Two processes are available to produce an order of size 8000 products. The stati
ID: 3061970 • Letter: T
Question
Two processes are available to produce an order of size 8000 products. The statistical situations of these two processes are shown in the two figures be low. The complete order can produced on any of the processes SL-193USL-225 P -25 Po21 Check and answer the following questions: A- What is the target value of Process A (oa)? B- What is the amount of changes in the target values of process A, per unit of standard deviation, 8A C- What is the percentage of Rework in process A D- What is the percentage of Scrap in process B E- Assume that the cost of one reworked item is 15 QAR and the cost of one scrapped item is 10 QAR. What is the cost of producing the order on process A (in terms of scrap and rework cost only)? F Which of the following Production Plans is more expensive: Plan A: Plan B: Produce the whole order (8000) in Process A Produce 5000 products on A and 3000 on B
Explanation / Answer
Given, for process A,
LSL = 20, USL = 24, u = 25, variance = 9, std.dev (s)= 3 (sqrt root of variance)
a) Target value of Process A = the value desired to be attained = mid point of LCL and UCL
= 22
b) Amount of changes in process A of target A values from mean per standard deviation
= (X-u(target))/s
= (22-25)/3
= -1
c) percentage of rework in Process A
Z = (X-u)/s = (UCL-u)/s
=(24-25)/3=-0.33
According to Z table
fraction of values lying outside UCL is X>Z
= 0.1293+0.5 = 0.6293
percentage of rework
= 62.93%
d) In B, there are two process,
Assuming each process takes equal quantity (4500+4500)
for std.dev = 4
The fraction of items lying outside LCL and UCL
Let us calculate fraction of item lying outside UCL
Z = (22.5-21)/2 = 0.75
The fraction of value from u to UCL = 0.2734 (from Z table)
The fraction of values lying outside UCL = 0.5-0.2734 = 0.2266
total fraction lying outside limits = 0.2266*2 = 0.4532
For, std.dev = 3,
Z = (22.5-21)/3 = 0.5 = 0.1915
The fraction of values lying outside UCL = 0.5-0.1915
= 0.3085
total fraction = 0.617
Here rework are items which are produced above UCL
scrap are item below LCL
Hence altogether total scrap = 0.3085+0.2266 = 0.5351
e) cost of rework = 15
cost of scrap = 10
In Process A, 62.93% items undergo rework
total cost of rework = 8000*0.6293*15 = 75516
f) A : 75,516 (as calculated above)
B : 5000 on A
Total rework cost on A = 5000*0.6293*15 = 47,197.5
3000 on B
on std.dev = 2
cost of rewok and scrap = 1500*0.2296*10+1500*0.2296*15 = 8610
on std.dev = 3
cost of rework and scrap = 1500*0.3085*10+1500*0.3085*15 = 11,568.75
total = 20,178.75
Altogether on process B = 47,197.5+ 20,178.75 = 67,376.25
Process B to be choosen for less cost of rrework and scrap
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