠
μ60°C ≠ μ70°C
Ha: μ50°C =
μ60°C =
μ70°CH0:
μ50°C = μ60°C =
μ70°C
Ha: μ50°C ≠
μ60°C ≠
μ70°C H0:
μ50°C = μ60°C =
μ70°C
Ha: Not all the population means are
equal.H0: At least two of the population means
are equal.
Ha: At least two of the population means are
different.H0: Not all the population means are
equal.
Ha: μ50°C =
μ60°C = μ70°C
Find the value of the test statistic. (Round your answer to two decimal places.)
Find the p-value. (Round your answer to four decimal places.)
p-value =
State your conclusion.
Reject H0. There is sufficient evidence to conclude that the mean yields for the three temperatures are not equal.Reject H0. There is not sufficient evidence to conclude that the mean yields for the three temperatures are not equal. Do not reject H0. There is not sufficient evidence to conclude that the mean yields for the three temperatures are not equal.Do not reject H0. There is sufficient evidence to conclude that the mean yields for the three temperatures are not equal.
In: Statistics and Probability
On January 1, 2018, the general ledger of Big Blast Fireworks includes the following account balances:
| Accounts | Debit | Credit | ||||
| Cash | $ | 22,300 | ||||
| Accounts Receivable | 37,500 | |||||
| Inventory | 32,000 | |||||
| Land | 64,600 | |||||
| Allowance for Uncollectible Accounts | 3,500 | |||||
| Accounts Payable | 31,400 | |||||
| Notes Payable (9%, due in 3 years) | 32,000 | |||||
| Common Stock | 58,000 | |||||
| Retained Earnings | 31,500 | |||||
| Totals | $ | 156,400 | $ | 156,400 | ||
The $32,000 beginning balance of inventory consists of 320 units, each costing $100. During January 2018, Big Blast Fireworks had the following inventory transactions:
| January | 3 | Purchase 1,100 units for $117,700 on account ($107 each). | ||
| January | 8 | Purchase 1,200 units for $134,400 on account ($112 each). | ||
| January | 12 | Purchase 1,300 units for $152,100 on account ($117 each). | ||
| January | 15 | Return 110 of the units purchased on January 12 because of defects. | ||
| January | 19 | Sell 3,700 units on account for $555,000. The cost of the units sold is determined using a FIFO perpetual inventory system. | ||
| January | 22 | Receive $533,000 from customers on accounts receivable. | ||
| January | 24 | Pay $363,000 to inventory suppliers on accounts payable. | ||
| January | 27 | Write off accounts receivable as uncollectible, $2,700. | ||
| January | 31 | Pay cash for salaries during January, $116,000. |
The following information is available on January 31, 2018. At the end of January, the company estimates that the remaining units of inventory are expected to sell in February for only $100 each. At the end of January, $4,200 of accounts receivable are past due, and the company estimates that 40% of these accounts will not be collected. Of the remaining accounts receivable, the company estimates that 5% will not be collected. Accrued interest expense on notes payable for January. Interest is expected to be paid each December 31. Accrued income taxes at the end of January are $12,500.
Jounral Entries: 1.)Purchase 1,100 units for $117,700 on account ($107 each). 2.) Purchase 1,200 units for $134,400 on account ($112 each). 3.) Purchase 1,300 units for $152,100 on account ($117 each). 4.) Return 110 of the units purchased on January 12 because of defects. 5.) Sell 3,700 units on account for $555,000. 6.) Record the cost of the units sold, which is determined using a FIFO perpetual inventory system. 7.) Receive $533,000 from customers on accounts receivable. 8.) Pay $363,000 to inventory suppliers on accounts payable. 9.) Write off accounts receivable as uncollectible, $2,700. 10.) Pay cash for salaries during January, $116,000. 11.) Record the adjusting entry for inventory. 12.) Record the adjusting entry for uncollectible accounts. 13.) Record the adjusting entry for interest. 14.) Record the adjusting entry for income tax. 15.) Record the closing entry for revenue. 16.) Record the closing entry for expenses. 17.) Record the closing entry for income summary.
Additional: Prepare a multiple-step income statement for the period ended January 31, 2018
Additional: Prepare a classified balance sheet as of January 31, 201
In: Accounting
Gallatin Carpet Cleaning is a small, family-owned business operating out of Bozeman, Montana. For its services, the company has always charged a flat fee per hundred square feet of carpet cleaned. The current fee is $22.10 per hundred square feet. However, there is some question about whether the company is actually making any money on jobs for some customers—particularly those located on remote ranches that require considerable travel time. The owner’s daughter, home for the summer from college, has suggested investigating this question using activity-based costing. After some discussion, she designed a simple system consisting of four activity cost pools. The activity cost pools and their activity measures appear below:
| Activity Cost Pool | Activity Measure | Activity for the Year | |
| Cleaning carpets | Square feet cleaned (00s) | 7,000 | hundred square feet |
| Travel to jobs | Miles driven | 180,500 | miles |
| Job support | Number of jobs | 2,000 | jobs |
| Other (organization-sustaining costs and idle capacity costs) | None | Not applicable | |
The total cost of operating the company for the year is $352,000 which includes the following costs:
| Wages | $ | 140,000 |
| Cleaning supplies | 34,000 | |
| Cleaning equipment depreciation | 16,000 | |
| Vehicle expenses | 35,000 | |
| Office expenses | 56,000 | |
| President’s compensation | 71,000 | |
| Total cost | $ | 352,000 |
Resource consumption is distributed across the activities as follows:
| Distribution of Resource Consumption Across Activities | ||||||||||
| Cleaning Carpets | Travel to Jobs | Job Support | Other | Total | ||||||
| Wages | 75 | % | 14 | % | 0 | % | 11 | % | 100 | % |
| Cleaning supplies | 100 | % | 0 | % | 0 | % | 0 | % | 100 | % |
| Cleaning equipment depreciation | 73 | % | 0 | % | 0 | % | 27 | % | 100 | % |
| Vehicle expenses | 0 | % | 78 | % | 0 | % | 22 | % | 100 | % |
| Office expenses | 0 | % | 0 | % | 55 | % | 45 | % | 100 | % |
| President’s compensation | 0 | % | 0 | % | 32 | % | 68 | % | 100 | % |
Job support consists of receiving calls from potential customers at the home office, scheduling jobs, billing, resolving issues, and so on.
Required:
1. Prepare the first-stage allocation of costs to the activity cost pools.
|
2. Compute the activity rates for the activity cost pools. (Round your answers to 2 decimal places.)
|
3. The company recently completed a 400 square foot carpet-cleaning job at the Flying N Ranch—a 50-mile round-trip journey from the company’s offices in Bozeman. Compute the cost of this job using the activity-based costing system. (Round your intermediate calculations and final answer to 2 decimal places.)
|
4. The revenue from the Flying N Ranch was $88.40 (4 hundred square feet @ $22.10 per hundred square feet). Calculate the customer margin earned on this job. (Round your intermediate calculations and final answers to 2 decimal places.)
|
In: Accounting
Presented below are the balance sheets of Trout Corporation as of December 31, Year 1 and Year 2, and the income statement for the year ended December 31, Year 2. The statement of retained earnings for the year ended December 31, Year 2 is on the next page. All dollars are in thousands.
Trout Corporation
Balance Sheets
December 31, Year 1 and Year 2
Assets Year 1 Year 2
Cash $ 85 $ 127
Accounts receivable 245 253
Less: Allowance for doubtful accounts (9) (11)
Prepaid insurance 15 9
Inventory 225 234
Long-term investment 65 42
Land 160 160
Buildings and equipment 250 300
Less: Accumulated depreciation (75) (100)
Trademark 25 22
Total Assets $ 986 $1,036
Liabilities & Stockholders’ Equity
Accounts payable $ 50 $ 36
Salaries payable 9 6
Deferred tax liability 15 18
Lease liability -- 75
Bonds Payable 275 125
Less: Discount (26) (24)
Common Stock 250 280
Paid-In Capital –in excess of par 75 70
Preferred Stock - 105
Retained Earnings 338 345
Total Liabilities & Stockholders’ Equity $ 986 $ 1,036
Trout Corporation
Income Statement
For the Year Ended December 31, Year 2
Net sales revenue $ 380
Investment revenue 12
Operating Expenses:
Cost of Goods $ 150
Salaries expense 58
Depreciation expense 35
Trademark amortization 3
Bad debts expense 8
Insurance expense 20
Bond interest expense 45 319
Operating Income $ 73
Other Income (Expense):
Loss on building fir $(27)
Gain on sale of investments 4 (23)
Pre-Tax Income from Continuing Operations $ 50
Less: Income Tax Expense: 25
Net Income $ 25
Additional Information:
Shareholders were paid cash dividends of $18 million.
A building that originally cost $40 million, and which was one-fourth depreciated, was destroyed by fire. Some undamaged parts were sold for $3 million.
Investment revenue includes Trout Corporation's $7 million share of the net income of Bass Corporation, an equity method investee.
$30 million par value of common stock was sold for $60 million, and $70 million of preferred stock was sold at par.
A long-term investment in bonds, originally purchased for $30 million, was sold for $34 million.
Pretax accounting income exceeded taxable income causing the deferred income tax liability to increase by $3 million.
The right to use a building was acquired with a seven-year lease agreement; present value of lease payments, $90 million. Annual lease payments of $15 million are paid at January 1st of each year starting in Year 2.
$150 million of bonds were retired at maturity.
Required:
Use the EXCEL worksheet template provided. There are three tabs-
Direct Method Statement of Cash Flows (SCF)
Show your work
Cash flows from Operating Activities – CFOs Indirect Method
In: Accounting
Applying and Analyzing Inventory Costing Methods
At the beginning of the current period, Chen carried 1,000 units of
its product with a unit cost of $21. A summary of purchases during
the current period follows. During the period, Chen sold 2,800
units.
| Units | Unit Cost | Cost | |
|---|---|---|---|
| Beginning Inventory | 1,000 | $ 21 | $ 21,000 |
| Purchase #1 | 1,800 | 23 | 41,400 |
| Purchase #2 | 800 | 27 | 21,600 |
| Purchase #3 | 1,200 | 30 | 36,000 |
(a) Assume that Chen uses the first-in, first-out method. Compute
both cost of good sold for the current period and the ending
inventory balance. Use the financial statement effects template to
record cost of goods sold for the period.
Ending inventory balance $Answer
Cost of goods
sold
$Answer
Use negative signs with answers, when appropriate.
|
Balance Sheet |
||||||||||
|---|---|---|---|---|---|---|---|---|---|---|
| Transaction | Cash Asset | + |
Noncash Assets |
= | Liabilities | + |
Contributed Capital |
+ |
Earned Capital |
|
| Record FIFO cost of goods sold | Answer | Answer | Answer | Answer | Answer | |||||
|
Income Statement |
||||
|---|---|---|---|---|
Revenue |
- |
Expenses |
= |
Net Income |
| Answer | Answer | Answer | ||
(b) Assume that Chen uses the last-in, first-out method. Compute
both cost of good sold for the current period and the ending
inventory balance.
Ending inventory balance $Answer
Cost of goods
sold $Answer
(c) Assume that Chen uses the average cost method. Compute both
cost of good sold for the current period and the ending inventory
balance.
Ending inventory balance $Answer
Cost of goods
sold
$Answer
(d) Which of these three inventory costing methods would you choose
to:
| 1. Reflect what is probably the physical flow of goods? |
|
|||
| 2. Minimize income taxes for the period? |
|
|||
| 3. Report the largest amount of income for the period? |
|
Please answer all parts of the question.
In: Accounting
This problem is from 2008.
The US Open is an annual two week tennis event in Flushing NY in late August, early September.
In a year with no significant rain interruption, the US Open makes approximately $275 million in revenue and incurs expenses of approximately $225 million, for a profit of $50 million. Of the $275 million in revenue approximately $100 million is from ticket sales. As a non-profit organization, it incurs no tax.
The US Open can work around rain delays but if all play is suspended in either the afternoon or evening sessions, tickets are good for the same session in the following year, in which case the USTA foregoes revenue. The largest ticket prices are for the women’s and men’s finals so a rain-out on either of these days forgoes the most revenue.
The Open is interested in buying a contract to protect itself from foregone revenues from rain interruptions during the finals. Working with its insurance broker, it approaches the insurance market to see if it can buy a weather derivative or insurance policy.
The US Open estimates that between foregone ticket sales and lost margin on concessions and broadcasting rights, a rain out on either the men’s or women’s finals will mean $30 mil in lost profits.
The insurance broker is able to secure an insurance policy that will indemnify the US Open if rainfall occurs during the men’s or women’s finals. The policy treats each event separately, meaning there is coverage and a corresponding premium charged for postponement of either final. The insurer is willing to provide a policy covering each separate event that will indemnify the US Open with a limit of $30 million and a policy premium of $10 million for each. As with all insurance policies, the US Open can collect the insurance payments only once it demonstrates the losses.
The weather desks at three major reinsurance holding companies with broker/dealers supply the probabilities associated with significant rainfall (> ¼ inch) on days 13 and 14 of this calendar year, which is 20% for either day, and conditional on rain on the 13th day, the chance of rain on the 14th day is 30%.
Write out all possible rain/dry possibilities for the 13th and 14th days, with their associated probabilities.
Without insurance, what are the profits if there are rain postponements to either or both finals?
Without insurance, what are the expected profits?
With insurance, what are profits if there are rain postponements?
With insurance what are profits if there is no rain?
What are the expected profits if insurance is purchased?
Should the US Open explore including additional days into the policy?
Over a ten year period, assuming baseline revenue and costs are approximately the same amounts as today, what would the US Open expect to earn (i) in the absence of an insurance policy and (ii) with the insurance policy?
The weather desk is also willing to write two weather derivative contracts, one for day 13 and one for day 14, each with a payout of $30 million and a cost of $12 million. The derivative pays the US Open regardless of whether play is suspended or not. It pays based on measured rainfall within 24 hour period exceeding ¼ of an inch.
What is the best strategy for the US Open to manage its exposure to rain?
Explain.
Without insurance, what are the profits if there are rain postponements to either or both finals?
Without insurance, what are Expected profits?
With insurance, what are profits if there are rain postponements?
With insurance what are profits if there is no rain?
Should the US Open explore including additional days into the policy?
Over a ten year period, assuming baseline revenue and costs are approximately the same amounts as today, what would the US Open expect to earn (i) in the absence of an insurance policy and (ii) with the insurance policy?
What is the best strategy for the US Open to manage its exposure to rain?
This problem is from 2008.
The US Open is an annual two week tennis event in Flushing NY in late August, early September.
In a year with no significant rain interruption, the US Open makes approximately $275 million in revenue and incurs expenses of approximately $225 million, for a profit of $50 million. Of the $275 million in revenue approximately $100 million is from ticket sales. As a non-profit organization, it incurs no tax.
The US Open can work around rain delays but if all play is suspended in either the afternoon or evening sessions, tickets are good for the same session in the following year, in which case the USTA foregoes revenue. The largest ticket prices are for the women’s and men’s finals so a rain-out on either of these days forgoes the most revenue.
The Open is interested in buying a contract to protect itself from foregone revenues from rain interruptions during the finals. Working with its insurance broker, it approaches the insurance market to see if it can buy a weather derivative or insurance policy.
The US Open estimates that between foregone ticket sales and lost margin on concessions and broadcasting rights, a rain out on either the men’s or women’s finals will mean $30 mil in lost profits.
The insurance broker is able to secure an insurance policy that will indemnify the US Open if rainfall occurs during the men’s or women’s finals. The policy treats each event separately, meaning there is coverage and a corresponding premium charged for postponement of either final. The insurer is willing to provide a policy covering each separate event that will indemnify the US Open with a limit of $30 million and a policy premium of $10 million for each. As with all insurance policies, the US Open can collect the insurance payments only once it demonstrates the losses.
The weather desks at three major reinsurance holding companies with broker/dealers supply the probabilities associated with significant rainfall (> ¼ inch) on days 13 and 14 of this calendar year, which is 20% for either day, and conditional on rain on the 13th day, the chance of rain on the 14th day is 30%.
The weather desk is also willing to write two weather derivative contracts, one for day 13 and one for day 14, each with a payout of $30 million and a cost of $12 million. The derivative pays the US Open regardless of whether play is suspended or not. It pays based on measured rainfall within 24 hour period exceeding ¼ of an inch.
Explain.
please make sure the second part is answer.
In: Math