Advanced Operations Management
Solution Set 6
1.
A
B
1
20
10
2
0
30
Week
3
4
30
10
0
10
a. Product C: Lot-for-Lot Method
Week
1
2
3
GRt
70
90
60
OHIt (120)
50
10
0
SRt
0
50
50
NRt
0
0
50
PPt
0
50
50
5
20
5
6
0
10
4
50
0
50
50
55
5
55
0
55
55
30
6
30
0
30
30
X
4
220
0
0
220
120
5
120
0
0
120
X
6
X
0
0
X
X
b. Product D: Lot-for-Lot Method
Week
GRt
OHIt(160)
SRt
NRt
PPt
1
0
160
0
0
0
2
200
40
80
0
160
3
200
0
0
160
220
c. Product C: EOQ Lot-Sizing Method
EOQ = ((2*100*200)/44)^(1/2) = 100
Week
GRt
OHIt (120)
SRt
NRt
PPt
1
70
50
0
0
0
2
90
10
50
0
100
3
60
50
100
50
0
4
50
0
0
50
100
5
55
45
100
55
0
6
30
15
0
30
X
d. Product C: Part-Period Balancing Lot-Sizing Method
Week
1
2
3
4
5
6
GRt
70
90
60
50
55
30
OHIt(120)
50
10
50
0
30
0
SRt
0
50
100
0
85
0
NRt
0
0
50
50
55
30
PPt
0
100
0
85
0
X
Produced during week
2 for Week(s)
3
3, 4
Set Up Cost
Holding Cost
200
200
$0
4(50) = $200*
3, 4, 5
Produced during week
3 for Week(s)
5
5, 6
200
Set Up Cost
4(50+55+55) = $640
Holding Cost
200
200
$0
4(30) = $120*
2. Let I represent Indianapolis, C Chicago, and B Bloomington.
f4(I) = 12, f4(C) = 17-2 = 15, f4(B) = 16-5 = 11
f3(I) = Max(12-0+ f4(I) = 24, 12-5+ f4(B) = 18, 12-2+ f4(C) = 25*)
f3(B) = Max(16-5+ f4(I) = 23, 16-0+ f4(B) = 27*,16-7+ f4(C) = 24)
f3(C) = Max(17-2+ f4(I) = 27, 17-7+ f4(B) = 21, 17-0+ f4(C) = 32*)
f2(I) = Max(12-0+ f3(I) = 37, 12-5+ f3(B) = 34, f3(C) = 42*)
f2(B) = Max(16-5+ f3(I) = 36, 16-0+ f3(B) =43*, 16-7+ f3(C) = 41)
f2(C) = Max(17-2+ f3(I) = 40, 17-7+f3(B) = 37, 17-0+ f3(C) = 49*)
f1(B) = Max(-5+ f2(I) = 37, 0+ f2(B) = 43*, -7+ f2(C) = 42)
Thus the salesman should speak in Bloomington on Monday through Wednesday, and
go to Indianapolis on Thursday.
3. Let inventory be denoted by i. Clearly, 300 units is the most we would want to enter
the last month. Then
f3(300) = 0, f3(200) = 250+12(100) = 1450, f3(100) = 250+12(200) = 2650,
f3(0) = 250+12(300) = 3850
Similarly,
f2(0) = Min (3250+0+ f3(0) = 7100 (x2 = 300), 4250+150+ f3(100) =7050(x2 = 400),
5250+300+ f3(200) = 7000(x2 = 500), 06250+450+ f3(300) = 6700*(x2 = 600) )
f2(100) = Min (6100 (x2 = 200), 6050 (x2 = 300), 6000(x2 = 400), 5700* (x2 = 500))
f2(200) = Min ( 5100 (x2 = 100), 5050 (x2 = 200), 5000 (x2 = 300), 4700* (x2 = 400))
f2(300) = Min (3850 (x2 = 0), 4050 (x2 = 100), 4000 (x2 = 200), 3700* (x2 = 300))
f2(400) = Min (2800 (x2 = 0), 3000 (x2 = 100), 2700* (x2 =200))
f2(500) = Min (1750 (x2 = 0), 1700* (x2 =100))
f2(600) = 450 (x2 = 0).
f1 (0) = Min (8950* (x1 = 200), 9100 (x1 = 300), 9250 (x1 = 400), 9400 (x1 = 500),
9550 (x1 = 600), 9700 (x1 = 700), 9600 (x1 = 800))
Thus we produce 200 radios in month 1 and 600 in month 2.
4. Let the revenue remaining to be invested in period t be denoted by i. Then
f3(4) = 15, f3(3) = 13, f3(2) = 8, f3(1) = 7, f3(0) = 3.
f2(1) = Max (3+ f3(1) =10*, 6+ f3(0) = 9)
f2(2) = Max (3+ f3(2) = 11, 6+ f3(1) = 13*, 10+ f3(0) = 13*)
f2(3) = Max (3+ f3(3) = 16, 6+ f3(2) = 14, 10+ f3(1) = 17*, 12+ f3(0) = 15)
f2(4) = Max (3+ f3(4) = 18,6+ f3(3) = 19*,10+ f3(2) = 18,12+ f3(1) = 19*,14+ f3(0)= 17)
f1(4) = Max (4+ f2(4) = 23,7+ f2(3) = 24*,8+ f2(2) = 21,9+ f2(1) =19,11+ f2(0) = 17)
Thus site 1 and 3 should get 1 million dollars, site 2 2 million.
5. Cost of operation for keeping a phone n years.
N
Cost
1
40+20 =60
2
40+20+30=90
3
40+20+30+40=130
4
40+20+30+40+60=190
5
40+20+30+40+60+70=260
Let g(t) be the cost of keeping a phone to the end of year 5 assuming that we have a new
phone at the beginning of year t.
g(5) = 60
g(4) = Min (60+g(5) =120, 90*)
g(3) = Min( 60+g(4) = 150, 90+g(5) = 150, 130*)
g(2) = Min(60+g(3)= 190, 90+g(4)= 180*, 130+g(5)=190, 190)
g(1) = Min (60+g(2)=240, 90+g(3)=220*, 130+g(4)=220*, 190+g(5)=250, 260)
g(0) = Min (60 +g(1) = 280, 90+g(2)=270, 130+g(3)= 260*, 190+g94)=280,
260+g(5)= 320)
Thus a phone purchased at time 0 should be kept for three years, and then a new one
purchased and kept for three years.
6. Let NYC = City 3, LA = City 2, Miami = City1.
f3(0) = 0, f3(1) = 80, f3(2) = 150, f3(3) = 210, f3(4) = 250, f3(5) = 270, f3(6) = 280.
f2(0) = 0*
f2(1) = max (0+ f3(1) = 80, 100+ f3(0) = 100*)
f2(2) = max (0+ f3(2) = 150, 100+ f3(1) = 180, 195+ f3(0) = 195*)
f2(3) = max (0+ f3(3) = 210, 100+ f3(2) = 250, 195+ f3(1) = 275*, 275+ f3(0) = 275*)
f2(4) = max (0+ f3(4) = 250,100+ f3(3) = 310, 195+ f3(2) =345, 275+ f3(1) =355*,
325+ f3(0) = 325)
f2(5) = max (0+ f3(5) = 270, 100+ f3(4) = 350, 195+ f3(3) = 405, 275+ f3(2) = 425*,
325+ f3(1) = 405, 300+ f3(0) = 300)
f2(6) = max (0+ f3(6) = 280, 100+ f3(5) = 370, 195+ f3(4) = 445, 275+ f3(3) = 485*,
325+ f3(2) = 475, 300+ f3(1) = 380, 250+ f3(0) = 250)
f1(6) = max(0+ f2(6) = 485, 90+ f2(5) = 515, 180+ f2(4) = 535, 265+ f2(3) = 540*,
310+ f2(2) = 505, 350+ f2(1) = 450, 320+ f2(0) = 320)
Hence 3 flights are sent to Miami, and either 2 to LA and 1 to NY or 3 to LA and 0 to
NY. If only 4 flights are available:
f1(6) = max (0+ f2(4) = 355, 90+ f2(3) = 365, 180+ f2(2) = 375*, 265+ f2(1) = 365,
310+ f2(0) = 310)
Here 2 flights should be sent to Miami, and 2 to LA.