Solutions on Stoch Inventory Ch 5. (Nahmias)

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

=

σ2 =

∑xf(x) ∑x2f(x) σ =

= (0)(.05) + (5)(.10) +...+(35)(.05) = 18 - µ2 = 402.5 - (18)2 = 78.5 (2)(32)(1032) .36

=

8.86

The z value corresponding to a critical ratio of .84375 is 1.01. Hence, Q* = 5.9

cu c0

= =

σz + µ = (8.86)(1.01) + 18 = 26.95 ~ 27.

.65 -.50 = .15 .50

Critical ratio =

.15 = .231 .15 + .50

The Cumulative Distribution Function of Demand is: Quantity Sold

CDF

100,000-150,000

.10 < - - - Critical Ratio = .231

150,001-200,000

.25

•

•

•

•

•

•

Using linear interpolation, the optimal solution is: Q

5.10

*

⎡ .231 − .10 ⎤ 50,000 + 150,000 = 193,667. ⎢⎣ .25 − .10 ⎥⎦

=

a) A period is three months. Holding cost per year is $500, which means that the cost for a 3month period is: 500/4 = $125 = c0 cu = 250 (emergency shipment cost) 250 2 = 250 + 125 3

Critical ratio = Hence Q b)

*

Critical ratio

c0 = 125 *

F(Q ) = .5454 *

=

= .44.

= σz + µ = (6)(.44) + 60 = 62.64 ≈ 63 cars

cu = 150

Q

= .667 ⇒ z

σz +

µ

⇒ z

=

150 150 + 125

= .5454

= .11

= (6)(.11) + 60 = 60.66 ~ 61 cars.

c) This corresponds to an infinite horizon problem with lost sales. From part 3 of Appendix B at the end of the chapter. cu = lost profit = $3,500 c0 = holding cost = 125 Critical ratio =

3500 3500 + 125

= .9655

*

⇒ z = 1.82 which gives Q = (6)(1.82) + 60 = 70.92 ~ 71 cars.

5.11

a) X s2 b)

= 34.0 = 204.4 (s = 14.3) cu = 60 - 40 = 20 c0 = 40 - 29 = 11 Critical ratio = z = .37,

c) Value

20 20 + 11

Q* =

= .6452

σz + µ = (14.3)(.37) + 34 = 39.3 ≈ 39

#Times Observed

Relative Freq.

Cum Freq.

10

1

.1

.1

20

1

.1

.2

30

4

.4

.6 < - - - critical ratio = .6452

Q

*

40

2

.2

.8

50

1

.1

.9

60

1

.1

1.0

≈ 33 by linear interpolation.

d) The normal approximation is not very accurate since the order quantity it recommends is almost 20% too large. 5.12

co

= 28.50 - 20.00 + (.4)(28.50) = 19.90

cu

= 150 - 28.50 = 121.50. Critical Ratio

= cu/(cu + co) = 121.5/(121.5 + 19.9) = .86

a) Demand is assumed to be uniform from 50 to 250. We wish to find Q satisfying:

Area = 0.86

50

Q

250

From the picture it follows that (Q - 50)/200 = .86,

Q = (.86)(200) + 50 =

222.

b) In this case the demand distribution is assumed to be normal with mean 150 and standard deviation 20. We wish to find Q to solve: F(Q) = .86 From Table A-4 we have that z = 1.08, giving: Q = µ +

σz = 150 + (20)(1.08) = 172.

c) The uniform distribution in this case has a higher variance. The formula for the variance of a uniform variate on (a,b) is (b-a)2/12. Substituting b = 250 and a = 50 gives a variance in part a) of 3333. The variance in part b) is (20)2 = 400. 5.13

a)

h K p λ

= = = =

(1.50)(.28) = .42 100 12.80 (280)(12) = 3360

µ = (280)(5) = 1400 σ = 77 5 = 172.18

2Kλ + h

EOQ = Q0 =

1 - F(R1)

=

Qh (1265)(.42) = pλ (12.80)(3360)

z1 = 2.24, Q1

2λ [K + pn(R)] = h

=

= .0124 n(R1) = .75

(2)(3360) [100 + 12.80(.75)] .42

Qh (1324)(.42) = pλ 12.80(3360)

z2 = 2.23

=

= 1324

.0129

L(z2) = .004486

2(3360) [100 + 12.80(.77)] .42

=

= 1265

L(z1) = .0044,

1 - F(R2) =

Q2

(2)(100)(3360) .42

n(R2) = .77

= 1326

Close enough to stop. z = 2.23, R = σz + µ = (172.18)(2.23) + 1400 =1784 Optimal (Q,R) = (1326,1784)

b)

G(Q,R)

=

Kλ ⎡ Q ⎤ pλn(R) + h + R− µ + ⎣ 2 ⎦ Q Q

Kλ (100)(3360) = Q 1326

h

⎡ Q + R − µ ⎤ ⎣ 2 ⎦

=

=

$253.39

$439.74

pλn(R) (12.80)(3360)(.77) = Q 1326

= $24.97

c) σ = 0 ⇒ EOQ solution cost =

⇒

2Kλh = (2)(100)(3360)(.42)

= $531.26

G(Q,R) = $718.10 ⇒ cost of uncertainty = $718.10 - 531.26 = $186.84 yearly. 5.14

a)

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) h λ p K µ

= = = = =

3.5

= 15

Ic = (.3)(6) = 1.8 (28)(12) = 336 year 10 15 98

Q0 = EOQ = F(R1) =

(2)(336)(15) 1.8

(75)(1.8) (10)(336)

= 75

= .04 ⇒ z = 1.75 giving

R1 = σz + µ = 124 and n(R1) = σL(z) = .2426 Q1 = F(R2) +

(2)(336) (15 + (10)(.2426)) 1.8 (81)(1.8) = (10)(336)

= 81

.0434

z = 1.71 ⇒ R2 = σz + µ = 124. Conclude that (Q,R) = (81,124) b)

S = R - µ = 124 - 98 = 26 units.

Since R2 = R1, we stop.

5.15

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

⇒ 5.16

=

.0095

β = .9905 (99.05% service level)

Type 1 service of 95% Q = EOQ = 1265 ⎛ R − µ ⎞ F(R) = .95, ⎝ σ ⎠

5.17

= 1.645, which gives R = 1683.

Type 2 service of 95%. Requires iterative solution. Q0 = EOQ = 1265 n(R1) = (1 - β)Q = (.05)(1265) = 63.25 L(z1) =

n(R1 )

σ

=

63.25 172.18

= .3673

z1 = .065, 1 - F(R1) = .474 Q1 =

=

⎛ n(R) ⎞ n(R) + (EOQ) 2 + ⎜ ⎝ 1 − F(R)⎠ 1 − F(R)

⎛ n(R) ⎞ 63.25 2 + (1265) + ⎜ ⎝ 1 − F(R)⎠ .474

2

2

= 1405

n(R2) = (1 - β)Q1 = (.05)(1405) = 70.25 L(z2) =

70.25 172 .18

= .4080

z2 ≈ -.02

1 - F(R2) = .508 Q2

=

R2 = σz + µ = 1397

⎛ n(R) ⎞ 70.25 2 + (1265) + ⎜ ⎝ 1 − F(R) ⎠ .508

2

= 1411

n(R3) = (.05)(1411) = 70.54 L(z3) = .4097,

z3 ≈ -.02,

R3 = 1397

Same value. Stop. (Q,R) = (1411,1397) Imputed p =

5.18

Qh (1411)(.42) = λ (1 − F(R)) (3360)(.508)

= $ .35

Holding cost is h[Q/2 + R - µ]. Using a type 1 service objective, the policy obtained was (Q,R) = (1265, 1683) which results in an average annual holding cost of $384.51. Using a type 2 service objective, the policy was (Q,R) = (1411,1397) which results in an average annual holding cost of $295.05 The difference is $89.46

5.19

Weekly demand has mean 38 and standard deviation

130

LTD has µ = 38 x 3 = 114 and σ = a)

3

1 - F(R) = z = 2.59,

b)

130 =

19.75

Qh (500)(.40)(18.80) = pλ (400)(1976)

= .004757

R = σz + µ = 165

Must determine optimal Q by iteration EOQ = 198 = Q0 1 - F(R0) =

(198)(7.52) (400 )(1976)

= .0018838

z = 2.90, L(z) = .000542, n(R0) = σL(z) = .0107 Q =

2 λ [K + pn(R)] h

= 204

This value of Q turns out to be optimal. (must iterate once more to obtain R = 171). c) Use formula G(Q,R) = h[Q/2 + R - µ] + Kλ/Q + pλn(R)/Q Substitute (Q,R) = (500,165) from part (a) = (204,171) from part (b) Obtain G(500,165)

=

$2606.75 > Δ cost = $641.59 yearly

G(204,171)

=

$1965.16

d) Use the relationship n(R) = (1 - β)Q to determine R and the fact that n(R) = σ sL(z). Hence, L(z) =

n(R)

σ

=

(1 − β )Q

σ

=

(.01)(198) 19.75

= .10026

From Table B-4, z ≈ .90. Since R = σz + µ we obtain R = (19.75)(.90) + 114 = 132. The imputed shortage cost is: ^p

=

Qh λ ((1 − F(R))

= $4.09.