30
October 2011 1
October 2011
1
Little’s Law for Retail calls, May 2002: US Bank
: Throughput Rate, Retail, May 2002; US Bank Arrivals to offered Retail Total
May2002
0
5000
10000
15000
20000
25000
30000
35000
40000
45000
50000
0 5 10 15 20 25 30
days
Number of cases
W: Average Waiting Time, Retail, May 2002; US Bank (Wq)Average Waiting Time, Retail
May 2002; US Bank
0
5
10
15
20
25
30
35
40
0 5 10 15 20 25 30
days
Means, Seconds
L: Average Queue Length, Retail, May 2002; US Bank (Lq) # Customers in Queue (Average) Retail
May 2002; US Bank
0
1
2
3
4
5
6
0 5 10 15 20 25 30
days
Av
era
ge
Nu
mb
er
in Q
ue
ue
# Customers in Queue Little's Law
Date 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16
38476 36144 37414 14194 7107 38587 33572 33220 33349 34009 14807 7141 41293 37653 36872 35266
W 5.6 3.3 8.0 10.0 6.1 2.6 1.9 1.9 1.8 3.8 10.5 4.8 3.5 3.4 6.8 4.2
* W 2.49 1.39 3.48 1.65 0.50 1.14 0.74 0.74 0.69 1.50 1.79 0.40 1.69 1.47 2.91 1.71
L 2.50 1.40 3.49 1.66 0.50 1.15 0.74 0.74 0.71 1.51 1.81 0.40 1.71 1.47 2.94 1.73
Date 17 18 19 20 21 22 23 24 25 26 27 28 29 30 31
35338 15533 7530 40534 35493 34070 34005 32512 13100 5909 1558 43980 38163 38416 40284
W 5.4 27.6 8.1 3.0 2.1 2.1 3.2 3.5 2.3 6.4 37.0 6.5 2.7 2.5 3.2
* W 2.19 4.96 0.71 1.41 0.87 0.83 1.25 1.30 0.35 0.44 0.67 3.29 1.18 1.11 1.47
L 2.20 4.97 0.71 1.42 0.88 0.84 1.27 1.31 0.37 0.44 0.67 3.30 1.20 1.12 1.48
2
Little’s Law for Retail calls, August 16th, 2001: US Bank
: Throughput Rate, Retail, August 16th
, 2001; US Bank Arrivals to queue Retail
16 August 2001
0
500
1000
1500
2000
2500
7:00 9:00 11:00 13:00 15:00 17:00 19:00 21:00 23:00
Time (Resolution 30 min.)
Number of cases
W: Average Waiting Time, Retail, August 16th
, 2001; US Bank Average wait time(waiting) Retail
16 August 2001
0
10
20
30
40
50
60
70
7:00 9:00 11:00 13:00 15:00 17:00 19:00 21:00 23:00
Time (Resolution 30 min.)
Means, Seconds
L: Average Queue Length, Retail, August 16th
, 2001; US Bank # Customers in Queue (Average) Retail
August 16, 2001; US Bank
0
5
10
15
20
25
7:00 9:00 11:00 13:00 15:00 17:00 19:00 21:00 23:00
time (resolution 30 min)
Av
era
ge
Nu
mb
er
in Q
ue
ue
# Customers in Queue Little's Law
Time 7:00 7:30 8:00 8:30 9:00 9:30 10:00 10:30 11:00 11:30 12:00 12:30 13:00 13:30 14:00 14:30 15:00
443 639 987 1291 1998 2166 2278 2231 2158 2135 2000 1408 1311 1303 1323 1285 1340
W 1.7 3.2 1.2 1.5 2.4 2.8 2.4 2.6 2.0 1.3 1.3 0.8 1.0 1.0 0.8 0.8 1.5
* W 0.42 1.14 0.68 1.06 2.72 3.42 3.01 3.18 2.44 1.55 1.47 0.64 0.72 0.72 0.62 0.59 1.09
L 0.42 1.14 0.68 1.06 2.72 3.40 3.02 3.17 2.41 1.59 1.48 0.64 0.72 0.72 0.62 0.57 1.11
Time 15:30 16:00 16:30 17:00 17:30 18:00 18:30 19:00 19:30 20:00 20:30 21:00 21:30 22:00 22:30 23:00 23:30
1258 1235 1157 942 788 752 803 619 485 437 421 386 336 311 274 251 193
W 3.5 3.6 15.8 4.2 2.4 4.9 51.9 10.0 3.5 1.7 1.3 2.1 3.3 1.4 2.0 14.3 32.6
* W 2.422 2.45 10.2 2.173 1.06 2.05 23.16 3.43 0.95 0.41 0.314 0.44 0.62 0.24 0.30 2.00 3.50
L 2.37 2.49 10.17 2.16 1.07 1.94 23.11 3.59 0.95 0.40 0.31 0.45 0.62 0.24 0.30 1.83 3.63
3
Little’s Law for Private calls, May 4th, 2004: Israeli Telecom
: Throughput Rate, Private, May 4th
, 2004; Israeli TelecomArrivals to queue Private
4 May 2004
0
100
200
300
400
500
600
7:00 9:00 11:00 13:00 15:00 17:00 19:00 21:00 23:00
Time (Resolution 30 min.)
Number of cases
W: Average Waiting Time, Private, May 4th
, 2004; Israeli Telecom Average wait time(all) Private
4 May 2004
0
10
20
30
40
50
60
70
80
90
7:00 9:00 11:00 13:00 15:00 17:00 19:00 21:00 23:00
Time (Resolution 30 min.)
Means, Seconds
L: Average Queue Length, Private, May 4th
, 2004; Israeli Telecom # Customers in Queue (Average) Private
May 4, 2004; Israeli Telecom
0
5
10
15
20
25
7:00 9:00 11:00 13:00 15:00 17:00 19:00 21:00 23:00
time (resolution 30 min)
Av
era
ge
Nu
mb
er
in Q
ue
ue
# Customers in Queue Little's Law
Time 7:00 7:30 8:00 8:30 9:00 9:30 10:00 10:30 11:00 11:30 12:00 12:30 13:00 13:30 14:00 14:30 15:00
84 133 245 341 368 417 397 447 429 474 505 455 513 451 426 418 437
W 0.5 0.4 0.8 47.9 48.8 61.5 11.3 55.6 40.0 81.9 15.8 15.3 32.6 10.8 31.3 2.9 22.5
* W 0.02 0.03 0.11 9.08 9.98 14.24 2.50 13.81 9.52 21.57 4.44 3.86 9.29 2.70 7.40 0.67 5.46
L 0.02 0.03 0.11 8.47 10.59 14.24 1.85 14.40 9.42 21.57 4.61 3.86 8.98 2.92 7.49 0.67 5.46
Time 15:30 16:00 16:30 17:00 17:30 18:00 18:30 19:00 19:30 20:00 20:30 21:00 21:30 22:00 22:30 23:00 23:30
449 452 458 486 534 508 557 433 450 448 408 389 347 285 274 208 147
W 7.2 5.1 17.1 25.8 15.6 18.9 35.8 3.0 18.3 59.5 24.1 47.7 23.6 32.5 65.3 59.5 1.3
* W 1.78 1.28 4.36 6.96 4.62 5.34 11.07 0.723 4.572 14.8 5.45 10.31 4.543 5.14 9.93 6.879 0.10
L 1.78 1.22 4.42 6.96 4.62 5.27 11.09 0.78 3.89 15.32 5.39 10.44 4.64 5.14 9.82 6.99 0.10
4
Little’s Law for Telesales calls, October 10th, 2001: US Bank
: Throughput Rate, Telesales, October 10th
, 2001; US Bank
Arrivals to queue Telesales
10 October 2001
0
50
100
150
200
250
300
350
400
450
500
7:00 9:00 11:00 13:00 15:00 17:00 19:00 21:00 23:00
Time (Resolution 30 min.)
Number of cases
W: Average Waiting Time, Telesales, October 10th
, 2001; US Bank
Average wait time(all) Telesales
10 October 2001
0
100
200
300
400
500
600
700
7:00 9:00 11:00 13:00 15:00 17:00 19:00 21:00 23:00
Time (Resolution 30 min.)
Means, Seconds
L: Average Queue Length, Telesales, October 10th
, 2001; US Bank # Customers in Queue (Average) Telesales
October 10, 2001; US Bank
0
20
40
60
80
100
120
140
160
180
200
7:00 9:00 11:00 13:00 15:00 17:00 19:00 21:00 23:00
time (resolution 30 min)
Av
era
ge
Nu
mb
er
in Q
ue
ue
# Customers in Queue Little's Law
Time 7:00 7:30 8:00 8:30 9:00 9:30 10:00 10:30 11:00 11:30 12:00 12:30 13:00 13:30 14:00 14:30 15:00
76 102 182 262 379 464 440 433 410 431 422 418 401 439 453 432 373
W 109.8 123.8 383.5 403.7 503.5 522.5 607.9 602.1 552.4 521.1 508.6 468.8 442.1 467.3 545.9 483.1 442.1
* W 4.63 7.01 38.77 58.76 106.01 134.69 148.60 144.84 125.82 124.77 119.23 108.86 98.48 113.98 137.39 115.93 91.61
L 4.28 6.91 31.73 54.36 96.50 140.70 168.10 174.34 166.14 146.13 154.48 137.47 118.29 121.44 144.07 146.01 119.83
Time 15:30 16:00 16:30 17:00 17:30 18:00 18:30 19:00 19:30 20:00 20:30 21:00 21:30 22:00 22:30 23:00 23:30
405 427 298 242 182 134 132 134 112 105 105 87 80 55 45 28 30
W 419.2 442.2 458.8 387.9 415.1 357.1 121.6 179.8 267.9 445.7 536.0 416.9 403.9 326.0 463.6 187.3 0.9
* W 94.31 104.89 75.96 52.15 41.97 26.58 8.92 13.38 16.67 26.00 31.27 20.15 17.95 9.96 11.59 2.91 0.02
L 107.86 101.22 111.60 82.93 42.23 32.32 10.57 13.24 18.67 21.07 32.50 24.10 20.33 10.69 11.13 4.35 0.02
5
Little's Law for Telesales calls, October 10th, 2001: US Bank (Revisited)
(Unbounded waiting, as apposed to being truncated at 30 min.)
: Throughput Rate, Telesales, October 10th
, 2001; US Bank
Arrivals to queue, Telesales
10 October 2001
0
50
100
150
200
250
300
350
400
450
500
07:00 09:00 11:00 13:00 15:00 17:00 19:00 21:00 23:00
Time (Resolution 30 min.)
Num
ber
of
cases
W : Average Waiting Time - Untruncated, Telesales, October 10th, 2001; US Bank
Average wait time-Untruncated Telesales
10 October 2001
0
100
200
300
400
500
600
700
07:00 09:00 11:00 13:00 15:00 17:00 19:00 21:00 23:00
Time (Resolution 30 min.)
Means,
Seconds
L: Average Queue Length, Telesales, October 10th
, 2001; US Bank
# Customers in queue (Average), Telesales
10 October 2001
0
25
50
75
100
125
150
175
200
07:00 09:00 11:00 13:00 15:00 17:00 19:00 21:00 23:00
Time (Resolution 30 min.)
Avera
ge N
um
ber
in Q
ueue
# Customers in Queue Little's Law
Time 7:00 7:30 8:00 8:30 9:00 9:30 10:00 10:30 11:00 11:30 12:00 12:30 13:00 13:30 14:00 14:30 15:00
76 102 182 262 379 464 440 433 410 431 422 418 401 439 453 432 373
W 109.8 123.8 383.5 486.7 591.3 667.4 725.3 734.4 690.3 586.9 593.4 515.8 483.3 550.3 626.7 549.6 530.9
* W 4.63 7.01 38.77 70.84 124.51 172.05 177.30 176.65 157.23 140.54 139.12 119.77 107.66 134.21 157.73 131.90 110.01
L 4.28 6.91 31.73 54.36 96.50 140.70 168.10 174.34 166.14 146.13 154.48 137.47 118.29 121.44 144.07 146.01 119.83
Time 15:30 16:00 16:30 17:00 17:30 18:00 18:30 19:00 19:30 20:00 20:30 21:00 21:30 22:00 22:30 23:00 23:30
405 427 298 242 182 134 132 134 112 105 105 87 80 55 45 28 30
W 446.6 511.4 602.1 387.9 415.1 357.1 121.6 179.8 330.7 458.6 536.0 416.9 403.9 326.0 463.6 187.3 0.9
* W 100.49 121.32 99.67 52.15 41.97 26.58 8.92 13.38 20.58 26.75 31.27 20.15 17.95 9.96 11.59 2.91 0.02
L 107.86 101.22 111.60 82.93 42.23 32.32 10.57 13.24 18.67 21.07 32.50 24.10 20.33 10.69 11.13 4.35 0.02
5.1
Little’s Law for Russian calls, May 23rd, 2005: Israeli Telecom
: Throughput Rate, Russian, May 23rd
, 2005; Israeli Telecom
Arrivals to queue Russian
23 May 2005
0
10
20
30
40
50
60
70
80
90
7:00 9:00 11:00 13:00 15:00 17:00 19:00 21:00 23:00
Time (Resolution 30 min.)
Number of cases
W: Average Waiting Time, Russian, May 23rd
, 2005; Israeli Telecom Average wait time(all) Russian
23 May 2005
0
50
100
150
200
250
7:00 9:00 11:00 13:00 15:00 17:00 19:00 21:00 23:00
Time (Resolution 30 min.)
Means, Seconds
L: Average Queue Length, Russian, May 23rd
, 2005; Israeli Telecom # Customers in Queue (Average) Russian
May 23, 2005; Israeli Telecom
0
1
2
3
4
5
6
7
8
9
10
7:00 9:00 11:00 13:00 15:00 17:00 19:00 21:00 23:00
time (resolution 30 min)
Av
era
ge
Nu
mb
er
in Q
ue
ue
# Customers in Queue Little's Law
Time 7:00 7:30 8:00 8:30 9:00 9:30 10:00 10:30 11:00 11:30 12:00 12:30 13:00 13:30 14:00 14:30 15:00
12 12 22 46 59 59 36 52 43 56 81 61 80 46 67 56 50
W 16.9 1.3 11.4 166.0 148.9 88.7 7.9 27.4 0.7 3.9 20.3 74.9 125.4 46.3 41.4 47.9 37.7
* W 0.11 0.01 0.14 4.24 4.88 2.91 0.16 0.79 0.02 0.12 0.91 2.54 5.57 1.18 1.54 1.49 1.05
L 0.08 0.04 0.14 3.97 4.88 3.18 0.16 0.79 0.02 0.12 0.88 2.57 5.32 1.44 1.36 1.67 1.00
Time 15:30 16:00 16:30 17:00 17:30 18:00 18:30 19:00 19:30 20:00 20:30 21:00 21:30 22:00 22:30 23:00 23:30
57 52 62 70 75 79 68 68 60 55 55 56 56 43 27 14 6
W 65.7 22.4 79.2 156.6 118.6 139.3 143.6 150.2 179.2 151.3 209.5 219.5 224.7 88.4 33.7 107.0 0.7
* W 2.08 0.65 2.73 6.09 4.94 6.11 5.42 5.68 5.97 4.62 6.40 6.83 6.99 2.11 0.51 0.83 0.00
L 2.13 0.62 2.34 5.51 5.82 5.61 6.03 3.04 8.63 4.34 5.99 7.18 6.85 2.61 0.51 0.83 0.00
6
Motivation 1: λ customers/hour, each charged $1/hour while remaining in the system.Then λ × W is the rate at which the system generates cash which, in turn, “clearly”equals L.
Motivation 2: If there is always a single customer (L = 1) in the system, and everycustomer remains in the system W hours on average (customers arrive one after the other),then λ = 1/W is clear. When there are L in the system, on the average, λ = L/W is justone leap of faith.
Hint at a stochastic version: think of i.i.d sojourn times and use the Strong Law of LargeNumbers.
Motivation 3 (finite horizon): Consider a system that operates in a finite horizon(interval of time), and think of customers that arrive and leave (discrete units).Interval length is T .
Note: Little’s Law will work if the system is empty at time 0, and empty at time T .
Motivation 4 (work in cycles): Consider a system that operates in cycles of equaldurations and has the same statistical behavior during each cycle.Cycle length is T .
Note: Little’s Law will work if the system is at the same level (not necessarily 0) at thebeginning and at the end of the cycle, and if all the customers that are in the system atthe beginning of the cycle leave the system before the end of the cycle.This happens, for instance, if there is a moment during the cycle when the system becomesempty (see Example 10 on page 12, or Serfozo’s treatment on page 18).
Graphical representation. N customers flow though the system during a cycle.A customer is represented by a rectangle of unit height, whose length equals the time thecustomer spends in the system (see the figure below).
7
1
2
time
A(T)=N # customers
0
W7
1
2
time
L(t)
0
L(0)=0 L(T)=0
T0
0 T
Note: Vertical cut = number of customers in the system.
S = Shaded area (units: customer × hours), measures total waiting.
W =S
N, divides waiting among customers
(the customer’s view).
L =S
T, divides waiting over time
(the manager’s view).
λ =N
T, implies L = λ ·W
(which adds the server’s view).
8
More formally:
Area =∫ T
0L(t)dt =
A(T )∑
k=1
Wk (whenever L(0) = L(T ) = 0) .
DefineL = 1
T
∫ T0 L(t)dt
W = 1A(T )
∑A(T )k=1 Wk
λ = A(T )T
=⇒ L = λW
Examples
1. Management Strategy and Control: Only two out of the three λ, L, W deter-mine a strategy; the third is implicitly determined.
Scenario: λ = demand (projected), W = goal (set), L = means of monitoring W .
2. Inventory Management
L = average inventory;W = average time in inventory;λ = average throughput rate.
The quantity 1/W = λ/L is often referred to as the turnover ratio.
Scenario: A fast food restaurant processes on the average 5000 lbs. of hamburgerper week. The observed inventory level of raw meat, over a long period of time,averages 2500 lbs.Data:L = 2500 lbs., λ = 5000 lbs./week;W = L/λ = 2500/5000 = 1/2 week is the average time spent by a pound of meatin inventory; 1/W = 2 times per week is the inventory turnover ratio.
3. Services Management
L = average number of customers;W = average customer’s delay;λ = average customers’ throughput rate.
Scenario:3.1 A restaurant processes on average 1500 customers per day (=15 hours). Onaverage, there are 50 customers waiting to place an order, waiting for an order toarrive or eating.
9
λ = 1500 customers/day = 100 customers/hour;L = 50 customers;W = L/λ = 50/100 = 1/2 hours, average time in the restaurant.
3.2 Out of the 50 customers, 40 customers on the average are eating.
λ = 100, L = 50− 40 = 10 customers at the service counter;W = L/λ = 10/100 hours = 6 minutes average wait at the counter.
4. Workforce Management: A certain Japanese company has 36,000 employees,20% of whom are women. The average term of employment for a woman is 37months, whereas for men it is 200 months. Assume that the total employment leveland the mix of men and women are stable over time.Lw = average number of women in system = 36, 000× 0.2 = 7, 200 women.λw =
720037
= 194.6 women/month is the average number of new women employeeshired per month.Lm = 36, 000× 0.8 = 28, 800 men,λm =
28,800200
= 144 men/month.Thus, the company hires an average of 194.6 + 144 = 338.6 new employees permonth, or equivalently, 338.6× 12 = 4063.2 new employees per year.
4063.2
36, 000= 0.1128× 100% = 11.28% labor turnover during a year
= 2.82% turnover during a 3-month period
(compared with 40% at fast food, for example,
and about 100% in many Call Centers).
5. Little’s Law in Transportation Science
5.1 Cars flow through a highway. We wish to relate the 3 quantities: HighwayDensity, Flow Rate, Car Velocity.
System = 1 km of highway
L = avg. number of cars in system (1 km) = Density
λ = Flow, in avg. number of cars per hour (in = out = through)
W = avg. time to travel 1 km, say in hours
⇒ 1W
= Velocity, in km/hr; denote it V .
By Little’s Law:
Density =Flow
Velocity
10
5.2 Cars flow over a single-loop detector, that can measure Occupancy = % timethere is a car above the detector;Flow = avg. # cars per hour.
System = Detector
L = Occupancy (E [Indicator])
λ = Flow
W = ℓV
time to traverse one detector
where V = Velocity, ℓ = av. car length.
By Little’s Law:
Occupancy =Flow × car-length
Velocity× 100%
Note: Occupancy = Density × car-length.
5.3 Empirically, transportation flow reveals the following “flow vs. occupancy”relation (”flow vs. density” would look the same):
From “Causes and Cures of Highway Congestion”,Chao Chen, Zhanfeng Jia and Pravin Varaiya, 2001
Speed = 60 mph
Maximum
Flow
Free Flow,
100 %
Efficiency
Congestion,
Inefficient
Operation
Depth of
Congestion
Critical
Occupancy Level
5:30 am
6:45 am9:00 am
.
Figure 6: Flow vs. occupancy on a section at postmile 37.18 on I-10W, midnight to noon
on October 3, 2000.
11
From “The freeway congestion paradox”,Chao Chen and Pravin Varaiya, 2001.
6:10
12:00
11:00
7:00
6:00
5:30
5:25
5:104:00
Figure 1 Congestion begins at 5:20 am. By 7:00 am, both speed and flow
have dropped dramatically.
From “Causes and Cures of Highway Congestion”,Chao Chen, Zhanfeng Jia and Pravin Varaiya, 2001
Occupancy (%)
Flo
w (
VP
H)
Maximum
Flow
CongestionFree Flow
60 mph
Depth ofCongestionRecovery Phase
Critical Occupancy
Level
Figure 7: Model of congestion. If occupancy is maintained below critical level, section
12
The critical occupancy is the occupancy-level beyond which congestion startsbuilding up.
Note: For each point on the curve, the slope of the line connecting it with theorigin is proportional (equal) to the velocity; indeed:
Flow
Occupancy=
Velocity
Car-lenght;
Flow
Density= Velocity
This explains the (almost) straight line to the left of the critical occupancy: its slopeis the congestion-free velocity (60 miles/hr in California highways).
Note: with a single-loop detector covering N lanes, and assuming that traffic isevenly divided among the lanes (though typically this is not the case), the Occu-pancy should be calculated by using Flow/N , instead of merely Flow.
6. Abandonment: Calls arrive at a call center at rate α. A fraction Pab of themabandons due to impatience. Individual abandonment rate is θ.
Let Lq, Wq denote, respectively, the average number of customers waiting to beserved, and the average queueing time (waiting for service). Then
α · Pab = θ · Lq.
But Lq = αWq, hencePab = θ ·Wq.
Thus, the abandonment rate is proportional to the average waiting time. This hasbeen confirmed empirically for new (potential) customers. Indeed, (Pab, Wq) wereobserved and scatterplotted. The slope (via regression) can be used to estimatecustomers’ (average) patience.
13
The data is from a bank call center. Each point corresponds to a 15-minute period ofa day (Sunday to Thursday), starting at 7:00am, ending at midnight, and averagedover the whole year of 1999.
• Why a positive y-intercept?
• What about experienced customers?
7. Loan Application Flow from Managing Business Process Flows, by R.Anupindi,S.Chopra, S.Deshmukh, J.Van Mieghem, E.Zemel, Chapter 3. (In Recitation.)
8. Process Flow: A supermarket receives from suppliers 300 tons of fish over thecourse of a full year, which averages out to 25 tons per month. The average quantityof fish held in freezer storage is 16.5 tons.On average, how long does a ton of fish remain in freezer storage between the timeit is received and the time it is sent to the sales department?
W = L/λ = 16.5/25 = 0.66 months, on average, is the period that a ton of fishspends in the freezer.How does one get L = 16.5? This comes out of the following inventory build-updiagram by calculating the area below the graph:
Inventory/Queue Build-up Diagram.
0
2
4
6
8
10
12
14
16
18
20
22
24
26
28
30
0 1 2 3 4 5 6 7 8 9 10 11 12
time (months)
invento
ry L
(t)
17× 412
+ 24× 412
+ 12× 212
+ 5× 212
=
173
+ 8 + 2 + 56
= 16.5.
14
9. Shop Flow Control: JIT (Just-In-Time) principles advocate limiting the numberof active jobs (those that have been released to the shop floor).
Scenario: A job shop with Lold = 300 active jobs, Wold = 20 weeks, λold = 15jobs/week.
Management familiar with Little’s law and JIT principles imposes Lhope ≤ 150 active
jobs, in anticipation of λhope = 15 jobs/week, Whope =Lhopeλhope
≤ 10 weeks.
It turns out, however, that
Lactual ≤ 150, Wactual = 20 weeks, λactual =LactualWactual
≤ 7.5 jobs/week.
What is lacking? Congestion curves (Strategic Q-theory): later.
10. Assembly Lines
L = average WIP;W = average production time of a unit;λ = average production rate.
The quantity 1λ
is often called a cycle time.
Scenario: Watches are produced by an assembly line consisting of 20 workers. Theline produces 4 watches per minute.Data:L = 20 watches,λ = 4 watches/min = 4
60watches per second,
1/λ = 15 seconds cycle time is the average time between consecutive assemblycompletion (a watch is assembled every 15 seconds);W = L/λ = 20/4 = 5 minutes assembly time of a watch.
15
11. Insurance: An insurance company processes 10,000 claims per year. The averageprocessing time of a claim is 3 weeks. Assuming 50 weeks per year, we have
λ = 10,000 claims/year = 200 claims/week;W = 3 weeks;L = λW = 3× 200 = 600 claims backlog on the average.
12. Cash Flow (Accounts Receivable): A company sells 300M$ worth of finishedgoods per year. The average amount of accounts receivable is 45M$.
λ = 300M$/year;L = 45M$;W = L/λ = 45/300 = 0.15 years = 1.8 months.
So it takes, on average, 1.8 months from the time a customer is billed until the timepayment is received.
13. Cash Flow: A paper mill processes 40M$ of raw material per year. Direct conver-sion costs are 20M$ per year. Average inventory cost (raw material + conversion)is 5M$.
λ = 40+20 = 60M$ year;L = 5M$;W = 5/60 = 1/12 years = 1 month.
Thus, there is an average lag of one month between the time a dollar enters thesystem in the form of raw material (example: logs) or conversion cost (example:chemicals), and the time it leaves the system in the form of finished goods (example:paper).
14. Loss Queues: Customers arrive at a service facility at rate α. A fraction β of themare blocked (do not enter). The others join a queue and wait until being served.Assuming existence of averages and flow conservation, letτ = average service time,ν = long-run time-average number of customers in service. (Think G/G/S/N.)
Thenβ = 1−
ν
ατ· (Any three of (α, β, τ, ν) determine the fourth.)
By: system = servers, L = ν, W = τ, λ = α(1− β).
Alternative scenario: An Internet site. (Think G/G/s/s.)S = number of servers. Then ρ = ν/S is servers’ utilization and
β = 1−ρS
ατ,
where (S, τ) are known, ρ is measured, hence α and β determine each other. Onecould also use this to determine an appropriate S, given service level.
16
15. Little’s Law in the “Production of Justice”.
• 5 Judges “process” 3 types of files.
• System = “drawer” of a Judge.
Judges: Performance Analysis (λ, W )
3
001
3
0
01
01
0
3
01
3
0
03
01
0
1
2
3
4
5
6
7
8
9
10
0 5 10 15 20 25 30
(6.2, 7.4) (13.5, 7.4)
(26.3, 4.5)
(12, 4.9)
(7.2, 4.6)
.
.
.
.
.
Judges: Performance Analysis
Case Type 0 Judge1
Case Type 01 Judge2
Case Type 3 Judge3
Judge4
Judge5
Avg.
Month
s - W
Avg. Cases / Month -
Judges: Performance Analysis (L)
(6.2, 7.4) (13.5, 7.4)
(26.3, 4.5)
(12, 4.9)
(7.2, 4.6)
3
001
3
0
01
01
0
3
01
3
0
03
01
0
1
2
3
4
5
6
7
8
9
10
0 5 10 15 20 25 30
.
.
.
.
.
45 100
118
59
33
Case Type 0 Judge1
Case Type 01 Judge2
Case Type 3 Judge3
Judge4
Judge5
Avg.
Month
s - W
Avg. Cases / Month -
17
Little's Law over a Finite Horizon
Although Little's Law seems quite straightforward, one should actually pay close
attention to the way one measures the quantities that it relates. Indeed, when using
different methods for computing ,E W , Little's Law may well result in different L .
So far, we have used the following method to compute ,E W : for every specific
interval, a call was considered to arrive in that interval (thus included in the
computation of ) if it arrived to the queue during that interval, in which case its
entire waiting time (no matter how long) was considered to "belong" to that interval.
This method is also used in a software that we shall be using later in the course,
SEEStat.
The last two examples do not adhere to the assumptions of a finite-horizon-Little's-
law, in some of the intervals. Due to very heavy loads during these days, not all
incoming calls in an interval were answered in that interval (i.e. they were
"transferred" to the next interval, still pending to be answered). The result is adjacent
intervals that very much affect one another.
In order to adjust Little's law to hold under the previously mentioned conditions (or,
for that matter, under any condition) one reasonable approach is to conceptually force
the 0-to-0 assumption on the given interval. In other words, we should pretend the
system is indeed empty, both at the beginning of the interval and at its end.
Practically, this implies adjusting and WE . To elaborate:
Arrival Rate~
- We count all calls that spend some of their wait time within the
considered interval, i.e. both calls that arrived to the queue in the
interval and calls from previous intervals which still have not been
answered (instead of counting only the former, as done regularly by
SEEStat).
Average Waiting Time WE~
- Averaging over only the parts of the waiting time that
intersected the considered time-interval (instead of
considering the whole waiting time of calls that
arrived during the interval, as done regularly by
SEEStat).
Practically, we treat any call that entered queue prior to the interval's start as if it
arrived exactly at the interval's beginning; and any call that has not left the queue until
the interval's completion as if it left the queue exactly at interval's end.
Let us examine empirically the effects of this approach:
17.1
Note: Recall that waiting-times are not truncated to 30 minutes, the latter being
SEEStat standard.
~
WE
WE~
17.2
Little’s Law: Remaining Work in System
Consider a system in which entities are characterized by their arrival time and sojourn time.
Fix a finite time interval 1 2,t t , and focus on entities for which their sojourn time intersects
1 2,t t . These can be divided into 3 types:
1. Entities that both arrived and departed within the interval.
2. Entities that arrived within the interval and departed after it was over.
3. Entities that arrived prior to the interval's start but spent some time within the
interval.
Introduce a new function , 0V V t t , where V t is the remaining work in the
system at time t .
Formally, V t is the sum of all remaining sojourn times of entities which are in the system
at time t . It follows that a new arrival increases V t by the sojourn time of the arriving
entity, whereas with every passing time unit, the value of V t decreases by that time unit
multiplied by the number of entities in the system.
For example, let us examine the interval 1 2,t t in the following system:
Figure 1: The 3 types of entities, and V t over .
In Figure 1, every entity is represented by a rectangle with length equal to the entity's
sojourn time and width of 1 (i.e. each rectangle's area equals its length). A total of 6 entities
pass through the system during the interval. However, only two of the entities are in the
system at time 1t ; and only three of the entities are in the system at time 2.t Therefore only
these entities are considered when computing1V t and 2V t respectively.
Computing1 ,V t we see that at time 1t one of the entities has 3 time units of sojourn time
left and the other has 9 time units of sojourn time left. Hence 1 3 9 12,V t which is
represented by the dashed area. In the same manner,2 2 2 1 5,V t represented by
the dotted area.
Type 1 entity
Type 2 entity
Type 3 entity
2t1t
2
17.3
Claim:
Let 1 2t t .
a. If1 2V t V t then on the interval 1 2[ , ), .t t L W
b. The absolute difference between L and W is given by: 2 1
2 1
( ) ( )V t V tW L
t t
Explanation (for 'b' only as it is a generalization of 'a'):
Recall that in an interval which starts with an empty system and ends with an empty system,
.L E W
Let us force this condition on some interval [t1,t2) by treating all entities of Type 3 from
above (i.e. entities which arrive before t1) as if they arrived exactly at time t1; and treating all
entities of Type 2 from above (i.e. entities which depart after t2) as if they depart exactly at
time t2 ).
We now compute L according to Little's Law:
#entities that spent sometimein the systemL
1 2
2 1
,
#
sojourn times enclosed within t t
t t entities that spent sometimein the system
E W
Next, recall how SEEStat computes the arrival rate and average waiting time and finds their
product:
1 2# ,Arrivals during t tE W
1 2
2 1 1 2
,
# ,
sojourn times of entities toarriveduring t t
t t Arrivals during t t.
But:
1 2
1 2 1 2
,
,
sojourn times enclosed within t t
sojourn times of entities to arrive during t t V t V t
Hence:
1 2
2 1
V t V tL E W
t t
___________
Let us return to the example above:
4;
7
2 4 5 3 142
4 4E W E W
6 3 7 2 4 3 2 213
7 6 7L
And indeed:
2 1
2 1
5 121.
7
V t V t
t t
17.4
userTypewritten Text
userText Box(Abir Koren, 2011):
userRectangle
userLine
userText Box
Let us examine an empirical chart of V t :
V(t): Remaining Work in System, Retail, August
16th, 2001; US Bank
0
500
1000
1500
2000
2500
07:0
0
08:0
0
09:0
0
10:0
0
11:0
0
12:0
0
13:0
0
14:0
0
15:0
0
16:0
0
17:0
0
18:0
0
19:0
0
20:0
0
21:0
0
22:0
0
23:0
0
00:0
0
Time (resolution 30 sec.)
seconds
Figure 2: Remaining Work in System throughout a day
The above graph, repeated appropriately, enables one to identify the intervals over which
Little's Law prevails. For example, let us focus on the interval 18 : 00,18 : 30 . We then get:
V(t): Remaining Work in System, Retail,
August 16th, 2001, 18:00-18:30; US Bank
0
20
40
60
80
100
120
140
160
180
200
18:00 18:05 18:10 18:15 18:20 18:25 18:30
Time (resolution 30 sec.)
seconds
Figure 3: Remaining Work in System throughout a half-hour interval
From Figure 3 we get V(18:00)=5 and V(18:30)=189, so there is a slight difference between
the actual L and that calculated using Little's Law.
To be precise,
17.5
18 : 30 18 : 00 189 50.102
18 : 30 18 : 00 1800
V VL .
This matches the difference presented previously, when putting L against W .
Our analysis may well explain why, on this day, we see that Little's Law 'works properly'.
Finally, we examine an interval where we previously found Little's Law does NOT work:
Consider October 10th2001 in Telesales of US Bank call center, during 11:00 11:30. It is
possible to compute the remaining work at both the beginning and the end of the interval.
We get (note how loaded the call center is):
11: 00 170,670 sec,V
11: 30 154,634 sec.V
Therefore:
154,634 170,670 160368.91.
1800 1800W L
Comparing this result to the measured difference in the corresponding graph above (note
that you should compare it to the difference found in the graph with untruncated waiting
times), we get the exact same difference.
17.6
Little’s Formula. Deterministic Model.Infinite Horizon (Stidham’s formulation)
Averages: L = λ ·W
L = number of units in the system;λ = throughput rate;W = sojourn time.
Rigorous formulation
The system is characterized by {(An, Dn), n ≥ 1}, where
An – time of the nth arrival.Dn – departure-time of the nth arrival.
0 ≤ An ≤ An+1 An ≤ Dn
Extension (Brumelle)
Associate with every n a corresponding function fn(t), t ∈ [An, Dn].Assume that fn(t) = 0, if t 6∈ [An, Dn].
Interpret fn(t) as income-rate at time t (average income per unit of time).
Define
GN =1
N
N∑
n=1
∫ ∞
0fn(t)dt (average income per customer);
HT =1
T
∫ T
0
∞∑
n=1
fn(t)dt (average income per time unit).
Then,limN↑∞
GN = G exists ⇔ limT↑∞
HT = H exists,
in which case
H = λ · G
19
Stochastic example: M/M/1
Model
Birth-and-death process, birth rate λ, death rate µ.
Assumption
ρ = λµ
< 1, answers existence of stationary (limit) distribution π:
πk = (1− ρ)ρk, k = 0, 1, 2, . . . (geometric distribution).
L =∞∑
k=0
kπk =ρ
1− ρ=
λ
µ− λ.
Little: W =1
λL =
1
µ− λ=
1
µ
1
1− ρ.
Check out:
W = (PASTA) =∞∑
k=0
E[sojourn time/k customers in system] πk
= (memoryless property) =∞∑
k=0
[
k
µ+
1
µ
]
πk
=1
µ+
1
µL = · · · =
1
µ
1
1− ρ.
System = queue: Lq = λ Wq, Wq = W −1µ
= 1µ
ρ
1−ρ.
Lq – queue-length,Wq – waiting-time.
System = server:
L = λ ·1
µ,
L = ρ = probability that the system is not empty (customer waits)= proportion of time when the server is busy (traffic intensity).
20
Stochastic Model (à la Serfozo1 )
{(An, Dn), n ≥ 1} random variables; limits are a.s. (with probability 1)
e.g. λ = limt↑∞
1
tA(t) a.s. ;
1
T
∫ T
0L(t)dt→ L a.s., as T ↑ ∞, etc.
“Periodic” System (Serfozo, pg. 17)
A system is periodically empty if there exist strictly increasing random times τn ↑ ∞,such that
1. τn ∼ τn+1 i.e. limn↑∞
τn+1τn
= 1 a.s. (implied, for example, by τn/n→ c).
2. For all n, there exists t ∈ [τn, τn+1) such that A(t) = D(t), i.e. L(t) = 0.
Theorem. If a system is periodically empty, the existence of any two positive limits outof (L, λ, W ) implies existence of the third, as well as the relation L = λW .
Typical application: τn starts a “cycle” (eg. empty system; state 7), which gives riseto a regenerative structure (eg. Markovian).
1Introduction To Stochastic Networks, Springer 1999, Chapter 5
21
Application of H = λG : Brumelle’s formula (1971), in Whitt, pg. 257.
Framework: a single queue (think of G/G/1 ; G/G/S is o.k. as well).
Characteristics of customer n: An arrival time;Wn waiting time (before service);Sn service time (Dn = An + Wn + Sn).
Let
fn(t) =
Sn An ≤ t < An + WnSn + Wn + An − t An + Wn ≤ t ≤ An + Wn + Sn = Dn0 otherwise.
fn(t) – Remaining work associated with customer n:
fn(t) = Sn while customer is waiting, then decreases at rate 1 while she is served.
G = limN↑∞
1
N
N∑
n=1
∫ ∞
0fn(t)dt = lim
1
N
N∑
1
(
SnWn +1
2S2n
)
= E(SWq) +1
2E(S2)
(assuming SLLN-behavior, which is o.k. if steady state exists).
H = limT↑∞
1
T
∫ T
0
∞∑
n=1
fn(t)dt = lim1
T
∫ T
0V (t)dt = E(V )
V (t) – Work load process (under FIFO in G/G/1, it is equal to virtual waiting time).
22
Brumelle’s Formula E(V ) = λ[
E(SWq) +12
E(S2)]
If, as usually assumed in G/G/1, service times are independent of waiting times,
E(V ) = λ[
E(S) E(Wq) +1
2(S2)
]
.
If ASTA = arrivals see time averages, as in the case of Poisson arrivals, and if we have asingle-server queue with FIFO, then
Vd= Wq .
⇒ E(Wq) = λ[
E(S)E(Wq) +1
2E(S2)
]
, which yields
E(Wq) =1
1− λE(S)·
1
2λE(S2)
=1
1− ρ
1
2λE(S2) (using λE(S) = ρ)
=ρ
1− ρ
1
2
E(S2)
E(S)
Khinchine Pollatcheck E(Wq) = E(S)ρ
1−ρ
1+C2S
2
(
C2 = σ2
E2
)
Hall, Formula (5.64) (M/G/1)
Kingman’s bound (G/G/1),EWqES
≈ρ
1− ρ
C2a + C2S
2- Upper bound,
Allen-Cunneen approx. - Asymptotically correct, as ρ ↑ 1Hall, Formula (5.70) (in Heavy Traffic)
The following picture is based on call center data.
Service Level vs. Availability.
0
50
100
150
200
250
300
350
0.00 0.10 0.20 0.30 0.40 0.50 0.60 0.70 0.80 0.90 1.00
Utilization (Hourly Avg.)
Av
era
ge W
ait
, sec
Average Wait Trendline
23
