ISyE 2030 Practice Test 1

Transcription

1 1 NAME ISyE 2030 Practice Test 1 Summer 2005 This test is open notes, open books. You have exactly 90 minutes. 1. Some Short-Answer Flow Questions (a) TRUE or FALSE? One of the primary reasons why theoretical capacity in a plant is not achieved is due to variability. ANSWER: True. (b) TRUE or FALSE? If arrivals occur according to a Poisson process, then the interarrival times are exponentially distributed. ANSWER: True. (c) TRUE or FALSE? A bottleneck in a flow line is that operation with the smallest processing time. ANSWER: False. (It tends to have the largest processing time.) (d) A job shop has four areas: milling (M), turning (T), deburring (D), and painting (P). On a typical work day, the shop make 100 parts P1 with the routing of M-T-P, 50 parts of P2 with a routing of T-M-T-D-P, and 60 parts of P3 with a routing of M-P. What would the from-to chart have for the cell from M to T? ANSWER: In this case you have 100 of P1 plus 50 of P2 for a total of 150.

2 2 2. Suppose that we are simulating the sales of turkeys at a local market. The owner starts off with 15 turkeys. The daily demand for turkeys for the next 10 days is as follows Each turkey costs the owner $6 to buy, and can be sold to a customer for $13. It costs the owner $1 to store each turkey overnight (i.e., if it wasn t sold that day). If, at the end of a particular day, the owner has less than or equal to 8 turkeys in stock, he places an order to bring the inventory level back up to 15 the next morning. (He gets the turkeys the first thing in the morning before any customers arrive.) Compute his total profit (or loss) for this ten-day period. How many turkeys are left at the end of the last day? ANSWER: Day Beginning Demand End Order More? Totals Assuming we have to pay $90 to buy the first day s turkeys, we have Profit = 90 + (47 13) (81 1) (41 6) = $194. Finally, looking at the table, we see that 9 turkeys are left at the end of the 10th day.

3 3 3. The first eight customers at Joey s single-server ice cream shop have the following interarrival times and priorities: cust / priority arrival time The symbol denotes a high-priority customer, while denotes a low-priority guy. High-priorities get to go ahead of all low-priorities in line, but customers within a priority class get processed FIFO. Also, high-priority guys can t bump a lowpriority out of service once it starts. Further, if an arriving customer sees two or more people in the line, he will become disgusted and leave the shop without ever entering the line. Each customer requires exactly 4 time units to be served. (a) How many customers are served immediately? (b) When can Joey close the shop? (c) What is the average time-in-system for the eight customers? (d) What is the average number of customers in the system during the first 10 time units? ANSWER: Cust Arrl Time Start Serve Depart Time in Sys (a) From the table, we see that only 1 customer is served immediately (cust 1). (b) From the table, we see that the last guy leaves at time 32. (c) From the table, the average of the 8 times in the system is (d) To do this part, we need to draw a graph (like we did in class) or examine the following table:

4 4 Time Event L(t) Cust Order 0 1 arrives arrives 2 1, departs arrives 2 2, arrives 3 2, 3, departs 2 3, end of problem Then the average number of customers is 10 0 L(t) dt/10 = 1.69.

5 5 4. Short-Answer Probability Questions (a) The set of all outcomes of an experiment is called. Solution: The sample space. (b) Any subset of the above set is called. Solution: An event. (c) If A and B are disjoint, then Pr(A B) =? Solution: P(A) + P(B). (d) If Pr(A) = 0.7 and Pr(B) = 0.6, and A and B are independent, then i. Pr(A B) =? Solution: P(A)P(B) = ii. Pr(A B) =? Solution: P(A) + P(B) P(A B) = (e) TRUE or FALSE? Ā B = A B Solution: TRUE

6 6 5. Consider the continuous random variable Y having p.d.f. f(y) = { c y 3 if 1 y 1 0 otherwise. (a) What does p.d.f. mean? Solution: probability density function. (b) Find c. Solution: c = 2 (c) Find Pr( 1 Y 0). Solution: 1/2 (d) Find Pr(0 Y Y 1). Solution: 1/16 (e) Find Pr(0 Y Y 0.5). Solution: 0 (f) Find E[Y ]. Solution: 0 (g) Find Var(Y ). Solution: 2/3 (h) Find E[3Y 2]. Solution: 2 (i) Find Var(3Y 2). Solution: 6

7 7 6. TRUE-FALSE Questions. X and Y must be independent if (a) f(x y) = f Y (y) for all y. Solution: FALSE (b) Cov(X, Y ) = 0. Solution: FALSE (c) f(x, y) = cy, 0 < x < y < 1. Solution: FALSE (d) f(x, y) = cy 2 /(1 + x 3 ), 0 < x < 1, 1 < y < 3. Solution: TRUE (e) E(XY ) = E(X) E(Y ). Solution: FALSE 7. Suppose f(x, y) = cx, 0 < y < x < 1. (a) Find c. Solution: 3 (b) Find Pr(X < 0.5 and Y > 0.5). Solution: 0 (c) Find the p.d.f. of Y. Solution: f Y (y) = 3 2 (1 y2 ), 0 < y < 1. (d) Find the conditional p.d.f. of X given that Y = y. Solution: f(x y) = 2x, 0 < y < x < 1. 1 y 2 8. Suppose that E(X) = 3, E(Y ) = 2, Var(X) = 5, Var(Y ) = 4, and Cov(X, Y ) = 2. (a) Find E(2X + 3Y ). Solution: E(2X + 3Y ) = 2E(X) + 3E(Y ) = 12. (b) Find Var(2X + 3Y ). Solution: Var(2X + 3Y ) = 4Var(X) + 9Var(Y ) + 2(2)(3)Cov(X, Y ) = 32.

8 8 9. If the m.g.f. of X is M X (t) = e 2t2, find E(X). Solution: E(X) = d dt M X(t) = d t=0 dt e2t2 = d t=0 dt 4te2t2 = 0. t=0 10. Suppose that a light bulb has a lifetime that is exponentially distributed with a mean of 1000 hours. Suppose the bulb has already survived 3000 hours. What s the probability that it will survive another 1000 hours? Solution: By the memoryless property, P(X 4000 X 3000) = P(X 1000) = e λx = e 1 =