Chapter 6 Solution Ops Management Essay

Submitted By ilateem
Words: 2004
Pages: 9

CHAPTER 6
PROCESS DESIGN AND Facility LAYOUT

Answers to Problems
1.
Longest task = 2.4 minutes
Total task times = 18 minutes

OT = 450 minutes per day

a.
Minimum cycle time = length of longest task, which is 2.4 minutes.

a.
Maximum cycle time =  task times = 18 minutes.CT = 450 / 180 = 2.50 minutes per unit
N = 18 / 2.5 = 7.2, round to 8

b.b.
CT = 450 / 125 = 3.6 minutes per unit
Range of output:

e. c
i. output = 450 / 9 = 50 units per day ii. output = 450 / 15 = 30 units per day

2.
Desired output = 33.33 units per hour
Operating time = 60 minutes per hour

CT =
Operating time
=
60 minutes per hr.
= 1.80 82 minutes per unit

Desired output

33.33 units per hr.

a.
Station
Time left
Eligible
Will fit
Assign (time)
Idle

1
1.82
0.42 a b a -- a (1.4)

0.42

2
1.82
1.32
0.52
b c, d, e c, d b c, d, e
--
b (0.5) e* (0.8)

0.52

3
1.82
1.12
0.52
.02 c, d c, g g, f g c, d c, g f -- d* (0.7) c** (0.6) f (0.5)

.02

4
1.82
0.82
0.32
g h g h g (1.0) h (0.5)

0.32

1.28
* is tied in no. of followers, but is longer(longest)
** has more followers

b. Efficiency = 1 – [1.28 / 4(1.82)] = .82 or 82%.

3.

Desired output = 4 units per hour

Operating time = 56 minutes per hour

CT =
Operating time
=
56 minutes per hr.
= 14 minutes per unit

Desired output

4 units per hr.

a.
Station
Time left
Eligible
Will fit
Assign (time)
Idle

1
14
9
6
a, d, f a, d, g b, d, g a, d, f a, d, g b, g f* (5) a** (3) g* (6)

0

2
14
7
5
1 b, d b, e c, e c b, d b, e c, e
--
d* (7) b** (2) e*** (4)

1

3
14
10
1
c h i c h
--
c (4) h (9)

1

4
14
9 i i

i (5)

9

11
* is tied for no. of followers, but is longer (longest)
** has more (most) no. of followers
*** tied in no. of followers and time; choose randomly

b. Efficiency = 1 –
Total idle time
= 1 –
11
= 80.4%

CT x no. of stations

14(4)

1. CT = 1.3 minutes per unit
Time [no. followers]

.3 [3]
a. i

.2 [4] .4 [3] 1.3 [2] 1.2 [0]

.1[3] .8[2] .3[1]

a. ii
Station
Time left Eligible
Will fit
Assign (time)
Idle time

1
1.3
a, c, e a, c, e a* (.2)

1.1 b, c, e b, c, e b** (.4)

.7 c, e c, e c** (.3)

.4 d, e e e (.1)

.3 d, f
--

0.3

2
1.3
d, f d, f d** (1.3)

0.0 f --

0.0

3
1.3
f f f (.8)

.5 g g g (.3)

.2 h --

0.2

4
1.3
h h h (1.2)

0.1
--

0.1

0.6
* most followers
** tied in no. of followers, but longer (longest)

a. iii
Percentage idle time:
(idle time)
=
.6
= 11.5%

N × CT

4(1.3)

a. iv
Output:
OT
=
420 min./day
= 323.1 units/day

CT

1.3 min./unit

b.
i.
Total time = 4.6 min., CT =
Total time
=
4.6
= 2.3 minutes.

N

2

Assign a, b, c, d, and e to workstation 1: 2.3 minutes

Assign f, g, and h to workstation 2: 2.3 minutes

ii.
Percentage idle time = 0

iii.
Output =
OT
=
420
= 182.6 units per day.

CT

2.3

5. Output rate = 240 units per eight-hour day

a.

b. CT =
OT
=
480 min/day
= 2 minutes per unit

output

240 units/day

c. N =
t
=
4.6
= 2.3 (round up to 3) workstations

CT

2.0

d.
Station
Time Left
Eligible
Will Fit
Assign (time)
Idle time

1
2
a, c, e a, c, e a* (.2)

1.8 b, c, e b, c, e e** (1.2)

.6 b, c, f b, c b** (.4)

.2 c, f c c (.2)

0

.0

2
2
d, f d, f f** (1.2)…