ISYE 3104 Summer 2003 Homework 4 Solution

Chapter 9Layout Strategy

PROBLEM SET

The accompanying table lists the sequence of machine types required for nine part types produced by a manufacturing facility:

(a)Set up a part-machine processing indicator matrix for the data in the table.

part-machine incidence matrix is as follow:

Machine
1 / 2 / 3 / 4 / 5 / 6
1 / 1 / 1 / 1
2 / 1 / 1
3 / 1 / 1 / 1
4 / 1 / 1
Part / 5 / 1 / 1
6 / 1 / 1
7 / 1 / 1 / 1
8 / 1 / 1
9 / 1 / 1 / 1

(b)Determine machine groups and part families using

(i)the Row and Column Masking clustering algorithm presented in class (c.f. also the reading assignment posted on the reserves);

Machine
1 / 2 / 3 / 4 / 5 / 6
1 / 1 / 1 / 1
2 / 1 / 1
3 / 1 / 1 / 1
4 / 1 / 1
Part / 5 / 1 / 1
6 / 1 / 1
7 / 1 / 1 / 1
8 / 1 / 1
9 / 1 / 1 / 1

CELL 1: Machine group {2 3 5} with part family {1 2 7 8}

CELL 2: Machine group {1 4 6} with part family {3 4 5 6 9}

(ii)the Similarity Coefficient approach, with the clustering threshold set to 0.5.

Using Similarity Coefficient approach:

Remember that the Similarity Coefficient for any pair of machines Mi and Mj is defined by:

SCij = |P(Mi) ∩ P(Mj)| / |P(Mi) U P(Mj)|

where P(Mi) denotes the set of parts using machine Mi, and |X| denotes the cardinality – i.e., the number of elements - of any given set X.

Then, using this definition, we obtain the following similarity coefficients for the various machine pairs:

Machine / Parts / 1 / -
1 / 3 4 6 9 / 2 / 0 / -
2 / 1 7 / 3 / 0 / ½ / -
3 / 1 2 7 8 / / 4 / 3/5 / 0 / 0 / -
4 / 3 4 5 9 / 5 / 0 / ½ / 1 / 0 / -
5 / 1 2 7 8 / 6 / 3/5 / 0 / 0 / 3/5 / 0 / -
6 / 3 5 6 9 / Machine / 1 / 2 / 3 / 4 / 5 / 6

Setting the clustering threshold equal to ½, we obtain the following two machine groups:

{1 4 6}, {2 3 5}

1 / -
2 / 0 / -
3 / 0 / ½ / -
4 / 3/5 / 0 / 0 / -
5 / 0 / ½ / 1 / 0 / -
6 / 3/5 / 0 / 0 / 3/5 / 0 / -
Machine / 1 / 2 / 3 / 4 / 5 / 6

PROBLEMS 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 ti )/ (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.

PROBLEMS 9.12

(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[∑ti / Cycle time] = ceiling[13/5] = ceiling[2.6] = 3 workstations

A feasible layout

Efficiency = 13 minutes / (3 stations x 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[∑ti / Cycle time] = ceiling[13/6.675] = ceiling[1.95] = 2 workstations

Efficiency = 13 minutes / (3 stations / 6.67 minutes) = 0.649 or 64.9%

Problem 3:

a)Assuming that each workstation will eventually consist of a number of units form a single type, how many configurations are available?

Under the convention that a “configuration” is defined only by the machine types used at each workstation (and not the exact number of machine units used), then, there is a total of 3*2*2*2 = 24 different combinations of machines types.

However, if we look closer at the performances and prices of each machine type, we can rule out some machine types from further consideration, since they are dominated by some other machine candidate(s). For instance, among the machine candidates for station 1, MMOD, machine type 3 is obviously slower than type 1 and type 2 and it operates with high variance and with a much higher cost. A similar situation arises at station 3, HDBLD, where machine type 2 can be eliminated.

Thus, the viable combinations are now reduced to be 2*2*2*1 = 8.

b)Identify a good configuration that meets the demand and performance requirements while minimizing the deployment cost.

The easiest way to solve this problem is to build a spreadsheet. The tricky parts are converting rate in piece/hour to batch process times and computing the SCV for the batch processing.

Let:b= batch size

te= effective processing time for one piece

ts= time to perform a setup (to be taken equal to 0 in our case)

tb= effective processing time for a batch

Ce2= SCV for the processing time for a piece

Cb2= SCV for the processing time for a batch

ra= arrival rate

We compute the mean effective processing time for a batch as:

tb = b * te + ts = b * te = 50 * (1/ speed)

e.g., for machine type at station 1, tb = 50 / 42 = 1.19 hr/batch

Similarly, the fact that the batch processing time is the sum of the processing times of the b pieces in the batch, which are identically distributed, implies that b2 = b *e2. Since C2 = t2, we have that Cb2 = b2 / tb2= (b *e2) / (b * te)2 = Ce2/ b

e.g., for Type 1 machine at station 1, Ce2 = (2.0)2 / 50 = 0.08

We get the arrival rate in batches per hour as (assume that there are 8 working hrs/day):

ra = (1000 prt/day) / (50 prt/batch)(8 hr/day) = 2.5 batches / hour

We can now start our calculation as follow:

Min. machines required = ra*tb
Actual machine required, m = roundup (ra * tb)
Utilization, u = ra*tb / m
Average time in queue,

Total waiting time at the station,

Since the SCV for inter-departure time, Cd2 , of previous station is the SCV for the arrivals, Ca2, to the next station. The linking equation for multiple machine station is

We must also assume something about the SCV Ca2 for the inter-arrival times experienced at the first station. We will assume deterministic batch releases, which makes it equal to zero (This should be typical in repetitive, high throughput manufacturing).

We will start with a configuration, that selects the cheapest type of machine at each workstation, and utilizes the minimum number of machines that provides a machine utilization lower than one for the considered throughput.

The result is that we have 3, 3, 6, and 3 machines at station 1, 2, 3, and 4 respectively. Fro this configuration, we see that resulting total cycle time is over the cycle time constraint (less than 8 hrs) with a total cycle time of 21.71 hr.

The largest portion of cycle time is at station 2, 13.34 hr, so we will work on this first. By adding one more machine of the same type, the total cycle time goes down to 8.46 hrs and the cost goes up to $1,010, with a resulting benefit of 0.27 hr/$-added (You can check by editing the number of machine at station 2 in the spreadsheet in the previous page to be 4 machines). We now try to replace the machines at station 1 by picking a machine type 2, which has a smaller variability. This selection is entered in our spreadsheet by setting the corresponding part processing CV to be 1 and the corresponding cost per machine equal to $85. With four type-1 machines at station 2, the cycle time goes down to 6.95 hrs, i.e., our cycle time requirement is met. At the same time the total cost increases to $1,115.

Alternatively, we consider the possibility of adding one more machine of type 1 at station 1. The results for this configuration are given below:

We see that the total cycle time is equal to 6.34 hr and the resulting total cost is $1,060. Both of these numbers are better than the previously considered option of using three machines of type 2 for the station 1. Hence, we decide to stick with this configuration.

We note that the procedure described above is based on trial and error and it will not necessarily lead to an optimal solution, unless we perform an exhaustive search over all possible configurations. This is a typical use of queueing theory as a “design” tool; i.e., queueing theory essentially allows the systematic evaluation of the performance resulting from each considered alternative.

In the more specific context of our example, it is unclear without any further analysis and/or computation whether the generated solution is optimal; all we can say is that it is a configuration that meets the posed production requirements at a fairly low cost.