Question number (1):
a) For the part-machine matrix shown below, apply a clustering technique to form the GT cells of
machines.
b) Three machines from the seven above will constitute a GT cell. The From-To data for the machines
are shown in the table below.
FROM
A
B
D
A
0
0
70
To
B
10
0
0
D
0
85
0
1. Determine the most logical sequence of machines for this data, according to Hollier Method 1,
and construct the flow diagram for the data, showing where and how many parts enter and exit
the system.
Solution:
FROM
A
B
D
From sums
A
0
10
0
10
B
0
0
85
85
D
70
0
0
70
To sums
70
10
85
165
The minimum To sum is 10 for machine B and the same 10 is the minimum from sum for
machine A, So according to tie breaker 3rd rule, both machines with the minimum To ( B ) is first
and minimum From ( A) is last.
Then the sequence is B – D - A
10
75
B
85
D
70
A
60
15




From B 85 part exit and 10 enter from A, So the system starts with 75 parts enter directly
to the first machine B (85 exit – 10 enter = 75 should enter at B).
85 parts enter to D and only 70 exit to A, So Machine D ships 15 parts to the exit (70+15
exit=85enter)
Machine A has 70 parts enter from machine D and 10 exit to B, the remaining parts are
60 parts exit.
Note that the system has 75 parts enter and finally 75 parts exit to the shipping 60 from A
and 15 from D.
2. Repeat (1) using Hollier Method 2.
FROM
A
B
D
From sums
From/To
ratio
1/7
8.5
0.82
A
0
10
0
10
B
0
0
85
85
D
70
0
0
70
To sums
70
10
85
165
The sequence is B – D – A (the same as Hollier 1).
3. Compute the percentage of in-sequence moves and the percentage of backtracking moves in the
solution for the two methods. Which sequence is better, according to these measures?
For method 1 and 2 the same : percent of in-sequence moves
= [(85+70) / (85+70+10)] *100%
= 93.93%
Percent of backtracking moves
= [(10) / (85+70+10)] *100%
= 6.06%
Question number (2):
a) For the part-machine matrix shown below, apply the rank order clustering technique to form the GT
cells of machines.
Parts ‘Number’
1
2
A
Machines
1
5
6
1
1
1
1
D
E
4
1
B
C
3
1
1
1
1
1
1
b) If we suppose that the previous five machines will constitute a GT cell. The From-To data for the
machines are shown in the table below.
FROM
1
2
3
4
5
1
0
0
0
100
0
2
40
0
0
0
105
TO
3
110
0
0
50
0
4
0
115
0
0
50
5
0
0
0
0
0
1. Determine the most logical sequence of machines for this data, according to Hollier Method 1,
and construct the flow diagram for the data, showing where and how many parts enter and exit
the system.
2. Repeat (1) using Hollier Method 2.
3. Compute the percentage of in-sequence moves and the percentage of backtracking moves in the
solution for the two methods. Which sequence is better, according to these measures?
Solution for Hollier method:
FROM
1
2
3
4
5
SUM
1
0
0
0
70
0
70
2
10
0
0
0
75
85
TO
3
80
0
0
20
0
100
FROM
1
2
4
5
SUM
1
0
0
70
0
70
2
10
0
0
75
85
TO
4
0
85
0
20
105
4
0
85
0
0
20
105
5
0
0
0
0
0
0
5
0
0
0
0
0
SUM
10
85
70
95
TO
FROM
1
2
4
SUM
1
0
0
70
70
2
10
0
0
10
4
0
85
0
85
2
0
0
0
TO
4
85
0
85
SUM
85
0
SUM
10
85
90
2 4 1
FROM
2
4
SUM
Accordingly, the machines' flow will take the sequence: 5 _ 2 _ 4 _ 1 _ 3
SUM
90
85
0
90
95
20
10
95
5
75
2
85
70
4
20
80
1
3
100
20
15
(b) Repeat step (a) only using Hollier Method 2.
FROM
1
2
3
4
5
SUM
1
0
0
0
70
0
70
2
10
0
0
0
75
85
TO
3
80
0
0
20
0
100
4
0
85
0
0
20
105
Accordingly, the machines' flow will take the sequence: 5 _ 1 _ 2 _ 4 _ 3
5
0
0
0
0
0
0
SUM
90
85
0
90
95
Ratio
9/7
1
0
9/10.5
∞
(c) Compute the percentage of in-sequence moves and the percentage of backtracking moves in the
solution for the two methods. Which method is better, according to these measures?
In-sequence moves percentage considers moves between adjacent machines in the flow direction
implied by the machine sequence
So, for method 1: percent of in-sequence moves
= [(75+85+70+80) / (75+85+70+80+10+20+20)] *100%
= 86.1%
For method 2, percent of in-sequence moves
= [ (10+85+20) / ( 10+85+20+80+70+75+20) ]* 100%
= 32%
Similarly for the backtracking percentage, it is computed by dividing the values of backtracked moves
between adjacent points by the total number of moves.
In the two cases there is no backtracks between two adjacent nodes. That is, the percentage of
backtracking moves equal zero in both alternatives. However, there is another performance measure
called percentage of backward moves which consider all the backward moves regardless to cells'
adjacency. This percentage will equal 10/360 = 2.8% in method 1 and 70/360 = 19.4% in the second.
The percentage of in-sequence moves in the first alternative is grater than that of the second whereas
the percentage of the backward moves in the second is greater than the first. So, the machine sequence
resulted in Hollier method 1 is better than the sequence in the second method.