Arlington
460
810
1020
1270
Oakland
1500
1850
2080
10000
Baltimore
960
610
400
330
The optimal solution found using the QM for Windows assignment module is:
The Seattle crew will go to Detroit.
9-40. If the total distance is maximized, we assign a very low cost (miles) to the prohibited route
to prevent this assignment. A cost of 0 is used. The initial table is:
Kansas City
Chicago
Detroit
Toronto
Seattle
1500
1730
1940
2070
Arlington
460
810
1020
1270
Oakland
1500
1850
2080
0
Baltimore
960
610
400
330
9-41. Because this is a maximization problem, each number is subtracted from 95. The problem
is then solved using the minimization algorithm.
Assignment
Rating
Anderson—finance
95
Sweeney—economics
75
Williams—statistics
85
McKinney—management
80
9-42.
Assignment
Rating
Hawkins to cardiology
18
Condriac to urology
32
Bardot to orthopedics
24
Hoolihan to obstetrics
12
9-43. Each rating is subtracted from 27.1 because this is a maximization problem.
Assignment
Rating
1–2 P.M. on A
27.1
2–3 P.M. on C
17.1
3–4 P.M. on B
18.5
4–5 P.M. on independent
12.8
9-44.
Assignment
Rating
Adams to project 3
$ 6
Brown to project 2
10
Cooper to project 1
Davis to dummy
9-45. The following optimal assignments can be made:
Assignment
Cost
Component C53 to plant 6
0.06
Component C81 to plant 3
0.04
Component D5 to plant 4
0.30
Component D44 to plant 5
0.10
Component E2 to plant 2
0.07
Component E35 to plant 8
0.06
Component G99 to plant 1
0.55
9-46. This can be modeled as a transportation problem. Students should note the large numbers
used to block infeasible production plans (see Printout 1 on the next page).
a. The solution yields a cost of $2,591,200. The plan is shown in Printout 2. There are
multiple optimal solutions.
9-47. The schedules are found using software for the assignment problem. Here are the three
schedules.
Part a
Part b
Part c
Vincze
26-Mar
26-Mar
26-Mar
Veil
12-Apr
12-Apr
12-Apr
Anderson
26-Feb
1-May
1-May
Herbert
5-Feb
5-Feb
5-Feb
Schatz
12-Jan
12-Jan
26-Feb
9-Jun
9-Jun
9-Jun
19-Sep
26-Feb
12-Jan
1-May
19-Sep
19-Sep
20-Aug
20-Aug
20-Aug
27-Jan
27-Jan
27-Jan
9-48. This can be modeled as a maximization assignment problem. Using computer software
we obtain the following solution: Erica to Austin/San Antonio, Louis to El Paso/West Texas,
SOLUTIONS TO INTERNET HOMEWORK PROBLEMS
9-49. Jessie Cohen Clothing Group’s northwest corner solution:
TO
Factory
FROM
A
B
C
Availability
4
3
3
W
30
5
35
6
7
6
50
50
8
2
5
10
50
Demand
9-50. Optimal solution for Jessie Cohen Clothing Group’s problem:
TO
Factory
FROM
A
B
C
Availability
4
3
3
W
15
20
35
6
7
6
30
20
50
50
50
Demand
The total cost of this solution = 15($3) + 20($3) + 30($6) + 20($6) + 50($2) = $505.
9-51.
OFFICE
MAN
Omaha
Miami
Dallas
Jones
800
1,100
1,200
500
1,600
1,300
500
1,000
2,300
Row subtraction is done first.
OFFICE
MAN
Omaha
Miami
Dallas
Jones
0
300
400
Smith
0
800
0
500
Column subtraction is done next; and then draw the minimum number of lines to cover zeroes.
OFFICE
MAN
Omaha
Miami
Dallas
Jones
0
0
Smith
0
800
0
200
1,400
Subtract the smallest number (200) not covered by a line from all numbers not covered, and add
it to any number at the intersection of two lines. Then redraw minimum number of lines.
OFFICE
MAN
Omaha
Miami
Dallas
Jones
200
0
0
600
200
Optimal assignment: Jones to Dallas, Smith to Omaha, Wilson to Miami,
Cost = $1,200 + $500 + $1,000 = $2,700
9-52. Original problem:
CUSTOMER
SITE
A
B
C
D
1
7
3
4
8
2
6
5
3
9
6
4
7
4
Row subtraction is done first.
CUSTOMER
SITE
A
B
C
D
1
4
0
1
5
2
1
0
2
1
3
0
1
3
0
4
4
2
3
0
Column subtraction is done next.
CUSTOMER
SITE
A
B
C
D
1
4
0
0
5
2
1
0
1
1
3
0
1
2
0
4
4
2
2
0
9-53. Original problem.
CASE
SQUAD
A
B
C
D
E
1
14
7
3
7
27
2
20
7
6
30
3
10
3
4
5
21
5
13
25
26
8
CASE
SQUAD
A
B
C
D
E
1
11
4
0
4
24
2
14
1
6
0
24
3
7
0
1
2
18
5
5
17
16
18
0
Column subtraction is done next.
CASE
SQUAD
A
B
C
D
E
1
10
4
0
4
24
2
13
1
6
0
24
3
6
0
1
2
18
5
4
17
18
0
9-54. The optimal assignments are:
Assignment
Rating
C53 at plant 1
10 cents
C81 at plant 3
D5 at plant 4
30 cents
D44 at plant 2
14 cents
Total manufacturing cost 58 cents
9-55. A “northeast corner” rule would be directly analogous to the northwest corner rule, but it
would simply begin in the upper right-hand corner instead of the upper left-hand corner. We see
in the table that this initial solution is degenerate because only four squares (instead of the ex-
pected five) are occupied. The degeneracy condition, by the way, is just a peculiarity of the Ex-
ecutive Furniture Corporation data.
TO
Albu-
Factory
FROM
querque
Boston
Cleveland
Capacity
5
4
3
Des Moines
100
100
8
4
3
Evansville
200
100
300
Fort
9
7
5
Lauderdale
300
300
Requirements
300
200
200
700
SOLUTION TO ANDREW–CARTER, INC., CASE
This case presents some of the basic concepts of aggregate planning by the transportation meth-
od. The case involves solving a rather complex set of transportation problems. Four different
configurations of operating plants have to be tested. The solutions, although requiring relatively
few iterations to optimality, involve degeneracy if solved manually. The costs are:
Total
Total
Variable
Fixed
Total
Configuration
Cost
Cost
Cost
All plants operating
$179,730
$41,000
$220,730
1 and 2 operating, 3 closed
188,930
33,500
222,430
1 and 3 operating, 2 closed
183,430
34,000
217,430
2 and 3 operating, 1 closed
188,360
33,000
221,360
The lowest weekly total cost, operating plants 1 and 3 with 2 closed, is $217,430. This is
$3,300 per week ($171,600 per year) or 1.5% less than the next most economical solution, oper-
ating all three plants. Closing a plant without expanding the capacity of the remaining plants
means unemployment. The optimum solution, using plants 1 and 3, indicates overtime produc-
The optimum producing and shipping pattern is
From
To (Amount)
Plant 1 (R.T.)
W2 (13,000); W4 (14,000)
W4 (1,000); W5 (8,000)
Plant 3 (O.T.)
W1 (4,000)
SOLUTION TO OLD OREGON WOOD STORE CASE
1. The assignment algorithm can be utilized to yield the fastest time to complete a table with
each person assigned one task.
Time
Person
Job
(Minutes)
Tom
Preparation
100
Cathy
George
Leon
240
2. If Randy is used, the assignment problem becomes unbalanced and a dummy job must be
added. The optimum assignment would be
Time
Person
Job
(Minutes)
George
Preparation
80
Tom
Assembly
60
Leon
80
Randy
Total time
230
3. If Cathy is given the preparation task, the solution of the assignment with the remaining three
workers assigned to the remaining three tasks is
Time
Person
Job
(Minutes)
Cathy
Preparation
120
Tom
Assembly
60
George
60
Leon
Packaging
Total time
250
If Cathy is assigned to the finishing task, the optimum assignment is
Time
Person
Job
(Minutes)
George
Preparation
80
Tom
Assembly
60
Cathy
Leon
Total time
4. One possibility would be to combine the packaging operation with finishing. Then, George
2.09 tables—a total of almost 7 tables per day.
To utilize all five workers, George and Tom could each build entire tables, 2.09 and 1.75 per
day, respectively. Letting Randy do preparation (110 minutes), Cathy the assembly (70 minutes),
and Leon the finishing and packaging (90 minutes) allows an additional 4.36 tables per day for a
total of 8.2 per day.
INTERNET CASE STUDIES
Northwest General Hospital
Optimal Solution
Source
Destination
Number of Trays
From: Station 5A
To: Wing 5
60
5A
6
80
5A
3
60
3G
1
80
3G
3
90
3G
4
55
4
2
Optimal Cost: 4,825 minutes
SOLUTION TO CUSTOM VANS, INC. CASE
To determine whether the shipping pattern can be improved and where the two new plants
should be located, the total costs for the entire transportation system for each combination of
plants, as well as the existing shipping pattern costs, will have to be determined. In the headings
Cost Table for Custom Vans, Inc.
SHOP
PLANT
Chicago
Milwaukee
Minn.
Detroit
Capacity
Gary
10
20
40
25
300
20
30
50
15
150
Detroit*
26
36
56
1
22
37
10
30
35
*Since a plant at Detroit could purchase a gallon of fiberglass for $2 less than
any other plant, and one Shower-Rific takes 2 gallons of fiberglass, a systems
approach to transportation warrants that $2(2), $4, be deducted from each price
quoted in the case for shipments from Detroit.
The existing shipping pattern is shown in the table below.
Existing Shipping Pattern
Chicago
Milwaukee
Minn.
Detroit
Capacity
Gary
200
100
300
150
150
750
If the transportation algorithm were used to optimize the current shipments, the result
would be the solution shown in the table below. Note that Milwaukee and Minneapolis would
receive nothing.
Optimum Shipping Pattern with No New Plants
Chicago
Milwaukee
Minn.
Detroit
Capacity
Gary
300
300
Fort Wayne
150
150
300
200
750
Optimum Shipping Pattern with Detroit and Madison Plants
Chicago
Milwaukee
Minn.
Detroit
Capacity
Gary
200
100
300
Fort Wayne
100
50
150
Detroit
150
Madison
150
Demand
Optimum Shipping Pattern with Madison and Rockford Plants
Chicago
Milwaukee
Minn.
Detroit
Capacity
Gary
250
50
300
Fort Wayne
150
150
Madison
100
50
150
Rockford
50
100
150
Demand
300
100
150
200
750
Optimum Shipping Pattern with Detroit and Rockford Plants
Chicago
Milwaukee
Minn.
Detroit
Capacity
Gary
200
100
300
Fort Wayne
100
50
150
Detroit
150
Rockford
150
150
Demand
Since the total cost is lowest in the Gary–Fort Wayne–Detroit–Madison combination
($10,200), the new plants should be located in Detroit and Madison. This system is also an im-
provement over the existing pattern, which costs $9,000, on a cost-per-unit basis.
Status quo: $9,000/450 units = $20/unit
Proposed: $10,200/750 units = $13.60/unit