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Process Analysis III
Operations -- Prof. Juran
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Operations -- Prof. Juran
Outline
• Set-up times• Lot sizes• Effects on capacity• Effects on process choice
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Operations -- Prof. Juran
Set-up Times • Many processes can be described (at
least approximately) in terms of – a fixed set-up time and – a variable time per unit (a.k.a. cycle time)
• Capacity of a single activity is a function of lot size, set-up time, and cycle time
• Overall capacity of a system depends on these factors and the resulting bottlenecks across multiple activities
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Operations -- Prof. Juran
Example: Kristen
In general, a formula for the number of minutes to produce n one-dozen batches is given by this expression:
n1016Set-up time
Cycle time per 1-dozen batch
This views the cookie operation as a single activity. We arrived at these numbers through analysis of individual sub-activities at a more detailed level.
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Operations -- Prof. Juran
Example: Kristen
Note that Kristen’s effective cycle time is 10 minutes per 12 cookies, or 0.8333 minutes per cookie, assuming a lot size of 12 cookies.
We can determine the capacity of the system in a specific period of time T by solving for n:
Tn1016
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Operations -- Prof. Juran
Example 1 We can determine the capacity of the system in a specific period of time T by solving for n.
How many 1-dozen batches could Kristen produce in 4 hours?
n1016 T n1016 240 n10 224 n 22 one-dozen orders
In this situation, the capacity of the system is a linear function of the time available.
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Operations -- Prof. Juran
Capacity = f (Time)
-10
0
10
20
30
40
50
60
0 50 100 150 200 250 300 350 400 450 500
Time (Minutes)
Cap
acit
y (L
ots
)
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Operations -- Prof. Juran
12345678910
A B C D E FAvailable Time Capacity (Lots) Capacity (Cookies) Set-up Time Cycle Time Lot Size
0 -1.60 -19.20 16 0.833 125 -1.10 -13.20
10 -0.60 -7.2015 -0.10 -1.2020 0.40 4.8025 0.90 10.8030 1.40 16.8035 1.90 22.8040 2.40 28.80
=(A6-$D$2)/($E$2*$F$2)
=B9*$F$2
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Operations -- Prof. Juran
Example 2 This assumes that the set-up only needs to be done once.
What if there were a 16-minute set-up for every lot?
This effectively makes the set-up time zero, and the cycle time 26 minutes per 12-cookie lot.
Capacity is still a linear function of the time available.
n)1016( T n26 240 n 23.9 one-dozen orders
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Operations -- Prof. Juran
Capacity = f (Time)
0
10
20
30
40
50
60
0 50 100 150 200 250 300 350 400 450 500
Time (Minutes)
Cap
acit
y (L
ots
)
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Operations -- Prof. Juran
Let’s make some assumptions; a system similar (but not identical) to the Kristen system:• Produce individual units (cookies) • The cycle time is 0.8333 minutes per cookie• The set-up time is s minutes, and needs to be performed
again for every “lot” of 12 cookies
The capacity of this system (in “lots”) over 240 minutes is:
240/(s + 0.8333 * 12)
The capacity of this system (in “lots”) with a 16-minute set-up is:
240/(s + 0.8333 * 12) = 9.23
(or 9.23 * 12 = 110.77 cookies)
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Operations -- Prof. Juran
123456789
A B C D E FAvailable Time Capacity (Lots) Capacity (Cookies) Set-up Time Cycle Time Lot Size
0 0.00 0.00 16 0.833 125 0.19 2.31
10 0.38 4.6215 0.58 6.9220 0.77 9.2325 0.96 11.5430 1.15 13.8535 1.35 16.15
=A5/($D$2+$E$2*$F$2)
=B8*$F$2
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Operations -- Prof. Juran
Now let’s assume the time available is fixed at 240 minutes, and study the effect on capacity that results from changing the set-up time.The capacity of this system (in “lots”) with an s-minute set-up is:
240/(s + 0.8333 * 12)
(a nonlinear function of the set-up time)
Example 3
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Operations -- Prof. Juran
Capacity = f (Set-up Time)
0
5
10
15
20
25
30
0 20 40 60 80 100 120 140 160 180 200 220 240
s = Set-up Time (Minutes)
Cap
acit
y (L
ots
)
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Operations -- Prof. Juran
Capacity = f (Set-up Time)
0
50
100
150
200
250
300
350
0 20 40 60 80 100 120 140 160 180 200 220 240
s = Set-up Time (Minutes)
Cap
acit
y (C
oo
kies
)
Capacity could also be measured in “cookies” instead of “12-cookie lots”:
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Operations -- Prof. Juran
Extreme Case 1:
If the set-up time is zero, then the capacity of this system (in “lots”) over 240 minutes is:
240/(0 + 0.8333 * 12) = 24 lots
Extreme Case 2:
If the set-up time is 240, then the capacity of this system (in “lots”) over 240 minutes is zero (because all of the time is consumed by setting up)
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Operations -- Prof. Juran
12345678
A B C D E FSet-up Time Capacity (Lots) Capacity (Cookies) Available Time Cycle Time Lot Size
0 24.00 288.00 240 0.833 121 21.82 261.822 20.00 240.003 18.46 221.544 17.14 205.715 16.00 192.006 15.00 180.00
=$D$2/(A5+$E$2*$F$2)
=B7*$F$2
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Operations -- Prof. Juran
Now let’s assume the time available is fixed at 240 minutes, AND fix the set-up time at 16 minutes, to study the effect on capacity that results from changing the lot size.The capacity of this system (in “cookies”) with an s-minute set-up is:
240/(16 + 0.8333 * Q)
(another nonlinear function)
Example 4
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Operations -- Prof. Juran
Extreme Case 1:
240/16 = 15 gives an upper bound to the number of lots; in that case we would use up all of our time setting up, and never make any cookies.
Extreme Case 2:
If we assume only one set-up, then the capacity is
240 - 16/0.8333 = 268.8 cookies
The largest lot that can be completed in 240 minutes is 268.
Extreme Case 3:
If we assume no set-up, then the capacity is
240/0.8333 = 288 cookies
The largest lot that can be completed in 240 minutes is 288.
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Operations -- Prof. Juran
Capacity = f (Lot Size)
0
50
100
150
200
250
300
0 20 40 60 80 100 120 140 160 180 200 220 240
Q = Lot Size (Cookies)
Cap
acit
y (C
oo
kies
)
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Operations -- Prof. Juran
12345678
A B C D E FLot Size Capacity (Lots) Capacity (Cookies) Available Time Cycle Time Set-up Time
0 15.00 0.00 240 0.833 161 14.26 14.262 13.58 27.173 12.97 38.924 12.41 49.665 11.90 59.506 11.43 68.57
=$D$2/($F$2+$E$2*A5)
=B7*A7
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Operations -- Prof. Juran
Example 5
What if the lot size AND the set-up time are variables?
We can determine the capacity of the system in a specific period of time using this complicated function of lot size, cycle time, set-up time, and the time available for production:
Capacity in lots
sizelot *time cycletime up-setavailable time
Capacity in units
sizelot *time cycletime up-setavailable time*sizelot
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Operations -- Prof. Juran
Assume 240 minutes available, and 0.8333 minute cycle time:
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Operations -- Prof. Juran
123456789
1011121314
A B C D E F G H I J240 0.8333
1 12 24 36 48 600 288.0 288.0 288.0 288.0 288.0 288.02 84.7 240.0 261.8 270.0 274.3 276.94 49.7 205.7 240.0 254.1 261.8 266.76 35.1 180.0 221.5 240.0 250.4 257.1
Set-up Time 8 27.2 160.0 205.7 227.4 240.0 248.310 22.2 144.0 192.0 216.0 230.4 240.012 18.7 130.9 180.0 205.7 221.5 232.314 16.2 120.0 169.4 196.4 213.3 225.016 14.3 110.8 160.0 187.8 205.7 218.218 12.7 102.9 151.6 180.0 198.6 211.820 11.5 96.0 144.0 172.8 192.0 205.7
Lot Size
=E$2*$A$1/($B11+$B$1*E$2)
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Operations -- Prof. Juran
Why Do We Care?
• It might be on the exam• Drives major decisions regarding
operations strategy, technology choice, process design, and capital investment
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Operations -- Prof. Juran
Process Choice
• Sometimes we get to choose among several possible technologies
• One important factor is capacity: Which technology can meet demand fastest?
• This may depend on lot size• Similar to make-vs-buy decisions
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Operations -- Prof. Juran
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Operations -- Prof. Juran
Colarusso Confectioners needs to fill an order for 500 sfogliatelle (a famous Italian pastry) for one of their clients.
Colarusso has the in-house capability to produce sfogliatelle, but this is an unusually large order for them and they are considering whether to outsource the job to Tumminelli Industries, Inc. (a regional pastry supplier with equipment designed for greater volume).
The customer service rep from Tumminelli quotes a rate for sfogliatelle as follows: a fixed order cost of $135 plus $0.25 per sfogliatella. Colarusso’s in-house costs are $75.00 to set up production and $0.39 per unit.
Example: Make vs. Buy
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Operations -- Prof. Juran
What should Colarusso do with this order for 500 svogliatelle?
1234567
A B C DColarusso Tumminelli
Set-up cost 75.00$ 135.00$ Per unit 0.39$ 0.25$
500 270.00$ 260.00$
=B$2+$A5*B$3 =C$2+$A5*C$3
The total cost of the order will be lower if Colarusso outsources this job to Tumminelli.
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Operations -- Prof. Juran
Obviously Colarusso has an advantage for small lot sizes, and Tumminelli has an advantage for large lot sizes. What is the break-even point?
Make vs. Buy?
$-
$50
$100
$150
$200
$250
$300
$350
$400
$450
$500
0 100 200 300 400 500 600 700 800 900 1000
Lot Size
Tot
al C
ost
Colarusso
Tumminelli
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Operations -- Prof. Juran
Qcs CC Qcs TT
Q39.075 Q25.0135
Q25.039.0 75135
Q14.0 60
Q 429
Finding the break-even point algebraically:
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Operations -- Prof. Juran
Process Choice Example
Yamada Fukuda Setup Time (min) 5 60 Cycle Time (min./ unit) 2.2 2
All-American Industries is considering which of two machines to purchase:
1. If the typical lot size is 200 units, which machine should they buy?
2. What is the capacity of that machine in a 480-minute shift?
3. What is the break-even lot size for these two machines?
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Operations -- Prof. Juran
1. If the typical lot size is 200 units, which machine should they buy?
Total flow time for a 200-unit lot:
Yamada: Qcs YY 200*2.25
445
Fukuda: Qcs FF 200*260
460
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Operations -- Prof. Juran
123456
D E F G H IYamada Fukuda Shift
Setup Time (min) 5 60 480 minutesCycle Time (min./unit) 2.2 2
200 lot size
Flow Time per Lot 445 460=F2+F3*$H$4
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Operations -- Prof. Juran
2. What is the capacity of that machine in a 480-minute shift?
Capacity of Yamada machine in 480 minutes:
size lot*time cycletime up-setavailable time
200*2.25480
= about 1.1 Lots
size lot*time cycletime up-set
available time*size lot
200*2.25
480*200
= about 215.7 Units
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Operations -- Prof. Juran
1234567
D E F G H IYamada Fukuda Shift
Setup Time (min) 5 60 480 minutesCycle Time (min./unit) 2.2 2
200 lot size
Capacity in Lots 1.1Capacity in Units 215.7
=H2/(E2+E3*H4)=E6*H4
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Operations -- Prof. Juran
3. What is the break-even lot size for these two machines?
Break-even Lot Size:
Qcs YY Qcs FF
Q2.25 Q*260
Q2.0 55
Q 275
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Operations -- Prof. Juran
12345678
D E F G H IYamada Fukuda Shift
Setup Time (min) 5 60 480 minutesCycle Time (min./unit) 2.2 2
275 lot size
Flow Time 610 610
0=10000000*(E6-F6)
=F2+F3*$H$4
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Operations -- Prof. Juran
Summary
• Set-up times• Lot sizes• Effects on capacity• Effects on process choice