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Worked Examples for Chapter 3

Example for Section 3.1

The Apex Television Company has to decide on the number of 27- and 20-inch sets to beproduced at one of its factories. Market research indicates that at most 0 of the 27-inchsets and !0 of the 20-inch sets can be sold per month. The maximum number of "ork-hours available is #00 per month. A 27-inch set re$uires 20 "ork-hours and a 20-inch setre$uires !0 "ork-hours. %ach 27-inch set sold produces a profit of &!20 and each 20-inchset produces a profit of &'0. A "holesaler has a(reed to purchase all the television setsproduced if the numbers do not exceed the maxima indicated by the market research.

(a) Formulate a linear programming model for this problem.

The decisions that need to be made are the number of 27-inch and 20-inch T) sets to be

produced per month by the Apex Television Company. Therefore* the decision variablesfor the model are

x!+ number of 27-inch T) sets to be produced per month*x2+ number of 20-inch T) sets to be produced per month.

Also let

, + total profit per month.

The model no" can be formulated in terms of these variables as follo"s.

The total profit per month is , + !20 x! '0 x2.

The resource constraints are

/! 1umber of 27-inch sets sold per month x!0

/2 1umber of 20-inch sets sold per month x2!0

/ 3ork-hours availability 20 x! !0 x2#00.

1onne(ativity constraints on T) sets produced

x!0x20

3ith the ob4ective of maximi5in( the total profit per month* the 6 model for thisproblem is

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Maximi5e , + !20 x! '0 x2*sub4ect to

x! 0

x2 !0

20 x! !0 x2 #00

and x!0* x20.

(b) se the graphical method to sol!e this model.

The constraint* x! 0 * has a constraint boundary at x! + 0. 8imilarly* the

constraint* x2 !0* has a constraint boundary at x2 + !0.

The constraint boundary for the constraint* 20 x! !0 x2 #00* intercepts the

x!-axis at 20x! !0/0 + #00* so at x!+ 2#. 8imilarly* this constraint boundary interceptsthe x2-axis at 20/0 !0x2+ #00* so at x2+ #0. This constraint boundary lies "ell "ithin

the constraint boundary for the first constraint* x!

0* so the first constraint isredundant and can be i(nored.9or a sample value of ,* say* , + 2*00* the correspondin( ob4ective function line*

, + 2*00 + !20 x! '0 x2* intercepts the x!-axis at !20 x! '0/0 + 2*00* so at x!+20. :t intercepts the x2-axis at !20 /0 '0 x2+ 2*00* so at x2+ 0. The correspondin(ob4ective function lines for other values of , are parallel to this line. ushin( these linesup as much as possible "hile still passin( throu(h a point in the feasible re(ion revealsthat the optimal solution is /x!* x2 + /20* !0 "ith , + *200* as depicted in the follo"in((raph.

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Example for Section 3."

;"i(ht is an elementary school teacher "ho also raises pi(s for supplemental income. A - !> = ? 0.

1onne(ativity constraints A 0* = 0.

The resultin( linear pro(rammin( model for this problem is

Minimi5e , + 0. A &0.' =*sub4ect to

'00 A !000 = '000

!0 A 70 = 700

2> A - !> = 0

and

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A 0* = 0.

(b) se the graphical method to sol!e this model. What is the resulting dail$ cost per

pig'

As sho"n belo"* the optimal solution is /A* = + /20>7* 0>7. The resultin( daily cost perpi( is , + 0>7 + .7!.

Example for Sections 3." and 3.

The 9a(ersta 8teel"orks currently is "orkin( t"o mines to obtain its iron ore. This ironore is shipped to either of t"o stora(e facilities. 3hen needed* it then is shipped on to the

company@s steel plant. The dia(ram belo" depicts this distribution net"ork* "here M!and M2 are the t"o mines* 8! and 82 are the t"o stora(e facilities* and is the steelplant. The dia(ram also sho"s the monthly amounts produced at the mines and needed atthe plant* as "ell as the shippin( cost and the maximum amount that can be shipped permonth throu(h each shippin( lane.

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Mana(ement no" "ants to determine the most economic plan for shippin( the iron orefrom the mines throu(h the distribution net"ork to the steel plant.

(a) Formulate a linear programming model for this problem.

The decision variables are defined as follo"s

xm!-s! number of units /tons shipped from Mine ! to 8tora(e 9acility !*xm!-s2 number of units /tons shipped from Mine ! to 8tora(e 9acility 2*

xm2-s! number of units /tons shipped from Mine 2 to 8tora(e 9acility !* xm2-s2 number of units /tons shipped from Mine ! to 8tora(e 9acility 2* xs! -p number of units /tons shipped from 8tora(e 9acility ! to the lant* xs2 -p number of units /tons shipped from 8tora(e 9acility 2 to the lant.

The total shippin( cost is

, + 2000 xm!-s! !700 xm!-s2 !00 xm2-s! !!00 xm2-s2 00 xs!-p '00 xs2-p

The constraints "e need to consider are

/! 8upply constraint on M! and M2

xm!-s! xm!-s2+ 0 xm2-s! xm2-s2+ 0

/2 Conservation-of-flo" constraint on 8! and 82

xm!-s! xm2-s! - xs!-p + 0 xm!-s2 xm2-s2 - xs2-p + 0

/ ;emand constraint on

xs!-p xs2-p+ !00

M1

M2 S2

S1

P

$2,000/ton30 tons max.

40 tonsproduced

60 tonsproduced

$1,100/ton50 tons max.

$400/ton70 tons max.

$800/ton70 tons max.

$1,700/ton30 tons max.

$1,600/ton50 tons max.

100 tonsneeded

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/ Capacity constraints

xm!-s!0* xm!-s2 0

xm2-s!#0* xm2-s2 #0

xs!-p 70* xs2-p 70

/# 1onne(ativity constraints

xm!-s!0* xm!-s2 0

xm2-s!0* xm2-s2 0

xs!-p 0* xs2-p 0

The resultin( linear pro(rammin( model for this problem is

Maximi5e , + 2000 xm!-s! !700 xm!-s2 !00 xm2-s! !!00 xm2-s2 00 xs!-p '00 xs2-p*

sub4ect to xm!-s! xm!-s2+ 0

xm2-s! xm2-s2+ 0xm!-s! xm2-s! - xs!-p + 0

xm!-s2 xm2-s2 - xs2-p + 0 xs!-p xs2-p+ !00

xm!-s!0* xm!-s2 0

xm2-s!#0* xm2-s2 #0

xs!-p

70* xs2-p

70 and

xm!-s!0* xm!-s2 0

xm2-s!0* xm2-s2 0

xs!-p 0* xs2-p 0

(b) Sol!e this model b$ the simplex method.

This model can be solved by usin( the %xcel 8olver*"hich employs the simplex method.

This spreadsheet formulation and solution of the problem is sho"n next.

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