Process Analysis III
Outline
• Set-up times
• Lot sizes
• Effects on capacity
• Effects on process choice
Operations -- Prof. Juran
2
©The McGraw-Hill Companies, Inc., 2004
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
Operations -- Prof. Juran
3
©The McGraw-Hill Companies, Inc., 2004
Example: Kristen
In general, a formula for the number of minutes to produce n onedozen batches is given by this expression:
16  10n
Set-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.
Operations -- Prof. Juran
4
©The McGraw-Hill Companies, Inc., 2004
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:
16  10n  T
Operations -- Prof. Juran
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©The McGraw-Hill Companies, Inc., 2004
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?
16  10n  T
16  10n  240
10n  224
n  22 one-dozen orders
In this situation, the capacity of the system is a linear function of
the time available.
Operations -- Prof. Juran
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©The McGraw-Hill Companies, Inc., 2004
Capacity = f (Time)
60
50
Capacity (Lots)
40
30
20
10
0
0
50
100
150
200
250
300
350
400
450
-10
Time (Minutes)
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©The McGraw-Hill Companies, Inc., 2004
500
1
2
3
4
5
6
7
8
9
10
A
B
C
D
E
F
Available Time Capacity (Lots) Capacity (Cookies) Set-up Time Cycle Time Lot Size
0
-1.60
-19.20
16
0.833
12
5
-1.10
-13.20
10
-0.60
-7.20
15
-0.10
-1.20
=(A6-$D$2)/($E$2*$F$2)
20
0.40
4.80
25
0.90
10.80
30
1.40
16.80
=B9*$F$2
35
1.90
22.80
40
2.40
28.80
Operations -- Prof. Juran
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©The McGraw-Hill Companies, Inc., 2004
Example 2
This assumes that the set-up only needs to be done once.
What if there were a 16-minute set-up for every lot?
This effectively makes the set-up time zero, and the cycle time 26
minutes per 12-cookie lot.
(16  10)n
26n
n
T
 240
 9.23 one-dozen orders
Capacity is still a linear function of the time available.
Operations -- Prof. Juran
9
©The McGraw-Hill Companies, Inc., 2004
Capacity = f (Time)
60
Capacity (Lots)
50
40
30
20
10
0
0
50
100
150
200
250
300
350
400
450
Time (Minutes)
Operations -- Prof. Juran
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©The McGraw-Hill Companies, Inc., 2004
500
Let’s make some assumptions; a system similar (but not identical)
to the Kristen system:
•Produce individual units (cookies)
•The cycle time is 0.8333 minutes per cookie
•The set-up time is s minutes, and needs to be performed again for
every “lot” of 12 cookies
The capacity of this system (in “lots”) over 240 minutes is:
240/(s + 0.8333 * 12)
The capacity of this system (in “lots”) with a 16-minute set-up is:
240/(s + 0.8333 * 12) = 9.23
(or 9.23 * 12 = 110.77 cookies)
Operations -- Prof. Juran
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©The McGraw-Hill Companies, Inc., 2004
1
2
3
4
5
6
7
8
9
A
B
C
D
E
F
Available Time Capacity (Lots) Capacity (Cookies) Set-up Time Cycle Time Lot Size
0
0.00
0.00
16
0.833
12
5
0.19
2.31
10
0.38
4.62
=A5/($D$2+$E$2*$F$2)
15
0.58
6.92
20
0.77
9.23
25
0.96
11.54
=B8*$F$2
30
1.15
13.85
35
1.35
16.15
Operations -- Prof. Juran
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©The McGraw-Hill Companies, Inc., 2004
Example 3
Now let’s assume the time available is fixed at 240 minutes, and
study the effect on capacity that results from changing the set-up
time.
The capacity of this system (in “lots”) with an s-minute set-up is:
240/(s + 0.8333 * 12)
(a nonlinear function of the set-up time)
Operations -- Prof. Juran
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©The McGraw-Hill Companies, Inc., 2004
Capacity = f (Set-up Time)
30
Capacity (Lots)
25
20
15
10
5
0
0
20
40
60
80
100
120
140
160
180
200
220
s = Set-up Time (Minutes)
Operations -- Prof. Juran
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©The McGraw-Hill Companies, Inc., 2004
240
Capacity could also be measured in “cookies” instead of “12-cookie lots”:
Capacity = f (Set-up Time)
350
Capacity (Cookies)
300
250
200
150
100
50
0
0
20
40
60
80
100
120
140
160
180
200
220
s = Set-up Time (Minutes)
Operations -- Prof. Juran
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©The McGraw-Hill Companies, Inc., 2004
240
Extreme Case 1:
If the set-up time is zero, then the capacity of this system (in “lots”)
over 240 minutes is:
240/(0 + 0.8333 * 12) = 24 lots
Extreme Case 2:
If the set-up time is 240, then the capacity of this system (in “lots”)
over 240 minutes is zero (because all of the time is consumed by
setting up)
Operations -- Prof. Juran
16
©The McGraw-Hill Companies, Inc., 2004
1
2
3
4
5
6
7
8
A
B
C
D
E
F
Set-up Time Capacity (Lots) Capacity (Cookies) Available Time Cycle Time Lot Size
0
24.00
288.00
240
0.833
12
1
21.82
261.82
2
20.00
240.00
=$D$2/(A5+$E$2*$F$2)
3
18.46
221.54
4
17.14
205.71
=B7*$F$2
5
16.00
192.00
6
15.00
180.00
Operations -- Prof. Juran
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©The McGraw-Hill Companies, Inc., 2004
Example 4
Now let’s assume the time available is fixed at 240 minutes, AND
fix the set-up time at 16 minutes, to study the effect on capacity that
results from changing the lot size.
The capacity of this system (in “cookies”) with an s-minute set-up
is:
240/(16 + 0.8333 * Q)
(another nonlinear function)
Operations -- Prof. Juran
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©The McGraw-Hill Companies, Inc., 2004
Extreme Case 1:
240/16 = 15 gives an upper bound to the number of lots; in that
case we would use up all of our time setting up, and never make
any cookies.
Extreme Case 2:
If we assume only one set-up, then the capacity is
240 - 16/0.8333 = 268.8 cookies
The largest lot that can be completed in 240 minutes is 268.
Extreme Case 3:
If we assume no set-up, then the capacity is
240/0.8333 = 288 cookies
The largest lot that can be completed in 240 minutes is 288.
Operations -- Prof. Juran
19
©The McGraw-Hill Companies, Inc., 2004
Capacity = f (Lot Size)
300
Capacity (Cookies)
250
200
150
100
50
0
0
20
40
60
80
100
120
140
160
180
200
220
Q = Lot Size (Cookies)
Operations -- Prof. Juran
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©The McGraw-Hill Companies, Inc., 2004
240
1
2
3
4
5
6
7
8
A
Lot Size
0
1
2
3
4
5
6
B
C
D
E
F
Capacity (Lots) Capacity (Cookies) Available Time Cycle Time Set-up Time
15.00
0.00
240
0.833
16
14.26
14.26
13.58
27.17
=$D$2/($F$2+$E$2*A5)
12.97
38.92
12.41
49.66
=B7*A7
11.90
59.50
11.43
68.57
Operations -- Prof. Juran
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©The McGraw-Hill Companies, Inc., 2004
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

Capacity in units

Operations -- Prof. Juran
time
available
set - up time  cycle time * lot size 
lot size  * time
available
set - up time  cycle time * lot size 
22
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Assume 240 minutes available, and 0.8333 minute cycle time:
Operations -- Prof. Juran
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©The McGraw-Hill Companies, Inc., 2004
A
B
240 0.8333
1
2
3
4
5
6
7 Set-up Time
8
9
10
11
12
13
14
0
2
4
6
8
10
12
14
16
18
20
C
D
1
288.0
84.7
49.7
35.1
27.2
22.2
18.7
16.2
14.3
12.7
11.5
12
288.0
240.0
205.7
180.0
160.0
144.0
130.9
120.0
110.8
102.9
96.0
Operations -- Prof. Juran
E
F
G
H
I
Lot Size
24
36
48
60
288.0 288.0 288.0 288.0
261.8 270.0 274.3 276.9
240.0 254.1 261.8 266.7
221.5 240.0 250.4 257.1
205.7 227.4 240.0 248.3
192.0 216.0 230.4 240.0
180.0 205.7 221.5 232.3
169.4 196.4 213.3 225.0
=E$2*$A$1/($B11+$B$1*E$2)
160.0 187.8 205.7 218.2
151.6 180.0 198.6 211.8
144.0 172.8 192.0 205.7
J
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©The McGraw-Hill Companies, Inc., 2004
Why Do We Care?
• It might be on the quiz
• Needed for cases like Kristen
• Drives major decisions regarding
operations strategy, technology choice,
process design, and capital investment
Operations -- Prof. Juran
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©The McGraw-Hill Companies, Inc., 2004
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
Operations -- Prof. Juran
26
©The McGraw-Hill Companies, Inc., 2004
Operations -- Prof. Juran
©The McGraw-Hill Companies, Inc., 2004
Example: Make vs. Buy
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.
Operations -- Prof. Juran
©The McGraw-Hill Companies, Inc., 2004
What should Colarusso do with this order for 500
svogliatelle?
A
B
C
D
1
Colarusso Tumminelli
2 Set-up cost $ 75.00 $ 135.00
3 Per unit
$ 0.39 $
0.25
4
5
500 $ 270.00 $ 260.00
6
=B$2+$A5*B$3
=C$2+$A5*C$3
7
The total cost of the order will be lower if Colarusso
outsources this job to Tumminelli.
Operations -- Prof. Juran
©The McGraw-Hill Companies, Inc., 2004
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?
$500
$450
$400
Colarusso
Tumminelli
Total Cost
$350
$300
$250
$200
$150
$100
$50
$0
100
200
300
400
500
600
700
800
900
1000
Lot Size
Operations -- Prof. Juran
©The McGraw-Hill Companies, Inc., 2004
Finding the break-even point algebraically:
sC  cCQ
75  0.39Q
0.39  0.25Q
0.14Q
Q
 sT  cT Q
 135  0.25Q
 135  75
 60
 429
Operations -- Prof. Juran
©The McGraw-Hill Companies, Inc., 2004
Process Choice Example
All-American Industries is considering which of two machines to purchase:
Setup Time (min)
Cycle Time (min./unit)
Yamada
5
2.2
Fukuda
60
2
1.
If the typical lot size is 200 units, which machine should they buy?
2.
What is the capacity of that machine in a 480-minute shift?
3.
What is the break-even lot size for these two machines?
Operations -- Prof. Juran
©The McGraw-Hill Companies, Inc., 2004
1.
If the typical lot size is 200 units, which machine should they buy?
Total flow time for a 200-unit lot:
Yamada: sY  c Y Q
 5  2.2 * 200
 445
Fukuda:
sF  c F Q
 60  2 * 200
 460
Operations -- Prof. Juran
©The McGraw-Hill Companies, Inc., 2004
D
E
Yamada
Setup Time (min)
5
Cycle Time (min./unit)
2.2
1
2
3
4
5
6 Flow Time per Lot
445
F
Fukuda
60
2
460
G
H
Shift
480
I
minutes
200
lot size
=F2+F3*$H$4
Operations -- Prof. Juran
©The McGraw-Hill Companies, Inc., 2004
2.
What is the capacity of that machine in a 480-minute shift?
Capacity of Yamada machine in 480 minutes:
time available
set - up time   cycle time * lot size

480
5  2.2 * 200 
= about 1.1 Lots
lot size * time available
set - up time   cycle time * lot size

200 * 480
5  2.2 * 200 
= about 215.7 Units
Operations -- Prof. Juran
©The McGraw-Hill Companies, Inc., 2004
D
1
2
3
4
5
6
7
E
Yamada
Setup Time (min)
5
Cycle Time (min./unit)
2.2
Capacity in Lots
Capacity in Units
1.1
215.7
F
Fukuda
60
2
G
H
Shift
480
I
minutes
200
lot size
=H2/(E2+E3*H4)
=E6*H4
Operations -- Prof. Juran
©The McGraw-Hill Companies, Inc., 2004
3.
What is the break-even lot size for these two machines?
Break-even Lot Size:
sY  c Y Q
 sF  c F Q
5  2 .2 Q
 60  2 * Q
0 .2 Q
Q
 55
 275
Operations -- Prof. Juran
©The McGraw-Hill Companies, Inc., 2004
D
E
Yamada
Setup Time (min)
5
Cycle Time (min./unit)
2.2
1
2
3
4
5
6 Flow Time
7
8
610
0
F
Fukuda
60
2
610
G
H
Shift
480
I
minutes
275
lot size
=F2+F3*$H$4
=10000000*(E6-F6)
Operations -- Prof. Juran
©The McGraw-Hill Companies, Inc., 2004
Summary
• Set-up times
• Lot sizes
• Effects on capacity
• Effects on process choice
Operations -- Prof. Juran
39
©The McGraw-Hill Companies, Inc., 2004