1 BUS P301:01

CHAPTER 9: LAYOUT STRATEGIES – Suggested Solutions to

Selected Questions

Summer II, 2009

Question 9.1

Question 9.2

(a) Plan A movements = (20x6)+(12x18)+(8x2)+(6x4)+(10x2)+(4x18)

= 120+216+16+24+20+72

= 468 (in 100’s) = 46,800

Cost =46,800 x 0.50 = $23,400.00

(b) Plan B movements = (20x6)+(8x18)+(12x2)+(10x4)+(6x2)+(4x18)

= 120+144+24+40+12+72

= 412 (in 100’s) = 41,200

Cost =41,200 x 0.50 = $20,600.00

(c) Plan C movements = (20x6)+(10x18)+(6x2)+(8x4)+(12x2)+(4x18)

= 120+180+12+32+24+72

= 440 (in 100’s) = 44,000

Cost =44,000 x 0.50 = $22,000.00

(d) Plan B is the lowest cost, at $20,600.

Movements = (4 × 8) + (9 × 7) + (7 × 4) + (6 × 3) + (8 × 2) + (10 × 6)

= 32 + 63 + 28 +18 +16 + 60 = 217 (in 100s)

= 21,700

Cost = 21,700 × $1 = $21,700

2 BUS P301:01

Question 9.10

(a) Takt time Minutes available per day/Units demanded per day

420/250 1.68 minutes

(b) Number of cross-trained employees (1.1 1.1 1.7 3.1 1.0)/1.68 8.0/1.68

4.76 5

(c) The cleaning operation is substantially longer than the others so it warrants special

consideration to ensure that a smooth flow can be maintained. A machine constrained

task or lack of cross-training may suggest that more traditional assembly line balancing

techniques be used.

Question 9.11 Station 2 with B and D; and Station 3 with E.

(60)(60 sec)(a) Cycle time =

180 PLAs

3,600= = 20 seconds per PLA

180

task time(b) Theoretical minimum of workstations =

cycle time

60= = 3

20

(c) Yes, it is feasible. Station 1 with A and C;

3 BUS P301:01

Question 9.15

(a)

(b) Station 1 gets A, G, and B and has 0.5 minutes left over. Station 2 gets C, D, and E,

with no time left over. Station 3 gets F, H, I, and J and has 0.5 minutes left over.

Improvements in efficiency would seem impossible. The times are in 0.5 minute

increments and can’t be sub-divided to achieve exact balance. Total of 1 minute of

idle time/cycle.

(c) If stations 1 and 3 each had 0.5 minute more work to do, the line would be 100%

efficient; perhaps support tasks could be assigned to them.

Time Time Left Ready Station Task (minutes) (minutes) Tasks

1 A 5 5 B, G G 3 2 B B 1.5 0.5 C, E

2 C 3 7 E, D D 4 3 E E 3 0 F

3 F 2 8 H H 3.5 4.5 I I 2 2.5 J J 2 0.5

Summary Statistics

Cycle time 10 minutes

Time allocated (cyc sta) 30 minutes/cycle

Time needed (sum task) 29 minutes/unit

Idle time (allocated-needed) 1 minute/cycle

Efficiency (needed/allocated) 96.67%

Balance delay (1-efficiency) 3.333%

(d) Theoretical minimum no. of stations 3

4 BUS P301:01

Question 9.19

The assembly-line activities are:

Time Ready Time Station

Task (in minutes) Predecessors Tasks Left Assignment

A 3 None A, B 1 1

B 6 None C, D, E 1

C 7 A D, E, F 3 2

D 5 A, B 3

E 2 B F, G, H, I 3 3

F 4 C 4

G 5 F J, H, I, K 2 4

H 7 D, E 5

I 1 H K, J 2 5

J 6 E 6

K 4 G, I, J 0 6

50

(a)

=1 unit 60 min 24 hr

(c) 144unitsperday.10 min hr day

24hr 60min(b) 15min

96 units hr

50 min per unit(d) 5 stations

10 min per cycle

5 BUS P301:01

(f) Idle time = time allocated per unit - time needed per unit

= 60 - 50

= 10 min/cycle

(g) Best assignment is shown in part (a); efficiency is shown in part (e) (i.e., 83.33%)

Time needed per unit(e) Efficiency

Time allocated per unit

Total task time

(Cycle time) (Number of stations)

50

(10)(6)

50

60

.8333, or 83.33%