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Supplement to Chapter 11 Analytical queuing models 337 again, using Little’s law u WIP q = r a × W q = t e r a (1 − u) and since r a u = = r a t e r e u r a = t e then, u u WIP q = × t e × (1 − u) = u 2 (1 − u) t e For M/M/m systems When there are m servers at a station the formula for waiting time in the queue (and therefore all other formulae) needs to be modified. Again, we will not derive these formulae but just state them. W q = u 2(m+1)−1 m(1 − u) From which the other formulae can be derived as before. t e For G/G/1 systems The assumption of exponential arrival and processing times is convenient as far as the mathematical derivation of various formulae are concerned. However, in practice, process times in particular are rarely truly exponential. This is why it is important to have some idea of how a G/G/1 and G/G/m queue behaves. However, exact mathematical relationships are not possible with such distributions. Therefore some kind of approximation is needed. The one here is in common use, and although it is not always accurate, it is for practical purposes. For G/G/1 systems the formula for waiting time in the queue is as follows. W q = c 2 2 A a + c e D C 2 F A u D C(1 − u) F t e VUT formula There are two points to make about this equation. The first is that it is exactly the same as the equivalent equation for an M/M/1 system but with a factor to take account of the variability of the arrival and process times. The second is that this formula is sometimes known as the VUT formula because it describes the waiting time in a queue as a function of: V – the variability in the queuing system U – the utilization of the queuing system (that is demand versus capacity), and T – the processing times at the station. In other words, we can reach the intuitive conclusion that queuing time will increase as variability, utilization or processing time increases.
338 Part Three Planning and control For G/G/m systems The same modification applies to queuing systems using general equations and m servers. The formula for waiting time in the queue is now as follows. W q = c 2 2 A a + c e D C 2 F A u 2(m+1)−1 D Cm(1 − u) F t e Worked example 1 ‘I can’t understand it. We have worked out our capacity figures and I am sure that one member of staff should be able to cope with the demand. We know that customers arrive at a rate of around 6 per hour and we also know that any trained member of staff can process them at a rate of 8 per hour. So why is the queue so large and the wait so long? Have at look at what is going on there please.’ Sarah knew that it was probably the variation, both in customers arriving and in how long it took each of them to be processed, that was causing the problem. Over a two-day period when she was told that demand was more or less normal, she timed the exact arrival times and processing times of every customer. Her results were as follows. The coefficient of variation, c a of customer arrivals = 1 The coefficient of variation, c e of processing time = 3.5 The average arrival rate of customers, r a = 6 per hour therefore, the average inter-arrival time = 10 minutes The average processing rate, r e = 8 per hour therefore, the average processing time = 7.5 minutes Therefore the utilization of the single server, u = 6/8 = 0.75 Using the waiting time formula for a G/G/1 queuing system Also because, A 1 + 12.25D A 0.75 D W q = 7.5 C 2 F C 1 − 0.75F = 6.625 × 3 × 7.5 = 149.06 mins = 2.48 hours WIP q = cycle time × throughput time WIP q = 6 × 2.48 = 14.68 So, Sarah had found out that the average wait that customers could expect was 2.48 hours and that there would be an average of 14.68 people in the queue. ‘Ok, so I see that it’s the very high variation in the processing time that is causing the queue to build up. How about investing in a new computer system that would standardize processing time to a greater degree? I have been talking with our technical people and they reckon that, if we invested in a new system, we could cut the coefficient of variation of processing time down to 1.5. What kind of a different would this make?’ Under these conditions with c e = 1.5 A 1 + 2.25D A 0.75 D W q = 7.5 C 2 F C 1 − 0.75F = 1.625 × 3 × 7.5 = 36.56 mins = 0.61 hour
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sixth edition OperatiOns ManageMent
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Further reading in Operations Manag
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Pearson Education Limited Edinburgh
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Contents Guide to ‘operations in
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Contents ix Chapter 12 Inventory pl
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Guide to ‘operations in practice
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Guide to ‘operations in practice
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Making the most of this book and My
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Making the most of this book and My
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Preface xix Illustrations-based Ope
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To the Student . . . Making the mos
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About the authors Nigel Slack is th
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Publisher’s acknowledgements We a
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Part One INTRODUCTION This part of
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Chapter 1 Operations management 3 O
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88 Part Two Design Figure 4.2 The d
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90 Part Two Design Environmentally
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92 Part Two Design Figure 4.3 Diffe
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94 Part Two Design Professional ser
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96 Part Two Design Figure 4.4 Devia
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98 Part Two Design Figure 4.6 Proce
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100 Part Two Design Figure 4.8 Flow
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102 Part Two Design Given that an i
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104 Part Two Design From Little’s
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106 Part Two Design The relationshi
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108 Part Two Design Summary answers
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110 Part Two Design the grant appli
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Chapter 5 The design of products an
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114 Part Two Design hopeless lack o
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116 Part Two Design Core products a
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118 Part Two Design Figure 5.3 The
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120 Part Two Design Short case Squa
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122 Part Two Design Critical commen
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124 Part Two Design Reducing design
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126 Part Two Design Figure 5.7 A QF
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128 Part Two Design Purpose Basic f
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130 Part Two Design Interactive des
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132 Part Two Design Early conflict
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134 Part Two Design Summary answers
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136 Part Two Design similar. It was
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Chapter 6 Supply network design Key
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140 Part Two Design The supply netw
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142 Part Two Design a particularly
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144 Part Two Design and expanding o
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146 Part Two Design Figure 6.4 Offs
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148 Part Two Design Controversially
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150 Part Two Design Energy costs. O
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152 Part Two Design product and fas
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154 Part Two Design Figure 6.7 Cent
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156 Part Two Design The blue dotted
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158 Part Two Design Table 6.4 The a
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160 Part Two Design Figure 6.11 Rep
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162 Part Two Design Case study Disn
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164 Part Two Design plastic, and ap
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166 Part Two Design of a failing Di
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170 Part Two Design Figure S6.2 Som
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172 Part Two Design Figure S6.3 Tim
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174 Part Two Design Table S6.3 Expo
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176 Part Two Design a comparative s
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188 Part Two Design Table 7.2 The a
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190 Part Two Design Combinatorial c
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204 Part Two Design Table 7.4 Stand
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208 Part Two Design What is process
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214 Part Two Design E-business E-co
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216 Part Two Design Table 8.2 Examp
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218 Part Two Design Customer-proces
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220 Part Two Design ‘drive’ it.
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226 Part Two Design Table 8.5 Prese
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230 Part Two Design Case study Roch
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232 Part Two Design Problems and ap
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238 Part Two Design Table 9.2 Cause
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240 Part Two Design ● Group resou
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242 Part Two Design Job design Job
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244 Part Two Design ● ● ● ●
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246 Part Two Design Figure 9.7 A ty
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248 Part Two Design asked to contri
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250 Part Two Design ● ● ● Eff
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252 Part Two Design Table 9.3 Examp
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254 Part Two Design Qualified worke
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256 Part Two Design Case study Serv
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258 Part Two Design 3 4 5 Step 1 -
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260 Part Two Design them. This mean
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264 Part Two Design Figure S9.1 Tim
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Key operations questions Chapter 10
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Chapter 10 The nature of planning a
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270 Part Three Planning and control
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276 Part Three Make-to-stock Planni
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Part Four IMPROVEMENT Even the best
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Notes on chapters 653 3 Sources inc
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Notes on chapters 657 national Jour
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Glossary 659 Breakthrough improveme
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Glossary 667 Reliability: when appl
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Glossary 669 Transformation process
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‘blue sky’ research projects 45
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cumulative representation 317-22 cu
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failures (continued) estimates of o
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inventories (continued) de-coupling
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mitigation of ethical practice brea
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precedence diagrams 197-9, 481 prec
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esources 7-8 business continuity 59
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supply chains (continued) meaning 3
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