Solve the Woof&Meow pet food company problem. Suppose your job is to present your ndings and recommendations to the company CEO. Compose...
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4. Solve the Woof&amp;Meow pet food company problem. Suppose your job is to
present your findings and recommendations to the company CEO. Compose one-page executive summary for your employer outlining the company’s most
profitable use of overstock resources. Include any findings which may be of interest to the company’s decision makers. 11.4 Woof&amp;Meow Pet Food Company Woof&amp;Meow has decided to use year-end overstock to produce specialty test-market
canned food for cats and dogs. The food Will be composed of five ingredients: chicken, beef, wheat, rice, and water. The following table shows for each item, the quantity
in stock, the fraction (by weight) of protein and fat, and the calorie content. Ingredient stock protein fat calories
(a) (wt fraC) (W's I5mm) (per a) 31750 0.27 0.14 2.4 21780 0.26 0.15 2.5 15600 0.14 0.024 3.4 21000 0.027 0.0025 1.3
unlimited 0 0 0 Any food produced must meet certain nutritional requirements. Acceptable ranges
are shown in the table below. protein range fat range calorie range
(wt frac) (wt frac) (per g) cat food 0.07—0.12 0.04—0.07 1.00—1.25
dog food 0.09—0.13 0.03—0.08 0.75—1.15

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In addition, consumer appreciation and marketability requirement suggest that
any food produced must contain at least 30% meat by weight. Formulate a linear
program whose solution provides a recipe for a batch of dog food and a batch of cat
food that maximizes the total weight of overstock used in the process. Solution. We must determine how much of each ingredient is used in each type of
food. Let
xi,- = the grams of ingredient 1' used in food 3', where i = 1, 2, 3, 4, 5 corresponds to chicken, beef, wheat, rice and water, respectively;
and j = 1,2 corresponds to cat food and dog food, respectively. We can consider
1353' 6 1R. Because there are many constraints on various relative quantities, we will
also use the following dependent variables. W;- = the total weight of food 3'. Wm = the total protein weight in food 3'. ij = the total fat weight in food j.
C, = the total calories in food 3'. All together there are 18 decision variables.
The objective is to maximize the use of overstock ingredients (by weight). That
is, we seek to maximize the objective function z = $11 + .1321 + $31 + $41 + $12 + $22 + $32 + 3342. We begin enumerating the problem constraints by defining our dependent variables
in terms of our initial decision variables. W1 = $11 + $21 +3331 + $41 + 1'51- W2 = $12 + $22 + 3332 + $42 + 1‘52-
WP] = 0.27:1:11 + 0.26321 + 0.14:1:31 + 0.0273641.
Wfl = 0.1451311 + 0.159321 + 0.0243231 + 0.0025x41.
ng = 0.273312 -|— 0.263922 + 0.143332 + 0.0275642.

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ng = 0.143312 + 0.15.7522 + 0.0245332 + 0002511742. 01 = 2.42311 + 2.51221 + 3.411131 + 1.33341.
02 = 2.42212 + 2.55822 -|- 3.4.1332 + 1.3:842. Now, our inequality constraints are relatively simply expressed in terms of our
augmented set of decision variables. For example, the requirement that cat food be
at least 30% meat by weight is expressed as $11 + 3721 W 2 0.3 or equivalently — $11 — 1321 + 0.3W1 5 0.
1 Notice that the above equivalence assumes that W1 &gt; 0. For this particular problem
we can be reasonably assured that this will be the case at the optimal solution. That
is, we expect Wf &gt; 0. However, the denominator of any such constraint should be
checked once the problem is solved to make sure that the equivalency is met. The
corresponding constraint for dog food is $12 + $22 W 2 0.3 01' equivalently — 3512 — $22 + 0.3W2 S 0.
2 There are constraints on the weight fraction of protiens. For the cat food we have W 1
0.07 &lt; —p &lt; 0.12 which is equivalent to the two constraints 0.07W1 — Wpl S 0 and — 0.12W1 + Wpl S 0 Remark 11.4.1. Suppose a: and y are decision variables and b is a constant. A constraint such as x/y 2 b is equivalent to a; 2 by only if we can be sure that
y &gt; 0 for any feasible 3;.

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For dog food we have the corresponding constraints
0.09W2 - Wp2 _ 0 and - 0.13W2 + Wp2 5 0.
We also have similar constraints on the weight fraction of fats:
0.04W1 - W/1 5 0 and - 0.07W1 + W/1 0.
0.03W2 - W/2 &lt; 0 and - 0.08W2 + W/2 5 0.
Next, there are restrictions on the calorie density for each food. Using the same type
of construction, we have
W1 - C1 0 and - 1.25W1 + C1 5 0,
0.75W2 - C2 &lt; 0 and - 1.15W2 + C2 &lt; 0.
Finally, we cannot use more ingredients than are available:
X11 + X12 &lt; 31750,
21 + 22 - 21780,
31 + X32 &lt; 15600,
241 + 242 - 21000.
Thus, the LP:

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