Question

You are a project team manager, and your team members report each day to you to receive their primary assignments.   Not every team member is as efficient as another with particular kinds of tasks.   

		Time required (hours) to complete tasks
Task	Task complexity	Team member 1 - Jones	Team member 2 - Nguyen	Team member 3 - Walpita	Team member 4 - Manderas
Task A	Very high	3	5	4	3
Task A	High	2	1	3	2
Task A	Medium	3	4	2	2
Task A	Low	4	3	3	4

a. Given the task roster, complexity levels, and time estimates, assign team members to tasks to optimize efficiency.   Also, determine how long it will take for the team to complete their routine tasks.

b.You want to focus on efficiency, so you assign ratings to each team member’s performance of each task.   Reassign the tasks to obtain the highest efficiency possible (a perfect score is 100).

c. Compare the results with those from the first question for this problem. What is the average rating of the assignment?   What other criteria could be used to assign tasks to team members?

	Rating
Task	Team member 1 Jones	Team member 2 Nguyen	Team member 3 Walpita	Team member 4 Manderas
Task A	89	95	83	84
Task A	88	80	96	85
Task A	87	92	82	84
Task A	93	82	36	94

Answer 1

a. Here the problem is of Maximization type and converts it into minimization by subtracting it from maximum value 5

	JONES	NGUYEN	WALPITA	MANDERAS
VH	2	0	1	2
HIG	3	4	2	3
MED	2	1	3	3
LOW	1	2	2	1

Here given problem is balanced.
Step-1: Find out each row minimum element and subtract it from that row

	JONES	NGUYEN	WALPITA	MANDERAS
VH	2	0	1	2	(-0)
HIG	1	2	0	1	(-2)
MED	1	0	2	2	(-1)
LOW	0	1	1	0	(-1)

Step-2: Find out each column minimum element and subtract it from that column.

	JONES	NGUYEN	WALPITA	MANDERAS
VH	2	0	1	2
HIG	1	2	0	1
MED	1	0	2	2
LOW	0	1	1	0
	(-0)	(-0)	(-0)	(-0)

Step-3: Make assignment in the opportunity cost table

(1) Row-wise cell (VH,NGUYEN) is assigned, so column-wise cell (MED,NGUYEN) crossed off.
(2) Row-wise cell (HIG,WALPITA) is assigned
(3) Column-wise cell (LOW,JONES) is assigned, so row-wise cell (LOW,MANDERAS) crossed off.
Row-wise column-wise assignment shown in table

	JONES	NGUYEN	WALPITA	MANDERAS
VH	2	[0]	1	2
HIG	1	2	[0]	1
MED	1	0	2	2
LOW	[0]	1	1	0

Step-4: Number of assignments = 3, number of rows = 4
Which is not equal, so the solution is not optimal.

Step-5: Draw a set of horizontal and vertical lines to cover all the 0

Step-5: Cover the 0 with a minimum number of lines
(1) Mark(✓) row MED since it has no assignment
(2) Mark(✓) column NGUYEN since row MED has 0 in this column
(3) Mark(✓) row VH since column NGUYEN has an assignment in this row VH.
(4) Since no other rows or columns can be marked, therefore draw straight lines through the unmarked rows HIG,LOW and marked columns NGUYEN

Tick mark not allocated rows and allocated columns

	JONES	NGUYEN	WALPITA	MANDERAS
VH	2	[0]	1	2	✓(3)
HIG	1	2	[0]	1
MED	1	0	2	2	✓(1)
LOW	[0]	1	1	0
		✓ (2)

Step-6: Develop the new revised opportunity cost table

Step-6: Develop the new revised table by selecting the smallest element, among the cells not covered by any line (say k = 1)
Subtract k = 1 from every element in the cell not covered by a line.
Add k = 1 to every element in the intersection cell of two lines.

	JONES	NGUYEN	WALPITA	MANDERAS
VH	1	0	0	1
HIG	1	3	0	1
MED	0	0	1	1
LOW	0	2	1	0

Repeat steps 3 to 6 until an optimal solution is obtained.
Iteration: 1
Step-3: Make assignment in the opportunity cost table
(1) Row-wise cell (HIG,WALPITA) is assigned, so column-wise cell (VH,WALPITA) crossed off.

(2) Column-wise cell (LOW,MANDERAS) is assigned, so row-wise cell (LOW,JONES) crossed off.

(3) Row-wise cell (VH,NGUYEN) is assigned, so column-wise cell (MED,NGUYEN) crossed off.

(4) Row-wise cell (MED,JONES) is assigned

Row-wise & column-wise assignment shown in table

	JONES	NGUYEN	WALPITA	MANDERAS
VH	1	[0]	0	1
HIG	1	3	[0]	1
MED	[0]	0	1	1
LOW	0	2	1	[0]

Step-4: Number of assignments = 4, number of rows = 4
Which is equal, so the solution is optimal
So, the Optimal assignments are

	JONES	NGUYEN	WALPITA	MANDERAS
VH	1	[0]	0	1
HIG	1	3	[0]	1
MED	[0]	0	1	1
LOW	0	2	1	[0]

The Optimal solution is

Work Difficulty	Team Members	Time
VH	NGUYEN	5
HIG	WALPITA	3
MED	JONES	3
LOW	MANDERAS	4
	Total	15

b. and c.

Based on the rating out of 100, the problem can be designed as:

Here the problem is of Maximization type and converts it into minimization by substracting it from maximum value 96

	JONES	NGUYEN	WALPITA	MANDERAS
VH	7	1	13	12
HIG	8	16	0	11
MED	9	4	14	12
LOW	3	14	60	2

Here given problem is balanced.
Step-1: Find out each row minimum element and subtract it from that row

	JONES	NGUYEN	WALPITA	MANDERAS
VH	6	0	12	11	(-1)
HIG	8	16	0	11	(-0)
MED	5	0	10	8	(-4)
LOW	1	12	58	0	(-2)

Step-2: Find out each column minimum element and subtract it from that column.

	JONES	NGUYEN	WALPITA	MANDERAS
VH	5	0	12	11
HIG	7	16	0	11
MED	4	0	10	8
LOW	0	12	58	0
	(-1)	(-0)	(-0)	(-0)

Step-3: Make assignment in the opportunity cost table

Step-3: Make assignment in the opportunity cost table

(1) Row-wise cell (VH,NGUYEN) is assigned, so column-wise cell (MED,NGUYEN) crossed off.

(2) Row-wise cell (HIG,WALPITA) is assigned

(3) Column-wise cell (LOW,JONES) is assigned, sorow-wisee cell (LOW,MANDERAS) crossed off.

Row-wise & column-wise assignment shown in table

	JONES	NGUYEN	WALPITA	MANDERAS
VH	5	[0]	12	11
HIG	7	16	[0]	11
MED	4	0	10	8
LOW	[0]	12	58	0

Step-4: Number of assignments = 3, number of rows = 4
Which is not equal, so the solution is not optimal.
Step-5: Draw a set of horizontal and vertical lines to cover all the 0

Step-5: Cover the 0 with a minimum number of lines
(1) Mark(✓) row MED since it has no assignment
(2) Mark(✓) column NGUYEN since row MED has 0 in this column
(3) Mark(✓) row VH since column NGUYEN has an assignment in this row VH.
(4) Since no other rows or columns can be marked, therefore draw straight lines through the unmarked rows HIG,LOW and marked columns NGUYEN
Tick mark not allocated rows and allocated columns

	JONES	NGUYEN	WALPITA	MANDERAS
VH	5	[0]	12	11	✓(3)
HIG	7	16	[0]	11
MED	4	0	10	8	✓(1)
LOW	[0]	12	58	0
		✓ (2)

Step-6: Develop the new revised opportunity cost table

Step-6: Develop the new revised table by selecting the smallest element, among the cells not covered by any line (say k = 4)
Subtract k = 4 from every element in the cell not covered by a line.
Add k = 4 to every element in the intersection cell of two lines.

	JONES	NGUYEN	WALPITA	MANDERAS
VH	1	0	8	7
HIG	7	20	0	11
MED	0	0	6	4
LOW	0	16	58	0

Repeat steps 3 to 6 until an optimal solution is obtained.

Iteration: 1
Step-3: Make assignment in the opportunity cost table
(1) Row-wise cell (VH,NGUYEN) is assigned, so column-wise cell (MED,NGUYEN) crossed off.
(2) Row-wise cell (HIG,WALPITA) is assigned
(3) Row-wise cell (MED,JONES) is assigned, so column-wise cell (LOW,JONES) crossed off.
(4) Row-wise cell (LOW,MANDERAS) is assigned

Row-wise & column-wise assignment shown in table

	JONES	NGUYEN	WALPITA	MANDERAS
VH	1	[0]	8	7
HIG	7	20	[0]	11
MED	[0]	0	6	4
LOW	0	16	58	[0]

Step-4: Number of assignments = 4, number of rows = 4
Which is equal, so the solution is optimal
Optimal assignments are

	JONES	NGUYEN	WALPITA	MANDERAS
VH	1	[0]	8	7
HIG	7	20	[0]	11
MED	[0]	0	6	4
LOW	0	16	58	[0]

The Optimal solution is assigned based on ratings

Work Difficulty	Team Members	Ratings
VH	NGUYEN	95
HIG	WALPITA	96
MED	JONES	87
LOW	MANDERAS	94
	Total	372