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sz (Answers) (W‘W NV???) 1. Suppose that the farmer has $1 acres of arable land and :32 acres of grass. The
linear program Will be Math 482 Maximize z = 150271 + 100232 subject to 25171 +10% 3 2000,
331 +272 S 100,
33171;? Z The answer: z = 40000/3 is attained at $1 = %, $2 = %. 2. The equation solved for :51 is used to eliminate the 2 in the ﬁrst row: This corresponds to the basic solution 2:1 = x2 = $5 = 0, (133 = 1, (134 = 4. We see
from the ﬁrst row of the tableau that z = 4% + 5955 + 2, and since $2, m5 2 0, 2 has
a minimum value of 2, attained With the given basic solution. ? 3. We reformulate the LP in standard form: (t \a& Minimize z = ~5$1 — 5x2 —— 3x3
V v subject to
\ 1131 +3332 +173 +234 2 3
®\ ' M —C131 +3373 +335 2 2
2:61 ~$2 +2$3 +3136 — 4
“0‘7 2$1 +3532 —a:3 +5137 2 2
KW‘” m1, 272, 273, 1174, $5, 3:6, 237 Z 0 We represent this LP in the tableau I 0 5/2 —11/2 0 0 2 0 3/2 3/210 0 —1/2 0 3/2 and, since 2/3 < 3/(5/2) < 2/(3/2), we pivot on the position in bold above, obtaining thetableau 
26/3 0 —29/6 0 0 0 11/6 2/3
' O 7/2 0 1 0 ~1/2 0 (r5 0 29/6 0 0 1 —5/6 4/3 '
$3 0 —4/3 1 0 0 1/3 —1/3
931 1 5/6 0 0 0 1/6 1/3 Now the ratio (4/3)/(29/6) is the smallest of the ratios associated with the positive
entries in the :cg—column, so we pivot on the position in bold above and arrive at the following tableau: $4 $2 6/29 —5/29 8/29
$3 8/29 3/29 1/29
:51 —5/29 9/29 3/29 This corresponds to the basic solution 2:1 2 32/29, 232 = 8/29, x3 = 30/29,
:54 = 1/29, and 2:5 = 236 = 277 = 0. Since 225,336,237 2 0, we see that —z attains a
maximum possible value of 10 under this solution. Hence 2’s optimal value is ~10,
achieved when 231 = 32/29, 2:2 2 8/29, :53 = 30/29. 4. Since :134 is unrestricted, we solve for it in the ﬁrst equation and substitute
the result into the objective function and the remaining conditions, obtaining the equation
$4 = *‘21171— 5$2 + 3.113 + 2 and W rabbislle mama me A ~293 “iw‘ﬁ (33.010342 ...
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 Spring '08
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