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# Questions ch 4

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```Ise 216
Chapter 4
question hour
23.03.2011
Q 4.12
A large automobile repair shop installs about 1250 mufflers
per year, 18 percent pf which are for imported cars. All the
imported car mufflers are purchased from a single local
supplier at a cost of 18.50\$ each. The shop uses a holding
cost based on 25 percent annual interest rate. The set up
cost for placing an order is estimated to be 28\$.
a) Determine the optimal number of imported-car mufflers
the shop should purchase each time an order is placed,
and the time between placement of orders.
b) If the replenishment lead time is six weeks, what is the
reorder point based o the level of on-hand inventory?
c) The current reorder policy is to buy imported car mufflers
only once a year. What are the additional holding amd set
up costs incured by policy?
A 4.12
пЃ¬ = (1250)(.18) = 225
c = \$18.50
i = 0.25
h = i*c = (.25)(18.50) = 4.625
K = 28
a)
Q* =
2 KпЃ¬
h
=
( 2 )( 28 )( 225 )
пЂЅ 52
4 . 625
T =cycle time= Q*/пЃ¬ = 52/225
= .2311 yrs.
b) T = 0.2311*52=12.02 weeks
Hence, r = пЃ¬пЃґ = (225/52)(6) = 25.96
c)
п‚» 26 units
If Q = 225, the average inventory level is Q/2 = 225/2 = 112.5. The annual
holding cost is (112.5)(4.625) = \$520.31. At the optimal solution, the annual
holding cost is (52/2)(4.625) = \$120.25.
The excess holding cost is \$400.06 annually.
The annual holding and set-up cost incurred by this policy is \$520.31 + 28 =
\$548.31 since there is only one set-up annually.
The set-up cost at the optimal policy
Set up= 1/0.2311*28=121.16
The cost of optimal policy= 120.25+121.16=241.40
Therefore the annual difference
=
\$306.91
Q 4.22
A purchasing agent for a particular type of silicon wafer used in the
production of semiconductors must decide among three sources. Source A
will sell the silicon wafers for 2.5 per wafer, independently of the number
of wafers ordered. Source B will selll the wafers for 2.4 each but will not
consider an order for fewer than 3000 wafers. And source C will sell the
wafers 2.3 each but will not accept an order fewer than 4000 wafers.
Assume an order set up cost of \$100 and an annual requirement of 20000
wafers. Assume a 20 percent annual interest rate for holding cost
calculations.
a) Which source should be used, and what is the size of the standing order?
b) What is the optimal value of holding and set up costs for wafers when
the optimal source is used?
c) If the replenishment lead time for wafers is three months, determine the
reorder point based on the on-hand level of inventory of waters?
A 4.22
пЃ¬
K
i
=
20,000
=
100
= 0.20
c0
=
\$2.50
c1
=
\$2.40
c2
=
\$2.30
Q
Q
(0)
(1)
(2
Q
2 KпЃ¬
=
=
1c 0
2 KпЃ¬
1c 1
2 KпЃ¬
=
1c 2
пЂЅ
пЂЅ
All Units Discount
(2)( 100 )( 20 , 000 )
пЂЅ
= 2828
(. 20 )( 2.50 )
(2)( 100 )( 20 , 000 )
= 2887
(. 2)( 2.40 )
(2)( 100 )( 20 , 000 )
= 2949
(. 2)( 2.30 )
0
only Q is realizable.
Cost at Q = 4,000
(20,000)(2.30) +
(. 2)( 2.30 )( 4, 000 )
2
пЂЅ
(100 )( 20 , 000 )
= \$47,420
( 4, 000 )
Cost at Q = 3,000
(20,000)(2.40) +
(. 2)( 2.40 )( 3, 000 )
2
пЂ«
(100 )( 20 , 000 )
0
= \$49,386.67
3, 000
Cost at Q = Q
=
2828
(20,000)(2.50) +100*20000/2828+2828/2*0.2*2.5 =51414.214
It follows that the optimal order size is Q = 4,000
A 4.22
b)
4000/2*2.3*0.20=\$1420.
c)
T = Q/пЃ¬ = 4,000/20,000 = .2 years = 2.4 months.
Hence,
пЃ§ > T
.6
mos
.
i)
ii)
2.4 mos.
3 months
пЃ§/ T = 3 / 2 . 4 = 1 . 2 5
0.25*0.2*20000=1000
Q 4.20
Filter Systems produces air filters. One filter part number
JJ39877 is supplied on exclusive contract basis to Oil Changers
at a constant 200 units monthly. Filter systems can produce
this filter at a rate of 50 per hour. Set up time to change the
settings on the equipment is 1,5 hours. Worker time is
charged at the rate of 55\$ per hour. Idle time during set ups
cost 100\$ per hour in lost profit. Filter Systems has
established a 22 percent annual interest rate for determining
holding cost. Each filter costs 2.5\$ to produce. They are sold
for 5.5\$ each to Oil Changers. Assume 6-hour days, 20
working days per month, and 12 months per year for your
calculations.
A 4.20
a)
пЃ¬= 2 0 0 * 1 2 = 2 4 0 0
units/month
P=50*6*20*12=72000 (ignore set up times)
h' = h(1 -
Q =
пЃ¬/ P )
= (.22)(2.50)(1 - 2400/72,000) = .5317
( 2 )(( 100 пЂ« 50 ) * 1 . 5 )( 2400 )
= 1,449
0 . 5317
пЃ¬/ P )
b)
H = Q (1 -
c)
T = Q/пЃ¬ = 1449/2400 = .60375 years
= (1449)(1 - 2400/72,000) = 1,401.
T1 = Q/P = 1,449/72,000 = .0201 years
.0201/.60375 = .033 or 3.3% of each cycle is up-time.
Q 4.25
Parasol systems sells motherboards. For quantities up
through 25, the firm charges \$350 per board;
between 26 and 50, it charges \$315 for each
purchased beyond 25; and it charges \$285 each for
the additional quantities over 50. A large
communication firm expects to require these
motherboards for the next 10 years at a rate of at
least 140 per year. Order set up costs are based \$30
and holding costs are based on an 18 percent annual
interest rate. What should be the size of the standing
order?
A 4.25
Incremental discount schedule
пЃ¬ = 140
K
i
=
=
30
0.18
C(Q)
=
C(Q)/Q
=
350
for
Q
(350 )( 25 ) пЂ« 315 (Q пЂ­ 25 )
for
26
(350 )( 25 ) пЂ« (315 )( 25 ) пЂ« 285 (Q пЂ­ 50 )
for
51
for
Q
875 / Q пЂ« 315
for
26
п‚Ј
Q
51
п‚Ј
Q
2375 / Q пЂ« 285
=
пЃ¬* пѓЄ
G(Q)
=
(140)
G 1( Q )
for
K * lambda
пѓє пЂ«
Q
пѓ« Q
пѓ»
G(Q)
(0)
Q
п‚Ј
350
=
пѓ¦ C (Q )
пЂ« (. 18 ) пѓ§
пѓ§
Q
пѓЁ
25
=
(1)
Q
пѓ¦ C (Q ) пѓ¶пѓ¦ Q пѓ¶
( 30 )( 140 )
пѓ·
пЂ« (. 18 ) пѓ§
пѓЄ
пѓє пЂ«
пѓ§
пѓ·пѓ§ 2 пѓ·
Q
Q
пѓё
пѓ« Q
пѓ»
пѓЁ
пѓёпѓЁ
(2 )( 30 )(140 )
= 1 1 . 5 5 п‚» 1 2 . 0 0 (realizable)
(. 18 )( 350 )
140
=
50
пѓ¶пѓ¦ Q пѓ¶
пѓ·
пѓ·
пѓ·пѓ§
пѓёпѓЁ 2 пѓё
пѓ©пЂ 875 пѓ№пЂ (30 )( 140 )
пѓ№пЂ Q
пЂ« 315 пЂ«
пЂ« .18
пЂ« 315
пѓЄпЂ Q пѓ»пЂ пѓєпЂ пѓЄпЂ пѓєпЂ Q
пѓ«пЂ пѓ«пЂ Q
пѓ»пЂ 2
123 , 700
(56 .7 )Q
пЂ«
пЂ« 44 , 178 .75
Q
2
=
п‚Ј
(2 )(123 , 700 )
56 .7
пЂЅ 66
not
realizable
п‚Ј
п‚Ј
п‚Ј
25
Q
Q
п‚Ј
50
A 4.25(2)
пѓ№ ( 30 )(140 )
пѓ№пѓ¦ Q пѓ¶
G 2( Q ) = 1 4 0 пѓЄ
пЂ« 285 пѓє пЂ«
пЂ« . 18 пѓЄ
пЂ« 2 8 5пѓє пѓ§ пѓ·
Q
пѓ« Q
пѓ»
пѓ« Q
пѓ»пѓЁ 2 пѓё
333 , 700
=
Q
(2)
Q =
пЂ«
51 .3Q
2
пЂ« 40 ,113 .75
(2 )( 333 , 700 )
51 .3
пЂЅ 114 r e a l i z a b l e
(0)
(2)
Compare average annual costs at Q and Q .
(0)
G 0 (Q ) = ( 1 4 0 ) ( 3 5 0 ) + (30 )( 140 ) пЂ« (. 18 )( 350 )(12 ) пЂЅ \$49 , 728
12
(2)
G 2 (Q
) =
333 , 700
114
пЂ«
51 .3(114 )
2
2
пЂ« 40 ,113 .75 пЂЅ \$45, 965
(2)
Hence, the optimal solution is Q which requires maintaining a standing order of 114.
Q 4.26
A local outdoor vegetable stand has exactly 1000 sq feet of space to
display three vegetables. Tomatoes, lettuce, zucchini. The
appropriate data for these items are:
Annual demand (in
pounds)
Cost per pound
tomatoe
850
lettuce
1280
Zucchini
630
0,29
0,45
0,25
The set up cost for replenishment of vegetables is \$100. The space
consumed by each vegetable is proportional to its costs. With
tomatoes requiring 0,5 sq foot per pound. The annual int. rate 0,25.
What are the optimal quantities that should be purchased of these
three vegetables?
A 4.26
The space consumed by lettuce and zucchini:
0.45/0.29 = 1.55 пѓћ lettuce consumes (0.5)(1.55) = 0.775
0.25/0.29 = .862 пѓћ zucchini consumes (0.5)(0.862) = 0.431
Compute the respective EOQs:
EOQtom = (2 )(100 )( 850 ) пЂЅ 1531
(, .25 )(. 29 )
EOQlettuce
=
(2 )(100 )(1280 )
(. 25 )(. 45 )
EOQzucchini =
(2 )(100 )( 630
(. 25 )(. 25 )
multiplier m:
W
m=
пЂЅ
пѓҐ w EOQ
i
i
пЂЅ 1509
пЂЅ 1420
1000
(. 5)(1531 ) пЂ« (. 775 )(1509 ) пЂ« (. 431 )( 1420 )
Hence, the optimal order quantities:
Qtomatoes = (1531)(.3926) = 601
Qlettuce = (1509)(.3926) = 592
Qzucchini = (1420)(.3926) = 558
пЂЅ
1000
2547
= .3926
Q 4.29
A metal fabrication shop has a single punch press. There are currently three parts that the
shop has aggreed to produce that require the press, and it appears that they will be
supplying these parts well into the future. You may assume that the press is critical
resource for these parts, so that we need not to worry about the interaction of the
press with the other machines in the shop. The relevant information here is:
Part number
Demand
Setup cost
Cost (per unit)
Production rate
(per year)
1
2500
80
16
45000
2
5500
120
18
40000
3
1450
60
22
26000
Holding costs are based on an 18 percent annual interest rate, and the products are to be
produced in sequence on a rotation cycle. Set up times can be considered negligible.
a)
What is the optimal time between set ups for part number 1?
b)
What percantage of the time is the punch press idle, assuming an optimal rotation
cycle policy?
c)
What are the opt,mal lot sizes of each part put through the press at an optimal
solution?
d)
What is the total annual cost of holding and set up for these items on the punch
press, assuming an optimal rotation cycle?
A 4.29
пЃ¬
2500
5500
1450
Input:
P
45000
40000
26000
a) compute T*: T* =
h
2.88
3.24
3.96
hп‚ў
2.7200
2.7945
3.7392
K
80
120
60
( 2 )( 80 пЂ« 120 пЂ« 60 )
( 0 . 18 (1 пЂ­ 2500 / 45000 ) * 16 )( 2500 ) пЂ« ( 2 . 7945 )( 5500 ) пЂ« ( 3 . 7392 )( 1450 )
= .1373 years
b) and c) The order quantities for each item are given by the formula
Qj = пЃ¬Tj. Obtain:
Q1 = 343.21
Q2 = 755.05
Q3 = 199.06
The respective production times are given by Tj = Qj/Pj. Substituting, one obtains:
T1 = 343.21/45000= .007626
T2 =
.018876
T3 =
.007656
It follows that the total up time each cycle is the sum of these three quantities which gives: total up time = .03416. The
total idle time each cycle is .1373 - .0342 = .1031. The percentage of each cycle which is idle time is thus .1031/.1373 =
75%.
d) Using the formula
n
G(T) = пѓҐ ( K / T пЂ« h ' пЃ¬ T / 2 ) one obtains, G(T) = \$3787.82 annually
j
j пЂЅ1
j
j
```
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