ISyE 3104: Introduction to Supply Chain Modeling: Manufacturing and Warehousing
Instructor: Spyros Reveliotis
Summer 2002
Homework #4
Due Date: Monday, 7/1/02
Problem set:

From Chapter 9 in your textbook:

Discussion Questions: 2

Problems: 2, 4, 11, 12 and 13

Case Studies on

Des Moines National Bank

State Automobile License Renewals
DISCUSSION QUESTION

You need to collect statistics about the material flow in the facility. Therefore, you should study the recently issued move tickets and routing sheets, or track the orders currently run in the facility.
PROBLEMS
9.2 The following matrix indicates the bidirectional traffic (in number of trips) between any pair of processes:

M

W

D

L

G

B

M
W
D
L
G
B



125


75
0


0
75
0


50
0
150
20


60
0
20
0
0


Based on the above matrix, we can establish the following closeness ratings among the different processes:

M

W

D

L

G

B

M
W
D
L
G
B



A


E
U


U
E
U


I
U
A
O


E
U
O
U
U


Two layouts that tend to satisfy the above adjacency requirements are as follows:
Room 1

Room 2

Room 3

B

M

W

D

G

L

Room 4

Room 5

Room 6

Or
Room 1

Room 2

Room 3

W

M

B

L

G

D

Room 4

Room 5

Room 6


MovementDistance Calculations

M W:
M D:
M G:
M B:
W L:
D G:
D B:
L G:

125 x 20 =
75 x 40 =
50 x 20 =
60 x 20 =
75 x 20 =
150 x 20 =
20 x 20 =
20 x 20 =

2,500
3,000
1,000
1,200
1,500
3,000
400
400



13,000 = Minimum distance movement for both (symmetrical layouts)

To compute the optimal solution of this problem, in a strict sense, one would employ the following mathematical programming formulation:
Parameters:
d_{ij }: distance between available areas (rooms) i and j, i,j{1,…,6}
f_{kl} : number of trips between processes ik and l, k,l {1,…,6}
Variables:
x_{ki}^{ }binary variable with a value of 1 indicating that process k is assigned to location i, k,i {1,…,6}
Objective: min _{i,j,k,l} d_{ij}*f_{kl}*x_{ki}*x_{lj}
Constraints
_{i} x_{ki} = 1, k (i.e., every process k must be assigned to one and only one location)
_{k} x_{ki} = 1, i (i.e., every location i must be assigned to one and only one process)
x_{ki}{0,1}, i, k (i.e., the decision variables are all binary)
The above formulation is known as a quadratic assignment problem. (it is an assignment because as it is stated by the problem constraints, we are essentially trying to match the processes with the available areas/rooms, and it is quadratic because the objective is a quadratic function of the decision variables). However, the nonlinear structure (quadratic) of the objective, combined with the binary nature of the decision variables, makes it a computationally hard problem. i.e., there are no readily available efficient algorithms for its solution.
9.4 Layout 1:

Refrig.
(1)

Counter
(2)

Sink
(3)

Storage
(4)

Stove
(5)


Trip Matrix


Distance Matrix


1

2

3

4

5



1

2

3

4

5

1
2
3
4
5

0
5
3
3
0

8
0
12
0
8

13
3
0
0
4

0
3
4
0
10

0
8
0
5
0


1
2
3
4
5

0
4
8
12
16

4
0
4
8
12

8
4
0
4
8

12
8
4
0
4

16
12
8
4
0

∑ ∑ T_{ij} x D_{ij} = 600
Layout 2:


Refrig.
(1)


Counter
(2)


Stove
(5)


Trip Matrix


Distance Matrix


1

2

3

4

5



1

2

3

4

5

1
2
3
4
5

0
5
3
3
0

8
0
12
0
8

13
3
0
0
4

0
3
4
0
10

0
8
0
5
0


1
2
3
4
5

0
7
8
12
14

7
0
5
6
7

8
5
0
4
9

12
6
4
0
6

14
7
9
6
0

∑ ∑ T_{ij} x D_{ij} = 602
9.11
Task

Performance Time
(in minutes)

Task Must Follow
This Task

A
B
C
D
E
F
G
H
I

1
1
2
1
3
1
1
2
1
13


A
A
C
C
C
D, E, F
B
G, H

Since we need to produce 60 boats over a period of 200 minutes, we must produce one boat every 200/60 = 3.333 min; this defines the target cycle time for this line. Furthermore, the information provided above implies the precedence relationships among the various tasks expressed by the following precedence graph:
The above drawing indicates also the organization of the various tasks into stations, in a way that respects (i) the precedence relationship, and (ii) the target cycle time: notice that the maximum workload per cycle established by the above configuration is 3 min, at workstations 1, 2 and 4. The overall line efficiency can be computed as follows:
Efficiency = 13 minutes / (5 stations x 3.33 minutes) = 0.78 or 78%
Finally, notice that a lower bound for the number of workstations for this particular set of data is
N = ceiling [(_{i} t_{i} )_{ }/ (cycle time)] = ceiling[13/3.333]=ceiling[3.9] = 4.
However, you can convince yourselves that due to the task indivisibility, this lower bound is not achievable.

(a) Resolving Problem 9.11 with a production time of 300 minutes per day:
Cycle time = 300 minutes / 60 units = 5 minutes/unit
Lower bound to the number of stations = ceiling[∑t_{i} / Cycle time] = ceiling[13/5] = ceiling[2.6] = 3 workstations
A feasible layout
Efficiency = 13 minutes / (3 stations / 5 minutes) = 0.867 or 86.7%
(b) Resolving Problem 9.11 with a production time of 400 minutes per day:
Cycle time = 400 minutes / 60 units = 6.67 minutes/unit
Lower bound to the number of stations = celing[∑t_{i} / Cycle time] = ceiling[13/6.675] = ceiling[1.95] = 2 workstations
Efficiency = 13 minutes / (3 stations / 6.67 minutes) = 0.649 or 64.9%

Output = total operating time / cycle time
Based on the above formula, the daily output is maximized by minimizing the line cycle time. However, a hard lower bound to the line cycle time is 3 minutes, which is the time requested for the execution of task E. Hence, the maximum possible output for this case is: 400/3 = 133.33 133 boats.
CASE STUDIES : DES MOINES NATIONAL BANK
This case study is based on an analysis at the Federal Reserve Bank in El Paso. During the analysis, the author recognized the need for multiplecriteria facility layout procedure. An improved (at least for this application) multiplecriteria model was developed during this analysis and can be found in Timothy L. Urban, “A multiple Criteria Model for the Facilities Layout Problem,” International Journal of Production Research, 25, no. 12, (December 1987): 18051812
1. Develop a layout that minimized the total workflow.
Using techniques similar to those presented in the solution of Problem 9.2 above, one can develop the following layout when focusing on the minimization of the total workflow:
Check crediting

Check reconcilement

Check sorting

Credit adjustment

Returned checks

Check distribution

Offices

Government checks


2. Develop a layout using the relationships defined by the closeness ratings.
In this case, the proposed layout must satisfy as much as possible the closeness rating identified in the lower part of Table 9.5. A suggested solution is as follows:
Check crediting

Check adjustment

Check reconcilement

Offices

Returned checks

Check distribution

Check sorting

Government checks



Develop a layout that considers both the workflow and closeness relationships between departments.
As it was discussed in class, a systematic way to address this problem is through the revising of the closeness ratings presented in Table 9.5, in order to take into consideration the adjacency requirements that would facilitate the workflow taking place in this facility. Hence, for instance, the closeness rating for the pair of (1,4) should be upgraded to an A (from the current U value). Similarly, the existing E relationship between pairs (4,5) and (4,6) already takes care of the workflow between these units (40 and 60 trips, respectively), expressed in the upper part of Table 9.5. On the other hand, the X rating between pairs (1,2) and (1,3) is a much stronger requirement than any adjacency requirement that would be based on the facilitation of the material flow; hence, these pairs of units should be kept apart from each other, in spite of the relatively significant workflow between them.
Check reconcilement

Government checks

Check sorting

Check crediting

Returned checks

Check distribution

Offices

Credit adjustment



Comment on the various layouts developed.
The first layout is very efficient in terms of minimizing the total workflow movement. The only flows that do not travel between adjacent departments are the flows from check crediting to the offices (10 units) and from check distribution to government checks (40 units). However, because the analysis did not take into consideration the closeness ratings, the layout obviously does a poor job of addressing those concerns. For example, check sorting is located next to the check reconcilement area and, as stated in the case, it is desirable to keep these areas separate due to the noise from the sorting equipment. Also, government checks and check sorting are at opposite ends of the building even though they use the same equipment, which needs to be kept together.
The second layout then addresses the closeness ratings, but the total workflow travel more than doubles. This is evident by the movement of materials from check distribution to check sorting to check reconcilement to check crediting – these departments are at opposite corners of the building even though they use the same equipment, which needs to be kept together.
The third layout can be thought of as a compromise between the two objectives. The workflow is not as low as the first layout (an increase of about 7.5 percent) but is much better than the second. It does not address the closeness ratings quite as well as the second layout but does much better than the first (e.g., the government check department is now located next to check sorting). This is a good illustration of how managers frequently need to deal with multiplecriteria situations; we may not optimize any of the objectives, but try to “satisfice” all of them.
5. Discuss any other factors that should be considered when developing a layout of the check processing division.
Any number of suggestions could be identified at this point. Some of the more obvious may be as follows:

Offices may want to be placed near the windows, so a corner of the building may be most suitable. Also, the view (perhaps looking out on a park as opposed to the busy street) may lead us to place the office areas on a particular side of the building.

The location of the electrical wiring necessary for the sorting equipment may determine the location of that department.

The location of restrooms, stairs, etc., may affect the placement of the office areas.

The customers come onto the floor, aesthetic considerations (e.g., do not expose them to the sorting equipment or material handling) would need to be analyzed.

Personnel and material flow between floors (other than the checks coming up on the service elevator) should be considered by placing those departments near the stairs or elevators.

It may be beneficial to have department shapes that deviate from the 75’ x 75’ square. For example, 150’ x 37.5’ departments may benefit the workflow.

Historical workflow was used for the analysis. If there are forecasts of significant changes in workflow or in relative workflow (e.g., increase in government checks but not commercial checks), we should use the expected workflow.

This analysis focuses on the relative location of departments; it does not suggest how we arrange the equipment and personnel within the individual departments. These considerations may impact the layout, as the work may not flow between the centers of the departments.

We could extend this layout analysis to the multistory case. Perhaps this is not the best combination of departments to be placed on the same floor.
LICENSE RENEWALS

What is the maximum number of applications per hour that can be handled by the present configuration of the process?
Since, in the present configuration, each task is executed by a different person, the maximum output of renewals will be limited to 60 renewals/hour (3600 sec/hr / 60 sec/renewal), a limit established by task #3 which presents the longest process time, and therefore, it is characterized as the “bottleneck” task.
Every other station will be waiting for the clerk who checks the file for violations. In particular, the clerk and expensive equipment for the photographic step will be idle approximately 1/3 of the time (20 seconds / 60 seconds).
A (perfectly) balanced line process is one in which the process times of each step are the same. In practice, it is hard to achieve perfect balancing, but one should still strive to achieve comparable process times among the different stations. An obvious way to balance the line is to add stations to the bottleneck activity, reducing thus its effective processing time. Specifically, the effective process time of the activity is defined as its original/nominal process time (the time required to perform the activity) divided by the number of stations or locations performing the activity. However, this may not be the most efficient solution. As we showed in class, in many cases it is possible to combine activities creatively and end up with a significantly smaller number of workers, while making efficient use of them. In any case, the cycle time is the longest process time and dictates the rate at which the whole process produces output.

How many applications can be processed per hour if a second clerk is added to check for violations?
Based on the above discussion, if a second file clerk is added to the activity of checking files, the process time for this activity is reduced to 30 sec/location (60 seconds / 2 locations). The bottleneck now becomes the eye test. The maximum output of renewals become 90 renewals/hour (3600 sec/hour / 40 sec/renewal).

Assuming the addition of one more clerk, what is the maximum number of applications the process can handle?
If activities 1, 2, and 3 can be successfully combined to form a new activity taking 105 seconds that is accomplished by the same three people, the process time of the new combined task is 35 seconds. So without adding any personnel, as was done above, it is possible to process up to 90 renewals per hour.
Creative rearranging and combining of tasks can produce other cycle times. The only limitations are: (1) the tasks must be performed in a logical sequence, and (2) the facilities and equipment must be available for the tasks.

How would you suggest modifying the process in order to accommodate 120 applications per hour?
This question requires trialanderror creation of proposed solutions. Presented below are some proposed solutions that each provides the necessary capacity for handling 120 renewals per hour. Solution A was achieved by simply expanding the number of stations performing each job so that at least 120 licenses are processed per hour. Solution B combines jobs such that the process time at most stations equals the bottleneck process time (or cycle time). Although this reduced the number of employees from 8 to 7, one of these is an additional photographer with another camera. So the total costs are increased. Solutions C and D produce the same costs per renewal, $0.4333, and both employ 7 persons.
How do solutions C and D compare to each other? Some managers would argue that the five people who each perform jobs 1, 2, 3 and 4 have an enriched job. Others would argue that enlarging a job is not the same as enriching a job. How difficult will it be to monitor the performance of each of these five people working independently as a line process? How difficult will it be to teach each of the five employees all four jobs rather than teaching each person one or two jobs. This is where the quantitative analysis of the type pursued above ends and analysis related to human resources and job design takes over (c.f Chapter 10 for more on these issues).
PROPOSED SOLUTIONS – 120 RENEWAL/HOUR:
Solution A





Maximum



Time


Process

Output


Job

(sec)

Station

Time (sec)

(per hr)

Cost/Hour

1

15

1

15

240

$6.00

2

30

1

30

120*

6.00

3

60

2

30

120*

12.00

4

40

2

20

180

12.00

5

20

1

20

180

8.00 + 5.00

6

30

1

30

120*

9.00

Cost per renewal = $58.00 / 120 = $0.4833



$58.00

* Indicates a bottleneck step.
















Solution B





Maximum



Time


Process

Output


Job

(sec)

Station

Time (sec)

(per hr)

Cost/Hour

1 + 2 + 3

105

4

26.25

137.14

$24.00

4 + 5

60

2

30

120*

16.00 + 10.00

6

30

1

30

120*

9.00

Cost per renewal = $59.00 / 120 = $0.4916



$59.00

* Indicates a bottleneck step.
















Solution C





Maximum



Time


Process

Output


Job

(sec)

Station

Time (sec)

(per hr)

Cost/Hour

1 + 2 + 3 + 4

145

5

29

124.1

$30.00

5

20

1

20

180

8.00 + 5.00

6

30

1

30

120*

9.00

Cost per renewal = $52.00 / 120 = $0.4333



$52.00

* Indicates a bottleneck step.
















Solution D





Maximum



Time


Process

Output


Job

(sec)

Station

Time (sec)

(per hr)

Cost/Hour

1 + 4

55

2

27.5

131

12.00

2

30

1

30

120*

6.00

3

60

2

30

120*

$12.00

5

20

1

20

180

8.00 + 5.00

6

0

1

30

120*

9.00

Cost per renewal = $52.00 / 120 = $0.4333



$52.00

* Indicates a bottleneck step.










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