Lingnan University
Department of Computing and Decision Sciences
The First Term, 2015–2016
BUS 2211: Operations Management
Solution to Assignment 1
Question 1 (Productivity) [10 marks]
Answer:
1. The labor productivity of the desktop computers in January is computed as
Labor productivity of the
desktop computers in January
=
Output
$3; 000
=
Labor Cost
11 $60
500
8 8
35:51.
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2. The calculation of the productivity for the laptops is given as follows:
(a) The labor productivity of the laptops in January is computed as
Labor productivity
of the laptops in January
=
Output
=
Labor Cost
11
$60
$3; 200
12 8
650
1 $60
1
33:08.
8
(b) The productivity was higher for the desktop computers in January.
3. For the multi-factor problem, the computation is given as follows:
(a) The multi-factor productivity of the manufacturer in January is
650) + $500
20
is
Multi-factor productivity in January
Output
=
Labor+Materials + Overhead
$3; 000 500 + $3; 200 650
=
(11 $60 20 8 1 $60 1 8) + ($1; 800 500 + $2; 000
1:55.
Th
(b) This multi-factor productivity can be explained as follow: For every $1 input of labor,
material, and overhead, the manufacturer can generate $1:55 of output (sales revenue).
sh
4. The labor productivity of the desktop computers in February is computed as
Labor productivity of the
desktop computers in February
=
Output
$3; 000
=
Labor Cost
11 $60
550
9 8
34:72.
Thus, the productivity growth— the percentage change in labor productivity of desktop computers—
from January to February can be found as
Productivity growth of desktop
computers from January to February
=
34:72 35:51
35:51
1 10-26-2021 23:03:23 GMT -05:00
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2:23%,
BUS 2211: Operations Management
Solution to Assignment 1
which implies that, from January to February, the manufacturer experienced a decline in the
labor productivity of desktop computers.
Question 2 (Layout) [20 marks]
Answer:
1. The precedence graph is plotted as follows.
A
B
C
D
G
H
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E
F
2. The minimum possible number of workstations (Nmin ) is obtained as follows:
P
38
task time
:
=
= 3:45 = 4 workstations.
cycle time
11
Nmin =
The task assignment is made in the following table:
Station
Time
Remaining
1
Th
3
Assign
Revised Time
Fit
(Task Time)
Remaining
A, C
A, C
A (5)
6
6
B, C
B, C
C (4)
2
2
B, E, F
–
–
2
11
B, E, F
B, E, F
F (7)
4
4
B, E
B
B (4)
0
11
D, E
D, E
E (6)
5
5
D, G
D, G
G (4)
1
1
D
–
–
1
11
D
D
D (3)
8
H
H
H (5)
3
sh
4
Will
11
is
2
Eligible
8
Idle
2
0
1
3
3. Based on the result obtained from the above table, we plot the following …gure to show the
product layout.
4. The e¢ ciency of the process is obtained as follows:
E¢ ciency = 1
total idle time
=1
N cycle time
2+0+1+3
= 86:36%.
4 11
2 10-26-2021 23:03:23 GMT -05:00
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BUS 2211: Operations Management
A
Solution to Assignment 1
B
D
E
C
G
H
F
Workstation 2
Workstation 3
Workstation 4
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Workstation 1
5. The rate of output is computed as follows:
480
OT
=
= 43:6 cell phones/day.
CT
11
sh
Th
is
Output rate =
3 10-26-2021 23:03:23 GMT -05:00
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