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Little's LawWORK IN PROGRESS | WIP = TH*CT | Cycle Time | CT | WIP/TH | Throughput | TH | WIP/CT | Bottleneck Rate | rb | =1/Max Avg Processing Time | Raw Processing Time | To | Sum of Avg Processing Time | Critical WIP | Wo | rb*To |
BEST CASE PERFORMANCECT BEST | if w<=Wo | To | | | otherwise | W/rb | TH BEST | if w<=Wo | w/To | | | otherwise | rb |
PRACTICAL WORST CASECTpwc | To+((w-1)/rb) | THpwc | (w/(Wo+w-1))rb |
Sample Midterm Qsle Midterm Qs (Cont)b) A company supplying seats to an auto assembly plant sends trucks to its customer at an average rate of 6 trucks per day. Given the travel time to the customer is an average of three days, what is the average number of trucks in transit at any given time?
TH = 6 trucks/day
CT = 3 days
WIP = TH x CT = 18 trucks |
| | PREEMPTIVE ONLYNatural Proc. Time | to | STD of Nat. Proc. Time | σo | SCV of Nat. Proc. Time | co2 | σo2/to2 | STD of Nat. Proc. Time | mf | Mean Time to Repair | mr | STD of Time to Repair | σr | Mean Availability | A | mf/(mf+mr) | SCV of Time to Repair | cr2 | σr2/mr | Mean Eff. Time with Preemptive Outages | te(Po) | to/A | SCV of Eff. Time with Preemptive Outages | ce(PO)2 | co2+(1+cr2)A(1-A)*mr/to |
PREEMPTIVE PLUS NON PREEMPTIVEMean batch size | Ns | Mean batch size | ts | STD of Setup Time | σs | Mean Eff. Time with Preemptive Outages | te | te(PO)+ts/Ns | std. dev. Squared of eff. Time | σe2 | te(po)2 x cd(po)2+(rs2/Ns) + (Ns-1/Ns2) x ts2 | SCV of Eff. Time with Preemptive Outages | ce2 | σe2/te2 | Mean Utilization | u | te/ta | SCV of interarrival times | ca2 | SCV of interdeparture times | cd2 | 1 + (1-u2) x (ca2-1) + u2(ce2-1) | CTQ | | (ca2 + ce2)/2 x u/(1-u) x te | CT | | CTQ + te | WIP | | CT/ta | WIP | | CT/ta |
Sample Midterm Qsle Midterm Qs (Cont)b) A company supplying seats to an auto assembly plant sends trucks to its customer at an average rate of 6 trucks per day. Given the travel time to the customer is an average of three days, what is the average number of trucks in transit at any given time?
TH = 6 trucks/day
CT = 3 days
WIP = TH x CT = 18 trucks |
| | Sample Midterm QsSCV of Effective processing time | ce2 = ((σo2 + (σs2/N) + (Ns-1)/Ns2 x ts2)/te | Utilization | u = (te/ta) = (te(np)/ta) | Utilization (coffee sho) | u = ra/re=te/ta | WIP (M/M/1) | u/(1-u) | WIP (M/M/1) | u/(1-u) | CT (M/M/1) | te/(1-u) |
Sample Midterm Qs (Cont)3. a) Compute the average cycle time at machine 1.
CTq1 =( Ca2 + Ce2)/2 x (u/(1-u)) x te
u = (te/ta)
b) Compute the mean and coefficient of variation of the time between departures from Machine 1.
ta(2) = td(1) =ta(1)
cd2 = 1 + (1-u2) x (ca2 -1) + u2(ce2 - 1)
u2 = te2/ta2 = 18/22
c) Compute the average cycle time at machine 2
CT(2) = CTq(2) + te(2) (note use u2 to calculate CTq(2)
d) Calculate the total CT and WIP of the system (combining machine 1 and 2).
Total CT = CT(1) + CT(2)
From Little's Law
Total WIP = CT x TH = CT/ta
e) Now suppose the line must produce both products in equal proportion, i.e., one unit of Product 1 for each unit of Product 2. Estimate the bottleneck rate and raw process time of the line under this product mix.
Hint: Think about the what the average processing time will be at each machine.
Processing time at (M1 + M2)/#of machines (2 calculations ..1 for each product)
rb = 1/largest processing
to = to1+to2 (answer from processing time) |
Kendall NotationM | Memoryless or exponential | D | Deterministic | G | General | G/G/1 | not exponential, gives approx CT and CTq | M/M/1 | exponential, infinite source population unlimited queue length |
M/M/1 QueingWIP = u/(1-u)
CT = WIP/ra = te/(1-u)
CTq = CT-te =(u x te)/(1-u)
WIPq =raCTq=u2/(1-u) |
G/G/1 QUECT = ((ca2 + ce2)/2)(u/(1-u) x te |
M/M/1/bWIP = u/(1-u) - ((b+1)ub+1) / 1-ub+1
TH = ((1-ub) / (1-ub+1) x ra
• Smaller buffer sizes bring greater losses relative to uncapacitated system |
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