MAt 540

Week8-LinearProgramming-Continued.xls

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Lecture notes


WEEK 8 - LINEAR PROGRAMMING CONTINUED - APPLICATIONS
IN THIS CHAPTER, VARIOUS APPLICATIONS OF THE USE OF LINEAR PROGRAMMING
	ARE ILLUSTRATED. WE WILL DISCUSS FIVE OF THESE EXAMPLES.
I. PRODUCT MIX - PAGE 111 - 116
			PROCESS.	COST	PROFIT
	PRODUCTS		TIME (HR.)	PER	PER
	TO BE MIXED	VARIABLE	PER DOZEN	DOZEN	DOZEN
	Sweatshirt - Front	X1	0.10	36	90
	Sweatshirt - Both	X2	0.25	48	125
	T-Shirt - Front	X3	0.08	23	45
	T-Shirt - Both	X4	0.21	35	65
OBJECTIVE - MAXIMIZE PROFITS
CONSTRAINTS
1) PROCESSING TIME - MAXIMUM OF 72 HOURS
2) SHIPPING CAPACITY - MAXIMUM OF 1,200 BOXES
	[a sweatshirt is 3 times the size of a t-shirt]
3) COST BUDGET - MAXIMUM OF $25,000
4) SWEATSHIRTS - MAXIMUM OF 500 DOZENS
5) T-SHIRTS - MAXIMUM OF 500 DOZENS
THE LINEAR PROGRAMMING FORMULATION IS AS FOLLOWS:
QM FOR WINDOWS SOLUTION
INITIAL SCREEN
		X1	X2	X3	X4		RHS	Equation form
	Maximize	90.0	125.0	45.0	65.0			Max 90X1 + 125X2 + 45X3 + 65X4
	Constraint 1	0.1	0.25	0.08	0.21	<=	72	.1X1 + .25X2 + .08X3 + .21X4 <= 72
	Constraint 2	3	3	1	1	<=	1200	3X1 + 3X2 + X3 + X4 <= 1200
	Constraint 3	36	48	25	35	<=	25000	36X1 + 48X2 + 25X3 + 35X4 <= 25000
	Constraint 4	1	1	0	0	<=	500	X1 + X2 <= 500
	Constraint 5	0	0	1	1	<=	500	X3 + X4 <= 500
FINAL SOLUTION SCREEN
	Variable	Status	Value
	X1	Basic	175.5556		176
	X2	Basic	57.7778		58
	X3	Basic	500		234
	X4	NONBasic	0
	slack 1	NONBasic	0
	slack 2	NONBasic	0
	slack 3	Basic	3406.667
	slack 4	Basic	266.6667
	slack 5	NONBasic	0
	Optimal Value (Z)		45522.22
2. DIET PROBLEM PAGE 116 - 119 - TABLE OF DATA ON PAGE 116
OBJECTIVE - MINIMIZE COSTS
CONSTRAINTS
1) CALORIES - AT LEAST 420 CALORIES
2) IRON - AT LEAST 5 MILLIGRAMS
3) CALCIUM - AT LEAST 400 MILLIGRAMS
4) PROTEIN - AT LEAST 20 GRAMS
5) FIBER - AT LEAST 12 GRAMS
6) FAT - MAXIMUM OF 20 GRAMS
7) CHOLESTEROL - MAXIMUM OF 30 MILLIGRAMS
THE LINEAR PROGRAMMING FORMULATION IS AS FOLLOWS:
Subject to:
QM FOR WINDOWS SOLUTION
INITIAL SCREEN
		X1	X2	X3	X4	X5	X6	X7	X8	X9	X10		RHS	Equation form
	Minimize	0.18	0.22	0.1	0.12	0.1	0.09	0.4	0.16	0.5	0.07			Min .18X1 + .22X2 + .1X3 + .12X4 + .1X5 + .09X6 + .4X7 + .16X8 + .5X9 + .07X10
	Constraint 1	90	110	100	90	75	35	65	100	120	65	>=	420	90X1 + 110X2 + 100X3 + 90X4 + 75X5 + 35X6 + 65X7 + 100X8 + 120X9 + 65X10 >= 420
	Constraint 2	6	4	2	3	1	0	1	0	0	1	>=	5	6X1 + 4X2 + 2X3 + 3X4 + X5 + X7 + X10 >= 5
	Constraint 3	20	48	12	8	30	0	52	250	3	26	>=	400	20X1 + 48X2 + 12X3 + 8X4 + 30X5 + 52X7 + 250X8 + 3X9 + 26X10 >= 400
	Constraint 4	3	4	5	6	7	2	1	9	1	3	>=	20	3X1 + 4X2 + 5X3 + 6X4 + 7X5 + 2X6 + X7 + 9X8 + X9 + 3X10 >= 20
	Constraint 5	5	2	3	4	0	0	1	0	0	3	>=	12	5X1 + 2X2 + 3X3 + 4X4 + X7 3X10 >= 12
	Constraint 6	0	2	2	2	5	3	0	4	0	1	<=	20	2X2 + 2X3 + 2X4 + 5X5 + 3X6 + 4X8 + X10 <= 20
	Constraint 7	0	0	0	0	270	8	0	12	0	0	<=	30	270X5 + 8X6 + 12X8 <= 30
FINAL SOLUTION SCREEN
	Variable	Status	Value
	X1	NONBasic	0
	X2	NONBasic	0
	X3	Basic	1.0246
	X4	NONBasic	0
	X5	NONBasic	0
	X6	NONBasic	0
	X7	NONBasic	0
	X8	Basic	1.2414
	X9	NONBasic	0
	X10	Basic	2.9754
	surplus 1	NONBasic	0
	surplus 2	Basic	0.0246
	surplus 3	NONBasic	0
	surplus 4	Basic	5.2217
	surplus 5	NONBasic	0
	slack 6	Basic	10.0099
	slack 7	Basic	15.1035
	Optimal Value (Z)		0.5094
3. INVESTMENT EXAMPLE
			RETURN
	INVESTMENT		FROM
	OPTIONS	VARIABLE	INVESTMENT
	Municipal bonds	X1	0.085
	Cert. of deposit	X2	0.050
	Treasury bills	X3	0.065
	Growth stock fund	X4	0.130
OBJECTIVE: MAXIMIZE TOTAL RETURN ON INVESTMENTS
CONSTRAINTS
1. TOTAL INVESTMENT IN MUNICIPAL BONDS - MAXIMUM OF 20% OF $70,000
2. AMOUNT INVESTED IN C.D's - MAXIMUM = SUM OF OTHER 3 ALTERNATIVES
3. TOTAL INVESTMENT IN TREASURY BILLS AND CD's - AT LEAST 30% OF $70,000
4. TOTAL INVESTMENT IN TREASURY BILLS AND CD's - AT LEAST 1.2 TIMES
	THE INVESTMENT IN MUNICIPAL BONDS AND GROWTH STOCK FUNDS.
5. TOTAL INVESTMENT = $70,000
THE LINEAR PROGRAMMING FORMULATION IS AS FOLLOWS:
NOTE: THE FOURTH CONSTRAINT IS DETERMINED AS FOLLOWS:
QM FOR WINDOWS SOLUTION
	INITIAL SCREEN
		X1	X2	X3	X4		RHS	Equation form
	Maximize	0.085	0.05	0.065	0.13			Max .085X1 + .05X2 + .065X3 + .13X4
	Constraint 1	1	0	0	0	<=	14000	X1 <= 14000
	Constraint 2	1	1	-1	-1	<=	0	X1 - X2 - X3 - X4 <= 0
	Constraint 3	0	1	1	0	>=	21000	X2 + X3 >= 21000
	Constraint 4	-1.2	1	1	-1.2	>=	0	--1.2X1 + X2 + X3 --1.2X4 >= 0
	Constraint 5	1	1	1	1	=	70000	X1 + X2 + X3 + X4 = 70000
FINAL SOLUTION SCREEN
	Variable	Status	Value
	X1	NONBasic	0
	X2	NONBasic	0
	X3	Basic	38181.82
	X4	Basic	31818.18		38181.816
	slack 1	Basic	14000		38181.816
	slack 2	Basic	70000
	surplus 3	Basic	17181.82
	surplus 4	NONBasic	0
	artfcl 5	NONBasic	0
	Optimal Value (Z)		6618.182
4. MEDIA EXAMPLE (MARKETING EXAMPLE)
	MEDIA		AUDIENCE
	OPTIONS	VARIABLE	EXPOSURE	COST
	Television Comm.	X1	20,000	$ 15,000
	Radio Comm.	X2	12,000	6,000
	Newspaper Ads	X3	9,000	4,000
OBJECTIVE: MAXIMIZE AUDIENCE EXPOSURE
CONSTRAINTS
1. BUDGET LIMIT - MAXIMUM OF $100,000
2. TV STATION AVAILABILITY - MAXIMUM OF 4 COMMERCIALS
3. RADIO STATION AVAILABILITY - MAXIMUM OF 10 COMMERCIALS
4. NEWSPAPER AVAILABILITY - MAXIMUM OF 7 ADS
5. AD AGENCY TIME AVAILABILITY - MAXIMUM OF 15 COMMERCIALS OR ADS.
THE LINEAR PROGRAMMING FORMULATION IS AS FOLLOWS:
QM FOR WINDOWS SOLUTION
	INITIAL SCREEN
		X1	X2	X3		RHS	Equation form
	Maximize	20000	12000	9000			Max 20000X1 + 12000X2 + 9000X3
	Constraint 1	15000	6000	4000	<=	100000	15000X1 + 6000X2 + 4000X3 <= 100000
	Constraint 2	1	0	0	<=	4	X1 <= 4
	Constraint 3	0	1	0	<=	10	X2 <= 10
	Constraint 4	0	0	1	<=	7	X3 <= 7
	Constraint 5	1	1	1	<=	15	X1 + X2 + X3 <= 15
FINAL SOLUTION SCREEN
	Variable	Status	Value
	X1	Basic	1.8182		40000	2	30000
	X2	Basic	10		120000	10	60000
	X3	Basic	3.1818		27000	3	12000
	slack 1	NONBasic	0		187000		102000
	slack 2	Basic	2.1818
	slack 3	NONBasic	0
	slack 4	Basic	3.8182
	slack 5	NONBasic	0
	Optimal Value (Z)		185000
5. TRANSPORTATION EXAMPLE
		TRANSPORTATION COSTS
			DESTINATION
	SOURCE	New york	Dallas	Detroit	TOTALS
	Cincinnati	$ 16	$ 18	$ 11	300
	Atlanta	$ 14	$ 12	$ 13	200
	Pittsburgh	$ 13	$ 15	$ 17	200
					700
	TOTALS	150	250	200	600
OBJECTIVE: MINIMIZE TOTAL COST OF TRANSPORTATION
CONSTRAINTS
1. AVAILABILITY FROM CINCINNATI - MAXIMUM OF 300
2. AVAILABILITY FROM ATLANTA - MAXIMUM OF 200
3. AVAILABILITY FROM PITTSBURGH - MAXIMUM OF 200
4. REQUIREMENTS FOR NEW YORK = 150
5. REQUIREMENTS FOR DALLAS = 250
6. REQUIREMENTS FOR DETROIT = 200
QM FOR WINDOWS SOLUTION - USING THE TRANSPORTATION MODEL INSTEAD OF THE
		LINEAR PROGRAMMING MODEL
INITIAL SCREEN
		New York	Dallas	Detroit	SUPPLY
	Cincinnati	16	18	11	300
	Atlanta	14	12	13	200
	Pittsburgh	13	15	17	200
	DEMAND	150	250	200
FINAL SOLUTION SCREEN
	Optimal solution value = $7300	New York	Dallas	Detroit	Dummy
	Cincinnati			200	100
	Atlanta		200
	Pittsburgh	150	50		0
	Optimal solution value = $7300

Problems


PROBLEMS 1 - 4 PAGES 147 - 148
I.	REVISED PRODUCT MIX PROBLEM
1a)			PROCESS.	COST	PROFIT
	PRODUCTS		TIME (HR.)	PER	PER
	TO BE MIXED	VARIABLE	PER DOZEN	DOZEN	DOZEN
	Sweatshirt - Front	X1	0.090	39.60	86.40
	Sweatshirt - Both	X2	0.225	52.80	120.20
	T-Shirt - Front	X3	0.072	25.30	42.70
	T-Shirt - Both	X4	0.189	38.50	61.50
	INITIAL SCREEN
		X1	X2	X3	X4		RHS	Equation form
	Maximize	86.4	120.2	42.7	61.5			Max 86.4X1 + 120.2X2 + 42.7X3 + 61.5X4
	Constraint 1	0.09	0.225	0.072	0.189	<=	72	.09X1 + .225X2 + .072X3 + .189X4 <= 72
	Constraint 2	3	3	1	1	<=	1200	3X1 + 3X2 + X3 + X4 <= 1200
	Constraint 3	36	48	25	35	<=	25000	36X1 + 48X2 + 25X3 + 35X4 <= 25000
	Constraint 4	1	1	0	0	<=	500	X1 + X2 <= 500
	Constraint 5	0	0	1	1	<=	500	X3 + X4 <= 500
	FINAL SOLUTION SCREEN
	Variable	Status	Value
	X1	Basic	122.2222
	X2	Basic	111.1111
	X3	Basic	500
	X4	NONBasic	0
	slack 1	NONBasic	0
	slack 2	NONBasic	0
	slack 3	Basic	2766.667
	slack 4	Basic	266.6667
	slack 5	NONBasic	0
	Optimal Value (Z)		45265.55
	THIS VALUE IS LOWER THAN THE ORIGINAL
1b)	TO MAKE ALL QUANTITIES EQUAL WE NEED TO ADD THE FOLLOWING
		CONSTRAINTS:
	6: X1 = X2 or X1 - X2 = 0
	7: X1 = X3 or X1 - X3 = 0
	8: X1 = X4 or X1 - X4 = 0
	INITIAL SCREEN
		X1	X2	X3	X4		RHS	Equation form
	Maximize	90	125	45	65			Max 90X1 + 125X2 + 45X3 + 65X4
	Constraint 1	0.1	0.25	0.08	0.21	<=	72	.1X1 + .25X2 + .08X3 + .21X4 <= 72
	Constraint 2	3	3	1	1	<=	1200	3X1 + 3X2 + X3 + X4 <= 1200
	Constraint 3	36	48	25	35	<=	25000	36X1 + 48X2 + 25X3 + 35X4 <= 25000
	Constraint 4	1	1	0	0	<=	500	X1 + X2 <= 500
	Constraint 5	0	0	1	1	<=	500	X3 + X4 <= 500
	NEW Constraint 6	1	-1	0	0	=	0	X1 - X2 = 0
	NEW Constraint 7	1	0	-1	0	=	0	X1 - X3 = 0
	NEW Constraint 8	1	0	0	-1	=	0	X1 - X4 = 0
	FINAL SOLUTION SCREEN
	Variable	Status	Value
	X1	Basic	112.5
	X2	Basic	112.5
	X3	Basic	112.5
	X4	Basic	112.5
	slack 1	NONBasic	0
	slack 2	Basic	300
	slack 3	Basic	8800
	slack 4	Basic	275
	slack 5	Basic	275
	artfcl 6	NONBasic	0
	artfcl 7	NONBasic	0
	artfcl 8	NONBasic	0
	Optimal Value (Z)		36562.5
II.	REVISED DIET PROBLEM
2a	CHANGE THE MINIMUM CALORIE REQUIREMENT FOR THE BREAKFAST
		TO 500 CALORIES
	FINAL SOLUTION SCREEN
	Variable	Status	Value
	X1	NONBasic	0
	X2	NONBasic	0
	X3	Basic	2.9951
	X4	NONBasic	0
	X5	NONBasic	0
	X6	NONBasic	0
	X7	NONBasic	0
	X8	Basic	1.3517
	X9	NONBasic	0
	X10	Basic	1.0049
	surplus 1	NONBasic	0
	surplus 2	Basic	1.9951
	surplus 3	NONBasic	0
	surplus 4	Basic	10.1557
	surplus 5	NONBasic	0
	slack 6	Basic	7.598
	slack 7	Basic	13.7793
	Optimal Value (Z)		0.5861
2b	CHANGE THE MINIMUM CALORIE REQUIREMENT FOR THE BREAKFAST
		TO 600 CALORIES
	FINAL SOLUTION SCREEN
	Variable	Status	Value
	X1	NONBasic	0
	X2	NONBasic	0
	X3	Basic	4.6218
	X4	NONBasic	0
	X5	NONBasic	0
	X6	NONBasic	0
	X7	NONBasic	0
	X8	Basic	1.3782
	X9	NONBasic	0
	X10	NONBasic	0
	surplus 1	NONBasic	0
	surplus 2	Basic	4.2437
	surplus 3	NONBasic	0
	surplus 4	Basic	15.5126
	surplus 5	Basic	1.8655
	slack 6	Basic	5.2437
	slack 7	Basic	13.4622
	Optimal Value (Z)		0.6827
2c	CHANGE THE MINIMUM CALORIE REQUIREMENT FOR THE BREAKFAST
		TO 700 CALORIES
	FINAL SOLUTION SCREEN
	Variable	Status	Value
	X1	NONBasic	0
	X2	NONBasic	0
	X3	Basic	5.6723
	X4	NONBasic	0
	X5	NONBasic	0
	X6	NONBasic	0
	X7	NONBasic	0
	X8	Basic	1.3277
	X9	NONBasic	0
	X10	NONBasic	0
	surplus 1	NONBasic	0
	surplus 2	Basic	6.3445
	surplus 3	NONBasic	0
	surplus 4	Basic	20.3109
	surplus 5	Basic	5.0168
	slack 6	Basic	3.3445
	slack 7	Basic	14.0672
	Optimal Value (Z)		0.7797
III.	REVISED INVESTMENT PROBLEM
3a	CHANGE THE $70,000 INVESTMENT AMOUNT FROM AN EQUALITY
		TO A MAXIMUM AMOUNT.
	FINAL SOLUTION SCREEN
	Variable	Status	Value
	X1	NONBasic	0
	X2	NONBasic	0
	X3	Basic	38181.82
	X4	Basic	31818.18
	slack 1	Basic	14000
	slack 2	Basic	70000
	surplus 3	Basic	17181.82
	surplus 4	NONBasic	0
	slack 5	NONBasic	0
	Optimal Value (Z)		6618.182
	THE CHANGE HAS NO EFFECT, SINCE THE ENTIRE $70,000
	IS INVESTED ANYWAY.
3b	IF THE AMOUNT AVAILABLE IS INCREASED TO $80,000,
	WHAT WOULD BE THE NEW SOLUTION?
	FINAL SOLUTION SCREEN
	Variable	Status	Value
	X1	NONBasic	0
	X2	NONBasic	0
	X3	Basic	43636.36
	X4	Basic	36363.64		43636.368
	slack 1	Basic	14000
	slack 2	Basic	80000
	surplus 3	Basic	22636.37
	surplus 4	NONBasic	0
	slack 5	NONBasic	0
	Optimal Value (Z)		7563.636
	ALL $80,000 WILL BE INVESTED.
	THE OPTIMAL VALUE WILL INCREASE FROM
	$6,618.18 TO $7,563.64, AND INCREASE OF			945.45
	THE $10,000 INCREASE WILL BE SPLIT BETWEEN
	X3 AND X4.
IV.	REVISED MARKETING PROBLEM
4a	INCREASE THE BUDGET BY $20,000.
	FINAL SOLUTION SCREEN
	Variable	Status	Value
	X1	Basic	3.6364
	X2	Basic	10
	X3	Basic	1.3636
	slack 1	NONBasic	0
	slack 2	Basic	0.3636
	slack 3	NONBasic	0
	slack 4	Basic	5.6364
	slack 5	NONBasic	0
	Optimal Value (Z)		205000
	THE OPTIMAL VALUE INCREASES BY 20,000, FROM 185,000
		TO 205,000.
4b	IF THE STORE WANTS THE SAME NUMBER OF PEOPLE TO BE
	EXPOSED TO EACH TYPE OF ADVERTISMENT, WHAT CHANGES
	SHOULD BE MADE IN THE FORMULATION, AND WHAT WOULD BE
	THE NEW SOLUTION, ASSUMING THE NEW BUDGET REMAINS AT
		$120,000
	TWO NEW CONSTRAINTS ARE NEEDED:
	6) 20000X1 = 12000X2; OR 20000X1 - 12000X2 = 0
	7) 20000X1 = 9000X3; OR 20000X1 - 12000X3 = 0
	FINAL SOLUTION SCREEN
	Variable	Status	Value
	X1	Basic	3.0682
	X2	Basic	5.1136
	X3	Basic	6.8182
	slack 1	Basic	16022.73
	slack 2	Basic	0.9318
	slack 3	Basic	4.8864
	slack 4	Basic	0.1818
	slack 5	NONBasic	0
	artfcl 6	NONBasic	0
	artfcl 7	NONBasic	0
	Optimal Value (Z)		184090.9

Key Points


	WEEK 8 - CHAPTER 4 - KEY POINTS TO REMEMBER
1	When formulating a linear programming model on a spreadsheet,
	the measure of performance is located in the target cell.
2	Determining the production quantities of different products manufactured
	by a company based on resource constraints is a product mix linear programming problem.
3	When using linear programming model to solve the "diet" problem,
	the objective is generally to minimize cost.
4	The standard form for the computer solution of a linear programming problem requires
	all variables to the left and all numerical values to the right of the inequality or equality sign.
5	A constraint for a linear programming problem can have a zero as its right-hand-side value.
6	Media selection is an important decision that advertisers have to make.
	In most media selection decisions, the objective of the decision maker is not to minimize cost.
7	In a media selection problem, instead of having an objective of maximizing profit or
	minimizing cost, generally the objective is to maximize the audience exposure.
8	Linear programming model of a media selection problem is used to determine
	the allocation of the advertising budget to various advertising media.
9	In a balanced transportation model, supply equals demand such that all
	constraints can be treated as equalities.
10	In an unbalanced transportation model, supply does not equal demand
	and supply constraints have signs.
11	Feasible solutions are ones that satisfy all the constraints simultaneously.
12	In a media selection problem, the estimated number of customers reached by
	a given media would generally be specified in the objective function.
	Even if these media exposure estimates are correct, using media exposure
	as a surrogate does not lead to maximization of profits.
13	Profit is maximized in the objective function by subtracting costs from revenues.
14	In a multi-period scheduling problem the production constraint usually takes the form of :
	beginning inventory - demand + production = ending inventory.

Key Points 2


	WEEK 8 - CHAPTER 4 - KEY POINTS TO REMEMBER
1	The standard form for the computer solution of a linear programming problem
		requires all variables to the left and all numerical values to the right
		of the inequality or equality sign.
2	Fractional relationships between variables are not permitted
		in the standard form of a linear program.
3	A constraint for a linear programming problem will sometimes
		have a zero as its right-hand-side value.
4	A systematic approach to model formulation does not require constructing
		the objective function before determining the decision variables
5	Product mix problems can have "greater than or equal to" (≥) constraints.
6	When using linear programming model to solve the "diet" problem,
	the objective is generally to minimize cost.
7	In a balanced transportation model, supply equals demand such that
		all constraints can be treated as equalities.
8	In a transportation problem, a demand constraint (the amount of product
		demanded at a given destination) is an equal-to constraint (=).
9	In a media selection problem, instead of having an objective of maximizing
		profit or minimizing cost, generally the objective is to maximize
		the audience exposure.
10	In formulating a typical diet problem using a linear programming model, we
		would not expect most of the constraints to be related to calories.
11	When systematically formulating a linear program, the first step is
		Identify the decision variables.
12	Compared to blending and product mix problems, transportation problems
		are unique because the solution values are always integers.
13	Balanced transportation problems have the equal (=) type of constraints.

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=++

++£

123123123123123:20,00012,0009,000:1)15,0006,0004,000100,0002)43)104)75)15,,0MaximizeZxxxSubjectToxxxxxxxxxxxx

1234

2134

1234

:0.0850.0500.0650.130

1)14,000

3)21,000

4)1.21.20

5)70,000

,,,0

MaximizeZxxxx

SubjectTo

xxxx

=+++

£++

---£

+³

-++-³

+++=

123412134213423123412341234:0.0850.0500.0650.130:1)14,0002)03)21,0004)1.21.205)70,000,,,0MaximizeZxxxxSubjectToxxxxxORxxxxxxxxxxxxxxxxxx

MAt 540

Lecture notes

Problems

Key Points

Key Points 2

MBD09EABA1F.unknown

MBD0A68B70E.unknown

MBD0A6C9B22.unknown

MBD0A686A48.unknown

MBD09E9E25C.unknown

MBD09E9E5D3.unknown

MBD09E87141.unknown