Department

ManagementCourse Code

MGOC20H3Professor

Vinh QuanStudy Guide

FinalThis

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Department of Management, UTSC

MGOC20 Operations Management: A Mathematical Approach – L01, Midterm – Solution

Q1.

(a)

p4 = $18.75, H = $18.75(.10) = $1.875/unit/year

875.1

)460)(40(22

4== H

SD

EOQ

= 140.095, EOQ4 is not admissible → set Q = 400

From

2

)( HQ

pD

Q

SD

QTC ++=

→ TC(400) = (40)(460)/(400) + 18.75(460) + (1.88)(400)/2 = $9047

p3 = $19.00, H = $19.00(.10) = $1.90/unit/year

90.1

)460)(40(2

3=EOQ

= 139.17, EOQ3 is not admissible → set Q = 200

TC(200) =

(40)(460) (1.90)(200)

(19.00)(460)

200 2

++

= $9022.00

p2 = $19.50, H = $19.50(.10) = $1.95/unit/year

95.1

)460)(40(2

2=EOQ

=137.37, EOQ2 is admissible so stop → set Q = 137.37

TC(q2*) =

(40)(460) (1.95)(137.37)

(19.55)(460)

137.37 2

++

= $9237.88

Therefore it is best to order 200 units each time since it has lowest cost. Total annual cost will be

$9022.00

(b) # of orders = D/Q* = 460/200 =2.3,

# of days between orders = Q*/D = 200/460 = .434 * 365 = 158.7 days

Q2.

E[L] = (5 + 2)/2 = 3.5 weeks, VAR[L] = (5/50 – 2/50)2/12 = 0.0003 years

700

1

)5000)(49(2][2

*== H

DSE

Q

boxes

lead time demand X is normally dist. with mean = E[X]= E[L]* E[D] = (3.5/50)(5000) = 350

var[X] = E[L]*var[D] + E[D]2 * var[L] = (3.5/50)(45)2 + 50002(0.0003) = 7641.75

75.7641=

X

= 87.4171, CB = $5/box/year

][

*

*][DEC

hq

rXP

B

=

,

)5000(5

)700)(1(

]

][*][

[=

−

−

XX

XErXEX

P

,

028.0]

4171.87

]350*

[=

−

r

ZP

,

972.0028.1]

4171.87

]350*

[=−=

−

r

ZP

From Z-Table, P[Z ≤ 1.91] = .9719 close enough, solve

4171.87

350*

91.1 −

=r

get r* = 516.97 boxes.

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