武汉理工大学研究生生产计划与控制(2)作业2 及对应答案

Homeworks1. A food store has an average of 220 customers arriving per hour during peak shoppinghours . During these peak periods all eight checkout counters will be open and operating with a capacity of serving an average of 35 customers per hour per counter . All checkout counters are identical .a. What proportion of the time would all of the checkout counters and waiting lines beempty of customers?b. How long would customers wait in line on the average?c. How many customers would be waiting in each line on the average? Solution:From the question, we know: λ=220,μ=35, N=8,a. compute 0P , probability that the system would be empty: n=0,()()01n 01P //!!1n NN n N N λμλμλμ-==⎡⎤+⎢⎥⎛⎫⎢⎥⎣⎦- ⎪⎝⎭∑=()()881n 01220/35220/35220!8!135*8n n -=⎡⎤+⎢⎥⎛⎫⎢⎥⎣⎦- ⎪⎝⎭∑ =()()()87n 0144/744/7!8!3/14n n =⎡⎤+⎢⎥⎢⎥⎣⎦∑= 10.0015377.831685208282.048=+ b. Compute 1t :()()801222200.0015352203588!1!1358N P t N N N λμλμμ⎡⎤⎡⎤⎢⎥⎢⎥⎛⎫⎛⎫⎢⎥⎢⎥== ⎪ ⎪⎢⎥⎢⎥⎝⎭⎝⎭⎛⎫⎛⎫⎢⎥⨯--⎢⎥ ⎪ ⎪⨯⎝⎭⎢⎥⎣⎦⎝⎭⎣⎦ 82200.00150.007135518469⎛⎫=⨯= ⎪⎝⎭ c. compute 1n :()()()()()81022/22035220/350.0015 1.55131!7!835220N n P N N λμλμμλ⎡⎤⎡⎤⨯⨯==⨯=⎢⎥⎢⎥--⨯-⎢⎥⎢⎥⎣⎦⎣⎦There are about 1.5513 customers would be waiting in each line on the average.Use POM software to solve this problem. The solution is as follows: MODEL: Multiple Channels Arrival Rate (lambda) = 220Service Rate (mu) = 35 Number of Channels = 8Average Number of Units in Waiting Line = 1.5672 Average Number of Units in System = 7.8529 Average Waiting Time in Line = 0.0071 Average Time in System = 0.0357Probability of Idle System = 0.00152. The Financial Aid Center at Rhode Island State University would like to improve itsresponsiveness to students who come to the center for assistance. One idea is to hire additional financial aid counselors . A recent study of students using the Financial Aid Center found that an average of 17 students come to the center each hour . The study also found that each of the center's three financial aid counselors saw 4 students per hour on average . Try to determine how many total counselors it would take so that the average waiting time for students before they see a counselor is less than 10 minutes .Solution: Compute 1t : λ= 17, μ=4,From the question, we need to hire N counselors to server the students:1μ=N μ, ρ=μλN =N417<1, N>4.25 If N=5, then()()()()0514n 0n 011P //17/417/417!!5!1!154nNnN n n N N λμλμλμ-====⎡⎤⎡⎤++⎢⎥⎢⎥⎛⎫⎛⎫⎢⎥⎢⎥⎣⎦⎣⎦-- ⎪ ⎪⨯⎝⎭⎝⎭∑∑=10.008577.032240.66986276=+()()50122170.0085417455!1!145N P t N N N λμλμμ⎡⎤⎡⎤⎢⎥⎢⎥⎛⎫⎛⎫⎢⎥⎢⎥== ⎪ ⎪⎢⎥⎢⎥⎝⎭⎝⎭⎛⎫⎛⎫⎢⎥⨯--⎢⎥ ⎪ ⎪⨯⎝⎭⎢⎥⎣⎦⎝⎭⎣⎦ 0.00851386.57910.21813.1m i n10m i n54hour =⨯==> if N=6, then()()()()0615n 0n 011P //17/417/417!!6!1!164nNnN n n N N λμλμλμ-====⎡⎤⎡⎤++⎢⎥⎢⎥⎛⎫⎛⎫⎢⎥⎢⎥⎣⎦⎣⎦-- ⎪ ⎪⨯⎝⎭⎝⎭∑∑=10.012552.22425828.0617199=+()()60122170.0125417466!1!146N P t N N N λμλμμ⎡⎤⎡⎤⎢⎥⎢⎥⎛⎫⎛⎫⎢⎥⎢⎥== ⎪ ⎪⎢⎥⎢⎥⎝⎭⎝⎭⎛⎫⎛⎫⎢⎥⨯--⎢⎥ ⎪ ⎪⨯⎝⎭⎢⎥⎣⎦⎝⎭⎣⎦()662170.0125170.01250.0501 3.0066min 10min 44147017466!146hour ⎡⎤⎢⎥⎛⎫⎛⎫⎢⎥==⨯==< ⎪ ⎪⎢⎥⎝⎭⎝⎭⎛⎫⨯-⎢⎥⎪⨯⎝⎭⎣⎦So the Financial Aid Center needs 6 counselors in the system.Use POM software to solve this problem. The solution is as follows: Number of Channels > = 17/4=4.25 If Number of Channels = 5, then MODEL: Multiple ChannelsArrival Rate (lambda) = 17 Service Rate (mu) = 4 Number of Channels = 5Average Number of Units in Waiting Line = 3.7087 Average Number of Units in System = 7.9587Average Waiting Time in Line = 0.2182>10min Average Time in System = 0.4682Probability of Idle System = 0.0085 If Number of Channels = 6, then MODEL: Multiple ChannelsArrival Rate (lambda) = 17 Service Rate (mu) = 4 Number of Channels = 6Average Number of Units in Waiting Line = 0.8488Average Number of Units in System = 5.0988Average Waiting Time in Line = 0.0499<10minAverage Time in System = 0.2999Probability of Idle System = 0.0125So the Financial Aid Center needs 6 counselors in the system.Precision Calibration Services CompanyThe Precision Calibration Services Company provides on-site calibration services for electronic medical equipment. When a customer calls Precision for service, a calibrator travels to the customer's medical center, observes operation of the equipment, and calibrates the equipment. Calibration means a precise measurement and adjustment of the electronic characteristics of the equipment so that it gives precise performance. Quick and dependable service has been the cornerstone of Precision's growth and success.Precision has had a few complaints about the time required for calibrators to respond to customers' calls. At a recent staff meeting, the calibrators said that they were working as hard and fast as humanly possible, but that on some days of the week the number of calls for service was so great that it could take a few days to get to some of these requests. All calibrators now work five days per week, 8 hours per day, Monday through Friday. A different scheduling arrangement is being considered for the calibrators so that they would be available to respond to calls six days per week. Each calibrator would work four days per week and 10 hours per day, and have two consecutive days off per week not including Sunday (Sunday is never worked). Such an arrangement would allow the number of adjustors available to answer requests to better match the daily pattern of the volume of requests.These averages seem to be reasonable estimates of the number of daily calls for service for the next several weeks:Mon. Tues. Wed. Thurs. Fri. Sat. Total Number of requests 29 25 19 30 33 40 176Each request is expected to require an average of about2 hours of a calibrator's time. Precision wants to plan work shifts for the calibrators to approximately match the requests for service, but is concerned about how this will affect the utilization of calibrators, group morale, and the time required to perform customer service.Assignment1.What is the minimum number of calibrators required?Minimum number of calibrators17628.89410⨯==≈⨯calibratorsIn the diagram, the red words and “0”mean the calibrator has rest or is idle, others mean the calibrator is working. The number of the calibrators should be at least 10.3. How much slack per week is present in your work-shift schedules?The 9th calibrator is idle on Friday and 10th calibrator only works on Tuesday. So there are four days (40 hours) slack per week in my work-shift schedules.4. What ways could be used to reduce the amount of slack in your work-shift schedules?Way 1: We can see from work-shift schedules that if the 9th calibrator has rest on Monday and Friday, not on Tuesday, the 10th calibrator can be reduced.Way 2: They hire a part-time calibrator who only works on Tuesday, they can also reduce the amount of slack in my work-shift schedules by reducing the 10th calibrator.5. What factors should be considered in changes in work-shift schedules such as the one under consideration at Precision'? Which of these factors is most important? Can you suggest how Precision should go about making the schedule changes being considered?Factors: the time required to perform customer service in each day, number of rest days in the regulation, the rest period should be consecutive or not, the utilization of calibrators.The time required to perform customer service in each day may be the most important one among these factors.I think Precision should set an information system to collect the number of daily calls for service and the number of daily calls for service that have finished the service, then they can predict the possible number of daily calls for service more precisely.Cincinnati Trucking Inc.Cincinnati Trucking Inc. (CTI) ships personal and small-business computers by company trucks to regional warehouses in the eastern half of the United States. Trucks return from the warehouses to the plant for loading on the average of five per eight-hour day. The plant uses one loading crew in the shipping department that assembles customers' orders for personal and small-business computers and loads outgoing orders onto the trucks. The loading crew does warehouse workwhen no outgoing trucks are at the plant and loads the outgoing trucks on a first-come, first-served basis. The loading crew can load seven trucks per eight-hour day on the average . Each truck occupies about 280 square feet of parking space, each truck driver is paid $30 per hour including fringe benefits, and the loading crew is paid a total of $200 per hour including fringe benefits. CTI ’s union contract with its truck drivers does not allow drivers to assist in loading or unloading trucks.a. On the average, how much parking space should be necessary for trucks waiting to be loaded? Use POM software to solve this problem. The solution is as follows: MODEL: Single Channel Arrival Rate (lambda) = 5 Service Rate (mu) = 7Average Number of Units in Waiting Line = 1.7857 Average Number of Units in System = 2.5000 Average Waiting Time in Line = 0.3571 Average Time in System = 0.5000 Probability of Idle System = 0.2857The parking space should be contain 2.5 trucks because of the average number of units in system is 2.5, so 280*2.5=700 square feet should be necessary as parking space.b. How much does the union contract clause barring drivers from doing loading work cost CTIper year if the Atlanta plant works 250 days per year and if we assume that drivers' idle time could be put to a use that would be of equal value to their present pay? Average time arrivals are in system:11t 0.575s μλ===-- d Each day work eight hours, t s =0.5×8=4hThe cost of CTI per day is about ： 2.5*(4*30) =300$ CTI per year is about ：300*250=75000$So the union contract clause barring drivers from doing loading work costs CTI about $7500 per year.The warehouse manager at CTI has established a policy that requires truck drivers to take their trucks to be serviced at the plant's maintenance center if one truck is waiting to be loaded, but each truck may be serviced a maximum of once per trip. Assume that the serviced trucks do not affect the arrival rate of five trucks per eight-hour day and that the truck-loading terminal operates much like a single-channel, limited-queue-length waiting-line system. If each truck makes a trip every 10 days on the average, how often will the trucks be serviced?Use POM software to solve this problem. The solution is as follows: MODEL: Single Channel with Limited Waiting Line LengthArrival Rate (lambda) = 5Service Rate (mu) = 7Maximum # in System = 2Average Number of Units in Waiting Line = 0.2976Average Number of Units in System = 0.7798Probability of Idle System = 0.4495Probability of 1 units in the system = 0.3211Probability of 2 units in the system = 0.2294Because when there are two trucks in the system, the third should be serviced at the plant's maintenance center. So a truck will be serviced in 2.294day (10day*0.2294=2.294day).The truck-loading dock described above has this arrival distribution:（1）（2）（3）Minutes betweenTrucks Arriving Arriving Trucks Relative Frequencyin an 8-Hour Day [1/( 1 )×480] (percent)1 480 5%2 240 103 160 154 120 205 96 256 80 157 69 10The time that it takes the single loading crew to load the trucks is normally distributed with a mean of 69 minutes and a standard deviation of 15 minutes. Use these uniform random numbers to establish the number of trucks arriving in 10 eight-hour days: 73, 22, 51, 45, 37, 59, 82, 08, 60, and 00. Use the normally distributed Z scores to establish loading times for each truck. (Begin in the third row of the table.) To determine:a.The number of trucks arriving in each day for 10 days.b.The number of minutes required to load trucks in each day and the total for the 10 days.c.The number of minutes of driver idle time in each day and the total for the 10 days.d. The amount of loading crew idle time in each day and the total for the 10 days.Because of a large increase in sales, CTI has doubled the number of trucks that are hauling personal and small-business computers to regional warehouses. The trucks are now arriving at the rate of 10 trucks per eight-hour day on the average. A second loading crew has been added, and the two crews work separate loading docks and each can load 7 trucks per eight-hour day on the average. With the extra trucks, should the drivers expect to spend more time at the plant now than previously?Solutions：a.Establishing Random Number Ranges for Each Truck in Discrete ArrivalDistribution:Random Numbers:73, 22, 51, 45, 37, 59, 82, 08, 60, 00;From the third row of the table of the normally distributed Z scores, we can get the ten number for ten days:Chart 3ti= μ + zi(σ) ti=69+zi(15)( minutes)Day1:t1=69+1.40×(15)=90；t2=69+（-0.49）×(15)=61.65; t3=69+0.56×(15)=77.4；t4=69+0.10×(15) =70.5;t5=69+（-1.05）×(15)=53.25;The number of minutes required to load trucks: 352.8 (minutes);Day2:t1=69+0.48×(15)=76.2；t2=69+1.00×(15)=84;t3=69+( -0.35)×(15)=63.75;The number of minutes required to load trucks: 223.95 (minutes);Day3:t1=69+（-0.04）×(15)=68.4；t2=69+1.21×(15)=87.15;t3=69+1.80×(15)=96；t4=69+（-0.21）×(15) =65.85; t5=69+（-1.58）×(15)=45.3;The number of minutes required to load trucks: 362.7(minutes);Day4:t1=69+0.15×(15)=71.25；t2=69+（-2.75）×(15)=27.75; t3=69+ 0.45×(15)=75.75；t4=69+ 0.47×(15) =76.05; The number of minutes required to load trucks: 250.8(minutes);Day5:t1=69+（-0.28）×(15)=64.8；t2=69+2.02×(15)=99.3;t3=69+3.00×(15)=114；t4=69+ 1.14 ×(15) =86.1; The number of minutes required to load trucks: 364.2(minutes);Day6:t1=69+（-0.54）×(15)=60.9；t2=69+1.72×(15)=94.8;t3=69+ 0.60×(15)=78；t4=69+ 0.11×(15) =70.65;t5=69+ 0.77×(15)=80.55;The number of minutes required to load trucks: 384.9(minutes);Day7:t1=69+1.14×(15)=86.1；t2=69+ 0.46×(15)=75.9;t3=69+1.01×(15)=84.15；t4=69+ 0.04×(15) =69.6;t5=69+（-1.05）×(15)=53.25；t6=69+（-0.11）×(15)=67.35;The number of minutes required to load trucks: 436.35(minutes);Day8:t1=69+0.22×(15)=72.3；t2=69+ 1.94×(15)=98.1;The number of minutes required to load trucks: 170.4(minutes);Day9:t1=69+（-0.11）×(15)=67.35；t2=69+ 1.02×(15)=84.3;t3 =69+ (-0.79)×(15)=57.15；t4 =69+（-0.24）×(15)=65.4;t5=69+0.52×(15)=76.8;The number of minutes required to load trucks: 351(minutes);Day10:t1=69+1.66×(15)=93.9;The number of minutes required to load trucks: 93.9(minutes);of one truck’s load time pluses the number of minutes of the last truck’s waiting time and then minuses Minutes between these two arriving trucks. The number of minutes of driver waitingBecause of a large increase in sales ： λ=10，μ=7，N=2；()()()()0211n 0n 01111P 17256//10/710/77710!!2!1!172n Nn N n n N N λμλμλμ-======⎡⎤⎡⎤+++⎢⎥⎢⎥⎛⎫⎛⎫⎢⎥⎢⎥⎣⎦⎣⎦-- ⎪⎪⨯⎝⎭⎝⎭∑∑()()220122*********.14887761610722!1!172N P t N N N λμλμμ⎡⎤⎡⎤⎢⎥⎢⎥⎛⎫⎛⎫⎛⎫⎢⎥⎢⎥===⨯⨯= ⎪ ⎪ ⎪⎢⎥⎢⎥⎝⎭⎝⎭⎝⎭⎛⎫⎛⎫⎢⎥⨯--⎢⎥ ⎪ ⎪⨯⎝⎭⎢⎥⎣⎦⎝⎭⎣⎦s 111t t 0.14880.29177μ⎛⎫=+=+= ⎪⎝⎭d5.05711't s =-=-=λμBecause't t s s <, With the extra trucks, the drivers should not expect to spend moretime at the plant now than previously.。