Planeación y control de la producción
Dr. Omar Romero Hernández
Tarea3
SOLUCIÓN TAREA 3
5.8
a) c0
cu
= .08 - .03 = .05
= .35 - .08 = .27
Critical ratio =
.27
.05 + .27
= .84375
From the given distribution, we have:
Q
f(Q)
F(Q)
0
.05
.05
5
.10
.15
10
.10
.25
15
.20
.45
20
.25
.70
< - - - - .84375
25
.15
.85
30
.10
.95
35
.05
1.00
Since the critical ratio falls between 20 and 25 the optimal is Q = 25 bagels.
b) The answers should be close since the given distribution appears to be close to the normal.
c) µ
= ∑xf(x) = (0)(.05) + (5)(.10) +...+(35)(.05) = 18
σ2 = ∑x2f(x) - µ2 = 402.5 - (18)2 = 78.5
(2)(32)(1032)
.36
σ =
=
8.86
The z value corresponding to a critical ratio of .84375 is 1.01.
Hence,
Q* =
5.14
a)
σz + µ = (8.86)(1.01) + 18 = 26.95 ~ 27.
Note: We assume 4 weeks/month and 48 weeks/year.
Monthly demand is normal (µ = 28, σ = 8)
γ = 14 weeks = 3.5 months ⇒ LTD ~ normal
with µ = (28)(3.5) = 98
σ = (8)
3.5
= 15
Planeación y control de la producción
Dr. Omar Romero Hernández
h
λ
p
K
µ
=
=
=
=
=
Tarea3
Ic = (.3)(6) = 1.8
(28)(12) = 336 year
10
15
98
Q0 = EOQ =
(2)(336)(15)
1.8
F(R1) = (75)(1.8)
(10)(336)
= 75
= .04 ⇒ z = 1.75 giving
R1 = σz + µ = 124
and n(R1) = σL(z) = .2426
Q1 =
(2)(336)
(15 + (10)(.2426))
1.8
F(R2) + (81)(1.8) =
= 81
.0434
(10)(336)
z = 1.71 ⇒ R2 = σz + µ = 124.
Conclude that (Q,R) = (81,124)
b)
5.15
Since R2 = R1, we stop.
S = R - µ = 124 - 98 = 26 units.
a) Type 1 service of 90%
EOQ = 75 (from 13 (a))
F(R) = .10, z = 1.28, R = σz + µ = (15)(1.28) + 98 = 117
Q,R) = (75,117)
b) Find Type II service level achieved in part (a).
n( R )
σL( z) (15)(.0475)
=1−β =
=
Q
Q
75
⇒
5.20
=
.0095
β = .9905 (99.05% service level)
λ
µ
σ
h
K
p
=
=
=
=
=
=
(12)(52) = 624 units per year
(12)(3) = 36 units per lead time
4 3 = 6.9282
(.2)(4) = $0.80
$75
$25
EOQ =
(2)(75)(624)
0.8
= 342.
Planeación y control de la producción
Dr. Omar Romero Hernández
a)
Tarea3
1 - F(R0) = (342)(.8)/(25)(624) = .0174.
From Table A-4, z = 2.11, L(z) = .0063, n(R) = σL(z) = .04365.
Q1 =
(2)(624)[75 + (25)(.04365)]
.8
=
345.
1 - F(R1) = (345)(.8)/(25)(624) = .0177.
Table A-4 gives z = 2.105 which is close enough to the previous z value to stop.
R = σz + µ = (6.9282)(2.105) + 36 = 51
Q = 345.
Hence the optimal solution to part a) is (Q,R) = (345,51).
b) The type 1 service is the value of F(R) at the final solution.
This is 1 - .0177 = .9823 (98.23%).
c)
The type 2 service (or fill rate) is the value of 1 - n(R)/Q at the optimal solution. This is 1 .04365/345 = .99987 (99.99%).
d) Solving for F(R) = .95, one obtains z = 1.645, giving R = (6.9282)(1.645) + 36 = 47. The
optimal Q is the EOQ of 342.
5.23
e)
Set Q = 342 and solve n(R)/Q = 1 - β for R. One obtains: n(R) = (342)(.05) = 17.1. L(z) =
17.1/6.9282
=
2.4682.
From
Table
A-4,
one
finds
z
=
-2.465, giving R = (-2.465)(6.9282) + 36 = 19.
f)
This is similar to e) except that β = .99 and Q = 500. In this case one obtains n(R) =
(500)(.01) = 5. L(z) = 5/6.9282 = .7217 and from Table A-4, z = -.535. Hence, R = (.535)(6.9282) + 36 = 32.
a)
From the solution of problem 14
(Q,R) = (81,124)
Set s = R = 124
S = R + Q = 205
b) x1 = 26 < 124 ⇒ order to 205 in January.
(order quantity = 205 - 26 = 179)
x2 = 205 - 37 = 168.
do not order in February.
x3 = 168 - 33 = 135.
do not order in March.
x4 = 135 - 26 = 109 < 124.
order to 205 in April.
(order quantity = 205 - 109 = 96)
x5 = 205 - 31 = 174.
x6 = 174 - 14 = 160.
(x7 = 160 - 40 = 120
do not order in May.
do not order in June.
⇒ order 85 units in July.)
Planeación y control de la producción
Dr. Omar Romero Hernández
5.25
Tarea3
a)
Cum
Item
Volume
Profit
Fraction
Annual
Annual
of
Profit
Profit
Total
Costume Jewelry
1900
$15.00
28500.00
28500.00
0.41
Novelty Gifts
2050
$12.25
25112.50
53612.50
0.76
Earrings
1285
$3.50
4497.50
58110.00
0.83
Children's Clothes
575
$6.85
3938.75
62048.75
0.88
Men's Jewelry
875
$4.50
3937.50
65986.25
0.94
Tee Shirts
1550
$1.25
1937.50
67923.75
0.97
Greeting Cards
3870
$0.40
1548.00
69471.75
0.99
Chocolate Cookies
7000
$0.10
700.00
70171.75
1.00
A
B
C
b) The cookies attract customers to the store. (In retail parlance, it would be known as a loss
leader.)
5.38
a) Exponential smoothing equations are:
Dt = αDt + (1 - α) Dt-1
MADt = α|Dt - Dt-1| + (1 - α)MADt-1
D-3 = (.2)(126) + (.8)(135) = 133.2
MAD-3 = (.2) |126 - 135| + (.8)(18.5) = 16.6
D-2 = (.2)(138) + (.8)(133.2) = 134.2
MAD-2 = .2 |138 - 133.2| + .8(16.6) = 14.24
D-1 = (.2)(94) + (.8)(134.2) = 126.2
MAD-1 = (.2) |94 - 134.2| + (.8)(14.24) = 19.43
Answer:
Mean: 126.2
MAD:
19.43
b) Since the demand is tracked by month, we use 2.5 months for the lead time.
σmonth ≈ 1.25 MAD = (1.25)(19.43) = 24.29
µmonth = 126.2
µ = (2.5)(126.2) = 315.5
σ =
24.29 = 38.4
2.5
c) µ = λµτ where λ = monthly demand
µτ = expected lead time
σ2 = µv2 + λ2στ2
where v2 = var. of monthly demand
στ2 = var. of lead time
We are given that στ = 3.3 weeks. This is converted to months by dividing by 4 to obtain στ
= 3.3/4 = .825 months.
Hence:
µ = (126.2)(2.5) = 315.5
σ2 = (2.5)(24.29)2 + (126.22(0.825)2 = 12,314.93
σ = 111.0.
d) F(Q) = .98 gives z = 2.05
R = σz + µ = (211.7)(2.05) + 315.5 = 750.