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ISE 463 MID-TERM Cheat Sheet by

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+­cr­2A­(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+(r­s/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
 

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