Production and Operations Analysis, Fifth Edition
Solutions To Problems From Chapter 4

c
K
I
4.10
=
=
=
=
280
2.40
45
.20
2K

h
(2)(45)(280)
 229
(.2)(2.40)
a)
Q* =
b)
T = Q*/ = 229/280 = .8179 yrs. (= 9.81 months)
c)
G* = 2K h =
(2)(45)(280)(.2)(2.40) = $109.98
d)
229
3
wks
R
9.82 mos.
r =  = (280)(3/52) = 16.15
r = 16 units
4.12
 = (1250)(.18) = 225
c = $18.50
I = .25
h = Ic = (.25)(18.50) = 4.625
K = 28
a) Q* =
2K
(2)(28)(225)
 52
=
(.25)(18.50)
h
T = Q*/ = 52/225 = .2311 yrs.
b) T = 12.02 weeks
Solutions For Chapter 4
Hence, r =  = (225/52)(6) = 25.96  26 units
c) 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 average annual holding and set-up cost at the optimal policy
is
2Kh
=
(2)(28)(225)(4.65)
= $241.40.
Therefore the annual difference
4.13
True optimal Q =
(2)(15)(280)
.48
= $306.91
= 132
fractional error = 1/2[Q/Q* + Q*/Q]
= 1/2[229/132 + 132/229]
true optimal cost = 2Kh = (2)(15)(280)(.48)
a 15.6% error = $9.91 annually.
4.17
P

K
c
I
=
1.156
= 63.50
= 10,000/day
= .6 x 106/year = .6 x 106/250 = 2400/day
= 1500
= 3.50
= .22 + .12 = .34
h = Ic(1 - /P) = (.34)(3.50)(1 - 2400/10,000) = .9044
2K
(2)(1500)(2400)(250)

= 44.612 lbs
h'
.9044
a)
Q =
b)
T1 = Q/P = 44,612/10,000 = 4.46 days
T = Q/ = 44,612/2400 = 18.59 days
T2 = T - T1 = 14.13
Production and Operations Analysis, Fifth Edition
Up time = 4.46 days or 4.46/18.59 = .24 (24% of cycle is up time)
Down time = 14.13 days (76% of cycle is down time)
c)
Annual holding and set-up cost =
=
2Kh'
(2)(1500)(600, 000)(.9044) = $40,347.49 annually.
If the compound sells for $3.90 per lb., then the annual profit exclusive of
holding and set-up cost is
(3.90 - 3.50)(600,000) = $240,000.00
Hence the profit generated from this item is
$240,000.00 - 40,347.49 = $199,652.51 annually.
4.20
a)
h' = h(1 - /P) = (.22)(2.50)(1 - 2400/72,000) = .5317
(2)( 232.5)( 2400)
= 1,449
.5317
Q =
b)
H = Q (1 - /P) = (1449)(1 - 2400/72,000) = 1,401.
c)
T = Q/ = 1449/2400 = .60375 years
T1 = Q/P = 1,449/72,000 = .0201 years
.0201/.60375 = .033 or 3.3% of each cycle is up-time.
4.23
Incremental schedule for source B.
C(Q)
=
2.55Q for Q  3,000
C(Q)
=
=
(2.55)(3,000) + 2.25(Q - 3,000) for Q  3,000
7650 + 2.25Q - 6750 = 900 + 2.25Q for Q  3,000
C(Q)/Q
=
=
2.55 for Q ≤ 3,000
900/Q + 2.25 for Q ≥ 3,000
Solutions For Chapter 4
G(Q) = [20,000]
900
(100)(20, 000)
900
Q
 2.25 
 (.20)
 (2.25)
Q
Q
Q
2
=
18,000,000 2,000,000 (.20)(2.25)Q


 $45,090
Q
Q
2
=
20,000,000
 0.225Q  $45,090
Q
The minimizing Q occurs where Q =
cost =
20,000,000
.225
= 9428
20,000,000
+ (.225)(9428) + $45,090 = $49,332.64
9428
Since the annual cost for source A exceeds $50,000, now source B should be used.
4.25
Incremental discount schedule
 = 140
K = 30
I = .18
C(Q) =
C(Q)/Q =
350
for Q  25
(350)(25)  315(Q  25)
for 26  Q  50
(350)(25)  (315)(25)  285(Q  50)
for 51  Q
350
for Q  25
875/ Q  315
for 26  Q  50
2375 / Q  285
for 51  Q
 C (Q)  (30)(140)
 C (Q)  Q 
 
G(Q) = (140) 

 (.18)

Q
 Q 
 Q  2 
Production and Operations Analysis, Fifth Edition
Q(0) =
G1(Q) = 140
=
(2)(30)(140)
= 11.55  12.00 (realizable)
(.18)(350)
875 
(30)(140)
875
Q
 315 
 .18
 315



Q
 Q 

 Q
 2
123,700 (56.7)Q

 44,178.75
Q
2
Q(1) =
(2)(123,700)
 66 not realizable
56.7
 2375
 (30)(140)
 2375
 Q 
G2(Q) = 140 
 285 
 .18
 285 
Q
 Q

 Q
 2 
=
333, 700 51.3Q

 40,113.75
Q
2
Q(2) =
(2)(333,700)
 114 realizable
51.3
Compare average annual costs at Q(0) and Q(2).
G0(Q(0)) = (140)(350) +
G2(Q(2)) =
(30)(140) (.18)(350)(12)

 $49,728
12
2
333, 700 51.3(114)

 40,113.75  $45,965
114
2
Hence, the optimal solution is Q(2) which requires maintaining a standing order of 114.
4.29
The input data for this problem are:

2500
5500
1450
P
45000
40000
26000
h
2.88
3.24
3.96
h
2.7200
2.7945
3.7392
K
80
120
60
Solutions For Chapter 4
a) First we compute T*.
(2)(80  120  60)
(2.72)(2500)  (2.7945)(5500)  (3.7392)(1450)
T* =
= .1373 years
b) and c)
The order quantities for each item are given by the formula
Qj
= Tj.
Q1
Q2
Q3
= 343.21
= 755.05
= 199.06
Obtain:
The respective production times are given by Tj = Qj/Pj.
Substituting, one obtains:
T1
T2
T3
= .007626
= .018876
= .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
G(T) =  (K
n
j 1
j
/ T  hj '  j T / 2)
one obtains, G(T) = $3787.82 annually.
4.36
President used the following EOQ.
Q1 =
(2)(100)(1800)
(.3)(2.40)
= 707
"True" EOQ was:
Q2
=
(2)(40)(1800)
(,.2)(2.40)
= 548
Production and Operations Analysis, Fifth Edition
Cost error in percent =
1  Q Q*  1 707 548  =



2 
Q * Q 
 2 548 707 
1.033
Error is 3.3% only
Optimal cost =
= 263
(2)(40)(1800)(.2)(2.40)
Additional cost = 263(.033) = $8.68 annuall
4.41
a)
c = .35 + .15 + .05 = .55
h = Ic = .11
EOQ
b)
2K

h
Q
50,000

Q* 120,604
(2)(400)(2,000,000)
(.11)
= 120,604
= .41458
Error = 0.5[Q/Q* + Q*/Q] = 1.4133  41.33% additional cost
Optimal cost
=
2Kh 
=
(2)(400)(2, 000, 000)(.11)
13,266.5
~
$13,267 annually.
Hence, the additional cost of using suboptimal policy:
= (13,267)(.4133) = $5483 annually.