Cheatography

# ISE 463 MID-TERM Cheat Sheet by musikdr

### Little's Law

 WORK 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 PERFOR­MANCE

 CT BEST if w<=Wo To otherwise W/rb TH BEST if w<=Wo w/To otherwise rb

### WORST CASE PERFOR­MANCE

 CTworst =w*To THworst 1/To

### PRACTICAL WORST CASE

 CTpwc 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 ONLY

 Natural 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 PREEMPTIVE

 Mean 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 Qs

 SCV 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 Notation

 M 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 Queing

 WIP = 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 QUE

 CT = ((ca2 + ce2)/2)(u­/(1-u) x te

### M/M/1/b

 WIP = 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