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Stochastic Mfg&Serv Sys

by: Maryse Thiel

Stochastic Mfg&Serv Sys ISYE 3232

Maryse Thiel

GPA 3.82

Hayriye Ayhan

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Hayriye Ayhan
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This 0 page Class Notes was uploaded by Maryse Thiel on Monday November 2, 2015. The Class Notes belongs to ISYE 3232 at Georgia Institute of Technology - Main Campus taught by Hayriye Ayhan in Fall. Since its upload, it has received 17 views. For similar materials see /class/234201/isye-3232-georgia-institute-of-technology-main-campus in Industrial Engineering at Georgia Institute of Technology - Main Campus.


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Date Created: 11/02/15
lSyE 3232 Stochastic Manufacturing and Service Systems Fall 2011 Hi Ayhan Solutions to Homework 7 1 You must be careful in this problem because A is not the arrival rate in this problemi Because A is being used in the service rate distribution lwill use 77 read as eta to denote the arrival rater Because the arrival rate is given in terms of hours 1 will rst convert it into minutes to match the service distribution s unitsi You can also convert the service rate information into hours both methods will work so long as the units agree 1 hour 60 minutes 7 2 minutes lt30 customersgt lt 1hour gt 1 customer n 7 Because the arrival distribution is exponential the squared coef cient of variation is 1 so oi 1 W 39ch is always true for the exponential distributioni Now that the arrival information is calculated 1 need to determine the mean and variance for the arrival distribution which if you look carefully should see is not an exponential distribution The rst way to do this is to calculate the mean and variance directly from the de nitions Let S be a random variable representing the service time ES 000 sfs d3 000 84A28672 d3 2 minutes VarS ms 7 195 0 32163 d3 7 000 3163 13gt 2 00 00 2 3 1 1 9 2 2 72A 2 72A 70 34Ase dsiltO 34Ase 18gt 72A27A272A27 An alternative method for calculating the mean and variance is possible that avoids the ne cessity of calculating the above integrals but requires knowledge of the Erlang k distribution The pidifi for an Erlang k distribution with parameter a is 11631667045 for s gt 0 93 UH 0 otherwise By setting a 2A and k 2 and considering only 8 Z 0 1 can rewrite the given pidifi as follows 22 se aMS 0386 271 7 1671 1 have now shown that the service times have an Erlang 2 distribution with parameter a 2A1 What is useful about this result is that an Erlang k distribution with parameter 1 corresponds to the distribution of the sum of k iiiidi exponential random variables with rate Oh So let S X1 X2 where X1 and X2 are independent exponential random variables with rate 1 allowing the mean and variance to be calculated as fs 4A28672 195 EX1 X2 EX1 EX2 i g a VarS VarX1 X2 VarX1 VarX2 0 l 7 N Having calculated the mean and variance of the service times I can now calculate the service rate 1 and the squared coef cient of variation for the service distribution 05 1 2 l m s VarS 1 1952 2 The last piece of information I need to calculate is the traf c intensity cg PZ a Using Kingman s formula the average waiting time for each customer in the queue is Wc2c p 43 3 quot 2 w 2 373 3375 minutes b Because the arrival rate is less than the service rate the throughput of the system is 77 Using Little s Law the average number of customers in the queue is Lq an 3375 116875 customers c Since the mean service time is ES 32 minutes the average time spent at the site by a customer is W Wq m 31375 32 4875 minutes By using Little s Law again the average number of customers at the site is 1 L 77W 5 4875 214375 customers a To determine the state space for Xn you should consider how the system moves from state to state If the inventory drops below 3 units during a particular day an order is placed to return the inventory to 6 units by the next morning therefore the day can never start with fewer than 3 units in stock The stock is only ever replenished to at most 6 units so the day will never start with more than 6 units From these observations the state space can be written as S 3436 The initial state is deterministic which means the initial distribution is given by a prob ability distribution with only a single nonzero va ue PltX 771fori5 0 7 Z 7 0 otherwise or equivalently as a vector aT 0 0 1 0 Next we nd the transition matrix Note that PXn1 31Xn 3 PXn1 41X 3 PXn1 5m 3 0 A 0quot because whenever the inventory level goes below 3 we order up to 6 Therefore PXn1 6an 3 1 Now suppose the inventory at the beginning of the day is 4 units If the demand during the current day day n is 1 we end up with 3 units of inventory and because we do not order we will have 3 units of inventory at the beginning of the next day day n 1 Because the demand equals 1 with probability 16 the probability of transitioning from state 4 to state 3 is 16 1 6 1f the demand is 2 or 3 the inventory level drops below 3 so we order hence Pom1 3m 4 PD 1 PXn1 6 an 4 PD 2 PD 3 Proceeding by the same logic row by row the transition matrix is given by U H calm oalwmha O onlwoalH O O ma O O O O mmmm H where Pij PXn1 j ZlX i 2 Let Yn be the number of units of inventory in stock at the end of day mi The inventory at the end of a day can be any value from 0 to 5 It cannot be 6 because at the beginning of a day the maximum number of items we can have in inventory is 6 and the demand is strictly greater than zero with probability 1 So the state space in this case is S 0113415 If it still seems odd to exclude 6 include 6 in the transition matrix and see what happens You should notice in the construction involving state 6 that state 6 cannot be reached from any initial state including state 6 for this reason we do not include it in this formulation The initial state is deterministic and the initial distribution is given by PY021 ifz 2 0 otherw1se Observe that PYn1 SlY 0 PD 1 16 Similarly PYn1 4m 0 PD 2 36 PYn1 3Y 0 PD 3 26 Continuing in this fashion we arrive at the following transition matrix 0 0 O Omlm O O D 03100 cab A o mman calm NH O O onlwoalwoalw O O OmleM mlH O mlmmlw O O mlmmlwmlh O D where 11 PYn1 j7 111 i71i 3i Xn represents the number of consecutive days without injury on the morning of day ni 1f today the number of consecutive days without injury is m then it is possible with probability 99100 to have one more day without injury and begin the next morning with m1 consecutive days without injuryi Because this logic holds for every nonnegative integer the state space is equal to the nonnegative integers SZO12Hi To gain some intuition for how the transition probabilities are calculated consider the following table The rst row gives the number of consecutive days without injury on the morning of day mi The second row indicates whether an injury occurred later during the day but before the next morning The last row shows the number of consecutive days without injury on the following morning which is a result of the previous day s events n 1 2 3 4 5 6 7 8 Morning of day n Xn 0 1 2 0 1 2 3 0 Injury Today No No Yes No No No Yes Yes Morning of day n1 Xn1 1 2 0 1 2 3 0 0 This example serves to illustrate the progression of states for the Markov chain in one particular instance In general if at the start of the day there have been m 2 0 consecutive days without injury then one of two outcomes can occur If no injury occurs then the number of consecutive days without injury has increased by 1 so on the morning of the next day there will have been m 1 consecutive days without injuryi From this logic the rst transition probability can be determine 1 99 100 If an injury occurs then the streak has been broken and the number of consecutive days without injury resets to 0 This allows one more transition probability to be calculated 1 100 Notice that the sum of these two probabilities is 1 and they correspond to the same row of the transition matrix Because the sum of the elements of each row of the transition matrix must equal 1 all other entries must be 0 Also because m was generic this formula and logic applies to all m 2 0 The probabilities can be summarized by writing as fo owsi PXn1 m 11X m PNo Injury PXn1 01X m P1njury 99 m lfjil1 PMPXn1Jani L 100 1f 0 0 otherwise


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