Service Engineering Created: April 1997Last revision: April 2007

LITTLE’S LAW

A conservation law that applies to the following general setting:

Input: Continuous flow or discrete units (examples: granules of powder measured in tons,tons of paper, number of customers, $1000’s).

System: Boundary is all that is required (very general, abstract).

Output: Same as input, call it throughput.

Two possible scenarios:

• System during a “cycle” (empty → empty, finite horizon);

• System in steady state/in the long run (for example, over many cycles).

Quantities that are related via Little’s law (long-run averages, or time-averages):

• λ = rate at which units arrive(= long-run average rate at which units depart) = throughput-rate, whose units arequantity/time-unit or #/time-unit;

• L = inventory/quantity/number in the system(eg. WIP: Work-In-Process, customers);

• W = time a unit spends in the system = throughput time(eg. hours) = sojourn time.

Little’s Law L = λW

1

Little’s Law for Retail calls, May 2002: US Bank

1

Little’s Law for Retail calls, May 2002: US Bank

λ: Throughput Rate, Retail, May 2002; US Bank Arrivals to offered Retail Total

May2002

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W: Average Waiting Time, Retail, May 2002; US Bank (Wq)Average Waiting Time, Retail

May 2002; US Bank

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L: Average Queue Length, Retail, May 2002; US Bank (Lq) # Customers in Queue (Average) Retail

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Date 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16λ 38476 36144 37414 14194 7107 38587 33572 33220 33349 34009 14807 7141 41293 37653 36872 35266W 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

λ * 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 1.71L 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 31λ 35338 15533 7530 40534 35493 34070 34005 32512 13100 5909 1558 43980 38163 38416 40284W 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

λ * 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 1.47L 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

2

Little’s Law for Retail calls, August 16th, 2001: US Bank

λ: Throughput Rate, Retail, August 16th, 2001; US Bank Arrivals to queue Retail

16 August 2001

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W: Average Waiting Time, Retail, August 16th, 2001; US Bank Average wait time(waiting) Retail

16 August 2001

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L: Average Queue Length, Retail, August 16th, 2001; US Bank # Customers in Queue (Average) Retail

August 16, 2001; US Bank

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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 15:00λ 443 639 987 1291 1998 2166 2278 2231 2158 2135 2000 1408 1311 1303 1323 1285 1340W 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

λ * 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 1.09L 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 23:30λ 1258 1235 1157 942 788 752 803 619 485 437 421 386 336 311 274 251 193W 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

λ * 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 3.50L 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

3

Little’s Law for Private calls, May 4th, 2004: Israeli Telecom

λ: Throughput Rate, Private, May 4th, 2004; Israeli Telecom Arrivals to queue Private

4 May 2004

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W: Average Waiting Time, Private, May 4th, 2004; Israeli Telecom Average wait time(all) Private

4 May 2004

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L: Average Queue Length, Private, May 4th, 2004; Israeli Telecom # Customers in Queue (Average) Private

May 4, 2004; Israeli Telecom

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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 15:00λ 84 133 245 341 368 417 397 447 429 474 505 455 513 451 426 418 437W 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

λ * 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 5.46L 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 23:30λ 449 452 458 486 534 508 557 433 450 448 408 389 347 285 274 208 147W 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

λ * 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 0.10L 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

4

Little’s Law for Telesales calls, October 10th, 2001: US Bank

λ: Throughput Rate, Telesales, October 10th, 2001; US Bank Arrivals to queue Telesales

10 October 2001

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W: Average Waiting Time, Telesales, October 10th, 2001; US Bank Average wait time(all) Telesales

10 October 2001

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L: Average Queue Length, Telesales, October 10th, 2001; US Bank # Customers in Queue (Average) Telesales

October 10, 2001; US Bank

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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 15:00λ 76 102 182 262 379 464 440 433 410 431 422 418 401 439 453 432 373W 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

λ * 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 91.61L 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 23:30λ 405 427 298 242 182 134 132 134 112 105 105 87 80 55 45 28 30W 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

λ * 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 0.02L 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 Russian calls, May 23rd, 2005: Israeli Telecom

5

Little’s Law for Russian calls, May 23rd, 2005: Israeli Telecom

λ: Throughput Rate, Russian, May 23rd, 2005; Israeli Telecom Arrivals to queue Russian

23 May 2005

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W: Average Waiting Time, Russian, May 23rd, 2005; Israeli Telecom Average wait time(all) Russian

23 May 2005

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L: Average Queue Length, Russian, May 23rd, 2005; Israeli Telecom # Customers in Queue (Average) Russian

May 23, 2005; Israeli Telecom

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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 15:00λ 12 12 22 46 59 59 36 52 43 56 81 61 80 46 67 56 50W 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

λ * 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 1.05L 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 23:30λ 57 52 62 70 75 79 68 68 60 55 55 56 56 43 27 14 6W 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

λ * 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 0.00L 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).

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1

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A(T)=N # customers

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L(t)

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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

3

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.

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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 = 7200

37= 194.6 women/month is the average number of new women employees

hired per month.Lm = 36, 000× 0.8 = 28, 800 men,λm = 28,800

200= 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 = `V

time to traverse one detector

where V = Velocity, ` = 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,InefficientOperation

Depth ofCongestion

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 noonon October 3, 2000.

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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 (%)

Flow

(VPH

)

MaximumFlow

CongestionFree Flow

60 mph

Depth ofCongestionRecovery Phase

Critical OccupancyLevel

Figure 7: Model of congestion. If occupancy is maintained below critical level, sectionoperates at 100 % efficiency and speed is at 60 mph.

a dynamic model that exhibits such transitions. Measurements of the recovery phase showerratic fluctuations as in Fig. 6. The model in turn supports the following hypothesis abouttraffic behavior:

If a metering policy keeps occupancy below its critical level in every section,efficiency will be 100 %, speed will be maintained at 60 mph, and highwaycongestion will be prevented. A consequence of the metering is that vehicleswill be stopped at the ramps for some time.

We call this the ideal ramp metering (IMP) principle. The IMP feedback strategy is tomonitor the occupancy downstream of each on-ramp and to throttle the flow from the on-ramp whenever the occupancy exceeds its critical value. For a control-theoretic discussionof this policy, see [6].

The figures in Table 1 are computed as follows. For each highway section we calculatethe critical occupancy level from PeMS data, as in Fig. 6. We assume that the pattern ofdemand is unchanged. We now simulate the traffic flow using the model of Fig. 7 andthe IMP feedback strategy. In the simulation, vehicles will be held back at some ramps.We calculate the total time spent by the vehicles at the ramps. The ramp delay is 124,000vehicle-hours. There is a net savings of 280,000 vehicle-hours.

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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.

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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.

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).

10. 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.

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0 1 2 3 4 5 6 7 8 9 10 11 12

time (months)

inve

nto

ry L

(t)

17× 412

+ 24× 412

+ 12× 212

+ 5× 212

= 16.5

7

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 =Lhope

λhope≤ 10 weeks.

It turns out, however, that

Lactual ≤ 150, Wactual = 20 weeks, λactual =Lactual

Wactual

≤ 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

001

01

0

3

01

3

00

3

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

. Mon

ths -

W

Avg. Cases / Month - λ

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

001

01

0

3

01

3

00

3

01

0

1

2

3

4

5

6

7

8

9

10

0 5 10 15 20 25 30

.

.

.

..

45 100

118

59

33

Judges: Performance Analysis

Case Type 0 Judge1 Case Type 01 Judge2 Case Type 3 Judge3 Judge4 Judge5

Avg

. Mon

ths -

W

Avg. Cases / Month - λ

17

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 < ∞ .

Define:A(t) = number of arrivals until t;D(t) = number of departures until t;L(t) = number of units such that An ≤ t < Dn, i.e., number of units in the system;Wn = Dn − An, sojourn time of the nth unit in the system.

Theorem. Assume that

λ = limt↑∞

1

tA(t) = lim

t↑∞

1

tD(t), 0 < λ < ∞ .

Then

limN↑∞

1

N

N∑n=1

Wn = W exists ⇔ limT↑∞

1

T

∫ T

0L(t)dt = L exists,

in which case L = λW .

18

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

ρ1−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 (a 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+1

τn= 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 + Wn

Sn + Wn + An − t An + Wn ≤ t ≤ An + Wn + Sn = Dn

0 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

2S2

n

)= 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) + 1

2E(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) ρ1−ρ

1+C2S

2

(C2 = σ2

E2

)Hall, Formula (5.64) (M/G/1)

Kingman’s bound (G/G/1),EWq

ES≈ ρ

1− ρ

C2a + C2

S

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.)

Ave

rage

Wai

t, se

c

Average Wait Trendline

23