Question

TASKS		Predecessor	Task Time
A	Warm tortilla	-none	20 sec
B	Rice	-A	10 sec
C	Beans	-B	10 sec
D	Meat	-B	10 sec
E	Salsa	-C, D	10 sec
F	Cheese	-E	5 sec
G	Sour Cream	-E	10 sec
H	Lettuce	-E	10 sec
I	Wrap	-F, G, H	15 sec
J	Bag	-I	5 sec
K	Checkout/pay	-J	25 sec

1. What is the total task time (sum of tasks).
2. What is the minimum C.T., and the maximum C.T.?
3. Draw a precedence diagram:
4. If the operations manager wants to achieve 100 burritos per hour, what is the cycle time?
5. Achieving the above cycle time and output, what is the theoretical number of workstations?
6. Assign tasks to workstations:
7. Efficiency of your burrito line:
8. Actual output per hour:
9. What if demand were such that 200 / hour were needed?
10. What is the maximum # of burritos that can be produced in an hour with these specific times/tasks?

Answer 1

The precedence diagram, Critical Path analysis are as follows :

From the analysis, the 4 critical paths are :

(1) 1-2-3-4-5-6-8-9-10-11-12 and critical activities : A,B,C,d,E,G,d,I,J,K

(2) 1-2-3-4-5-6-9-10-11-12 and critical activities : A,B,C,d,E,H,I,J,K

(3) 1-2-3-5-6-8-9-10-11-12 and critical activities : A,B,D,E,G,d,I,J,K

(4) 1-2-3-5-6-9-10-11-12 and critical activities : A,B,D,E,H,I,J,K

The total project time is 105 seconds

Cycle Time = 105 / 100

= 1.05 units per hour

Positional Weights for Each Process :

A , B , C ,D, E ,F, G ,H, I , J , K = 20/105 , 10/105, 10/105, 10/105, 10/105, 15/105, 5/105, 25/105

Sum of these weights = 130/105 =1.23

Number of Workstations = Sum of Weights / Cycle Time

= 1.23/1.05

= 1.17 = 2 workstations

Efficiency = Sum of Task Times / Actual No. of Workstations x Cycle Time

= 105 / (2 x 1.05) = 50%

Actual output = 50 burritoes

If manager wanted 200 burritoes per hour,

then number of workstations = 4 because cycle time is halved.

Efficiency = 105 / (4 x 1.05) = 25%

Actual Ouput the becomes 25 burritoes.