Solution. For each \(x_1 = 0, \ldots , 7\) take the least \(x_2\) with \(320x_1 + 240x_2 \geq 2000\). The costs are: \(x_1 = 7, x_2 = 0\): \(481.6\); \(6, 1\): \(480.8\); \(5, 2\): \(480.0\); \(4, 3\): \(479.2\); \(3, 5\): \(546.4\); \(2, 6\): \(545.6\); \(1, 7\): \(544.8\); \(0, 9\): \(612\). The best is \(4\) Grade 1 and \(3\) Grade 2 inspectors at Rs. 479.2, which inspect exactly \(2000\) pieces. Rounding the LP solution up gives Rs. 481.6; the integer optimum mixes the grades, which the LP never does. The gap between Rs. 430 and Rs. 479.2 is the cost of indivisibility.

Solution. Maximise \(0.383x_1 + 0.367x_2\) subject to \(2x_1 + x_2 \leq 3000\), \(x_1 + 2x_2 \leq 3600\), \(x \geq 0\). The two constraints meet where \(x_1 = 800\), \(x_2 = 1400\), with profit \(0.383 \times 800 + 0.367 \times 1400 = 820\). The other vertices give \(0.383 \times 1500 = 575\) at \((1500, 0)\) and \(0.367 \times 1800 = 660\) at \((0, 1800)\), so the optimum is \(800\) kg of Fertilex and \(1400\) kg of Multiplex, using all \(1000\) kg of wheat and all \(1200\) kg of barley.

Solution. Interest for three months is \(0.20 \times \frac 3{12} = 0.05\) per rupee. Maximise \(Z = 4x_1 + 3x_2 - 0.05x_3\) subject to \(0.5x_1 + 0.3x_2 \leq 500\), \(0.3x_1 + 0.4x_2 \leq 400\), \(0.2x_1 + 0.1x_2 \leq 200\) (capacity); \(10x_1 + 8x_2 \leq 300\,000 + x_3\) (funds); \(0 \leq x_3 \leq 200\,000\). Here the capacity constraints bind long before the funds do (at most \(1000\) units of product 1 cost only Rs. 10 000), so \(x_3 = 0\) at the optimum: borrowing is worth while only when money, not capacity, is scarce. The optimum is \(x_1 = 727.3\), \(x_2 = 454.5\), where departments \(A\) and \(C\) are fully used, with profit Rs. 4272.7.

Solution. Introduce \(e_1, e_2 \geq 0\): minimise \(e_1 + e_2\) subject to \(-e_1 \leq x_1 - 2 \leq e_1\), \(-e_2 \leq x_2 + 1 \leq e_2\), \(x_1 + x_2 \geq 3\). For standard form, write \(x_j = x_j^+ - x_j^-\) and add a slack or surplus variable to each of the five inequalities, giving \(5\) equations in \(4 + 2 + 5 = 11\) non-negative variables. The optimal value is \(2\), attained for example at \(x = (2, 1)\).

Solution. Let \(x_j\) be the number of nurses whose five-day spell starts on day \(j\) (\(j = 1\) for Monday). A nurse works on day \(i\) if the spell started on day \(i\) or one of the four days before, cyclically. Minimise \(\sum _{j = 1}^7x_j\) subject to, for Monday, \(x_1 + x_4 + x_5 + x_6 + x_7 \geq 17\), for Tuesday \(x_1 + x_2 + x_5 + x_6 + x_7 \geq 13\), and so on cyclically, with \(x_j \geq 0\) and integer. Summing the seven constraints, each \(x_j\) appears five times, so \(5\sum x_j \geq 105\) and at least \(21\) nurses are needed.

Solution. (a) Proportionality: the cost is \(F + cx\) for \(x > 0\) and \(0\) for \(x = 0\). Repair with a binary variable \(y\), cost \(Fy + cx\) and \(x \leq My\) (see the section on integer programming). (b) Proportionality: revenue \(p(x)x\) is non-linear; if it is concave it can be approximated piecewise linearly for maximisation, otherwise use non-linear programming. (c) Certainty: the coefficient is random; use expected values with sensitivity analysis, or a stochastic model with scenarios.