MS&E 260 Final Exam Stanford University Summer 2018 Name: SUNet ID: Question Points Available Points earned 1 25 2 25 3 25 4 25 5 25 6 25 Total 150 Instructions: 1. This examination contains 17 pages,...

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MS&E 260 Final Exam Stanford University Summer 2018 Name: SUNet ID: Question Points Available Points earned 1 25 2 25 3 25 4 25 5 25 6 25 Total 150 Instructions: 1. This examination contains 17 pages, not including this page and the table page. 2. You have 24 hours to complete the examination. Please submit your exam by 5:00 pm (PDT) on Aug 18th, 2018. No late exams will be accepted. 3. You may use any resource you wish (notes, calculators, computers), except for other students. No collaboration is permitted. 4. Please sign the below Honor Code statement. On my honor, I have neither given nor received unauthorized assistance on this examination, and I have abided by all other provisions of the honor code in taking this examination. Signature: MS&E 260 Final Exam Stanford University Summer 2018 Problem 1 [25 pts] Consider the following queueing problems: 1. (10 pts) A network file server is accessed by users according to a Poisson process with rate 8 users/ hour. Each user establishes a communication session with the server and downloads files. The durations of the sessions are exponentially distributed with rate 15 users/ hour. Assume that the size of the waiting queue is infinite. (a) (5 pts) What is the average number of users in the system? What is the average waiting time in the system? (b) (5 pts) What is the average number of users in the queue? What is the average waiting time in the queue? 1 Summer 2018 MS&E 260 Final Exam Problem 1 (continued) 2. (5 pts) Now consider the situation where the inter-arrival time is a constant and is given by 5 minutes. The service time required by each user is also constant and equal to 4 minutes. What is the mean waiting time per user in this case? 3. (10 pts) Suppose now that instead of one single-server queue with infinite bu↵er space, we have two single-server queues, each with infinite bu↵er space. Customers are randomly dispatched to each queue with an equal probability. The service times are exponentially distributed with rate µ0 = 12 users/hour at each server. What is the average waiting time in the system? 2 Summer 2018 MS&E 260 Final Exam Problem 2 Problem 2 [25 pts] MusicForTheMasses organizes a small music festival and has one ticket to sell and two days re- maining. On the first day, MusicForTheMasses expects one person to arrive whose willingness to pay is drawn independently from U [0, 100]. On the second day, a di↵erent passenger is expected to show up whose willingness to pay is drawn independently from U [0, x]. (a) (10 pts) How do you price the ticket for each day? (b) (8 pts) For what value of x is the price on the first day lower than the price on the second day? Problem 2 continued on next page. . . 3 Summer 2018 MS&E 260 Final Exam Problem 2 (continued) Suppose that MusicForTheMasses organizes a second music festival and has one VIP ticket to sell and three days remaining. On the first day, MusicForTheMasses expects one person to arrive whose willingness to pay is drawn independently from U [0, 100]. On the second day, a di↵erent passenger is expected to show up whose willingness to pay is drawn independently from U [0, 300]. Finally, on the third day, a di↵erent passenger is expected to show up whose willingness to pay is drawn independently from U [0, 200]. If you can’t find a buyer after 3 days, you can sell it as a regular ticket at the price of pr = $30. (c) (7 pts) How do you price the ticket in this case? It is enough to give the formula. Do NOT try to solve the optimization problem numerically. 4 Summer 2018 MS&E 260 Final Exam Problem 3 Problem 3 [25 pts] Swift Travels is a distributor of Flying Bikes, bikes equipped with expandable wings to enable flying. Each Flying Bike is imported from Germany. Swift Travels buys each bike for $350. There is a lead time of 1 month. The holding cost is calculated with an annual interest rate of 15%. The cost of order processing is $2500 per order. Annual demand for Flying Bikes follows a normal distribution with mean 170 bikes and variance of 450 bikes (standard deviation of ⇠ 21.213 bikes). Assume that if a bike is demanded when the distributor is out of stock, then the demand is backordered and the cost associated with each backordered case is estimated to be $120. (a) (7 pts) Swift Travels manager uses a (Q,R) policy. What are the optimal values of the order quantity and reorder level? Problem 3 continued on next page. . . 5 Summer 2018 MS&E 260 Final Exam Problem 3 (continued) (b) (4 pts) Determine the safety stock. (c) (6 pts) What are the expected annual holding, setup and penalty costs associated with the bike? Problem 3 continued on next page. . . 6 Summer 2018 MS&E 260 Final Exam Problem 3 (continued) (d) (4 pts) What is the proportion of order cycles in which no stock-outs occur? (e) (4 pts) What is the proportion of demand that is met? 7 Summer 2018 MS&E 260 Final Exam Problem 4 Problem 4 [25 pts] The Mad Hatter is a retailer of luxury hats in Wonderland. The demand over the lifespan of these luxury hats is normally distributed with mean 250 and standard deviation 80. The Mad Hatter is considering buying hats from the Crazy Hat supplier instead of making them himself. Production costs for Crazy Hat to make each hat is $40. Crazy Hat sells each hat wholesale for $60 to retailers, including the Mad Hatter. Mad Hatter then plans to sell the hats at the suggested retail price of $180 per hat. Assume that the salvage value of the product is $10 per hat for the Mad Hatter. (a) (8 pts) What is the optimal order quantity from the retailers perspective? What are the expected profits for the Mad Hatter and for Crazy Hat? Problem 4 continued on next page. . . 8 Summer 2018 MS&E 260 Final Exam Problem 4 (continued) (b) (8 pts) What would be the optimal order quantity if the Mad Hatter decides to make the hats himself at the same cost it takes for Crazy Hat to produce each hat (i.e. vertical integration)? What would be the expected profit for the Mad Hatter then? Problem 4 continued on next page. . . 9 Summer 2018 MS&E 260 Final Exam Problem 4 (continued) (c) (9 pts) Assume now that the Mad Hatter and Crazy Hat are considering a buyback contract. What would be the buyback price that perfectly coordinates the supply chain, if wholesale price remains at $60 per hat? What is the expected profit for the Mad Hatter and Crazy Hat? 10 Summer 2018 MS&E 260 Final Exam Problem 5 Problem 5 [25 pts] John is the chief operating o�cer of Alpha Corp, which produces fuel lines for commercial automobiles (call this Product A). John is considering a total shift in Alpha’s current business into a new business line: producing fuel lines for commercial aircraft (call this Product B). This will require capital investment to build new manufacturing infrastructure and decommission the old machinery. John believes his prospects are as follows: Suppose that the demand uncertainty can only assume two levels: low or high. Currently, John assigns a probability of 0.3 of low market demand for Product B, and a probability of 0.6 of high market demand for Product A. (a) (10 pts) Let x denote the dollar worth of the prospect of shifting the product lines and high market demand for Product B. What is the value of x such that John is indi↵erent between shifting to Product B or staying with Product A? For part (a), assume that John is risk neutral. Problem 5 continued on next page. . . 11 Summer 2018 MS&E 260 Final Exam Problem 5 (continued) (b) (15 pts) Suppose John now believes that the value of the prospect of Product B with high demand is $40M . Also, suppose that a test is available on the future demand for Product A (the results only apply to Product A), with the following characteristics: Probability of false high demand signals = 0.3 Probability of false low demand signals = 0.1 What is the maximum value that John should be willing to pay for this test? For part (b), assume that John is risk neutral. Problem 5 continued on next page. . . 12 Summer 2018 MS&E 260 Final Exam Problem 5 (continued) Problem 5 continued on next page. . . 13 Summer 2018 MS&E 260 Final Exam Problem 5 (continued) Problem 5 continued on next page. . . 14 Summer 2018 MS&E 260 Final Exam Problem 5 (continued) 15 Summer 2018 MS&E 260 Final Exam Problem 6 Problem 6 [25 pts] Answer briefly the following questions. (a) (8 pts) List the 5 principles of Lean Manufacturing. (b) (8 pts) Describe how you would move your furniture to a new house using a Flow process. Problem 6 continued on next page. . . 16 Summer 2018 MS&E 260 Final Exam Problem 6 (continued) (c) (9 pts) When evaluating the tradeo↵s between the charge-coupled device (CCD) imagers versus com- plementary metal-oxide-semiconductor (CMOS) imagers for space imaging, the di↵erences in the CCD versus CMOS industrial base result in relative di↵erences in both fixed cost and marginal unit cost. What are those relative di↵erences? 17 Normal Loss Function Normal z- Table
Answered Same DayAug 03, 2021

Answer To: MS&E 260 Final Exam Stanford University Summer 2018 Name: SUNet ID: Question Points Available Points...

Ishmeet Singh answered on Aug 07 2021
138 Votes
Expansion of Commonwealth Bank in Asia
Managerial Accounting
problem-1:
1.
(a) Assumption: All hours are significant no priority given.
Secondly, no regression or corre
lation between any hour.
(b) Both done together.
Data Given:
Poisson process rate, ƛ = 8 users/hr.
Service rate, µ = 15 users/hr.
No. of servers, m=1
Therefore,
Utilization,
· 8/ 1* 15 = 0.53
Now, no. of users in queue, Lq = Utilization^2/1- Utilization
· 0.53^2/1-0.53 = 0.597
Average waiting time in queue, Wq = Lq/ ƛ
· 0.597 * 60/8 mins = 4.47 mins approx.
Now, average waiting time in system, Ws = Wq+ 1/ µ
· 4.47 + 4 = 8.47 mins approx.
Now, no. of users in system, Ls = Ws* ƛ = 8 * 0.14 = 1.12 users
(2)
Arrival rate = 12 users/ hr.
Service rate = 15 users/ hr.
Utilization = 12/15 = 0.8
Waiting time in queue, Wq = Lq/ ƛ = 16 mins/user
&, In system = 16 +5 = 21 mins./user
(3)
Data Given:
Poisson process rate, ƛ = 8 users/hr.
Service rate, µ = 12 users/hr.
No. of servers, m=2
Therefore,
Utilization,
· 8/ 2* 12 = 0.33
Now, no. of users in queue, Lq = Utilization^2/1- Utilization
· 0.33^2/1-0.33 = 0.165
Average waiting time in queue, Wq = Lq/ ƛ
· 0.165 * 60/8 mins = 1.24 mins approx.
Now, average waiting time in system, Ws = Wq+ 1/ µ
· 1.24 + 5 = 6.24 mins approx.
problem-2:
Assuming there are 2 prices on each day p1 & p2
Valuation of buyers to buy each day [0,100] & [0,x]
p1 > p2
Expected Revenue: max p1p2 p1*(1-p1/100) + p1*p2/100*(1-p2/100)
= Max. p2 p1/100*p2*(1-p2/100)
· p1^2 = p1^2*(1-p1/100) + p1*p2/100*(1-p2/100)
· 1 = 1-p1/100 + p2/p1(1-p2/100)
· p1/100 = p2/p1(1-p2/100)
· p1^2 = 100p2 – p2^2
· p1^2 = p2 (100-p2) gives both p1 = p2 =$50 Therefore, keeping x =100 both days we can keep prices $50
(b) However any value x < 100 => p2> p1
(c) Data Given:
Prices: [p1 & p2]
Probabilities:
U [0,100]
U [0,300]
U [0,200]
Expected Revenue: max p1p2p3 * P (U1 >= p1) + P (U1 <= p1) * p2 * P (U2 >= p2) + P (U1 <= p1) * p2 * P (U2 <= p2) * p3 * P (U3 <= p3) =
Max. p1p2p3...
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