Process Analysis III

Operations -- Prof. Juran

Operations -- Prof. Juran

Outline

• Set-up times• Lot sizes• Effects on capacity• Effects on process choice

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

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.

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

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.

Operations -- Prof. Juran

Capacity = f (Time)

-10

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0 50 100 150 200 250 300 350 400 450 500

Time (Minutes)

Cap

acit

y (L

ots

)

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

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

Operations -- Prof. Juran

Capacity = f (Time)

0

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0 50 100 150 200 250 300 350 400 450 500

Time (Minutes)

Cap

acit

y (L

ots

)

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)

Operations -- Prof. Juran

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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

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

Operations -- Prof. Juran

Capacity = f (Set-up Time)

0

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0 20 40 60 80 100 120 140 160 180 200 220 240

s = Set-up Time (Minutes)

Cap

acit

y (L

ots

)

Operations -- Prof. Juran

Capacity = f (Set-up Time)

0

50

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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”:

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)

Operations -- Prof. Juran

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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

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

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.

Operations -- Prof. Juran

Capacity = f (Lot Size)

0

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0 20 40 60 80 100 120 140 160 180 200 220 240

Q = Lot Size (Cookies)

Cap

acit

y (C

oo

kies

)

Operations -- Prof. Juran

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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

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

Operations -- Prof. Juran

Assume 240 minutes available, and 0.8333 minute cycle time:

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)

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

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

Operations -- Prof. Juran

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

Operations -- Prof. Juran

What should Colarusso do with this order for 500 svogliatelle?

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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.

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?

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0 100 200 300 400 500 600 700 800 900 1000

Lot Size

Tot

al C

ost

Colarusso

Tumminelli

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:

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?

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

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

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

Operations -- Prof. Juran

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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

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

Operations -- Prof. Juran

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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

Operations -- Prof. Juran

Summary

• Set-up times• Lot sizes• Effects on capacity• Effects on process choice