1. (20 points) Consider the following process diagram. Defect rate for the entire process is uniform, at 2%.

A B C D E F G 50,000

Determine the number of parts to be manufactured if:

2. (20 points) Consider the following process diagram.

Derive a general formula that will provide the number of outputs after the process, and compute for the total number of outputs
if 10,000 units were the inputs. Defect rates in percent are as follows: 3%, 2%, 5%, 4%, 1%, and 7%.

1 3 4 6

2 5
 d =
0.02

06=50 ooo units

a) DEFD performed once

Process A B C D E F D G
sequence
-

:
- -
- - - -

06
IA =

( i -

da ) ( I
-

d.) Ctdcctd ,D2(td⇒( I

d.) ( told

since d is uniform ,

In = 06 = 50000 =
58 770.77 a 58 771 unit
8
( i .

d) ( i
.

0.0278

b) DEFD
performed twice

Process A D E F D E F D G
sequence B C
- -
-
- -
- - -
:
- -

IA = 06

(1 -

data -

dB)a -

dd ( I .
d ,o)3( 1- di⇒2( 1- d.) 24 -

do )

IA = 50 000 =
62 442.92 a
64443 units
"
(1-0.02)
06
=
In (I -

da ) Given :

Oi 03
Ice =
04 t 05 d ,
=
d4= 0.04

04
=

I4 (I -

did dz =
0 .

02
dg
=
0.01

05 =
If (1 -

dot dg
= Oi 05 dg
=
0.07

Is =
I4d4

Ice =
I4 (I -d4) +
I4d¢( 1- ds)
If =
Iq [(l -

d 4) + d4 (I -

do )]

I4 =
03 =
I3 (1 .
DD
I3 0 02
=
t
,

Is =
Ii (l -
di ) + Iz ( l -

d2)
I3 = Ii (I -

di ) t Iid , ( I -

dz )
I3 =
I ,
[ ( I -

on + di ( I
-

dod ]

Iq
= Ii [ ( I
-

di ) + d ,
(I -

DD ] (, -
dg )

06 = Ii (I -

d3)( I -d6)[ ( I .

d.) + d ,
Ct d2)][ a -d4) + da ( tds )]

06 =
( 0.93 )[ ( 097 101<(0-95)
) +0.03 ( 0.98 ) ][ ( 0.96 ) +0.04 ( o.gg
) ]

=
8826 .
17
a
8826 units