Cheatography
https://cheatography.com
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 |
|
Created By
Metadata
Comments
No comments yet. Add yours below!
Add a Comment