# Recitation 5, Part 1 Queue Build-Up Diagram

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```					    Thus, the company hires an average of 194.6 + 144 = 338.6 new employees per
month, or equivalently, 338.6 × 12 = 4063.2 new employees per year.
4063.2
= 0.1128 × 100% = 11.28% labor turnover during a year
36, 000
= 2.82% turnover during a 3-month period
Service Engineering                                              Winter, 2009
(compared with 40% at fast food, for example
and about 100% in many Call Centers).
Recitation 5, Part 1 : Queue Build-Up Diagram
Process Flow: A supermarket receives suppliers 300 tons            ﬁsh over the
10. Process Flow: A supermarket receives from from suppliers 300oftons of ﬁsh
over the course which averages out to averages out to 25 tons per quantity
course of a full year,of a full year, which 25 tons per month. The averagemonth.
of ﬁsh held in freezer storage is 16.5 tons.
The average quantity of ﬁsh held in freezer storage is 16.5 tons.
On average, how long does a ton of ﬁsh remain in freezer storage between the time
On average, how long does a ton of ﬁsh remain in freezer storage between
it is received and the time it is sent to the sales department?
the time it 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 ﬁsh
W = L/λ freezer.
spends in the = 16.5/25 = 0.66 months, on average, is the period that a ton
of does one get = freezer.
How ﬁsh spends inLthe 16.5? This comes out of the following inventory build-up
How by calculating the 16.5? This comes out of the following inventory
diagramdoes one get L = area below the graph:
build-up diagram by calculating the area below the graph:

Inventory/Queue Build-up Diagram.
Inventory/Queue Build-up Diagram.
30
28
26
24
22
20
18
inventory L(t)

16
14
12
10
8
6
4
2
0
0   1   2   3    4       5       6        7    8   9   10   11   12

time (months)

4                       4                        2                 2
17 ×             12          + 24 ×      12   + 12 ×              12       + 5×     12   =
4                    4                        2                 2
×
17 17                        + 24 ×
+    8           + 12 ×
+    2                       + 5×
+   5         = 16.5
= 16.5.
3 12                                  12                       12           6    12

1
7
Service Engineering                                                  Winter, 2009

Recitation 5 - Fluid Models. Part 2.

Deterministic view

• λ(t) - instantaneous arrival rate at time t
• c(t) - instantaneous capacity of the system, at time t (maximal po-
tential processing rate)
• δ(t) - instantaneous processing rate at time t. Holds δ(t) ≤ c(t).
• Q(t) - total amount of material in the system, (being processed +
queued) at time t.

Assume that Q(0), λ(t) and δ(t) are given for all t ∈ [0, T ].

Then Q(t) is the solution of the nonlinear diﬀerential equation
d
Q(t) = λ(t) − δ(t),   Q(0) = Q0 ,   t ∈ [0, T ].
dt

2
Solving the Equation

The general solution may be very complicated.

How to ”create” or plot Q(t) = (Q(0), Q(t1 ), Q(t2 ), ..., Q(T ))?

Start with Q(0). Then for tn = tn−1 + ∆t,        n = 1, 2, ...

Q(tn ) = Q(tn−1 ) + λ(tn−1 ) · ∆t − δ(tn−1 ) · ∆t.

EXCEL software can be used.

Queueing Models. Staffing.

Each little tube (on the picture above) can be viewed as server.

Consider a queueing system with one class of customers and one service
station.

• λ(t) - instantaneous arrival rate at time t
• µ - service rate (service capacity of each server), constant in time.
• N (t) - number of servers in the system at time t
• Q(t) - total number of customers in the system, (being served +
queued) at time t.

The above implies
• c(t) = µ · N (t)
• δ(t) = µ · (Q(t) ∧ N (t)), where a ∧ b = min(a, b)
• Thus
d
Q(t) = λ(t) − µ · (Q(t) ∧ N (t)),    Q(0) = Q0 ,       t ∈ [0, T ].
dt

3
Adding the possibility to abandon queue

• θ - abandonment rate for each customer in queue.

• The equation for Q transforms to
d                                                       +
Q(t) = λ(t) − µ · (Q(t) ∧ N (t)) − θ · (Q(t) − N (t)) ,                                Q(0) = Q0 .
dt

Example.

Tele-SHOP is a commercial channel dealing with online sales, which is
operated by a call-center.
In order to increase proﬁts it was decided to place a short (30 sec) daily
It turns out that the arrivals to the call-center are well approximated by an
inhomogeneous Poisson process, with an arrival rate function (customers
per hour) given by the ﬁgure below:
1000

900

800

700
Arrival Rate

600

500

400

300

200

100

0
16:00     17:00   18:00   19:00   20:00   21:00   22:00   23:00   0:00
Time

Figure 1: Arrival rate function (customers per hour).

• The call center operates from 16:00-24:00.
• The service duration of each incoming call has an average of 12 min-
utes.

4
• The number of servers in the system is ﬁxed and equals N .
• Assume that no abandonment takes place.

Plot Q(t) for N = 0, 150, 200, .

5
Dynamics of Q(t) for N = 150
300

250

200

150

100

50

0
16:00   17:00   18:00   19:00     20:00     21:00      22:00   23:00   0:00
number in system   Staffing

Dynamics of Q(t) for N = 200
250

200

150

100

50

0
16:00   17:00   18:00   19:00     20:00     21:00      22:00   23:00   0:00
number in system   Staffing

On the second picture, (for N = 200), Q actually satisﬁes
d
Q(t) = λ(t) − µ · Q(t),                 Q(0) = 0.
dt

6
Dynamics of Q(t) for N = 0
2500

2000

1500

1000

500

0
16:00   17:00   18:00     19:00     20:00     21:00       22:00   23:00   0:00
number in system    Staffing

On the last picture, (for N = 0), Q(t) actually satisﬁes
t
Q(t) =            λ(t)dt.
0

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Now assume that waiting customers can abandon from the system and the
abandonment rate equals θ = 6 (i.e. each waiting customer abandons after
an average of 1/θ hours, if he was not admitted to service before).

Plot Q(t) for N = 150.

Dynamics of Q(t) for N = 150 with abandonment
200

180

160

140

120

100

80

60

40

20

0
16:00   17:00   18:00   19:00     20:00     21:00      22:00   23:00   0:00
number in system   Staffing

8
Introducing costs. Profit maximization.

Deﬁne also

• c - staﬃng cost rate 1 .

• r - service completion reward per customer.

• s - abandonment penalty per customer.
2
• h - waiting cost rate

The total proﬁt is
T
C (N ) =        [rµ(Qt ∧ Nt ) − (sθ + h)(Qt − Nt )+ − cNt ]dt.
0

It depends on the staﬃng function N = {N (t), 0 ≤ t ≤ T }.

How to choose the optimal staffing N = N (t) in order to
maximize the profit C (N ) of running the call center?

Exact solution is usually not available.
Excel Solver can be used in order to compute an approximate solution.

1
The units of c are monetary units per one unit of work (salary).
2
The units of h are monetary units, eg.               shekels, per unit of waiting time,   i.e.
shekels          .
waiting customer×time unit

9
Numerical examples.
1. Minimize the cost of running the call center by assuming N ﬁxed and

• an hour work of a server costs 48 shekel, i.e. c = 0.8.
• minute waiting of a customer costs 1 shekel, i.e. h = 1.
• no abandonments (θ = 0) and no rewards (r = 0).
∗
Solution: The optimal staﬃng N ∗ = 149 and the cost C N = 63, 534.
300

250

200

150

100

50

0
16:00   17:00   18:00   19:00     20:00     21:00      22:00   23:00       0:00
number in system   Staffing

2. In addition, assume that the staﬃng can be changed at each hour.
∗
Solution : The optimal cost C N = 23, 133.
200

180

160

140

120

100

80

60

40

20

0
16:00   17:00   18:00   19:00     20:00     21:00      22:00   23:00       0:00
number in system   Staffing

10
V-model

Assume that there are two classes of customers:
• VIP (class 1), with arrival rate λ1 (t),
• Regular(class 2), with arrival rate λ2 (t),
• Each server serves both classes, with rates µi for class i, (i = 1, 2)
• Deﬁne Qi (t) to be the total number of class - i customers in the system,
i = 1, 2.

1                           2

1                     2

N

Figure 2: The structure of the system.

Let Q(t) = Q1 (t) + Q2 (t). Assume Q(0) = 0.

• Is it possible to write the diﬀerential equation for Q(t) without any

Answer: No, since the routing policy is not speciﬁed.

11
A Routing Policy for V-model

Assume that the call center works in the preemptive-resume regime:

• at every moment a service to a customer can be interrupted (in this
case a customer goes back to queue of its class) and resumed at a later
time.
• VIP customers are high priority customers, which means that no reg-
ular customer can be in service while VIP customer is waiting.

Write the diﬀerential equation for Qi (t), i = 1, 2.

dQ1
= λ1 (t) − µ1 (Q1 ∧ N ),      Q1 (0) = 0,
dt
dQ2
= λ2 (t) − µ2 (Q2 ∧ (N − Q1 )+ ),      Q2 (0) = 0.
dt

12

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