(b) Given the BOQ what would be the average inventory? What woulkd be the annual
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Question
(b) Given the BOQ what would be the average inventory? What woulkd be the annual inventory holding cost (o Given the BOQ, how many orders would be made each year What would be the annual order cost? (d) Given the EOQ, what is the total annual inventory cost? (e) What is the time between orders D What is the reorder point (ROP) 4. (10%) Consider the following order requirements: Period 11213 4 S Gross requirements040 Stockouts are not acceptable and we need a safety stoek of t0 (a) Develop the FOQ solution with a lead time of 2 periods b) Develop the POQ solution with a lead time of 1 week and P- (c) Develop the LAL. solution 5, (15%) The following tasks are required for the manufacture of a product with a desired production rate of 6 inished proslucts per hour The task eleent network with time-to-complete in minutes is as follows (a) Find the total time to complete a inished prosduct (b) Find the cycle time c) Find the smallest number of required workstations (d) Assign tasks to each workstation (e) What is the utidization rate? (1) What is the idle time? Page 7 of RExplanation / Answer
a) Total time to complete a finished product= 12 mins
b) Given,
Demand = 6 units / 60 mins
Average throughput rate = 60/6 = 10 mins
So, cycle rate = 10 mins
c) Minimum number of workstations= sum of total task times / cycle time
= (2+3+2+3+2+1+4+3)/10= 20/10= 2 workstations
Task Table
Task
Task time
Predecessor
A
2
-
B
3
A
C
2
B
D
3
B
E
2
C
F
1
C
G
4
D
H
3
E,F,G
The total station time is nothing but the cycle time. So, total station time of each station is 10 mins
In order of precedence relationship (primary rule) and shortest performance time (secondary rule), tasks: A->B->C->D->E->F->G->H
Workstation 1:
First task =A
Time left= 10-2=8 mins
Second task =B
Time left= 8-3=5 mins
Third task =C
Time left= 5-2=3 mins
Fourth task =D
Time left= 3-3=0 mins
So, workstation 1: A->B->C->D
Workstation 2:
First task =E
Time left= 10-2=8 mins
Second task =F
Time left= 8-1=7 mins
Third task =G
Time left= 7-4=3 mins
Fourth task =H
Time left= 3-3=0 mins
So, workstation 2: E->F->G->H
Work Station
Tasks
Total task Time
Idle Time
1
A->B->C->D
10
0
2
E->F->G->H
10
0
Were you able to assign all the tasks to the theoretical minimum no of workstations: Yes
e) Efficiency with 2 workstations = (sum of all tasks) / (no of workstations * Cycle time)
= (20)/ (10*2) = 1 or 100 %
e) Total idle time = 0 mins
Task
Task time
Predecessor
A
2
-
B
3
A
C
2
B
D
3
B
E
2
C
F
1
C
G
4
D
H
3
E,F,G
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