Cost minimization model

QUANTITATIVE ANALYSIS
FOR MANAGEMENT
PREPARED BY: ROWELL BALALA

TC = SC + WC
Where:
TC – Total cost per hour
SC – Service cost per hour
WC – Waiting cost per hour
General Expression for the total cost per unit time
to operate a queuing system
COST MINIMIZATION MODEL

Algebraic Expression for WC
cw = Cost of having one customer wait one
hour in the queue
Wcw = The average waiting cost per customer
λ = Arrival of customers per hour, in steady
state, we use L = λW to obtain the total waiting
cost per hour
WC = λ(Wcw) = (λW) cw = L cw

EXAMPLE #1: Freight Unloading
Workfor Metals purchases metals to be recycled
from several sources. These metals consist of steel
bars and sheets, aluminum shapes and iron plates.
Truckload shipments of scrap metals arrive at work
for in a more or less random pattern at an average
rate of one truckload per day. These trucks are
owned and operated by Workfor Metals. Each
shipment must be unloaded by a crew of workers. A
combination of unloading operations is used: simple
dumping use of crane and direct manual labor. A
crew of n workers can unload 0.8n trucks per day.

For each day that a truck is detained in unloading,
a cost of P300 is incurred (due to lost productive use
of the truck and hourly wages of the truck driver and
his assistant). Each worker in the unloading crew is
paid P105 per day.
Determine the optimal number of crew member to
hire to minimize the total cost per day. Assuming that
arrival and service rates are Poisson.

SOLUTION
Given:
λ = 1 (Average rate of truckload per day)
μ = 0.8n (Service rate)
n = The number of crew member
If n workers are on the crew, the service cost is given by:
SC = 105n (P/day wages)
The waiting cost is given:
cw = 300 (P/day/truck)
We can determine L (Expected average number of customers in the system)
with λ = 1 (Arrival of truck per day) and μ = 0.8n (Number of trucks a worker
could unload per day working alone)
L =
λ
μ −λ
=
1
0.8𝑛 −1
, recalling the algebraic expression of WC is L cw , therefore
WC = 300 x
1
0.8𝑛 −1
=
300
0.8𝑛 −1

Thus the total cost equation becomes
TC = 105𝑛 +
300
0.8𝑛 −1
If n = 2
TC = 105(2) +
300
0.8(2) −1
= P710/day
If n = 3
TC = 105(3) +
300
0.8(3) −1
= P529/day
If n = 4
TC = 105(4) +
300
0.8(4) −1
= P556/day
n = 3 is optimal, and the decision is to hire three crew members
only to minimize the total cost per day.

Trade – off between Service Cost and
Waiting cost
SC = 105n
WC =
𝟑𝟎𝟎
𝟎.𝟖𝒏 −𝟏
Cost (P/day)
TC = SC + WC
Number of crew
members, n
n = 3
P520
 As n is increased, the service
cost increases and the waiting
cost decreases. The optimal
trade - off point occurs at n = 3

NON – POISSON SERVICE RATES:
A USEFUL MODEL (M/G/1)
 We use M/G/1 as queuing model where M stand for Poisson arrival, G means
general service time and 1 means the system has one channel.
 Suppose the time required to serve a customer has mean 1/μ and the variance
σ2, and assume that ρ = λ/μ < 1. then the following steady – state equations
can be used.
ρο = 1 - ρ (Idle time)
Lq =
λ2
σ2 +ρ2
2 (1 −ρ)
Wq =
Lq
λ
W = Wq +
1
μ
L = λW

EXAMPLE #2: Normal Service Rate Density
 Consider an M/G/1 system in which arrivals occurs according to a Poisson
process at a rate of 5 per hour. The time it takes to service a customer is
normally distributed with a mean of 10 minutes and a standard deviation of 2
minutes. Determine the steady – state statistics of this system.
The mean service time is 1/6 hour per customer.
ρ =
λ
μ
=
𝟓
𝟔
= 0.833 <1
The proportion of time the server is idle is calculated below.
ρο = 1 - ρ (Idle time) = 1 – 0.833 = 0.167
(μ = 6 and σ = 2/60 hours)
Lq =
λ2
σ2 +ρ2
2 (1 −ρ)
= (5)2
(2/60)2 +(0.833)2
2 (1 −0.833)
= 2.167
Wq =
Lq
λ
=
2.167
5
= 0.433
W = Wq +
1
μ
= 0.433 +
1
6
= 0.600
L = 5 (0.600) = 3.000
S
O
L
U
T
I
O
N

EXAMPLE #3: An Empirical Distribution of
service time
Arrivals at AAA Transmission Repair are observed to be
random at the Poisson rate of 1.5 per day. The time
required to repair transmissions varies, and a frequency
distribution of this time is given in TABLE 1. The frequencies
are converted to probability by dividing by the total
number of observations (365). The mean service time is
computed by assuming the products of each service
time multiplied by its probability and is given in TABLE 1 as
0.492 day. The variance is also estimated by summing
the weighted squares of the deviations from the mean
and dividing by 365.

Thus the mean service time
and its variance are
1
μ
= 0.492 day
σ2 = 14.98336/365
= 0.041 day
Queuing Analysis
We model the system as an M/G/ 1
λ = 1.5 cars/ day
μ =
1
0.492
= 2.03 cars/day
And check to see that then steady –
state condition holds:
ρ =
λ
μ
=
𝟏.𝟓
𝟐.𝟎𝟑
= 0.739 <1
ρο = 1 - ρ (Idle time) = 1 – 0.739 = 0.261
Lq =
λ2
σ2 +ρ2
2 (1 −ρ)
= (1.5)2
(14.98336/365)2 +(0.739)2
2 (1 −0.739)
= 5.358 (cars)
Wq =
Lq
λ
=
5.358
1.5
= 3.572 (days)
W = Wq +
1
μ
= 3.572 +
1
0.492
= 4.064 (days)
L = λW = 1.5 (4.064) = 6.096 (cars)
1 2 3 4 5 6 7
Service
time
(days)
Number
of times
observed
Probability (1) x (3)
(1) Minus
the mean
(5)2
(2) x (6)
0.25 100 0.274 0.068 -0.242 0.058564 5.85640
0.50 200 0.548 0.274 0.008 0.000064 0.01280
0.75 40 0.110 0.082 0.258 0.066564 2.66256
1.00 25 0.068 0.068 0.508 0.258064 6.45160
365 1.000 0.492 0.383256 14.98336
TABLE 1: An Emipirical Service - Time Distribution

DIFFERENT TYPES OF CUSTOMER CLASSES
EXAMPLE #4. Different Customer Classes in a supermarket.
 Supershopper Supermarket operates two types of checkout
lanes: Express lane (for customer purchasing eight items or less
and a regular lane. The person at the express lane cash register
can check out fifteen customers per hour, and a person working
at the regular check out lane can service eleven customers per
hour on the average. The average arrival rate of customers at
the check out area is fifteen per hour. Sixty percent of these
have over eight grocery items. Assume the arrival and service
rates are Poisson distributed, and analyze the steady – state
system behavior.

SOLUTION:
 From the data of the problem, 60 percent or nine customers arrive at the regular
lane per hour and six customers arrive at the express lane per hour.
1. Regular lane system : λ = 9, μ = 11.
2. Express lane system: λ = 6, μ = 15.
Figure 1 below uses two separate M/M/1 system
15 customers
per hour
arrive to
supermarket
λ1 =
Regular arrival
rate = 9 per hour
λ2 =
Express arrival
rate = 6 per hour
Regular lane: μ = 11
Express lane: μ = 15

Table 2 uses the basic M/M/1 Steady state equation:
ρ =
λ
μ
and Wq =
λ
μ(μ−λ)
λ μ ρ ρo L Lq W (in hours) Wq (in hours)
Regular lane 9 11 0.818 0.182 4.500 3.682 0.500 0.409
Express lane 6 15 0.400 0.600 0.667 0.267 0.111 0.044
5.167 3.949
TABLE 2: Supershopper Supermarket Analysis
The proportion of time that there are no customers at the check
out area is
ρο (regular lane)x ρο (Express lane) = 0.182(0.600) = 0.109
Table 2 show that the long waiting period is 0.5 hour in the regular
lane subsystem. Thus the Supershopper should consider the
following options: (1) hire a faster checkher for this lane; (2)
provide more help for this checker (such as bagger); (3) add a
second regular lane check out register.

SELF – SERVICE FACILITIES (M/M/∞)
 Uses M/M/ ∞ model where ∞ is “infinite servers” or “no
waiting is necessary”
 The steady – state result for this system are:
ρ =
𝜆
𝜇
(we do not require that ρ be less than 1)
ρ n =
𝑒 − ρρ 𝑛
𝑛!
n = 0, 1, 2, 3…
L = ρ
W =
1
𝜇
(Recall that 𝑒 = 2.71828)

EXAMPLE #5. A Self – service System
 Supershopper Supermarket has fifteen customers entering per
hour on the average. Each customer spends an exponentially
distributed amount of time shopping before approaching the
checkout counter. The average time spent shopping in twenty
minutes.
1. On the average how many customers can be found shopping
in the aisles?
2. What is the average length of time a customer spends
shopping in the aisles?
3. Find the probability of there being six or more customers
shopping in the aisles?

 We have λ = 15, μ = 3, (since 1/μ = 20/60)
1. L = ρ =
𝜆
𝜇
=
15
3
= 5 means that five people can be found shopping in the
aisles on the average.
2. W =
1
𝜇
=
1
3
hour (or 20 minutes) is the length of time spent in the system.
3. The required probability can be found by computing one minus the
probability of five or fewer customers. We determine the probability of n
customers in the system as follows: (Answer = 1 – 0.61596 = 0.38404)
ρn =
𝑒 − ρρ 𝑛
𝑛!
= ρn =
𝑒 − 55 𝑛
𝑛!
(For #3, refer to TABLE 3)
SOLUTION:

TABLE 3
n ρn
0 0.00674
1 0.03369
2 0.08422
3 0.14037
4 0.17547
5 0.17547
0.61596
n = 0
ρn =
𝑒 − ρρ 𝑛
𝑛!
=
2.71828 − 550
0!
=
0.00674 (0)
1
= 0.00674
e = 2.71828 (constant value)
For n! recall fractional sign
0! = 1
1! = 1
2! = (2)(1) = 2
3! = (3)(2)(1) = 6

QUEUING NETWORKS
Consist of different queuing models
Types of Network
1. Series System – Consist one subsystem followed by
another subsystem (A customer must go through one
subsystem and then through the succeeding subsystem).
Subsystem 2Subsystem 1

1. Parallel System – Consist of two or more subsystems
arranged so that a customer must only go through one
of the subsystem.
Subsystem 3
Subsystem 1
Subsystem 2

EXAMPLE #6. Student registration System
 The student registration system of a local college consists of the
following three phases.
1. The student must first go to table 1 for a preliminary checking of
registration status.
2. The student then goes to any of tables 2, 3, or 4 to pay tuition.
3. The student then goes to table 5 for final checking, and departs
The students arrive at a Poisson rate of 30 per hour. The person at the
table 1 can service students at the Poisson rate of 40 students per
hour, each of the workers at table 2, 3, and 4, can service 20
students per hour, and the person working at table 5 can service 50
students per hour.

The diagram of registration system below consist of 3
models: M/M/1 followed by M/M/3, followed by M/M/1 .
SOLUTION:
System 2 (M/M/3)
λ = 30, μ = 20
λ = 30
System 3 (M/M/1)
λ = 30, μ = 50
TABLE 1
TABLE 2, 3, 4
TABLE 5
System 1 (M/M/1)
λ = 30, μ = 40
λ = 30 λ = 30

Cost minimization model

More Related Content

Similar to Cost minimization model

Recently uploaded

Cost minimization model