6-49.
Demand
Probability
P(Demand _____)
50
0.05
1.00
75
0.10
0.95
100
0.20
0.85
125
0.30
0.65
150
0.20
0.35
175
0.10
0.15
Stock 150
b. ML = 35; MP = 80 – 35 = 45
MR/(ML + MP) = 35/(35 + 45) = 0.4375
Stock 125
c.
Demand
Probability
P(Demand _____)
50
0.25
1.00
75
0.25
0.75
100
0.25
0.50
125
0.25
0.25
6-50. ML = 1.20; MP = 4.00 – 1.20 = 2.80
ML/(ML + MP) = 1.20/(1.20 + 2.80) = 0.30
Using the normal distribution table, the z-value associated with the upper 30% of the normal
distribution is between 0.52 and 0.53. We will choose 0.52 since P(Z0.52) is slightly more
than 0.30
6-51. ML = 35; MP = 15
ML/(ML + MP) = 35/(35 + 15) = 0.70
6-52. a. We are given = 400. While is not given, it can be found since P(350 X 450) =
0.85, which means there is a 15% chance it is outside this range. Note that 350 is 50 units below
the mean and 450 is 50 units above the mean. We know P(X 350) = 15%/2 = 7.5% and P(X
450) = 7.5%. From this,
P(X 450) = 0.85 + 0.075 = 0.925.
1.44 = 50
= 50/1.44 = 34.7
6-53. We are given = 3000. While is not given, it can be found since P(2900 X 3100) =
0.70, which means there is a 30% chance it is outside this range. Note that 2900 is 100 units be-
low the mean and 3100 is 100 units above the mean. We know that P(X 2900) = 30%/2 = 15%
1.04 = 100
= 100/1.04 = 96.15
6-54. Annual demand = D = 8,000; daily production rate = p = 200
a. daily demand = d = 8,000/250 = 32 units
b. number of days of production = 400/200 = 2 days
6-55.
( )
*2 8,000 120
2213.81 units
32
50 1
1200
s
h
DC
Qd
Cp
= = =
−
−
For this optimal value of Q,
Thus, we could save 10,800 – 8,980 = $1,820 per year.
6-56. Material structure tree for item A.
Part B: 1 × number of As = 1 × 50 = 50
Part C: 3 × number of As = 3 × 50 = 150
6-57. Gross material requirements plan for 50 units of A:
WEEK
LEAD
TIME
1
2
3
4
5
6
(WEEKS)
A
Required
date
50
1
Order re-
lease
50
Required
date
50
Order re-
lease
Required
date
150
Order re-
lease
150
D
Required
date
100
1
Order re-
lease
100
E
Order re-
lease
150
150
Required
date
300
Order re-
lease
300
Required
date
150
150
6-58. Net Material Requirements Plan
WEEK
Item
LEAD
TIME
1
2
3
4
5
6
(WEEKS)
A
Gross
50
1
On-hand
10
10
Net
40
Order
receipt
40
Order
release
40
B
Gross
40A
2
On-hand
15
15
Net
25
Order
receipt
25
Order
release
C
Gross
120A
1
On-hand
20
20
Net
100
Order
receipt
100
Order
release
100
D
Gross
50B
1
On-hand
10
Net
Order
release
40
Gross
75B
On-hand
10
Net
Order
receipt
100
6-59. a, b.
Units needed:
Gross Material Requirements Plan
WEEK
1
2
3
4
5
6
7
Lead
Time
(WEEKS)
S
Required
date
100
2
Order
release
100
T
Required
date
1
Order
release
100
U
Required
date
50
2
Order
release
50
V
Required
date
100
2
Order
release
100
W
Required
date
200
3
Order
release
200
Required
date
100
X
1
Order
release
100
Y
Required
date
25
2
Order
release
25
6-59. c. Net material requirements plan:
WEEK
LEAD
TIME
Item
1
2
3
4
5
6
7
(WEEKS)
Net
80
Order
release
80
On–
hand
20
Gross
100
On–
hand
20
T
1
Net
60
Order
receipt
60
Order
release
60
U
Gross
40S
Order
receipt
30
Gross
On–
hand
30
Net
On–
hand
10
Net
30
Gross
On–
hand
30
Net
W
3
Order
receipt
90
Order
release
90
Gross
1
On–
hand
25
Net
Order
receipt
35
Order
release
35
Gross
15U
Y
2
On–
hand
15
Net
0
Order
receipt
0
Order
release
0
1
Net
Order
receipt
80
Order
release
Gross
90U
On–
hand
10
6-60. Material structure tree for SL72
6-61.
1
2
3
4
Lead
time
SL72
Required
Date
800
1
On-Hand
0
Net
800
Order Re-
ceipt
800
Order Re-
lease
800
A
Required
Date
800SL72
1
On-Hand
0
Net
800
Order Re-
ceipt
800
Order Re-
lease
800
B
Required
Date
800 SL72
2
On-Hand
0
Net
800
Order Re-
ceipt
800
Order Re-
lease
800
On-Hand
0
Net
800
Order Re-
ceipt
800
On-Hand
Net
1600
Order Re-
ceipt
1600
Order Re-
lease
1600
E
Required
Date
3200 C
1
On-Hand
0
Net
3200
Order Re-
ceipt
3200
lease
On-Hand
Net
1600
Order Re-
1600
Order Re-
lease
1600
Order Re-
3200